Fluid Mechanics solution

Chapter 5 Mass, Bernoulli, and Energy Equations Solutions Manual for Fluid Mechanics: Fundamentals and Applications Second Edition Yunus A. Çengel & John M. Cimbala McGraw-Hill, 2010 CHAPTER 5 MASS, BERNOULLI, AND ENERGY EQUATIONS PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and protected by copyright and other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw-Hill: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of McGraw-Hill. 5-1 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations Conservation of Mass 5-1C Solution We are to discuss whether the flow is steady through a given control volume. Analysis No, a flow with the same volume flow rate at the inlet and the exit is not necessarily steady (even if the density is constant – see Discussion). To be steady, the mass flow rate through the device must remain constant in time, and no variables can change with time at any specified spatial position. Discussion If the question had stated that the two mass flow rates were equal, then the answer would still be not necessarily. As a counter-example, consider the steadily increasing flow of an incompressible liquid through the device. At any instant in time, the mass flow rate in must equal the mass flow rate out since there is nowhere else for the liquid to go. However, the mass flow rate itself is changing with time, and hence the problem is unsteady. Can you think of another counter-example? 5-2C Solution We are to discuss mass and volume flow rates and their relationship. Analysis Mass flow rate is the amount of mass flowing through a cross-section per unit time whereas volume flow rate is the amount of volume flowing through a cross-section per unit time. Discussion Mass flow rate has dimensions of mass/time while volume flow rate has dimensions of volume/time. 5-3C Solution We are to discuss the mass flow rate entering and leaving a control volume. Analysis The amount of mass or energy entering a control volume does not have to be equal to the amount of mass or energy leaving during an unsteady-flow process. Discussion If the process is steady, however, the two mass flow rates must be equal; otherwise the amount of mass would have to increase or decrease inside the control volume, which would make it unsteady. 5-4C Solution We are to discuss steady flow through a control volume. Analysis Flow through a control volume is steady when it involves no changes with time at any specified position. Discussion This applies to any variable we might consider – pressure, velocity, density, temperature, etc. 5-5C Solution We are to name some conserved and non-conserved quantities. Analysis Mass, energy, momentum, and electric charge are conserved, and volume and entropy are not conserved during a process. Discussion Students may think of other answers that may be equally valid. 5-2 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-6 Solution A house is to be cooled by drawing in cool night time air continuously. For a specified air exchange rate, the required flow rate of the fan and the average discharge speed of air are to be determined. Assumptions Flow through the fan is steady. Analysis The volume of the house is given to be V house = 450 m 3 . Noting that this volume of air is to be replaced every Δt = 5 min , the required volume flow rate of air is V = = V room Δt 450 m 3 ⎛ 1 min ⎞ 3 ⎟ = 1.50 m /s ⎜ 5 min ⎝ 60 s ⎠ House 450 m3 For the given fan diameter, the average discharge speed is determined to be V = V Ac = V πD 2 /4 = 1.50 m 3 /s π (1.20 m) 2 /4 = 1.33 m/s AIR Discussion Note that the air velocity and thus the noise level is low because of the large fan diameter. 5-7E Solution A garden hose is used to fill a water bucket. The volume and mass flow rates of water, the filling time, and the discharge velocity are to be determined. Assumptions splashing. 1 Water is an incompressible substance. 2 Flow through the hose is steady. 3 There is no waste of water by Properties We take the density of water to be 62.4 lbm/ft3. Analysis (a) The volume and mass flow rates of water are V = AV = ( π D 2 / 4 )V = [ π ( 1 / 12 ft) 2 / 4 ]( 8 ft/s) = 0.04363 ft 3 /s ≅ 0.0436 ft 3 /s  = ρV = (62.4 lbm/ft 3 )(0.04363 ft 3 /s) = 2.72 lbm/s m (b) The time it takes to fill a 20-gallon bucket is Δt = ⎞ ⎛ 1 ft 3 20 gal V ⎟ = 61.3 s ⎜ = 3 ⎜ 7.4804 gal ⎟  V 0.04363 ft /s ⎝ ⎠ (c) The average discharge velocity of water at the nozzle exit is Ve = V Ae = V πDe2 /4 = 0.04363 ft 3 /s [π (0.5 / 12 ft) 2 / 4] = 32 ft/s Discussion Note that for a given flow rate, the average velocity is inversely proportional to the square of the velocity. Therefore, when the diameter is reduced by half, the velocity quadruples. 5-3 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-8E Solution The ducts of an air-conditioning system pass through an open area. The inlet velocity and the mass flow rate of air are to be determined. Assumptions Flow through the air conditioning duct is steady. Properties The density of air is given to be 0.082 lbm/ft3 at the inlet. Analysis The inlet velocity of air and the mass flow rate through the duct are V1 = V1 A1 = V1 π D2 / 4 = π (16/12 ft )2 / 4 450 ft 3 /min = 322 ft/min = 26.9 ft/s 450 ft3/min AIR D = 16 in m = ρ1V1 = (0.082 lbm/ft 3 )(450 ft 3 / min) = 36.9 lbm/min = 0.615 lbm/s Discussion The mass flow rate though a duct must remain constant in steady flow; however, the volume flow rate varies since the density varies with the temperature and pressure in the duct. 5-9 Solution A rigid tank initially contains air at atmospheric conditions. The tank is connected to a supply line, and air is allowed to enter the tank until the density rises to a specified level. The mass of air that entered the tank is to be determined. Properties The density of air is given to be 1.18 kg/m3 at the beginning, and 7.20 kg/m3 at the end. Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. The mass balance for this system can be expressed as Mass balance: min − m out = Δmsystem → mi = m 2 − m1 = ρ 2V − ρ 1V Substituting, mi = ( ρ 2 − ρ 1 )V = [(7.20 - 1.18) kg/m 3 ](1 m 3 ) = 6.02 kg Therefore, 6.02 kg of mass entered the tank. Discussion V1 = 1 m3 ρ1 =1.18 kg/m3 Tank temperature and pressure do not enter into the calculations. 5-10 Solution The ventilating fan of the bathroom of a building runs continuously. The mass of air “vented out” per day is to be determined. Assumptions Flow through the fan is steady. Properties The density of air in the building is given to be 1.20 kg/m3. Analysis The mass flow rate of air vented out is  mair = ρVair = (1.20 kg/m 3 )(0.050 m 3 /s) = 0.060 kg/s Then the mass of air vented out in 24 h becomes m = m air Δt = (0.060 kg/s)(24 × 3600 s) = 5184 kg Discussion Note that more than 3 tons of air is vented out by a bathroom fan in one day. 5-4 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-11 Solution A desktop computer is to be cooled by a fan at a high elevation where the air density is low. The mass flow rate of air through the fan and the diameter of the casing for a given velocity are to be determined. Assumptions Flow through the fan is steady. Properties The density of air at a high elevation is given to be 0.7 kg/m3. Analysis The mass flow rate of air is m air = ρVair = (0.700 kg/m3 )(0.400 m3 /min) = 0.280 kg/min = 0.00467 kg/s If the mean velocity is 110 m/min, the diameter of the casing is V = AV = π D2 4 V → D= 4V 4(0.280 m3 /min) = = 0.0569 m πV π (110 m/min) Therefore, the diameter of the casing must be at least 5.69 cm to ensure that the mean velocity does not exceed 110 m/min. Discussion This problem shows that engineering systems are sized to satisfy given imposed constraints. 5-12 Solution A smoking lounge that can accommodate 15 smokers is considered. The required minimum flow rate of air that needs to be supplied to the lounge and the diameter of the duct are to be determined. Assumptions Infiltration of air into the smoking lounge is negligible. Properties The minimum fresh air requirements for a smoking lounge is given to be 30 L/s per person. Analysis The required minimum flow rate of air that needs to be supplied to the lounge is determined directly from Vair = Vair per person ( No. of persons) = (30 L/s ⋅ person)(15 persons) = 450 L/s = 0.45 m 3 /s The volume flow rate of fresh air can be expressed as V = VA = V (πD 2 / 4 ) Solving for the diameter D and substituting, 4(0.45 m 3 /s) 4V = = 0.268 m D= πV π (8 m/s) Smoking Lounge 15 smokers 30 L/s person Therefore, the diameter of the fresh air duct should be at least 26.8 cm if the velocity of air is not to exceed 8 m/s. Discussion Fresh air requirements in buildings must be taken seriously to avoid health problems. 5-5 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-13 Solution The minimum fresh air requirements of a residential building is specified to be 0.35 air changes per hour. The size of the fan that needs to be installed and the diameter of the duct are to be determined. Analysis The volume of the building and the required minimum volume flow rate of fresh air are V room = (2.7 m)(200 m 2 ) = 540 m 3 V = V room × ACH = (540 m 3 )(0.35/h ) = 189 m 3 / h = 189,000 L/h = 3150 L/min The volume flow rate of fresh air can be expressed as V = VA = V (πD 2 / 4 ) House D= 4V = πV 200 m2 0.35 ACH Solving for the diameter D and substituting, 4(189 / 3600 m 3 /s) = 0.116 m π (5 m/s) Therefore, the diameter of the fresh air duct should be at least 11.6 cm if the velocity of air is not to exceed 5 m/s. Discussion Fresh air requirements in buildings must be taken seriously to avoid health problems. 5-14 Solution Air is accelerated in a nozzle. The mass flow rate and the exit area of the nozzle are to be determined. Assumptions Flow through the nozzle is steady. Properties The density of air is given to be 2.21 kg/m3 at the inlet, and 0.762 kg/m3 at the exit. Analysis (a) The mass flow rate of air is determined from the inlet conditions to be m = ρ1 A1V1 = (2.21 kg/m 3 )(0.008 m 2 )(45 m/s) = 0.796 kg/s 1 = m 2 = m . (b) There is only one inlet and one exit, and thus m Then the exit area of the nozzle is determined to be m = ρ 2 A2V2 ⎯ ⎯→ A2 = Discussion V1 = 45 m/s A1 = 80 cm2 AIR V2 = 150 m/s 0.796 kg/s m = = 0.00696 m 2 = 69.6 cm 2 ρ 2V2 (0.762 kg/ m 3 )(150 m/s) Since this is a compressible flow, we must equate mass flow rates, not volume flow rates. 5-6 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-15 Solution Air is expanded and is accelerated as it is heated by a hair dryer of constant diameter. The percent increase in the velocity of air as it flows through the drier is to be determined. Assumptions Flow through the nozzle is steady. Properties The density of air is given to be 1.20 kg/m3 at the inlet, and 1.05 kg/m3 at the exit. Analysis 1 = m 2 = m  . Then, There is only one inlet and one exit, and thus m m 1 = m 2 ρ1 AV1 = ρ 2 AV2 V2 ρ1 1.20 kg/m3 = = = 1.14 V1 ρ 2 1.05 kg/m3 V2 V1 (or, an increase of 14%) Therefore, the air velocity increases 14% as it flows through the hair drier. Discussion It makes sense that the velocity increases since the density decreases, but the mass flow rate is constant. Mechanical Energy and Efficiency 5-16C Solution We are to discuss mechanical energy and how it differs from thermal energy. Analysis Mechanical energy is the form of energy that can be converted to mechanical work completely and directly by a mechanical device such as a propeller. It differs from thermal energy in that thermal energy cannot be converted to work directly and completely. The forms of mechanical energy of a fluid stream are kinetic, potential, and flow energies. Discussion It would be nice if we could convert thermal energy completely into work. However, this would violate the second law of thermodynamics. 5-17C Solution We are to define and discuss mechanical efficiency. Analysis Mechanical efficiency is defined as the ratio of the mechanical energy output to the mechanical energy input. A mechanical efficiency of 100% for a hydraulic turbine means that the entire mechanical energy of the fluid is converted to mechanical (shaft) work. Discussion No real fluid machine is 100% efficient, due to frictional losses, etc. – the second law of thermodynamics. 5-7 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-18C Solution We are to define and discuss pump-motor efficiency. Analysis The combined pump-motor efficiency of a pump/motor system is defined as the ratio of the increase in the mechanical energy of the fluid to the electrical power consumption of the motor, W pump E mech,out − E mech,in ΔE mech,fluid η pump-motor = η pumpη motor = = = W W W elect,in elect,in elect,in The combined pump-motor efficiency cannot be greater than either of the pump or motor efficiency since both pump and motor efficiencies are less than 1, and the product of two numbers that are less than one is less than either of the numbers. Discussion Since many pumps are supplied with an integrated motor, pump-motor efficiency is a useful parameter. 5-19C Solution We are to define and discuss turbine, generator, and turbine-generator efficiency. Analysis Turbine efficiency, generator efficiency, and combined turbine-generator efficiency are defined as follows: W shaft,out Mechanical energy output = η turbine = Mechanical energy extracted from the fluid | ΔE | mech,fluid η generator = Electrical power output W elect,out = Mechanical power input W shaft,in η turbine-gen = η turbineη generaor = Discussion W elect,out − E E mech,in mech,out = W elect,out | ΔE mech,fluid | Most turbines are connected directly to a generator, so the combined efficiency is a useful parameter. 5-8 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-20 Solution Wind is blowing steadily at a certain velocity. The mechanical energy of air per unit mass, the power generation potential, and the actual electric power generation are to be determined. Assumptions 1 The wind is blowing steadily at a constant uniform velocity. 2 The efficiency of the wind turbine is independent of the wind speed. Properties The density of air is given to be ρ = 1.25 kg/m3. Analysis Kinetic energy is the only form of mechanical energy the wind possesses, and it can be converted to work entirely. Therefore, the power potential of the wind is its kinetic energy, which is V2/2 per unit mass, and m V 2 / 2 for a given mass flow rate: emech = ke = V 2 (8 m/s) 2 ⎛ 1 kJ/kg = ⎜ 2 2 ⎝ 1000 m 2 /s 2 m = ρVA = ρV πD 2 4 ⎞ ⎟ = 0.032 kJ/kg ⎠ = (1.25 kg/m 3 )(8 m/s) π (50 m) 2 4 = 19,635 kg/s W max = E mech = m emech = (19,635 kg/s)(0.032 kJ/kg) = 628 kW The actual electric power generation is determined by multiplying the power generation potential by the efficiency, W elect = η wind turbineW max = (0.30)(628 kW) = 188 kW Therefore, 283 kW of actual power can be generated by this wind turbine at the stated conditions. Wind Wind turbine 8 m/s 50 m Discussion The power generation of a wind turbine is proportional to the cube of the wind velocity, and thus the power generation will change strongly with the wind conditions. 5-9 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-21 Solution The previous problem is reconsidered. The effect of wind velocity and the blade span diameter on wind power generation as the velocity varies from 5 m/s to 20 m/s in increments of 5 m/s, and the diameter varies from 20 m to 80 m in increments of 20 m is to be investigated. Analysis The EES Equations window is printed below, followed by the tabulated and plotted results. D1=20 "m" D2=40 "m" D3=60 "m" D4=80 "m" Eta=0.30 rho=1.25 "kg/m3" m1_dot=rho*V*(pi*D1^2/4); W1_Elect=Eta*m1_dot*(V^2/2)/1000 "kW" m2_dot=rho*V*(pi*D2^2/4); W2_Elect=Eta*m2_dot*(V^2/2)/1000 "kW" m3_dot=rho*V*(pi*D3^2/4); W3_Elect=Eta*m3_dot*(V^2/2)/1000 "kW" m4_dot=rho*V*(pi*D4^2/4); W4_Elect=Eta*m4_dot*(V^2/2)/1000 "kW" D, m 20 V, m/s 5 10 15 20 5 10 15 20 5 10 15 20 5 10 15 20 40 60 80 m, kg/s 1,963 3,927 5,890 7,854 7,854 15,708 23,562 31,416 17,671 35,343 53,014 70,686 31,416 62,832 94,248 125,664 Welect, kW 7 59 199 471 29 236 795 1885 66 530 1789 4241 118 942 3181 7540 8000 D = 80 m 7000 6000 WElect 5000 D = 60 m 4000 3000 D = 40 m 2000 1000 0 4 D = 20 m 6 8 10 12 14 16 18 20 V, m/s Discussion Wind turbine power output is obviously nonlinear with respect to both velocity and diameter. 5-10 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-22E Solution A differential thermocouple indicates that the temperature of water rises a certain amount as it flows through a pump at a specified rate. The mechanical efficiency of the pump is to be determined. Assumptions 1 The pump is adiabatic so that there is no heat transfer with the surroundings, and the temperature rise of water is completely due to frictional heating. 2 Water is an incompressible substance. Properties We take the density of water to be ρ = 62.4 lbm/ft3 and its specific heat to be C = 1.0 Btu/lbm⋅°F. Analysis The increase in the temperature of water is due to the conversion of mechanical energy to thermal energy, and the amount of mechanical energy converted to thermal energy is equal to the increase in the internal energy of water, m = ρV = (62.4 lbm/ft 3 )(1.5 ft 3/s) = 93.6 lbm/s ΔT =0.072°F E mech, loss = ΔU = m cΔT 1 hp ⎞ ⎛ = (93.6 lbm/s)(1.0 Btu/lbm ⋅ °F)(0.072°F)⎜ ⎟ = 9.53 hp ⎝ 0.7068 Btu/s ⎠ The mechanical efficiency of the pump is determined from the general definition of mechanical efficiency, η pump = 1 − E mech,loss 9.53 hp = 1− = 0.647 or 64.7%  27 hp W Pump 27 hp mech, in Discussion Note that despite the conversion of more than one-third of the mechanical power input into thermal energy, the temperature of water rises by only a small fraction of a degree. Therefore, the temperature rise of a fluid due to frictional heating is usually negligible in heat transfer analysis. 5-11 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-23 Solution Water is pumped from a lake to a storage tank at a specified rate. The overall efficiency of the pump-motor unit and the pressure difference between the inlet and the exit of the pump are to be determined. Assumptions 1 The elevations of the tank and the lake remain constant. 2 Frictional losses in the pipes are negligible. 3 The changes in kinetic energy are negligible. 4 The elevation difference across the pump is negligible. Properties We take the density of water to be ρ = 1000 kg/m3. Analysis (a) We take the free surface of the lake to be point 1 and the free surfaces of the storage tank to be point 2. We also take the lake surface as the reference level (z1 = 0), and thus the potential energy at points 1 and 2 are pe1 = 0 and pe2 = gz2. The flow energy at both points is zero since both 1 and 2 are open to the atmosphere (P1 = P2 = Patm). Further, the kinetic energy at both points is zero (ke1 = ke2 = 0) since the water at both locations is essentially stationary. The mass flow rate of water and its potential energy at point 2 are m = ρV = (1000 kg/m 3 )(0.070 m 3/s) = 70 kg/s ⎛ 1 kJ/kg ⎞ pe1 = gz1 = (9.81 m/s 2 )(18 m)⎜ ⎟ = 0.177 kJ/kg ⎝ 1000 m 2 /s 2 ⎠ Then the rate of increase of the mechanical energy of water becomes ΔE mech,fluid = m (emech,out − emech,in ) = m ( pe2 − 0) = m pe2 = (70 kg/s)(0.177 kJ/kg) = 12.4 kW The overall efficiency of the combined pump-motor unit is determined from its definition, η pump-motor = ΔE mech,fluid 12.4 kW = = 0.606 or 60.6% 20.4 kW W elect,in (b) Now we consider the pump. The change in the mechanical energy of water as it flows through the pump consists of the change in the flow energy only since the elevation difference across the pump and the change in the kinetic energy are negligible. Also, this change must be equal to the useful mechanical energy supplied by the pump, which is 12.4 kW: ΔE mech,fluid = m (e mech,out − e mech,in ) = m Solving for ΔP and substituting, ΔP = P2 − P1 ρ = VΔP 2 Storage tank 18 m Pump 1 ΔE mech,fluid 12.4 kJ/s ⎛⎜ 1 kPa ⋅ m 3 ⎞⎟ = = 177 kPa V 0.070 m 3 /s ⎜⎝ 1 kJ ⎟⎠ Therefore, the pump must boost the pressure of water by 177 kPa in order to raise its elevation by 18 m. Discussion Note that only two-thirds of the electric energy consumed by the pump-motor is converted to the mechanical energy of water; the remaining one-third is wasted because of the inefficiencies of the pump and the motor. 5-12 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-24 Solution A river is flowing at a specified velocity, flow rate, and elevation. The total mechanical energy of the river water per unit mass, and the power generation potential of the entire river are to be determined. Assumptions 1 The elevation given is the elevation of the free surface of the river. 2 The velocity given is the average velocity. 3 The mechanical energy of water at the turbine exit is negligible. Properties We take the density of water to be ρ = 1000 kg/m3. Analysis Noting that the sum of the flow energy and the potential energy is constant for a given fluid body, we can take the elevation of the entire river water to be the elevation of the free surface, and ignore the flow energy. Then the total mechanical energy of the river water per unit mass becomes V2 emech = pe + ke = gh + 2 4 m/s River 2 ⎛ (4 m/s) ⎞⎟⎛ 1 kJ/kg ⎞ = ⎜⎜ (9.81 m/s 2 )(70 m) + ⎟⎜⎝ 1000 m 2 /s 2 ⎟⎠ 2 ⎠ ⎝ 70 m = 0.695 kJ/kg The power generation potential of the river water is obtained by multiplying the total mechanical energy by the mass flow rate, m = ρV = (1000 kg/m 3 )(500 m 3 /s) = 500,000 kg/s W max = E mech = m emech = (500,000 kg/s)(0.695 kg/s) = 347,350 kW ≅ 347 MW Therefore, 347 MW of power can be generated from this river as it discharges into the lake if its power potential can be recovered completely. Discussion Note that the kinetic energy of water is negligible compared to the potential energy, and it can be ignored in the analysis. Also, the power output of an actual turbine will be less than 347 MW because of losses and inefficiencies. 5-13 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-25 Solution A hydraulic turbine-generator is generating electricity from the water of a large reservoir. The combined turbine-generator efficiency and the turbine efficiency are to be determined. Assumptions negligible. 1 The elevation of the reservoir remains constant. 2 The mechanical energy of water at the turbine exit is Analysis We take the free surface of the reservoir to be point 1 and the turbine exit to be point 2. We also take the turbine exit as the reference level (z2 = 0), and thus the potential energy at points 1 and 2 are pe1 = gz1 and pe2 = 0. The flow energy P/ρ at both points is zero since both 1 and 2 are open to the atmosphere (P1 = P2 = Patm). Further, the kinetic energy at both points is zero (ke1 = ke2 = 0) since the water at point 1 is essentially motionless, and the kinetic energy of water at turbine exit is assumed to be negligible. The potential energy of water at point 1 is ⎛ 1 kJ/kg ⎞ pe1 = gz1 = (9.81 m/s 2 )(70 m)⎜ ⎟ = 0.687 kJ/kg 1 ⎝ 1000 m 2 /s 2 ⎠ Then the rate at which the mechanical energy of the fluid is supplied to the turbine become ΔE mech,fluid = m (e mech,in − e mech,out ) = m ( pe1 − 0) = m pe1 = (1500 kg/s)(0.687 kJ/kg) = 1031 kW The combined turbine-generator and the turbine efficiency are determined from their definitions, η turbine-gen = η turbine = 750 kW 70 m Turbine W elect,out 750 kW = = 0.727 or 72.7% | ΔE mech,fluid | 1031 kW Generator 2 W shaft,out 800 kW = = 0.776 or 77.6%  | ΔE mech,fluid | 1031 kW Therefore, the reservoir supplies 1031 kW of mechanical energy to the turbine, which converts 800 kW of it to shaft work that drives the generator, which generates 750 kW of electric power. Discussion This problem can also be solved by taking point 1 to be at the turbine inlet, and using flow energy instead of potential energy. It would give the same result since the flow energy at the turbine inlet is equal to the potential energy at the free surface of the reservoir. Bernoulli Equation 5-26C Solution We are to define streamwise acceleration and discuss how it differs from normal acceleration. Analysis The acceleration of a fluid particle along a streamline is called streamwise acceleration, and it is due to a change in speed along a streamline. Normal acceleration (or centrifugal acceleration), on the other hand, is the acceleration of a fluid particle in the direction normal to the streamline, and it is due to a change in direction. Discussion In a general fluid flow problem, both streamwise and normal acceleration are present. 5-14 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-27C Solution We are to express the Bernoulli equation in three different ways. Analysis The Bernoulli equation is expressed in three different ways as follows: (a) In terms of energies: (b) In terms of pressures: (c) in terms of heads: ρ P + V2 + gz = constant 2 P+ρ V2 + ρgz = constant 2 P V2 + + z = H = constant ρg 2 g Discussion You could, of course, express it in other ways, but these three are the most useful. 5-28C Solution We are to discuss the three major assumptions used in the derivation of the Bernoulli equation. Analysis The three major assumptions used in the derivation of the Bernoulli equation are that the flow is steady, there is negligible frictional effects, and the flow is incompressible. Discussion If any one of these assumptions is not valid, the Bernoulli equation should not be used. Unfortunately, many people use it anyway, leading to errors. 5-29C Solution We are to define and discuss static, dynamic, and hydrostatic pressure. Analysis Static pressure P is the actual pressure of the fluid. Dynamic pressure ρV 2/2 is the pressure rise when the fluid in motion is brought to a stop isentropically. Hydrostatic pressure ρgz is not pressure in a real sense since its value depends on the reference level selected, and it accounts for the effects of fluid weight on pressure. The sum of static, dynamic, and hydrostatic pressures is constant when flow is steady and incompressible, and when frictional effects are negligible. Discussion The incompressible Bernoulli equation states that the sum of these three pressures is constant along a streamline; this approximation is valid only for steady and incompressible flow with negligible frictional effects. 5-30C Solution We are to define stagnation pressure and discuss how it can be measured. Analysis The sum of the static and dynamic pressures is called the stagnation pressure, and it is expressed as 2 Pstag = P + ρV / 2 . The stagnation pressure can be measured by a Pitot tube whose inlet is normal to the flow. Discussion Stagnation pressure, as its name implies, is the pressure obtained when a flowing fluid is brought to rest isentropically, at a so-called stagnation point. 5-15 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-31C Solution We are to define and discuss pressure head, velocity head, and elevation head. Analysis The pressure head P/ρg is the height of a fluid column that produces the static pressure P. The velocity head V 2/2 is the elevation needed for a fluid to reach the velocity V during frictionless free fall. The elevation head z is the height of a fluid relative to a reference level. Discussion It is often convenient in fluid mechanics to work with head – pressure expressed as an equivalent column height of fluid. 5-32C Solution We are to define hydraulic grade line and compare it to energy grade line. Analysis The curve that represents the sum of the static pressure and the elevation heads, P/ρg + z, is called the hydraulic grade line or HGL. The curve that represents the total head of the fluid, P/ρg + V 2/2g + z, is called the energy line or EGL. Thus, in comparison, the energy grade line contains an extra kinetic-energy-type term. For stationary bodies such as reservoirs or lakes, the EL and HGL coincide with the free surface of the liquid. Discussion The hydraulic grade line can rise or fall along flow in a pipe or duct as the cross-sectional area increases or decreases, whereas the energy grade line always decreases unless energy is added to the fluid (like with a pump). 5-33C Solution We are to discuss the hydraulic grade line in open-channel flow and at the outlet of a pipe. Analysis For open-channel flow, the hydraulic grade line (HGL) coincides with the free surface of the liquid. At the exit of a pipe discharging to the atmosphere, HGL coincides with the elevation of the pipe outlet. Discussion We are assuming incompressible flow, and the pressure at the pipe outlet is atmospheric. 5-34C Solution We are to discuss the effect of liquid density on the operation of a siphon. Analysis The lower density liquid can go over a higher wall, provided that cavitation pressure is not reached. Therefore, oil may be able to go over a higher wall than water. Discussion However, frictional losses in the flow of oil in a pipe or tube are much greater than those of water since the viscosity of oil is much greater than that of water. When frictional losses are considered, the water may actually be able to be siphoned over a higher wall than the oil, depending on the tube diameter and length, etc. 5-35C Solution We are to explain how and why a siphon works, and its limitations. Analysis Siphoning works because of the elevation and thus pressure difference between the inlet and exit of a tube. The pressure at the tube exit and at the free surface of a liquid is the atmospheric pressure. When the tube exit is below the free surface of the liquid, the elevation head difference drives the flow through the tube. At sea level, 1 atm pressure can support about 10.3 m of cold water (cold water has a low vapor pressure). Therefore, siphoning cold water over a 7 m wall is theoretically feasible. Discussion In actual practice, siphoning is also limited by frictional effects in the tube, and by cavitation. 5-16 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-36C Solution We are to discuss and compare the operation of a manometer. Analysis As the duct converges to a smaller cross-sectional area, the velocity increases. By Bernoulli’s equation, the pressure therefore decreases. Thus Manometer A is correct since the pressure on the right side of the manometer is obviously smaller. According to the Bernoulli approximation, the fluid levels in the manometer are independent of the flow direction, and reversing the flow direction would have no effect on the manometer levels. Manometer A is still correct if the flow is reversed. Discussion In reality, it is hard for a fluid to expand without the flow separating from the walls. Thus, reverse flow with such a sharp expansion would not produce as much of a pressure rise as that predicted by the Bernoulli approximation. 5-37C Solution We are to discuss and compare two different types of manometer arrangements in a flow. Analysis Arrangement 1 consists of a Pitot probe that measures the stagnation pressure at the pipe centerline, along with a static pressure tap that measures static pressure at the bottom of the pipe. Arrangement 2 is a Pitot-static probe that measures both stagnation pressure and static pressure at nearly the same location at the pipe centerline. Because of this, arrangement 2 is more accurate. However, it turns out that static pressure in a pipe varies with elevation across the pipe cross section in much the same way as in hydrostatics. Therefore, arrangement 1 is also very accurate, and the elevation difference between the Pitot probe and the static pressure tap is nearly compensated by the change in hydrostatic pressure. Since elevation changes are not important in either arrangement, there is no change in our analysis when the water is replaced by air. Discussion Ignoring the effects of gravity, the pressure at the centerline of a turbulent pipe flow is actually somewhat smaller than that at the wall due to the turbulent eddies in the flow, but this effect is small. 5-38C Solution We are to discuss the maximum rise of a jet of water from a tank. Analysis With no losses and a 100% efficient nozzle, the water stream could reach to the water level in the tank, or 20 meters. In reality, friction losses in the hose, nozzle inefficiencies, orifice losses, and air drag would prevent attainment of the maximum theoretical height. Discussion In fact, the actual maximum obtainable height is much smaller than this ideal theoretical limit. 5-39C Solution We are to compare siphoning at sea level and on a mountain. Analysis At sea level, a person can theoretically siphon water over a wall as high as 10.3 m. At the top of a high mountain where the pressure is about half of the atmospheric pressure at sea level, a person can theoretically siphon water over a wall that is only half as high. An atmospheric pressure of 58.5 kPa is insufficient to support a 8.5 meter high siphon. Discussion In actual practice, siphoning is also limited by frictional effects in the tube, and by cavitation. 5-17 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-40 Solution In a power plant, water enters the nozzles of a hydraulic turbine at a specified pressure. The maximum velocity water can be accelerated to by the nozzles is to be determined. Assumptions 1The flow of water is steady, incompressible, and irrotational with negligible frictional effects (so that the Bernoulli equation is applicable). 2 Water enters the nozzle with a low velocity. Properties We take the density of water to be ρ = 1000 kg/m3. Analysis We take points 1 and 2 at the inlet and exit of the nozzle, respectively. Noting that V1 ≅ 0 and z1 = z2, the application of the Bernoulli equation between points 1 and 2 gives P V2 P1 V12 + + z1 = 2 + 2 + z2 ρg 2 g ρg 2 g → P1 P V2 = atm + 2 ρg ρg 2 g → V2 = 2( P1 − Patm ) ρ 100 kPa Water V Substituting the given values, the nozzle exit velocity is determined to be V1 = 2(800 − 100) kPa ⎛⎜ 1000 N/m 2 1000 kg/m 3 ⎜⎝ 1 kPa ⎞⎛ 1 kg ⋅ m/s 2 ⎟⎜ ⎟⎜ 1N ⎠⎝ ⎞ ⎟ = 37.4 m/s ⎟ ⎠ 1 800 kPa Turbine nozzzle 2 Discussion This is the maximum nozzle exit velocity, and the actual velocity will be less because of friction between water and the walls of the nozzle. 5-41 Solution The velocity of an aircraft is to be measured by a Pitot-static probe. For a given differential pressure reading, the velocity of the aircraft is to be determined. Assumptions 1 The air flow over the aircraft is steady, incompressible, and irrotational with negligible frictional effects (so that the Bernoulli equation is applicable). 2 Standard atmospheric conditions exist. 3 The wind effects are negligible. Properties The density of the atmosphere at an elevation of 3000 m is ρ = 0.909 kg/m3. Analysis We take point 1 at the entrance of the tube whose opening is parallel to flow, and point 2 at the entrance of the tube whose entrance is normal to flow. Noting that point 2 is a stagnation point and thus V2 = 0 and z1 = z2, the application of the Bernoulli equation between points 1 and 2 gives V12 P2 − P1 V12 Pstag − P1 P V2 P1 V12 + + z1 = 2 + 2 + z2 → = → = ρg 2 g ρg 2 g ρg ρ 2g 2 Solving for V1 and substituting, V1 = 2( Pstag − P1 ) ρ = 2(3000 N/m 2 ) ⎛ 1 kg ⋅ m/s 2 ⎞ ⎟ = 81.2 m/s = 292 km/h ⎜ ⎟ 0.909 kg/m 3 ⎜⎝ 1 N ⎠ since 1 Pa = 1 N/m2 and 1 m/s = 3.6 km/h. Discussion Note that the velocity of an aircraft can be determined by simply measuring the differential pressure on a Pitot-static probe. 5-18 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-42 Solution The bottom of a car hits a sharp rock and a small hole develops at the bottom of its gas tank. For a given height of gasoline, the initial velocity of the gasoline out of the hole is to be determined. Also, the variation of velocity with time and the effect of the tightness of the lid on flow rate are to be discussed. Assumptions 1 The flow is steady, incompressible, and irrotational with negligible frictional effects (so that the Bernoulli equation is applicable). 2 The air space in the tank is at atmospheric pressure. 3 The splashing of the gasoline in the tank during travel is not considered. Analysis This problem involves the conversion of flow, kinetic, and potential energies to each other without involving any pumps, turbines, and wasteful components with large frictional losses, and thus it is suitable for the use of the Bernoulli equation. We take point 1 to be at the free surface of gasoline in the tank so that P1 = Patm (open to the atmosphere) V1 ≅ 0 (the tank is large relative to the outlet), and z1 = 0.3 m and z2 = 0 (we take the reference level at the hole. Also, P2 = Patm (gasoline discharges into the atmosphere). Then the Bernoulli equation simplifies to P V2 P1 V12 + + z1 = 2 + 2 + z2 ρg 2 g ρg 2 g → z1 = Gas Tank V22 2g 1 Solving for V2 and substituting, V 2 = 2 gz1 = 2(9.81 m/s 2 )(0.4 m) = 2.80 m/s 40 cm Therefore, the gasoline will initially leave the tank with a velocity of 2.80 m/s. 2 V2 The Bernoulli equation applies along a streamline, and streamlines generally do not make sharp turns. The Discussion velocity will be less than 2.80 m/s since the hole is probably sharp-edged and it will cause some head loss. As the gasoline level is reduced, the velocity will decrease since velocity is proportional to the square root of liquid height. If the lid is tightly closed and no air can replace the lost gasoline volume, the pressure above the gasoline level will be reduced, and the velocity will be decreased. 5-19 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-43E [Also solved using EES on enclosed DVD] Solution The drinking water needs of an office are met by large water bottles with a plastic hose inserted in it. The minimum filling time of an 8-oz glass is to be determined when the bottle is full and when it is near empty. Assumptions 1 The flow is steady, incompressible, and irrotational with negligible frictional effects (so that the Bernoulli equation is applicable). 2 All losses are neglected to obtain the minimum filling time. Analysis We take point 1 to be at the free surface of water in the bottle and point 2 at the exit of the tube so that P1 = P2 = Patm (the bottle is open to the atmosphere and water discharges into the atmosphere), V1 ≅ 0 (the bottle is large relative to the tube diameter), and z2 = 0 (we take point 2 as the reference level). Then the Bernoulli equation simplifies to P1 V12 P V2 + + z1 = 2 + 2 + z2 ρg 2 g ρg 2 g → z1 = V22 2g → V2 = 2 gz1 Substituting, the discharge velocity of water and the filling time are determined as follows: (a) Full bottle (z1 = 3.5 ft): 0.25 in V2 = 2(32.2 ft/s 2 )(3.5 ft) = 15.0 ft/s A = πD 2 / 4 = π (0.25 / 12 ft) 2 / 4 = 3.41 × 10−4 ft 2 V V 0.00835 ft 3 Δt =  = = = 1.6 s AV2 (3.41× 10− 4 ft 2 )(15 ft/s ) V 1.5 ft 1 (b) Empty bottle (z1 = 2 ft): V2 = 2(32.2 ft/s 2 )(2 ft) = 11.3 ft/s 2 ft V V 0.00835 ft Δt =  = = = 2.2 s AV2 (3.41× 10 − 4 ft 2 )(11.3 ft/s ) V 3 2 Discussion The siphoning time is determined assuming frictionless flow, and thus this is the minimum time required. In reality, the time will be longer because of friction between water and the tube surface. 5-44 Solution The static and stagnation pressures in a horizontal pipe are measured. The velocity at the center of the pipe is to be determined. Assumptions The flow is steady, incompressible, and irrotational with negligible frictional effects in the short distance between the two pressure measurement locations (so that the Bernoulli equation is applicable). Analysis We take points 1 and 2 along the centerline of the pipe, with point 1 directly under the piezometer and point 2 at the entrance of the Pitot-static probe (the stagnation point). This is a steady flow with straight and parallel streamlines, and thus the static pressure at any point is equal to the hydrostatic pressure at that point. Noting that point 2 is a stagnation point and thus V2 = 0 and z1 = 35 cm 26 cm z2, the application of the Bernoulli equation between points 1 and 2 gives P V2 P1 V12 + + z1 = 2 + 2 + z2 ρg 2 g ρg 2 g → V12 P2 − P1 = ρg 2g Water 1 2 V Substituting the P1 and P2 expressions give V12 P2 − P1 ρg (hpitot + R ) − ρg (hpiezo + R ) ρg (hpitot − hpiezo ) = = = = hpitot − hpiezo 2g ρg ρg ρg Solving for V1 and substituting, V1 = 2 g (hpitot − hpiezo ) = 2(9.81 m/s 2 )[(0.35 − 0.26) m] = 1.33 m/s Discussion Note that to determine the flow velocity, all we need is to measure the height of the excess fluid column in the Pitot-static probe. 5-20 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-45 Solution A water tank of diameter Do and height H open to the atmosphere is initially filled with water. An orifice of diameter D with a smooth entrance (no losses) at the bottom drains to the atmosphere. Relations are to be developed for the time required for the tank to empty completely and half-way. Assumptions 1 The orifice has a smooth entrance, and thus the frictional losses are negligible. 2 The flow is steady, incompressible, and irrotational with negligible frictional effects (so that the Bernoulli equation is applicable). Analysis We take point 1 at the free surface of the tank, and point 2 at the exit of orifice. We take the reference level at the orifice (z2 = 0), and take the positive direction of z to be upwards. Noting that the fluid at both points is open to the atmosphere (and thus P1 = P2 = Patm) and that the fluid velocity at the free surface is very low (V1 ≅ 0), the Bernoulli equation between these two points simplifies to P V2 P1 V12 + + z1 = 2 + 2 + z2 ρg 2 g ρg 2 g → z1 = → V22 2g V2 = 2 gz1 For generality, we express the water height in the tank at any time t by z, and the discharge velocity by V2 = 2 gz . Note that water surface in the tank moves down as the tank drains, and thus z is a variable whose value changes from H at the beginning to 0 when the tank is emptied completely. We denote the diameter of the orifice by D, and the diameter of the tank by Do. The flow rate of water from the tank is obtained by multiplying the discharge velocity by the orifice cross-sectional area, πD 2 V = AorificeV 2 = 2 gz 1 Tank Water 4 Then the amount of water that flows through the orifice during a differential time interval dt is H πD 2 D0 dV = Vdt = 2 gz dt (1) 4 D which, from conservation of mass, must be equal to the decrease in the volume of water in the tank, πD 2 2 (2) dV = Atank (−dz ) = − 0 dz 4 where dz is the change in the water level in the tank during dt. (Note that dz is a negative quantity since the positive direction of z is upwards. Therefore, we used –dz to get a positive quantity for the amount of water discharged). Setting Eqs. (1) and (2) equal to each other and rearranging, πD 2 D2 D02 1 πD 2 −1 2 gz dt = − 0 dz dt = − 02 dz = − z 2 dz → 2 4 4 2 gz D D 2g The last relation can be integrated easily since the variables are separated. Letting tf be the discharge time and integrating it from t = 0 when z = zi = H to t = tf when z = zf gives ∫ tf dt = − t =0 D02 D2 2g ∫ zf z = z1 z −1 / 2 dz → tf = - 1 D02 z2 D2 2 g 1 2 zf = ( 2g 2 D02 D2 ) zi − z f = z1 Then the discharging time for the two cases becomes as follows: (a) The tank empties halfway: zi = H and zf = H/2: tf = D02 D2 (b) The tank empties completely: zi = H and zf = 0: tf = D02 D2 D02 D2 ⎛ 2z 2 z f ⎞⎟ i ⎜ − ⎜ g g ⎟ ⎝ ⎠ ⎛ 2H H ⎞⎟ ⎜ − ⎜ g g ⎟⎠ ⎝ 2H g Discussion Note that the discharging time is inversely proportional to the square of the orifice diameter. Therefore, the discharging time can be reduced to one-fourth by doubling the diameter of the orifice. 5-21 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-46E Solution A siphon pumps water from a large reservoir to a lower tank which is initially empty. Water leaves the tank through an orifice. The height the water will rise in the tank at equilibrium is to be determined. Assumptions 1 The flow is steady, incompressible, and irrotational with negligible frictional effects (so that the Bernoulli equation is applicable). 2 Both the tank and the reservoir are open to the atmosphere. 3 The water level of the reservoir remains constant. Analysis We take the reference level to be at the bottom of the tank, and the water height in the tank at any time to be h. We take point 1 to be at the free surface of reservoir, point 2 at the exit of the siphon, which is placed at the bottom of the tank, and point 3 at the free surface of the tank, and point 4 at the exit of the orifice at the bottom of the tank. Then z1 = 20 ft, z2 = z4 = 0, z3 = h, P1 = P3 = P4 = Patm (the reservoir is open to the atmosphere and water discharges into the atmosphere) P2 = Patm+ρgh (the hydrostatic pressure at the bottom of the tank where the siphon discharges), and V1 ≅ V3 ≅ 0 (the free surfaces of reservoir and the tank are large relative to the tube diameter). Then the Bernoulli equation between 1-2 and 3-4 simplifies to P1 V12 P V2 + + z1 = 2 + 2 + z2 ρg 2 g ρg 2 g → P3 V32 P V2 + + z3 = 4 + 4 + z 4 ρg 2 g ρg 2 g → Patm P + ρgh V22 + z1 = atm + 2g ρg ρg h= → V42 2g 1 2 in Reservoir 20 ft 3 h Water Tank 2 4 → V2 = 2 gz1 − 2 gh = 2 g ( z1 − h ) V4 = 2 gh Noting that the diameters of the tube and the orifice are the same, the flow rates of water into and out of the tank will be the same when the water velocities in the tube and the orifice are equal since V2 = V4 → AV2 = AV4 → V2 = V4 Setting the two velocities equal to each other gives V2 = V4 → 2 g ( z1 − h) = 2 gh → z1 − h = h → h= z1 20 ft = = 10 ft 2 2 Therefore, the water level in the tank will stabilize when the water level rises to 10 ft. Discussion This result is obtained assuming negligible friction. The result would be somewhat different if the friction in the pipe and orifice were considered. 5-22 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-47 Solution Water enters an empty tank steadily at a specified rate. An orifice at the bottom allows water to escape. The maximum water level in the tank is to be determined, and a relation for water height z as a function of time is to be obtained. Assumptions 1 The orifice has a smooth entrance, and thus the frictional losses are negligible. 2 The flow through the orifice is steady, incompressible, and irrotational with negligible frictional effects (so that the Bernoulli equation is applicable). Analysis (a) We take point 1 at the free surface of the tank, and point 2 at the exit of orifice. We take the reference level at the orifice (z2 = 0), and take the positive direction of z to be upwards. Noting that the fluid at both points is open to the atmosphere (and thus P1 = P2 = Patm) and that the fluid velocity at the free surface is very low (V1 ≅ 0) (it becomes zero when the water in the tank reaches its maximum level), the Bernoulli equation between these two points simplifies to P V2 P1 V12 + + z1 = 2 + 2 + z2 ρg 2 g ρg 2 g → z1 = → V22 2g V2 = 2 gz1 m 1 Then the mass flow rate through the orifice for a water height of z becomes πD0 m out = ρVout = ρAorificeV 2 = ρ 4 2 2 gz 1 → z= 2g ⎛ 4m out ⎜ ⎜ ρπD 2 0 ⎝ ⎞ ⎟ ⎟ ⎠ Water Tank 2 Setting z = hmax and m out = m in (the incoming flow rate) gives the desired relation for the maximum height the water will reach in the tank, DT z D0 D0 ⎛ 4m in ⎞ ⎟ ⎜ ⎜ ρπD 2 ⎟ 0 ⎠ ⎝ (b) The amount of water that flows through the orifice and the increase in the amount of water in the tank during a differential time interval dt are πD02 dm out = m out dt = ρ 2 gz dt 4 πDT2 dm tank = ρAtank dz = ρ dz 4 The amount of water that enters the tank during dt is dmin = m in dt (Recall that m in = constant). Substituting them into the hmax = 2 2 1 2g conservation of mass relation dm tank = dm in − dm out gives dm tank = m in dt − m out dt → ρ ⎛ πD02 dz = ⎜ m in − ρ ⎜ 4 4 ⎝ πDT2 ⎞ 2 gz ⎟dt ⎟ ⎠ Separating the variables, and integrating it from z = 0 at t = 0 to z = z at time t = t gives 1 4 ρπDT2 dz m in − 14 ρπD02 2 gz = dt → ∫ z z =0 1 4 ρπDT2 dz m in − 14 ρπD02 2 gz ∫ = dt = t t t =0 Performing the integration, the desired relation between the water height z and time t is obtained to be 1 2 2 1 ⎛  ⎜ 1 ρπD 2 2 gz − m ln min − 4 ρπD0 2 gz 0 in 4 m in 2 g ) 2 ⎜⎝ ρπDT2 ( 14 ρπD02 ⎞ ⎟=t ⎟ ⎠ Discussion Note that this relation is implicit in z, and thus we can’t obtain a relation in the form z = f(t). Substituting a z value in the left side gives the time it takes for the fluid level in the tank to reach that level. Equation solvers such as EES can easily solve implicit equations like this. 5-23 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-48E Solution Water flows through a horizontal pipe that consists of two sections at a specified rate. The differential height of a mercury manometer placed between the two pipe sections is to be determined. Assumptions 1The flow through the pipe is steady, incompressible, and irrotational with negligible frictional effects (so that the Bernoulli equation is applicable). 2 The losses in the reducing section are negligible. The densities of mercury and water are ρHg = 847 lbm/ft3 and ρw = 62.4 lbm/ft3. Properties Analysis We take points 1 and 2 along the centerline of the pipe over the two tubes of the manometer. Noting that z1 = z2, the Bernoulli equation between points 1 and 2 gives P V2 P1 V12 + + z1 = 2 + 2 + z2 ρg 2 g ρg 2 g → P1 − P2 = ρ w (V22 − V12 ) 2 (1) We let the differential height of the mercury manometer be h and the distance between the centerline and the mercury level in the tube where mercury is raised be s. Then the pressure difference P2 – P1 can also be expressed as P1 + ρ w g ( s + h) = P2 + ρ w gs + ρ Hg gh → P1 − P2 = ( ρ Hg − ρ w ) gh h= ρ w (V 22 − V12 ) V 22 − V12 = 2 g ( ρ Hg − ρ w ) 2 g ( ρ Hg / ρ w − 1) Combining Eqs. (1) and (2) and solving for h, ρ w (V 22 − V12 ) 2 = ( ρ Hg − ρ w ) gh → Calculating the velocities and substituting, V1 = V2 = h= V A1 V A2 = = V πD12 /4 V πD 22 /4 = = ⎛ 0.13368 ft 3 ⎜ π (4/12 ft) 2 / 4 ⎜⎝ 1 gal 1 gal/s ⎛ 0.13368 ft 3 ⎜ π (2/12 ft) / 4 ⎜⎝ 1 gal 1 gal/s (6.13 ft/s) 2 − (1.53 ft/s) 2 2(32.2 ft/s 2 )(847 / 62.4 − 1) 2 ⎞ ⎟ = 1.53 ft/s ⎟ ⎠ ⎞ ⎟ = 6.13 ft/s ⎟ ⎠ = 0.0435 ft = 0.52 in (2) 1 2 4 in 2 in s h Therefore, the differential height of the mercury column will be 0.52 in. Discussion In reality, there are frictional losses in the pipe, and the pressure at location 2 will actually be smaller than that estimated here, and therefore h will be larger than that calculated here. 5-24 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-49 Solution An airplane is flying at a certain altitude at a given speed. The pressure on the stagnation point on the nose of the plane is to be determined, and the approach to be used at high velocities is to be discussed. Assumptions 1 The air flow over the aircraft is steady, incompressible, and irrotational with negligible frictional effects (so that the Bernoulli equation is applicable). 2 Standard atmospheric conditions exist. 3 The wind effects are negligible. Properties The density of the atmospheric air at an elevation of 12,000 m is ρ = 0.312 kg/m3. Analysis We take point 1 well ahead of the plane at the level of the nose, and point 2 at the nose where the flow comes to a stop. Noting that point 2 is a stagnation point and thus V2 = 0 and z1 = z2, the application of the Bernoulli equation between points 1 and 2 gives P V2 P1 V12 + + z1 = 2 + 2 + z2 ρg 2 g ρg 2 g → V12 P2 − P1 = ρg 2g Solving for Pstag, gage and substituting, Pstag, gage = ρV12 2 = (0.312 kg/m 3 )(300 / 3.6 m/s) 2 2 V12 Pstag − Patm Pstag, gage = = ρ ρ 2 → ⎛ 1N ⎜ ⎜ 1 kg ⋅ m/s 2 ⎝ ⎞ ⎟ = 1083 N/m 2 = 1083 Pa ⎟ ⎠ since 1 Pa = 1 N/m2 and 1 m/s = 3.6 km/h. Altitude 12,000 m 2 1 300 km/h Discussion A flight velocity of 1050 km/h = 292 m/s corresponds to a Mach number much greater than 0.3 (the speed of sound is about 340 m/s at room conditions, and lower at higher altitudes, and thus a Mach number of 292/340 = 0.86). Therefore, the flow can no longer be assumed to be incompressible, and the Bernoulli equation given above cannot be used. This problem can be solved using the modified Bernoulli equation that accounts for the effects of compressibility, assuming isentropic flow. 5-25 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-50 Solution A Pitot-static probe is inserted into the duct of an air heating system parallel to flow, and the differential height of the water column is measured. The flow velocity and the pressure rise at the tip of the Pitot-static probe are to be determined. Assumptions 1 The flow through the duct is steady, incompressible, and irrotational with negligible frictional effects (so that the Bernoulli equation is applicable). 2 Air is an ideal gas. Properties We take the density of water to be ρ = 1000 kg/m3. The gas constant of air is R = 0.287 kPa⋅m3/kg⋅K. Analysis We take point 1 on the side of the probe where the entrance is parallel to flow and is connected to the static arm of the Pitot-static probe, and point 2 at the tip of the probe where the entrance is normal to flow and is connected to the dynamic arm of the Pitot-static probe. Noting that point 2 is a stagnation point and thus V2 = 0 and z1 = z2, the application of the Bernoulli equation between points 1 and 2 gives P V2 P1 V12 + + z1 = 2 + 2 + z2 ρg 2 g ρg 2 g → P1 V12 P2 + = ρg 2 g ρg → where the pressure rise at the tip of the Pitot-static probe is ⎛ 1N P2 − P1 = ρ w gh = (1000 kg/m 3 )(9.81 m/s 2 )(0.024 m)⎜⎜ 2 ⎝ 1 kg ⋅ m/s = 235 N/m 2 = 235 Pa Also, ρ air = Substituting, V1 = V = 2( P2 − P1 ) ⎞ ⎟ ⎟ ⎠ P 98 kPa = = 1.074 kg/m 3 RT (0.287 kPa ⋅ m 3 /kg ⋅ K )(45 + 273 K ) 2(235 N/m 2 ) ⎛⎜ 1 kg ⋅ m/s 2 1N 1.074 kg/m 3 ⎜⎝ ρ air Air 2 V 1 h=2.4 cm ⎞ ⎟ = 20.9 m/s ⎟ ⎠ Note that the flow velocity in a pipe or duct can be measured easily by a Pitot-static probe by inserting the Discussion probe into the pipe or duct parallel to flow, and reading the differential pressure height. Also note that this is the velocity at the location of the tube. Several readings at several locations in a cross-section may be required to determine the mean flow velocity. 5-26 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-51 Solution The water in an above the ground swimming pool is to be emptied by unplugging the orifice of a horizontal pipe attached to the bottom of the pool. The maximum discharge rate of water is to be determined. Assumptions 1 The orifice has a smooth entrance, and all frictional losses are negligible. 2 The flow is steady, incompressible, and irrotational with negligible frictional effects (so that the Bernoulli equation is applicable). Analysis We take point 1 at the free surface of the pool, and point 2 at the exit of pipe. We take the reference level at the pipe exit (z2 = 0). Noting that the fluid at both points is open to the atmosphere (and thus P1 = P2 = Patm) and that the fluid velocity at the free surface is very low (V1 ≅ 0), the Bernoulli equation between these two points simplifies to P1 V12 P V2 + + z1 = 2 + 2 + z2 ρg 2 g ρg 2 g → z1 = V22 2g → V2 = 2 gz1 The maximum discharge rate occurs when the water height in the pool is a maximum, which is the case at the beginning and thus z1 = h. Substituting, the maximum flow velocity and discharge rate become V2,max = 2 gh = 2(9.81 m/s 2 )(3 m) = 7.67 m/s Vmax = ApipeV2,max = πD 2 4 V2,max = π (0.03 m) 2 4 (7.67 m/s) = 0.00542 m 3 /s = 5.42 L/s 1 Swimming pool 3m 3 cm 8m 2 m 25 m Discussion The result above is obtained by disregarding all frictional effects. The actual flow rate will be less because of frictional effects during flow and the resulting pressure drop. Also, the flow rate will gradually decrease as the water level in the pipe decreases. 5-27 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-52 Solution The water in an above the ground swimming pool is to be emptied by unplugging the orifice of a horizontal pipe attached to the bottom of the pool. The time it will take to empty the tank is to be determined. Assumptions 1 The orifice has a smooth entrance, and all frictional losses are negligible. 2 The flow is steady, incompressible, and irrotational with negligible frictional effects (so that the Bernoulli equation is applicable). Analysis We take point 1 at the free surface of water in the pool, and point 2 at the exit of pipe. We take the reference level at the pipe exit (z2 = 0). Noting that the fluid at both points is open to the atmosphere (and thus P1 = P2 = Patm) and that the fluid velocity at the free surface is very low (V1 ≅ 0), the Bernoulli equation between these two points simplifies to P1 V12 P V2 + + z1 = 2 + 2 + z2 ρg 2 g ρg 2 g → z1 = → V22 2g V2 = 2 gz1 For generality, we express the water height in the pool at any time t by z, and the discharge velocity by V2 = 2 gz . Note that water surface in the pool moves down as the pool drains, and thus z is a variable whose value changes from h at the beginning to 0 when the pool is emptied completely. We denote the diameter of the orifice by D, and the diameter of the pool by Do. The flow rate of water from the pool is obtained by multiplying the discharge velocity by the orifice cross-sectional area, πD 2 V = AorificeV 2 = 2 gz 4 Then the amount of water that flows through the orifice during a differential time interval dt is πD 2 dV = Vdt = 2 gz dt (1) 4 which, from conservation of mass, must be equal to the decrease in the volume of water in the pool, πD 2 (2) dV = Atank (−dz ) = − 0 dz 4 where dz is the change in the water level in the pool during dt. (Note that dz is a negative quantity since the positive direction of z is upwards. Therefore, we used –dz to get a positive quantity for the amount of water discharged). Setting Eqs. (1) and (2) equal to each other and rearranging, πD 2 D02 D2 πD 2 1 −1 z 2 dz dz = − → 2 gz dt = − 0 dz dt = − 02 4 4 2 gz D D 2 2g The last relation can be integrated easily since the variables are separated. Letting tf be the discharge time and integrating it from t = 0 when z = h to t = tf when z = 0 (completely drained pool) gives ∫ tf dt = − t =0 2g ∫ D02 D 2 0 −1 / 2 z = z1 z dz → tf =- D02 D 2 2g 1 0 = z2 1 2 2 D02 D 2 2g h= D02 D2 2h g z1 Substituting, the draining time of the pool will be (8 m) 2 2(3 m) = 55,600 s = 15.4 h tf = 2 (0.03 m) 9.81 m/s 2 1 Swimming pool 3m This is the minimum discharging time Discussion since it is obtained by neglecting all friction; the actual discharging time will be longer. Note that the discharging time is inversely proportional to the square of the orifice diameter. Therefore, the discharging time can be reduced to one-fourth by doubling the diameter of the orifice. D0= 8 m D=3 cm 2 m 25 m 5-28 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-53 Solution The previous problem is reconsidered. The effect of the discharge pipe diameter on the time required to empty the pool completely as the diameter varies from 1 to 10 cm in increments of 1 cm is to be investigated. The EES Equations window is printed below, followed by the tabulated and plotted results. Analysis g=9.81 "m/s2" rho=1000 "kg/m3" h=2 "m" D=d_pipe/100 "m" D_pool=10 "m" V_initial=SQRT(2*g*h) "m/s" Ac=pi*D^2/4 V_dot=Ac*V_initial*1000 "m3/s" t=(D_pool/D)^2*SQRT(2*h/g)/3600 "hour" Discussion Discharge time t, h 177.4 44.3 19.7 11.1 7.1 4.9 3.6 2.8 2.2 1.8 160 140 120 t, hour Pipe diameter D, m 1 2 3 4 5 6 7 8 9 10 180 100 80 60 40 20 0 1 2 3 4 5 6 7 8 9 10 D, cm As you can see from the plot, the discharge time is drastically reduced by increasing the pipe diameter. 5-29 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-54 Solution Air flows upward at a specified rate through an inclined pipe whose diameter is reduced through a reducer. The differential height between fluid levels of the two arms of a water manometer attached across the reducer is to be determined. Assumptions 1 The flow through the duct is steady, incompressible and irrotational with negligible frictional effects (so that the Bernoulli equation is applicable). 2 Air is an ideal gas. 3 The effect of air column on the pressure change is negligible because of its low density. 4 The air flow is parallel to the entrance of each arm of the manometer, and thus no dynamic effects are involved. Properties We take the density of water to be ρ = 1000 kg/m3. The gas constant of air is R = 0.287 kPa⋅m3/kg⋅K. Analysis We take points 1 and 2 at the lower and upper connection points, respectively, of the two arms of the manometer, and take the lower connection point as the reference level. Noting that the effect of elevation on the pressure change of a gas is negligible, the application of the Bernoulli equation between points 1 and 2 gives P V2 P1 V12 + + z1 = 2 + 2 + z 2 ρg 2 g ρg 2 g → P1 − P2 = ρ air ρ air = V 22 − V12 2 2 Air 1 P 110 kPa = = 1.19 kg/m 3 3 RT (0.287 kPa ⋅ m /kg ⋅ K )(50 + 273 K ) V V 0.045 m3/s V1 = = = = 15.9 m/s 2 A1 πD1 / 4 π (0.06 m)2 / 4 h 0.045 m 3 /s V V V2 = = = = 35.8 m/s A2 πD 22 / 4 π (0.04 m) 2 / 4 Substituting, ⎞ (35.8 m/s) 2 − (15.9 m/s) 2 ⎛⎜ 1N ⎟ = 612 N/m 2 = 612 Pa The differential height of P1 − P2 = (1.19 kg/m 3 ) 2 ⎜ ⎟ 2 ⋅ 1 kg m/s ⎝ ⎠ water in the manometer corresponding to this pressure change is determined from ΔP = ρ water gh to be where h= ⎛ 1 kg ⋅ m/s 2 P1 − P2 612 N/m 2 ⎜ = 3 2 ⎜ ρ water g (1000 kg/m )(9.81 m/s ) ⎝ 1 N ⎞ ⎟ = 0.0624 m = 6.24 cm ⎟ ⎠ When the effect of air column on pressure change is considered, the pressure change becomes ρ (V 2 − V12 ) P1 − P2 = air 2 + ρ air g ( z 2 − z1 ) 2 ⎞ ⎡ (35.8 m/s) 2 − (15.9 m/s) 2 ⎤⎛ 1N ⎟ = (1.19 kg/m 3 ) ⎢ + (9.81 m/s 2 )(0.2 m)⎥⎜ 2 ⎟ ⎜ 2 ⎣⎢ ⎦⎥⎝ 1 kg ⋅ m/s ⎠ Discussion = (612 + 2) N/m 2 = 614 N/m 2 = 614 Pa This difference between the two results (612 and 614 Pa) is less than 1%. Therefore, the effect of air column on pressure change is, indeed, negligible as assumed. In other words, the pressure change of air in the duct is almost entirely due to velocity change, and the effect of elevation change is negligible. Also, if we were to account for the Δz of air flow, then it would be more proper to account for the Δz of air in the manometer by using ρwater - ρair instead of ρwater when calculating h. The additional air column in the manometer tends to cancel out the change in pressure due to the elevation difference in the flow in this case. 5-30 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-55E Solution Air is flowing through a venturi meter with known diameters and measured pressures. A relation for the flow rate is to be obtained, and its numerical value is to be determined. Assumptions 1The flow through the venturi is steady, incompressible, and irrotational with negligible frictional effects (so that the Bernoulli equation is applicable). 2 The effect of air column on the pressure change is negligible because of its low density, and thus the pressure can be assumed to be uniform at a given cross-section of the venturi meter (independent of elevation change). 3 The flow is horizontal (this assumption is usually unnecessary for gas flow.). The density of air is given to be ρ = 0.075 lbm/ft3. Properties Analysis We take point 1 at the main flow section and point 2 at the throat along the centerline of the venturi meter. Noting that z1 = z2, the application of the Bernoulli equation between points 1 and 2 gives P1 V12 P V2 + + z1 = 2 + 2 + z 2 ρg 2 g ρg 2 g → P1 − P2 = ρ V 22 − V12 2 (1) The flow is assumed to be incompressible and thus the density is constant. Then the conservation of mass relation for this single stream steady flow device can be expressed as V1 = V2 = V → A1V1 = A2V 2 = V Substituting into Eq. (1), (V / A2 ) 2 − (V / A1 ) 2 ρV 2 P1 − P2 = ρ = 2 2 A22 → V1 = V A1 and ⎛ A22 ⎞ ⎟ ⎜1 − ⎜ A2 ⎟ 1 ⎠ ⎝ Solving for V gives the desired relation for the flow rate, V = A2 2( P1 − P2 ) ρ [1 − ( A2 / A1 ) 2 ] V2 = V A2 (2) 12.2 psia 11.8 psia 2 Air 1 2.6 in 1.8 in (3) The flow rate for the given case can be determined by substituting the given values into this relation to be V = πD 22 4 2( P1 − P2 ) ρ [1 − ( D 2 / D1 ) ] 4 = π (1.8 / 12 ft) 2 = 4.48 ft /s 4 ⎛ 144 lbf/ft 2 ⎜ (0.075 lbm/ft 3 )[1 - (1.8/2.6) 4 ] ⎜⎝ 1 psi 2(12.2 − 11.8) psi ⎞⎛ 32.2 lbm ⋅ ft/s 2 ⎟⎜ ⎟⎜ 1 lbf ⎠⎝ ⎞ ⎟ ⎟ ⎠ 3 Venturi meters are commonly used as flow meters to measure the flow rate of gases and liquids by simply Discussion measuring the pressure difference P1 - P2 by a manometer or pressure transducers. The actual flow rate will be less than the value obtained from Eq. (3) because of the friction losses along the wall surfaces in actual flow. But this difference can be as little as 1% in a well-designed venturi meter. The effects of deviation from the idealized Bernoulli flow can be accounted for by expressing Eq. (3) as V = C c A2 2( P1 − P2 ) ρ [1 − ( A2 / A1 ) 2 ] where Cc is the venturi discharge coefficient whose value is less than 1 (it is as large as 0.99 for well-designed venturi meters in certain ranges of flow). For Re > 105, the value of venturi discharge coefficient is usually greater than 0.96. 5-31 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-56 Solution The gage pressure in the water mains of a city at a particular location is given. It is to be determined if this main can serve water to neighborhoods that are at a given elevation relative to this location. Assumptions Properties Water is incompressible and thus its density is constant. We take the density of water to be ρ = 1000 kg/m3. Noting that the gage pressure at a dept of h in a fluid is given by Pgage = ρ water gh , the height of a fluid Analysis column corresponding to a gage pressure of 350 kPa is determined to be h= Pgage ρ water g = ⎛ 1 kg ⋅ m/s 2 ⎜ 1N (1000 kg/m 3 )(9.81 m/s 2 ) ⎜⎝ 350,000 N/m 2 ⎞ ⎟ = 35.7 m ⎟ ⎠ Water Main, 350 kPa which is less than 50 m. Therefore, this main cannot serve water to neighborhoods that are 50 m above this location. Discussion Note that h must be much greater than 50 m for water to have enough pressure to serve the water needs of the neighborhood. 5-57 Solution Water discharges to the atmosphere from the orifice at the bottom of a pressurized tank. Assuming frictionless flow, the discharge rate of water from the tank is to be determined. 1 Assumptions 1 The orifice has a smooth entrance, and thus the frictional losses are Air, 250 kPa negligible. 2 The flow is steady, incompressible, and irrotational with negligible frictional effects (so that the Bernoulli equation is applicable). Properties We take the density of water to be 1000 kg/m3. Analysis We take point 1 at the free surface of the tank, and point 2 at the exit of orifice, which is also taken to be the reference level (z2 = 0). Noting that the fluid velocity at the free surface is very low (V1 ≅ 0) and water discharges into the atmosphere (and thus P2 = Patm), the Bernoulli equation simplifies to P1 V12 P V2 + + z1 = 2 + 2 + z2 ρg 2 g ρg 2 g V22 P1 − P2 = + z1 2g ρg → 2( P1 − P2 ) ρ + 2 gz1 = 2(250 − 100) kPa ⎛⎜ 1000 N/m 2 1000 kg/m 3 ⎜⎝ 1 kPa 2.5 m 2 10 cm Solving for V2 and substituting, the discharge velocity is determined to V2 = Water Tank ⎞⎛ 1 kg ⋅ m/s 2 ⎟⎜ ⎟⎜ 1N ⎠⎝ ⎞ ⎟ + 2(9.81 m/s 2 )(2.5 m) = 18.7 m/s ⎟ ⎠ Then the initial rate of discharge of water becomes V = AorificeV2 = πD 2 4 V2 = π (0.10 m) 2 4 (18.7 m/s) = 0.147 m 3 /s Discussion Note that this is the maximum flow rate since the frictional effects are ignored. Also, the velocity and the flow rate will decrease as the water level in the tank decreases. 5-32 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-58 Solution The previous problem is reconsidered. The effect of water height in the tank on the discharge velocity as the water height varies from 0 to 5 m in increments of 0.5 m is to be investigated. The EES Equations window is printed below, followed by the tabulated and plotted results. Analysis g=9.81 "m/s2" rho=1000 "kg/m3" d=0.10 "m" P1=300 "kPa" P_atm=100 "kPa" V=SQRT(2*(P1-P_atm)*1000/rho+2*g*h) Ac=pi*D^2/4 V_dot=Ac*V 22.5 22 V, m/s 0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00 4.50 5.00 20.0 20.2 20.5 20.7 21.0 21.2 21.4 21.6 21.9 22.1 22.3 Discussion V , m3/s 0.157 0.159 0.161 0.163 0.165 0.166 0.168 0.170 0.172 0.174 0.175 21.5 V, m/s h, m 21 20.5 20 0 1 2 3 4 h, m Velocity appears to change nearly linearly with h in this range of data, but the relationship is not linear. 5-33 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 5 Chapter 5 Mass, Bernoulli, and Energy Equations 5-59 Solution A hand-held bicycle pump with a liquid reservoir is used as an atomizer by forcing air at a high velocity through a small hole. We are to explain how the liquid gets sucked up the tube. Assumptions 1 The flows of air and water are steady and nearly incompressible. 2 Air is an ideal gas. 3 The liquid reservoir is open to the atmosphere. 4 The device is held horizontally. 5 The water velocity through the tube is low (the water in the tube is nearly hydrostatic). Analysis At first glance, we are tempted to use the Bernoulli equation, thinking that the pressure in the air jet would be lower than atmospheric due to its high speed. However, as stated in the problem statement, the pressure through an incompressible jet exposed to the atmosphere is nearly atmospheric pressure everywhere. Thus, in the absence of the tube, the pressure in the air jet just above the tube would be nearly atmospheric. Meanwhile, the pressure at the liquid surface is also atmospheric. Applying hydrostatics from the liquid surface to the top of the tube reveals that the pressure at the top of the tube must be lower than atmospheric pressure by more than ρgh in order to suck the liquid up the tube. So, what causes the liquid to rise? It turns out that the answer has to do with streamline curvature. As the close-up sketch illustrates, the air streamlines must curve around the top of the tube. Since pressure decreases towards the center of curvature in a flow with curved streamlines, the pressure at the top of the tube must be less than atmospheric. At high enough air jet speed, the pressure is low enough not only to suck the liquid to the top of the tube, but also to break up the liquid at the top of the tube into small droplets, thereby “atomizing” the liquid into a spray of liquid droplets. Air Liquid rising Air jet streamlines Tube Atomized droplets Liquid Discussion If the geometry of the top of the tube were known, we could approximate the flow as irrotational and apply the techniques of potential flow analysis (Chap. 10) to estimate the pressure at the top of the tube. 5-34 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-60 Solution The water height in an airtight pressurized tank is given. A hose pointing straight up is connected to the bottom of the tank. The maximum height to which the water stream could rise is to be determined. 2 Assumptions 1 The flow is steady, incompressible, and irrotational with negligible frictional effects (so that the Bernoulli equation is applicable). 2 The friction between the water and air is negligible. Properties 1 We take the density of water to be 1000 kg/m . h Analysis We take point 1 at the free surface of water in the tank, and point 2 at the top of the water trajectory. Also, we take the reference level at the bottom of the tank. At the top of the water trajectory V2 = 0, and atmospheric pressure pertains. Noting that z1 = 20 m, P1,gage = 2 atm, P2 = Patm, and that the fluid velocity at the free surface of the tank is very low (V1 ≅ 0), the Bernoulli equation between these two points simplifies to P1 V12 P V2 + + z1 = 2 + 2 + z 2 ρg 2 g ρg 2 g Substituting, z2 = 3 atm 3 → ⎛ 101,325 N/m 2 ⎜ 1 atm (1000 kg/m 3 )(9.81 m/s 2 ) ⎜⎝ 3 atm P P1 + z1 = atm + z 2 → ρg ρg ⎞⎛ 1 kg ⋅ m/s 2 ⎟⎜ ⎟⎜ 1N ⎠⎝ z2 = 15 m P1,gage P1 − Patm + z1 = + z1 ρg ρg ⎞ ⎟ + 15 = 46.0 m ⎟ ⎠ Therefore, the water jet can rise as high as 46.0 m into the sky from the ground. Discussion The result obtained by the Bernoulli equation represents the upper limit, and should be interpreted accordingly. It tells us that the water cannot possibly rise more than 46.0 m (giving us an upper limit), and in all likelihood, the rise will be much less because of frictional losses. 5-35 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-61 Solution A Pitot-static probe equipped with a water manometer is held parallel to air flow, and the differential height of the water column is measured. The flow velocity of air is to be determined. Assumptions 1The flow of air is steady, incompressible, and irrotational with negligible frictional effects (so that the Bernoulli equation is applicable). 2 The effect of air column on the pressure change is negligible because of its low density, and thus the air column in the manometer can be ignored. Properties We take the density of water to be ρ = 1000 kg/m3. The density of air is given to be 1.25 kg/m3. Analysis We take point 1 on the side of the probe where the entrance is parallel to flow and is connected to the static arm of the Pitot-static probe, and point 2 at the tip of the probe where the entrance is normal to flow and is connected to the dynamic arm of the Pitot-static probe. Noting that point 2 is a stagnation point and thus V2 = 0 and z1 = z2, the application of the Bernoulli equation between points 1 and 2 gives P V2 P1 V12 + + z1 = 2 + 2 + z2 ρg 2 g ρg 2 g → P1 V12 P2 + = ρg 2 g ρg → V1 = 2( P2 − P1 ) ρ air (1) The pressure rise at the tip of the Pitot-static probe is simply the pressure change indicated by the differential water column of the manometer, P2 − P1 = ρ water gh (2) Combining Eqs. (1) and (2) and substituting, the flow velocity is determined to be V1 = 2 ρ water gh ρ air = 2(1000 kg/m 3 )(9.81 m/s 2 )(0.073 m) 1.25 kg/m 3 Air 1 = 33.8 m/s Pitot tube 7.3 cm 2 Manometer Discussion Note that flow velocity in a pipe or duct can be measured easily by a Pitot-static probe by inserting the probe into the pipe or duct parallel to flow, and reading the differential height. Also note that this is the velocity at the location of the tube. Several readings at several locations in a cross-section may be required to determine the mean flow velocity. 5-36 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-62E Solution A Pitot-static probe equipped with a differential pressure gage is used to measure the air velocity in a duct. For a given differential pressure reading, the flow velocity of air is to be determined. Assumptions The flow of air is steady, incompressible, and irrotational with negligible frictional effects (so that the Bernoulli equation is applicable). ΔP = 0.15 psi Properties The gas constant of air is R = 0.3704 psia⋅ft3/lbm⋅R. Analysis We take point 1 on the side of the probe where the entrance is parallel to flow and is connected to the static arm of the Pitot-static probe, and point 2 at the tip of the probe where the entrance is normal to flow and is connected to the dynamic arm of the Pitot-static probe. Noting that point 2 is a stagnation point and thus V2 = 0 and z1 = z2, the application of the Bernoulli equation between points 1 and 2 gives P V2 P1 V12 + + z1 = 2 + 2 + z2 ρg 2 g ρg 2 g → P1 V12 P2 + = ρg 2 g ρg → Air 1 70°F 13.4 psia V1 = Pitot tube 2 2( P2 − P1 ) ρ where ρ= 13.4 psia P = = 0.0683 lbm/ft 3 RT (0.3704 psia ⋅ ft 3 /lbm ⋅ R )(70 + 460 R ) Substituting the given values, the flow velocity is determined to be V1 = ⎛ 144 lbf/ft 2 ⎜ 0.0683 lbm/ft 3 ⎜⎝ 1 psi 2(0.15 psi ) ⎞⎛ 32.2 lbm ⋅ ft/s 2 ⎟⎜ ⎟⎜ 1 lbf ⎠⎝ ⎞ ⎟ = 143 ft/s ⎟ ⎠ Discussion Note that flow velocity in a pipe or duct can be measured easily by a Pitot-static probe by inserting the probe into the pipe or duct parallel to flow, and reading the pressure differential. Also note that this is the velocity at the location of the tube. Several readings at several locations in a cross-section may be required to determine the mean flow velocity. 5-37 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-63 Solution A water pipe bursts as a result of freezing, and water shoots up into the air a certain height. The gage pressure of water in the pipe is to be determined. Assumptions 1 The flow is steady, incompressible, and irrotational with negligible frictional effects (so that the Bernoulli equation is applicable). 2 The water pressure in the pipe at the burst section is equal to the water main pressure. 3 Friction between the water and air is negligible. 4 The irreversibilities that may occur at the burst section of the pipe due to abrupt expansion are negligible. Properties We take the density of water to be 1000 kg/m3. 2 Analysis This problem involves the conversion of flow, kinetic, and potential energies to each other without involving any pumps, turbines, and wasteful components with large frictional losses, and thus it is suitable for the use of the Bernoulli equation. The water height will be maximum under the stated assumptions. The velocity inside the hose is relatively low (V1 ≅ 0) and we take the burst section of the pipe as the reference level (z1 = 0). At the top of the water trajectory V2 = 0, and atmospheric pressure pertains. Then the Bernoulli equation simplifies to P V2 P1 V12 + + z1 = 2 + 2 + z2 ρg 2 g ρg 2 g → ρg P1 = Patm + z2 ρg → Solving for P1,gage and substituting, 42 m 1 Water Pipe P1 − Patm = z2 → ρg ⎛ 1 kPa P1,gage = ρgz 2 = (1000 kg/m 3 )(9.81 m/s 2 )(42 m)⎜ ⎝ 1 kN/m 2 P1,gage ρg 1 kN ⎞⎛⎜ ⎟⎜ 1000 kg ⋅ m/s 2 ⎠⎝ = z2 ⎞ ⎟ = 412 kPa ⎟ ⎠ Therefore, the pressure in the main must be at least 412 kPa above the atmospheric pressure. Discussion The result obtained by the Bernoulli equation represents a limit, since frictional losses are neglected, and should be interpreted accordingly. It tells us that the water pressure (gage) cannot possibly be less than 334 kPa (giving us a lower limit), and in all likelihood, the pressure will be much higher. 5-38 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-64 Solution We are to calculate the velocity measured by a Pitot-static probe for a range of velocities, and compare the analytical velocity to the actual velocity. Assumptions 1 The flow is steady, incompressible, and axisymmetric. 2 For the analytical analysis, we neglect irreversibilities such as friction so that the Bernoulli approximation can be used. Analysis 2 ( Pstag − P ) (a) We apply the equation given in the text for a Pitot-static probe, V= ρ In the CFD analysis, the stagnation pressure is reported as P1, and the static pressure as P2. For example, at an inlet velocity of 10 m/s, V= 2 ( P1 − P2 ) ρ = 2 ( 60.738 − 0.0244 ) N/m 2 ⎛ kg ⋅ m/s 2 ⎞ m ⎜ ⎟ = 9.956 3 1.225 kg/m N s ⎝ ⎠ We repeat the calculations for a range of inlet velocities, and show the results in the table below. Note that pressures P1 and P2 are outputs from the CFD code. The agreement is excellent, with less than 1% error at any particular velocity. The Bernoulli equation is apparently a good approximation for the flow around a Pitot-static probe, and that is why it makes for a simple but effective way to measure the fluid speed. Discussion When using a Pitot-static probe, it is critical that it be aligned properly. 5-39 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-65 Solution We are to calculate the velocity measured by a Pitot-static probe for a range of Reynolds numbers, and compare the analytical velocity to the actual velocity. Assumptions 1 The flow is steady, incompressible, and axisymmetric. 2 For the analytical analysis, we neglect irreversibilities such as friction so that the Bernoulli approximation can be used. Analysis 2 ( Pstag − P ) (a) We apply the equation given in the text for a Pitot-static probe, V= ρ In the CFD analysis, the stagnation pressure is reported as P1, and the static pressure as P2. For example, at Re = 1000, the inlet velocity is 0.055995 m/s, and using the two pressures we get V= 2 ( P1 − P2 ) ρ = 2 ( 0.0025537 − 0.0001738 ) N/m 2 ⎛ kg ⋅ m/s 2 ⎞ m ⎜ ⎟ = 0.06233 3 1.225 kg/m N s ⎝ ⎠ We repeat the calculations for a range of Reynolds numbers, and show the results in the table below. Note that pressures P1 and P2 are outputs from the CFD code. The agreement is horrible at the low Reynolds numbers, with the percentage error above 100% until Reynolds numbers above about 400. The Bernoulli equation is apparently a very poor approximation for the flow around a Pitot-static probe at low Reynolds numbers, but it does a good job at high Reynolds numbers. Why? The Bernoulli equation accounts for inertial effects, but not viscous effects. As will be discussed in Chapter 7, low Re means viscous effects are more important than inertial effects, and hence the Bernoulli approximation breaks down at very low Reynolds numbers. Discussion When using a Pitot-static probe at low speeds, you must be careful that the Reynolds number is not too small. Large Pitot-static tubes are available for low speeds, and there are also some correction factors that can be used. 5-40 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-66 Solution We are to generate expressions for inlet pressure and throat pressure and velocity in a converging-diverging duct for the case where Poutlet and Vinlet are known and irreversibilities are ignored. Assumptions 1 The flow is steady, incompressible, and two-dimensional. 2 We neglect irreversibilities such as friction. 3 4 The duct is horizontal – elevation differences do not play a role in this analysis. (a) We apply conservation of mass from the inlet to the outlet: Analysis ρVinlet Ainlet = ρVoutlet Aoutlet → Voutlet = Vinlet Ainlet Aoutlet where density has dropped out because of the incompressible flow approximation. We do a similar analysis at the throat. Thus, the average inlet velocity and average throat velocity are Voutlet = Vinlet Vthroat = Vinlet Ainlet Aoutlet Ainlet Athroat To estimate the pressure at inlet, we neglect irreversibilities and apply the Bernoulli equation along a streamline from the inlet to the outlet, Pinlet + 1 1 ρVinlet 2 + ρ gzinlet = Poutlet + ρVoutlet 2 + ρ gzoutlet 2 2 → Pinlet = Poutlet + 1 ρ (Voutlet 2 − Vinlet 2 ) 2 2 2 ⎞ 1 ⎛ ⎞ Ainlet ⎞ 1 ⎛⎛ 2 2 ⎛ Ainlet ⎞ ⎜ ⎟ ⎜ + ρ ⎜ Vinlet − Vinlet = ρVinlet ⎜ − 1⎟ ⎟ ⎟ ⎟ 2 ⎜ ⎝ Athroat ⎠ ⎟ 2 ⎜⎝ Aoutlet ⎠ ⎝ ⎠ ⎝ ⎠ and when we substitute the outlet velocity from conservation of mass we get Pinlet = Poutlet In like manner, we calculate the average pressure at any other axial location where the cross-sectional area is known. At the throat, for example, Pthroat = Poutlet 2 2 Ainlet ⎞ ⎛ Ainlet ⎞ ⎞ 1 1 ⎛⎛ 2 2 ⎜ + ρ (Voutlet − Vthroat ) = Poutlet + ρ ⎜ Vinlet ⎟ − ⎜ Vinlet ⎟ ⎟ Aoutlet ⎠ ⎝ Athroat ⎠ ⎟ 2 2 ⎜⎝ ⎝ ⎠ or, combining some terms, Pthroat = Poutlet ⎛ 1 ⎞2 ⎛ 1 ⎞2 ⎞ 1 2 ⎛ ⎟ + ρ (Vinlet Ainlet ) ⎜ ⎜ − ⎜ ⎝ Aoutlet ⎟⎠ ⎜⎝ Athroat ⎟⎠ ⎟ 2 ⎝ ⎠ (b) In this analysis, we have not accounted for any irreversibilities, such as friction, but in any real flow, friction would lead to higher pressure drop in the duct and thus the inlet pressure would have to be higher than predicted in order to overcome the additional losses due to friction. Discussion We must keep in mind that the Bernoulli equation is only an approximation. In Chap. 8 we learn how to approximate the additional pressure drop due to friction along the walls of a duct or pipe. 5-41 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-67 Solution We are to calculate and plot throat velocity and pressure drop from inlet to throat for converging-diverging duct flow at a given inlet velocity, both analytically and using CFD, and we are to compare the results. Assumptions 1 The flow is steady, incompressible, and two-dimensional. 2 For the analytical analysis, we neglect irreversibilities such as friction. 3 In the CFD analysis, the width (into the page) is unity, i.e., 1 m. 4 The duct is horizontal – elevation differences do not play a role in this analysis. Ainlet m ⎛ 0.043 m 2 ⎞ m =1 ⎜ = 2.65432 2 ⎟ Athroat s ⎝ 0.0162 m ⎠ s (a) We apply conservation of mass from the inlet to the throat: Analysis ρVinlet Ainlet = ρVthroat Athroat → Vthroat = Vinlet where density has dropped out because of the incompressible flow approximation. To estimate the pressure drop, we neglect irreversibilities and apply the Bernoulli equation along a streamline from the inlet to the throat, Pinlet + 1 1 ρVinlet 2 + ρ gzinlet = Pthroat + ρVthroat 2 + ρ gzthroat 2 2 → Pinlet − Pthroat = 1 ρ (Vthroat 2 − Vinlet 2 ) 2 2 2 ⎞ 1 ⎛ ⎞ Ainlet ⎞ 1 ⎛⎛ 2 2 ⎛ Ainlet ⎞ ⎟ ⎜ ⎟ V V − Pthroat = ρ ⎜ ⎜ Vinlet − = − ρ 1 ⎟ ⎜ ⎟ inlet ⎟ 2 inlet ⎜ ⎝ Athroat ⎠ ⎟ Athroat ⎠ 2 ⎜⎝ ⎝ ⎠ ⎝ ⎠ 2 2⎛ ⎞⎛ ⎞ N 1⎛ kg ⎞⎛ m ⎞ ⎛ 0.043 m 2 ⎞ N − 1⎟ ⎜ = 3017.269 2 = ⎜ 998.2 3 ⎟⎜ 1 ⎟ ⎜ ⎜ 2 ⎟ 2 ⎟ ⎜ ⎟ m 2⎝ m ⎠⎝ s ⎠ ⎝ 0.0162 m ⎠ kg ⋅ m/s ⎠ ⎝ ⎠⎝ and when we substitute the throat velocity from conservation of mass we get ΔP = Pinlet Thus, at an inlet velocity of 1 m/s, we predict the throat velocity and pressure drop to be 2.65 m/s and 3020 N/m2, respectively, to three significant digits. (b) We repeat the calculations for a range of inlet velocities, and show the results in the table below. (c) We run FlowLab for the same range of inlet velocities and plot ΔP as a function of Vinlet. On the same plot, we show the analytical predictions. The agreement is reasonable, with errors around 5%. Friction leads to higher pressure drop from the inlet to the throat; thus the CFD results (which account for irreversibilities) yield a higher ΔP. Discussion The Bernoulli equation is apparently a good approximation for the flow in the converging part of the duct. 5-42 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations Energy Equation 5-68C Solution We are to determine if frictional effects are negligible in the steady adiabatic flow of an incompressible fluid if the temperature remains constant. Analysis Yes, the frictional effects are negligible if the fluid temperature remains constant during steady, incompressible flow since any irreversibility such as friction would cause the entropy and thus temperature of the fluid to increase during adiabatic flow. Discussion Thus, this scenario would never occur in real life since all fluid flows have frictional effects. 5-69C Solution fluid. We are to analyze whether temperature can decrease during steady adiabatic flow of an incompressible Analysis It is impossible for the fluid temperature to decrease during steady, incompressible, adiabatic flow of an incompressible fluid, since this would require the entropy of an adiabatic system to decrease, which would be a violation of the 2nd law of thermodynamics. Discussion The entropy of a fluid can decrease, but only if we remove heat. 5-70C Solution We are to define and discuss irreversible head loss. Analysis Irreversible head loss is the loss of mechanical energy due to irreversible processes (such as friction) in piping expressed as an equivalent column height of fluid, i.e., head. Irreversible head loss is related to the mechanical e mech loss, piping E mech loss, piping = . energy loss in piping by h L = g m g Discussion hL is always positive. It can never be negative, since this would violate the second law of thermodynamics. 5-71C Solution We are to define and discuss useful pump head. Useful pump head is the useful power input to the pump expressed as an equivalent column height of w pump, u W pump, u = fluid. It is related to the useful pumping power input by hpump = . g m g Analysis Discussion Part of the power supplied to the pump is not useful, but rather is wasted because of irreversible losses in the pump. This is the reason that pumps have a pump efficiency that is always less than one. 5-43 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-72C Solution We are to define and discuss the kinetic energy correction factor. Analysis The kinetic energy correction factor is a correction factor to account for the fact that kinetic energy using average velocity is not the same as the actual kinetic energy using the actual velocity profile (the square of a sum is not equal to the sum of the squares of its components). The effect of kinetic energy factor is usually negligible, especially for turbulent pipe flows. However, for laminar pipe flows, the effect of α is sometimes significant. Discussion Even though the effect of ignoring α is usually insignificant, it is wise to keep α in our analyses to increase accuracy and so that we do not forget about it in situations where it is significant, such as in some laminar pipe flows. 5-73C Solution We are to analyze the cause of some strange behavior of a water jet. Analysis The problem does not state whether the water in the tank is open to the atmosphere or not. Let’s assume that the water surface is exposed to atmospheric pressure. By the Bernoulli equation, the maximum theoretical height to which the water stream could rise is the tank water level, which is 20 meters above the ground. Since the water rises above the tank level, the tank cover must be airtight, containing pressurized air above the water surface. In other words, the water in the tank is not exposed to atmospheric pressure. Discussion Alternatively, a pump would have to pressurize the water somewhere in the hose, but this is not allowed, based on the problem statement (only a hose is added). 5-74C Solution We are to analyze a suggestion regarding a garden hose. Analysis Yes. When water discharges from the hose at waist level, the head corresponding to the waist-knee vertical distance is wasted. When recovered, this elevation head is converted to velocity head, increasing the discharge velocity (and thus the flow rate) of water and thus reducing the filling time. Discussion If you are still not convinced, imagine holding the hose outlet really high up. If the outlet elevation is greater than the upstream supply head, no water will flow at all. If you are concerned about head losses in the hose, yes, they will increase as the volume flow rate increases, but not enough to change our answer. 5-75C Solution We are to analyze discharge of water from a tank under different conditions. Analysis (a) Yes, the discharge velocity from the bottom valve will be higher since velocity is proportional to the square root of the vertical distance between the hole and the free surface. (b) No, the discharge rates of water will be the same since the total available head to drive the flow (elevation difference between the ground and the free surface of water in the tank) is the same for both cases. Discussion Our answer to Part (b) does not change even if we consider head losses in the hose, because the hose is the same length in either case. Same hose, same length, same flow rate…yields the same head loss through the hose. Note: We are ignoring any effects of bends or curves in the hose – assume both cases have the same curves. 5-44 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-76E Solution In a hydroelectric power plant, the elevation difference, the power generation, and the overall turbinegenerator efficiency are given. The minimum flow rate required is to be determined. Assumptions 1 The flow is steady and incompressible. 2 The water levels at the reservoir and the discharge site remain constant. 3 We assume the flow to be frictionless since the minimum flow rate is to be determined, E mech,loss = 0. We take the density of water to be ρ = 62.4 Properties lbm/ft3. Analysis We take point 1 at the free surface of the reservoir and point 2 at the free surface of the discharge water stream, which is also taken as the reference level (z2 = 0). Also, both 1 and 2 are open to the atmosphere (P1 = P2 = Patm), the velocities are negligible at both points (V1 = V2 = 0), and frictional losses are disregarded. Then the energy equation in terms of heads for steady incompressible flow through a control volume between these two points that includes the turbine and the pipes reduces to 1 Water 240 ft 2 Turbine Generator P1 V2 P V2 + α 1 1 + z1 + hpump, u = 2 + α 2 2 + z 2 + h turbine, e + h L → hturbine, e = z1 2g 2g ρg ρg Substituting and noting that W turbine, elect = η turbine -gen m ghturbine, e , the extracted turbine head and the mass and volume flow rates of water are determined to be hturbine, e = z1 = 240 ft m = V = W turbine,elect η turbine-gen ghturbine m ρ = 370 lbm/s 62.4 lbm/ft 3 = ⎛ 25,037 ft 2 /s 2 ⎜ 0.83(32.2 ft/s 2 )(240 ft) ⎜⎝ 1 Btu/lbm 100 kW ⎞⎛ 0.9478 Btu/s ⎞ ⎟⎜ ⎟ = 370 lbm/s ⎟⎝ 1 kW ⎠ ⎠ = 5.93 ft 3 /s Therefore, the flow rate of water must be at least 5.93 ft3/s to generate the desired electric power while overcoming friction losses in pipes. Discussion In an actual system, the flow rate of water will be more because of frictional losses in pipes. 5-45 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-77E Solution In a hydroelectric power plant, the elevation difference, the head loss, the power generation, and the overall turbine-generator efficiency are given. The flow rate required is to be determined. Assumptions 1 The flow is steady and incompressible. 2 The water levels at the reservoir and the discharge site remain constant. We take the density of water to be ρ = 62.4 Properties lbm/ft3. Analysis We take point 1 at the free surface of the reservoir and point 2 at the free surface of the discharge water stream, which is also taken as the reference level (z2 = 0). Also, both 1 and 2 are open to the atmosphere (P1 = P2 = Patm), the velocities are negligible at both points (V1 = V2 = 0). Then the energy equation in terms of heads for steady incompressible flow through a control volume between these two points that includes the turbine and the pipes reduces to 1 Water 240 ft 2 Turbine Generator P1 V2 P V2 + α 1 1 + z1 + hpump, u = 2 + α 2 2 + z 2 + hturbine, e + h L → hturbine, e = z1 − h L ρg ρg 2g 2g Substituting and noting that W turbine, elect = η turbine -gen m ghturbine, e , the extracted turbine head and the mass and volume flow rates of water are determined to be hturbine, e = z1 − h L = 240 − 36 = 204 ft m = V = W turbine,elect η turbine-gen ghturbine m ρ = 435 lbm/s 62.4 lbm/ft 3 = ⎛ 25,037 ft 2 /s 2 ⎜ 0.83(32.2 ft/s )(204 ft) ⎜⎝ 1 Btu/lbm 100 kW 2 ⎞⎛ 0.9478 Btu/s ⎞ ⎟⎜ ⎟ = 435 lbm/s ⎟⎝ 1 kW ⎠ ⎠ = 6.98 ft 3 /s Therefore, the flow rate of water must be at least 6.98 ft3/s to generate the desired electric power while overcoming friction losses in pipes. Discussion Note that the effect of frictional losses in the pipes is to increase the required flow rate of water to generate a specified amount of electric power. 5-46 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-78 [Also solved using EES on enclosed DVD] Solution A fan is to ventilate a bathroom by replacing the entire volume of air once every 10 minutes while air velocity remains below a specified value. The wattage of the fan-motor unit, the diameter of the fan casing, and the pressure difference across the fan are to be determined. Assumptions 1 The flow is steady and incompressible. 2 Frictional losses along the flow (other than those due to the fan-motor inefficiency) are negligible. 3 The fan unit is horizontal so that z = constant along the flow (or, the elevation effects are negligible because of the low density of air). 4 The effect of the kinetic energy correction factors is negligible, α = 1. 1 Air 8 m/s D 2 The density of air is given to be 1.25 kg/m3. Properties Analysis (a) The volume of air in the bathroom is V = 2 m × 3 m × 3 m = 18 m3. Then the volume and mass flow rates of air through the casing must be V = V Δt = 18 m 3 = 0.03 m 3 /s 10 × 60 s m = ρV = (1.25 kg/m 3 )(0.03 m 3 /s) = 0.0375 kg/s We take points 1 and 2 on the inlet and exit sides of the fan, respectively. Point 1 is sufficiently far from the fan so that P 1 = Patm and the flow velocity is negligible (V1 = 0). Also, P2 = Patm. Then the energy equation for this control volume between the points 1 and 2 reduces to ⎞ ⎛P ⎞ ⎛P V2 V2 m ⎜ 1 + α 1 1 + gz1 ⎟ + W pump = m ⎜ 2 + α 2 2 + gz 2 ⎟ + W turbine + E mech,loss → ⎟ ⎜ ⎟ ⎜ρ 2 2 ⎠ ⎝ ρ ⎠ ⎝ V2 W fan, u = m α 2 2 2 since E mech, loss = E mech loss, pump in this case and W pump, u = W pump − E mech loss, pump . Substituting, V2 (8 m/s) 2 W fan, u = m α 2 2 = (0.0375 kg/s)(1.0) 2 2 and W fan, elect = W fan, u η fan-motor = ⎛ 1N ⎜ ⎜ 1 kg ⋅ m/s 2 ⎝ ⎞⎛ 1 W ⎞ ⎟⎜ ⎟⎝ 1 N ⋅ m/s ⎟⎠ = 1.2 W ⎠ 1.2 W = 2.4 W 0.5 Therefore, the electric power rating of the fan/motor unit must be 2.4 W. (b) For air mean velocity to remain below the specified value, the diameter of the fan casing should be 4(0.03 m 3 /s) 4V → D2 = = = 0.069 m = 6.9 cm V = A2V 2 = (πD 22 / 4 )V 2 πV 2 π (8 m/s) (c) To determine the pressure difference across the fan unit, we take points 3 and 4 to be on the two sides of the fan on a horizontal line. Noting that z3 = z4 and V3 = V4 since the fan is a narrow cross-section and neglecting flow loses (other than the loses of the fan unit, which is accounted for by the efficiency), the energy equation for the fan section reduces to W fan, u W fan, u P P → P4 − P3 = = m 3 + W fan, u = m 4 m / ρ ρ ρ V ⎛ 1 N ⋅ m/s ⎞ 2 ⎜ ⎟ = 40 N/m = 40 Pa 0.03 m /s ⎝ 1 W ⎠ Therefore, the fan will raise the pressure of air by 40 Pa before discharging it. Substituting , P4 − P3 = 1.2 W 3 Discussion Note that only half of the electric energy consumed by the fan-motor unit is converted to the mechanical energy of air while the remaining half is converted to heat because of imperfections. 5-47 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-79 Solution Water is pumped from a large lake to a higher reservoir. The head loss of the piping system is given. The mechanical efficiency of the pump is to be determined. 2 Assumptions 1 The flow is steady and incompressible. 2 The elevation difference between the lake and the reservoir is constant. Properties We take the density of water to be ρ = 1000 kg/m3. Reservoir 25 m Analysis We choose points 1 and 2 at the free surfaces of the lake and the reservoir, respectively, and take the surface of the lake as the reference level (z1 = 0). Both points are open to the atmosphere (P1 = P2 = Patm) and the velocities at both locations are negligible (V1 = V2 = 0). Then the energy equation for steady incompressible flow through a control volume between these two points that includes the pump and the pipes reduces to Pump 1 Lake ⎛P ⎞ ⎛P ⎞ V2 V2 m ⎜⎜ 1 + α 1 1 + gz1 ⎟⎟ + W pump = m ⎜⎜ 2 + α 2 2 + gz 2 ⎟⎟ + W turbine + E mech,loss → W pump, u = m gz 2 + E mech loss, piping 2 2 ⎝ρ ⎠ ⎝ ρ ⎠ since, in the absence of a turbine, E mech, loss = E mech loss, pump + E mech loss, piping and W pump, u = W pump − E mech loss, pump . Noting that E mech loss, piping = m gh L , the useful pump power is W pump,u = m gz 2 + m ghL = ρVg ( z 2 + hL ) ⎛ 1 kN = (1000 kg/m 3 )(0.025 m 3 /s)(9.81 m/s 2 )[(25 + 5) m]⎜⎜ 2 ⎝ 1000 kg ⋅ m/s = 7.36 kN ⋅ m/s = 7.36 kW ⎞ ⎟ ⎟ ⎠ Then the mechanical efficiency of the pump becomes η pump = W pump, u 7.36 kW = = 0.736 = 73.6% 10 kW W shaft Discussion A more practical measure of performance of the pump is the overall efficiency, which can be obtained by multiplying the pump efficiency by the motor efficiency. 5-48 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-80 Solution The previous problem is reconsidered. The effect of head loss on mechanical efficiency of the pump. as the head loss varies 0 to 20 m in increments of 2 m is to be investigated. Analysis The EES Equations window is printed below, followed by the tabulated and plotted results. g=9.81 "m/s2" rho=1000 "kg/m3" z2=25 "m" W_shaft=10 "kW" V_dot=0.025 "m3/s" W_pump_u=rho*V_dot*g*(z2+h_L)/1000 "kW" Eta_pump=W_pump_u/W_shaft Pumping power Wpump, u 6.13 6.38 6.62 6.87 7.11 7.36 7.60 7.85 8.09 8.34 8.58 8.83 9.07 9.32 9.56 9.81 ηpump Efficiency 0.613 0.638 0.662 0.687 0.711 0.736 0.760 0.785 0.809 0.834 0.858 0.883 0.907 0.932 0.956 0.981 1 0.95 0.9 0.85 ηpump Head Loss, hL, m 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0.8 0.75 0.7 0.65 0.6 0 2 4 6 8 10 12 14 16 hL, m Note that the useful pumping power is used to raise the fluid and to overcome head losses. For a given Discussion power input, the pump that overcomes more head loss is more efficient. 5-49 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-81 Solution A pump with a specified shaft power and efficiency is used to raise water to a higher elevation. The maximum flow rate of water is to be determined. Assumptions 1 The flow is steady and incompressible. 2 The elevation difference between the reservoirs is constant. 3 We assume the flow in the pipes to be frictionless since the maximum flow rate is to be determined, E mech loss, piping = 0. Properties We take the density of water to be ρ = 1000 kg/m3. Analysis We choose points 1 and 2 at the free surfaces of the lower and upper reservoirs, respectively, and take the surface of the lower reservoir as the reference level (z1 = 0). Both points are open to the atmosphere (P1 = P2 = Patm) and the velocities at both locations are negligible (V1 = V2 = 0). Then the energy equation for steady incompressible flow through a control volume between these two points that includes the pump and the pipes reduces to ⎛P ⎞ ⎛P ⎞ V2 V2 m ⎜⎜ 1 + α 1 1 + gz1 ⎟⎟ + W pump = m ⎜⎜ 2 + α 2 2 + gz 2 ⎟⎟ + W turbine + E mech,loss → W pump, u = m gz 2 = ρVgz 2 2 2 ⎝ρ ⎠ ⎝ ρ ⎠ since E mech, loss = E mech loss, pump in this case and W pump, u = W pump − E mech loss, pump . 2 The useful pumping power is W pump, u = η pump W pump, shaft = (0.82)(7 hp) = 5.74 hp Substituting, the volume flow rate of water is determined to be V = W pump, u ρgz 2 = ⎛ 745.7 W ⎞⎛ 1 N ⋅ m/s ⎞⎛ 1 kg ⋅ m/s 2 ⎟⎜ ⎜ ⎟⎜ 1N (1000 kg/m 3 )(9.81 m/s 2 )(15 m) ⎜⎝ 1 hp ⎟⎠⎝ 1 W ⎠⎜⎝ 5.74 hp = 0.0291 m 3 /s ⎞ ⎟ ⎟ ⎠ PUMP 15 m 1 Water This is the maximum flow rate since the frictional effects are ignored. Discussion In an actual system, the flow rate of water will be less because of friction in pipes. 5-50 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-82 Solution Water flows at a specified rate in a horizontal pipe whose diameter is decreased by a reducer. The pressures are measured before and after the reducer. The head loss in the reducer is to be determined. Assumptions 1 The flow is steady and incompressible. 2 The pipe is horizontal. 3 The kinetic energy correction factors are given to be α1 = α2 = α = 1.05. We take the density of water to be ρ = 1000 kg/m3. Properties Analysis We take points 1 and 2 along the centerline of the pipe before and after the reducer, respectively. Noting that z1 = z2, the energy equation for steady incompressible flow through a control volume between these two points reduces to P V2 P1 V2 + α 1 1 + z1 + h pump, u = 2 + α 2 2 + z 2 + hturbine, e + h L ρg ρg 2g 2g → hL = P1 − P2 α (V12 − V 22 ) + 2g ρg where V1 = V V2 = V A1 A2 = V πD12 / 4 = V πD22 / 4 = 0.035 m 3 /s π (0.15 m) 2 / 4 = 0.035 m 3 /s π (0.08 m) 2 / 4 = 1.981 m/s 1 = 6.963 m/s 480 kPa 15 cm Water 445 kPa 2 8 cm Reducer Substituting, the head loss in the reducer is determined to be hL = ⎛ 1 kN/m 2 ⎞⎛ 1000 kg ⋅ m/s2 ⎞ 1.05[(1.981 m/s) 2 − (6.963 m/s)2 ] (480 − 445) kPa ⎜ ⎟⎜ ⎟+ (1000 kg/m3 )(9.81 m/s 2 ) ⎝ 1 kPa ⎠⎝ 1 kN 2(9.81 m/s 2 ) ⎠ = 1.183 m ≅ 1.18 m Note that the 1.19 m of the head loss is due to frictional effects and 2.27 m is due to the increase in velocity. Discussion This head loss corresponds to a power potential loss of ⎛ 1N ⎞⎛ 1 W ⎞ E mech loss, piping = ρV ghL = (1000 kg/m3 )(0.035 m3 /s)(9.81 m/s 2 )(1.19 m) ⎜ ⎟ = 406 W 2 ⎟⎜ ⎝ 1 kg ⋅ m/s ⎠ ⎝ 1 N ⋅ m/s ⎠ 5-51 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-83 Solution A hose connected to the bottom of a tank is equipped with a nozzle at the end pointing straight up. The water is pressurized by a pump, and the height of the water jet is measured. The minimum pressure rise supplied by the pump is to be determined. Assumptions 1 The flow is steady and incompressible. 2 Friction between the water and air as well as friction in the hose is negligible. 3 The water surface is open to the atmosphere. We take the density of water to be ρ = 1000 kg/m3. Properties Analysis We take point 1 at the free surface of water in the tank, and point 2 at the top of the water trajectory where V2 = 0 and P1 = P2 = Patm. Also, we take the reference level at the bottom of the tank. Noting that z1 = 20 m and z2 = 27 m, hL =0 (to get the minimum value for required pressure rise), and that the fluid velocity at the free surface of the tank is very low (V1 ≅ 0), the energy equation for steady incompressible flow through a control volume between these two points that includes the pump and the water stream reduces to P V2 P1 V2 + α 1 1 + z1 + hpump, u = 2 + α 2 2 + z 2 + h turbine, e + h L ρg ρg 2g 2g → 2 hpump, u = z 2 − z1 Substituting, hpump, u = 27 − 20 = 7 m 27 m 1 A water column height of 7 m corresponds to a pressure rise of 20 m ⎞ ⎛ 1N ⎟ ΔPpump, min = ρghpump, u = (1000 kg/m 3 )(9.81 m/s 2 )(7 m)⎜⎜ 2⎟ 1000 kg m/s ⋅ ⎠ ⎝ = 68.7 kN/m 2 = 68.7 kPa Therefore, the pump must supply a minimum pressure rise of 68.7 kPa. Discussion The result obtained above represents the minimum value, and should be interpreted accordingly. In reality, a larger pressure rise will need to be supplied to overcome friction. 5-84 Solution The available head of a hydraulic turbine and its overall efficiency are given. The electric power output of this turbine is to be determined. Assumptions Properties 1 The flow is steady and incompressible. 2 The available head remains constant. We take the density of water to be ρ = 1000 kg/m3. Analysis When the turbine head is available, the corresponding power output is determined from W turbine = η turbine m ghturbine = η turbine ρVghturbine Substituting, Discussion Eff.=78% Turbine ⎛ 1N W turbine = 0.78(1000 kg/m 3 )(0.25 m 3 /s)(9.81 m/s 2 )(85 m)⎜⎜ 2 ⎝ 1 kg ⋅ m/s Generator ⎞⎛ 1 kW ⎞ ⎟⎜ ⎟⎝ 1000 N ⋅ m/s ⎟⎠ = 163 kW ⎠ The power output of a hydraulic turbine is proportional to the available turbine head and the flow rate. 5-52 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-85 Solution A pump is pumping oil at a specified rate. The pressure rise of oil in the pump is measured, and the motor efficiency is specified. The mechanical efficiency of the pump is to be determined. Assumptions 1 The flow is steady and incompressible. 2 The elevation difference across the pump is negligible. 3 All the losses in the pump are accounted for by the pump efficiency and thus hL = 0. 4 The kinetic energy correction factors are given to be α1 = α2 = α = 1.05. Properties The density of oil is given to be ρ = 860 kg/m3. Analysis We take points 1 and 2 at the inlet and the exit of the pump, respectively. Noting that z1 = z2, the energy equation for the pump reduces to P V2 P1 V2 + α 1 1 + z1 + h pump, u = 2 + α 2 2 + z 2 + hturbine, e + h L ρg ρg 2g 2g → hpump, u = P2 − P1 α (V 22 − V12 ) + ρg 2g where 25 kW V1 = V V2 = V A1 = V πD12 = A2 /4 V πD 22 / 4 = 0.1 m 3 /s π (0.08 m) / 4 = 2 0.1 m 3 /s π (0.12 m) 2 / 4 2 = 19.9 m/s Pump Motor = 8.84 m/s Substituting, the useful pump head and the corresponding useful pumping power are determined to be hpump, u = ⎛ 1 kg ⋅ m/s 2 ⎜ 1N (860 kg/m 3 )(9.81 m/s 2 ) ⎜⎝ 250,000 N/m 2 1 Oil ⎞ 1.05[ (8.84 m/s) 2 − (19.9 m/s) 2 ] ⎟+ = 29.6 − 17.0 = 12.6 m ⎟ 2(9.81 m/s 2 ) ⎠ ⎛ 1 kN W pump,u = ρVghpump, u = (860 kg/m 3 )(0.1 m 3 /s)(9.81 m/s 2 )(12.6 m)⎜ ⎜ 1000 kg ⋅ m/s 2 ⎝ ⎞⎛ 1 kW ⎞ ⎟⎜ = 10.6 kW ⎟⎝ 1 kN ⋅ m/s ⎟⎠ ⎠ Then the shaft pumping power and the mechanical efficiency of the pump become W pump,shaft = η motor W electric = (0.90)( 25 kW) = 22.5 kW η pump = W pump, u W pump, shaft = 10.6 kW = 0.471 = 47.1% 22.5 kW Discussion The overall efficiency of this pump/motor unit is the product of the mechanical and motor efficiencies, which is 0.9 × 0.471 = 0.42. 5-53 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-86 Solution Water flows through a horizontal pipe at a specified rate. The pressure drop across a valve in the pipe is measured. The corresponding head loss and the power needed to overcome it are to be determined. Assumptions 1 The flow is steady and incompressible. 2 The pipe is given to be horizontal (otherwise the elevation difference across the valve is negligible). 3 The mean flow velocities at the inlet and the exit of the valve are equal since the pipe diameter is constant. Properties We take the density of water to be ρ = 1000 kg/m3. Analysis We take the valve as the control volume, and points 1 and 2 at the inlet and exit of the valve, respectively. Noting that z1 = z2 and V1 = V2, the energy equation for steady incompressible flow through this control volume reduces to P1 V2 P V2 + α 1 1 + z1 + hpump, u = 2 + α 2 2 + z 2 + hturbine, e + h L ρg ρg 2g 2g Substituting, hL = ⎛ 1000 kg ⋅ m/s 2 kN/m ⎜ 3 2 ⎜ 1 kN (1000 kg/m )(9.81 m/s ) ⎝ 2 2 ⎞ ⎟ = 0.204 m ⎟ ⎠ hL = P1 − P2 ρg Valve Water 20 L/s The useful pumping power needed to overcome this head loss is W pump, u = m gh L = ρVgh L → ⎛ 1N = (1000 kg/m 3 )(0.020 m 3 /s)(9.81 m/s 2 )(0.204 m)⎜⎜ 2 ⎝ 1 kg ⋅ m/s 1 2 ΔP=2 kPa ⎞⎛ 1 W ⎞ ⎟⎜ ⎟⎝ 1 N ⋅ m/s ⎟⎠ = 40 W ⎠ Therefore, this valve would cause a head loss of 0.204 m, and it would take 40 W of useful pumping power to overcome it. Discussion The required useful pumping power could also be determined from ⎛ 1W ⎞ W pump = VΔP = (0.020 m 3 /s)(2000 Pa)⎜ ⎟ = 40 W ⎝ 1 Pa ⋅ m 3 /s ⎠ 5-54 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-87E Solution A hose connected to the bottom of a pressurized tank is equipped with a nozzle at the end pointing straight up. The minimum tank air pressure (gage) corresponding to a given height of water jet is to be determined. Assumptions 1 The flow is steady and incompressible. 2 Friction between water and air as well as friction in the hose is negligible. 3 The water surface is open to the atmosphere. 2 Properties We take the density of water to be ρ = 62.4 lbm/ft3. Analysis We take point 1 at the free surface of water in the tank, and point 2 at the top of the water trajectory where V2 = 0 and P1 = P2 = Patm. Also, we take the reference level at the bottom of the tank. Noting that z1 = 66 ft and z2 = 90 ft, hL = 0 (to get the minimum value for the required air pressure), and that the fluid velocity at the free surface of the tank is very low (V1 ≅ 0), the energy equation for steady incompressible flow through a control volume between these two points reduces to P1 1 90 ft 66 ft P1 V2 P V2 + α 1 1 + z1 + hpump, u = 2 + α 2 2 + z 2 + h turbine, e + h L 2g 2g ρg ρg or P1, gage P1 − Patm = z 2 − z1 → = z 2 − z1 ρg ρg Rearranging and substituting, the gage pressure of pressurized air in the tank is determined to be 1 lbf ⎛ ⎞ ⎛ 1 psi P1, gage = ρ g ( z2 − z1 ) = ( 62.4 lbm/ft 3 )( 32.2 ft/s 2 ) ( 90 − 66 ft ) ⎜ 2 ⎟⎜ 2 ⋅ 32 . 2 lbm ft/s ⎝ ⎠ ⎝ 144 lbf/ft ⎞ ⎟ = 10.4 psi ⎠ Therefore, the gage air pressure on top of the water tank must be at least 10.4 psi. Discussion The result obtained above represents the minimum value, and should be interpreted accordingly. In reality, a larger pressure will be needed to overcome friction. 5-88 Solution A water tank open to the atmosphere is initially filled with water. A sharp-edged orifice at the bottom drains to the atmosphere. The initial discharge velocity from the tank is to be determined. Assumptions 1 The flow is steady and incompressible. 2 The tank is open to the atmosphere. 3 The kinetic energy correction factor at the orifice is given to be α2 = α = 1.2. Analysis We take point 1 at the free surface of the tank, and point 2 at the exit of the orifice. Noting that the fluid at both points is open to the atmosphere (and thus P1 = P2 = Patm) and that the fluid velocity at the free surface of the tank is very low (V1 ≅ 0), the energy equation between these two points (in terms of heads) simplifies to P1 V2 P V2 + α 1 1 + z1 + hpump, u = 2 + α 2 2 + z 2 + h turbine, e + h L 2g 2g ρg ρg 1 which yields z1 + α 2 V 22 = z 2 + hL 2g Water 5m Solving for V2 and substituting, V2 = 2 g ( z1 − z 2 − hL ) / α = 2(9.81 m/s 2 )(5 − 0.3 m) / 1.2 = 8.77 m/s 2 10 cm Discussion This is the velocity that will prevail at the beginning. The mean flow velocity will decrease as the water level in the tank decreases. 5-55 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-89 Solution Water enters a hydraulic turbine-generator system with a known flow rate, pressure drop, and efficiency. The net electric power output is to be determined. Assumptions 1 The flow is steady and incompressible. 2 All losses in the turbine are accounted for by turbine efficiency and thus hL = 0. 3 The elevation difference across the turbine is negligible. 4 The effect of the kinetic energy correction factors is negligible, α1 = α2 = α = 1. We take the density of water to be 1000 kg/m3 and the density of mercury to be 13,560 kg/m3. Properties Analysis We choose points 1 and 2 at the inlet and the exit of the turbine, respectively. Noting that the elevation effects are negligible, the energy equation in terms of heads for the turbine reduces to P1 V2 P V2 + α 1 1 + z1 + hpump, u = 2 + α 2 2 + z 2 + hturbine, e + h L 2g 2g ρg ρg → P1 − P2 α (V12 − V 22 ) + 2g ρ water g hturbine, e = (1) where V1 = V V2 = V A1 A2 = = V πD12 0.6 m 3 /s / 4 π (0.30 m) / 4 2 V πD 22 / 4 = 3 0.6 m /s = 8.49 m/s π (0.25 m) 2 / 4 1 We Turbine = 12.2 m/s Generator The pressure drop corresponding to a differential height of 1.2 m in the mercury manometer is P1 − P2 = ( ρ Hg − ρ water ) gh ⎛ 1 kN = [(13,560 − 1000) kg/m 3 ](9.81 m/s 2 )(1.2 m)⎜⎜ 1000 kg ⋅ m/s 2 ⎝ = 148 kN/m 2 = 148 kPa 2 ⎞ ⎟ ⎟ ⎠ Substituting into Eq. (1), the turbine head is determined to be hturbine, e = ⎛ 1000 kg ⋅ m/s 2 ⎜ 1 kN (1000 kg/m 3 )(9.81 m/s 2 ) ⎜⎝ 148 kN/m 2 ⎞ (8.49 m/s ) 2 − (12.2 m/s) 2 ⎟ + (1.0) = 15.1 − 3.9 = 11.2 m ⎟ 2(9.81 m/s 2 ) ⎠ Then the net electric power output of this hydroelectric turbine becomes W turbine = η turbine-gen m ghturbine, e = η turbine-gen ρVghturbine, e ⎛ 1N = 0.83(1000 kg/m 3 )(0.6 m 3 /s)(9.81 m/s 2 )(11.2 m)⎜⎜ 1 kg ⋅ m/s 2 ⎝ ⎞⎛ 1 kW ⎞ ⎟⎜ ⎟⎝ 1000 N ⋅ m/s ⎟⎠ = 55 kW ⎠ Discussion It appears that this hydroelectric turbine will generate 55 kW of electric power under given conditions. Note that almost half of the available pressure head is discarded as kinetic energy. This demonstrates the need for a larger turbine exit area and better recovery. For example, the power output can be increased to 74 kW by redesigning the turbine and making the exit diameter of the pipe equal to the inlet diameter, D2 = D1. Further, if a much larger exit diameter is used and the exit velocity is reduced to a very low level, the power generation can increase to as much as 92 kW. 5-56 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-90 Solution The velocity profile for turbulent flow in a circular pipe is given. The kinetic energy correction factor for this flow is to be determined. Analysis The velocity profile is given by u (r ) = u max (1 − r / R ) 1 / n with n = 7 The kinetic energy correction factor is then expressed as 1 α= A ∫ ⎛ u (r ) ⎞ ⎜ ⎟ dA = 1 3 A ⎜ V avg ⎟ AV avg ⎝ ⎠ 3 where the average velocity is V avg = ∫ 1 1 u (r )dA = A A πR 2 From integral tables, ∫ ( a + bx ) n xdx = ∫ R ∫ ∫ u (r ) A 3 dA = r⎞ ⎛ u max ⎜1 − ⎟ r =0 ⎝ R⎠ R ( a + bx ) n + 2 b 2 ( n + 2) − 1 3 πR 2V avg ∫ R r =0 ( 2πr ) dr = 1/ n 3 ⎛ u max ⎜1 − ⎝ ∫ 3 2u max r ⎞n ⎟ (2πr )dr = 2 3 R⎠ R V avg 3 r⎞ ⎛ ⎜1 − ⎟ r =0 ⎝ R⎠ 2u max R2 R a ( a + bx ) n +1 b 2 ( n + 1) ∫ r ⎞n ⎛ ⎜1 − ⎟ rdr r =0 ⎝ R⎠ 3 R 1/ n rdr u(r) 2r Then, u (r )rdr = r =0 ∫ R ∫ r⎞ ⎛ ⎜1 − ⎟ r =0 ⎝ R⎠ R u (r ) 3 rdr = r =0 ∫ 1/ n r⎞ ⎛ ⎜1 − ⎟ r =0 ⎝ R⎠ R rdr = 3/ n (1 − r / R ) (1 − r / R ) − 1 1 1 1 ( + 2) ( n + 1) 2 n R R2 rdr = 1 +2 n 1 +1 n R = r =0 (1 − r / R ) (1 − r / R ) − 1 3 1 3 ( + 2) ( n + 1) 2 n R R2 3 +2 n 3 +1 n n2R2 ( n + 1)(2n + 1) R = r =0 n2R2 ( n + 3)( 2n + 3) Substituting, V avg = and 2u max R 2 2n 2 u max n2R2 = = 0.8167u max ( n + 1)( 2n + 1) ( n + 1)(2n + 1) 2u 3 ⎛ 2n 2 u max ⎞ ⎟ ⎜ α = max R 2 ⎜⎝ (n + 1)(2n + 1) ⎟⎠ −3 (n + 1) 3 (2n + 1) 3 (7 + 1) 3 (2 × 7 + 1) 3 n2R2 = 1.06 = = 4 (n + 3)(2n + 3) 4n (n + 3)(2n + 3) 4 × 7 4 (7 + 3)(2 × 7 + 3) Discussion Note that ignoring the kinetic energy correction factor results in an error of just 6% in this case in the kinetic energy term (which may be small itself). Considering that the uncertainties in some terms are usually more that 6%, we can usually ignore this correction factor in turbulent pipe flow analyses. However, for laminar pipe flow analyses, α is equal to 2.0 for fully developed laminar pipe flow, and ignoring α may lead to significant errors. 5-57 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-91 Solution Water is pumped from a lower reservoir to a higher one. The head loss and power loss associated with this process are to be determined. Assumptions 1 The flow is steady and incompressible. 2 The elevation difference between the reservoirs is constant. Properties We take the density of water to be 1000 kg/m3. Analysis The mass flow rate of water through the system is m = ρV = (1000 kg/m 3 )(0.030 m 3 /s) = 30 kg/s We choose points 1 and 2 at the free surfaces of the lower and upper reservoirs, respectively, and take the surface of the lower reservoir as the reference level (z1 = 0). Both points are open to the atmosphere (P1 = P2 = Patm) and the velocities at both locations are negligible (V1 = V2 = 0). Then the energy equation for steady incompressible flow for a control volume between 1 and 2 reduces to ⎞ ⎛P ⎞ ⎛P V2 V2 m ⎜ 1 + α 1 1 + gz1 ⎟⎟ + W pump, u = m ⎜⎜ 2 + α 2 2 + gz 2 ⎟⎟ + W turbine, e + E mech,loss ⎜ρ 2 2 ⎠ ⎝ ρ ⎠ ⎝ W pump, u = m gz 2 + E mech,loss → E mech,loss = W pump, u − m gz 2 Substituting, the lost mechanical power and head loss are calculated as ⎛ 1N E mech,loss = 20 kW − (30 kg/s)(9.81 m/s 2 )(45 m)⎜⎜ 2 ⎝ 1 kg ⋅ m/s ⎞⎛ 1 kW ⎞ ⎟⎜ ⎟⎝ 1000 N ⋅ m/s ⎟⎠ = 6.76 kW ⎠ Noting that the entire mechanical losses are due to frictional losses in piping and thus E mech, loss = E mech loss, piping , the irreversible head loss is determined to be hL = E mech loss, piping m g = ⎛ 1 kg ⋅ m/s 2 ⎜ 1N (30 kg/s )(9.81 m/s ) ⎜⎝ 6.76 kW 2 ⎞⎛ 1000 N ⋅ m/s ⎞ ⎟⎜ ⎟ = 23.0 m ⎟⎝ 1 kW ⎠ ⎠ The 6.76 kW of power is used to overcome the friction in the piping system. Note that the pump could raise Discussion the water an additional 23 m if there were no irreversible head losses in the system. In this ideal case, the pump would function as a turbine when the water is allowed to flow from the upper reservoir to the lower reservoir and extract 20 kW of power from the water. 5-58 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-92 Solution determined. Water from a pressurized tank is supplied to a roof top. The discharge rate of water from the tank is to be Assumptions 1 The flow is steady and incompressible. 2 The effect of the kinetic energy correction factor is negligible and thus α2 = 1 (we examine the effect of this approximation in the discussion). Properties We take the density of water to be 1000 kg/m3. Analysis We take point 1 at the free surface of the tank, and point 2 at the exit of the discharge pipe. Noting that the fluid velocity at the free surface of the tank is very low (V1 ≅ 0) and water discharges into the atmosphere (and thus P2 = Patm), the energy equation written in the head form simplifies to P1 V2 P V2 + α1 1 + z1 + hpump, u = 2 + α 2 2 + z2 + hturbine, e + hL ρg ρg 2g 2g → P1 − Patm V2 = α 2 2 + z2 − z1 + hL ρg 2g Solving for V2 and substituting, the discharge velocity is determined to V2 = = ⎤ 1 ⎡ 2 P1, gage − 2 g ( z2 − z1 + hL ) ⎥ ⎢ α2 ⎣ ρ ⎦ ⎤ 1 ⎡ 2 × (300 kPa) ⎛ 1000 N/m 2 ⎞⎛ 1 kg ⋅ m/s 2 ⎞ 2 ⎢ ⎟⎜ ⎟ − 2(9.81 m/s )(8 + 2 m) ⎥ 3 ⎜ 1 ⎣ 1000 kg/m ⎝ 1 kPa ⎠⎝ 1 N ⎠ ⎦ = 20.095 m/s ≅ 20.1 m/s 2 Then the initial rate of discharge of water becomes V = AorificeV2 = πD 2 V2 = π (0.025 m) h=8m 2 (20.095 m/s) 4 4 = 0.009864 m3 /s ≅ 0.00986 m3 /s = 9.86 L/s 300 kPa 1 Discussion This is the discharge rate that will prevail at the beginning. The mean flow velocity will decrease as the water level in the tank decreases. If we assume that the flow in the hose at the discharge is fully developed and turbulent, α2 ≈ 1.05, and the results change to V2 = 19.610 m/s ≈ 19.6 m/s, and V = 0.0096263 m3 /s ≅ 0.00963 m3 /s = 9.63 L/s , a decrease (as expected since we are accounting for more losses) of about 2.4%. 5-59 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-93 Solution An entrepreneur is to build a large reservoir above the lake level, and pump water from the lake to the reservoir at night using cheap power, and let the water flow from the reservoir back to the lake during the day, producing power. The potential revenue this system can generate per year is to be determined. Assumptions 1 The flow in each direction is steady and incompressible. 2 The elevation difference between the lake and the reservoir can be taken to be constant, and the elevation change of reservoir during charging and discharging is disregarded. 3 The given unit prices remain constant. 4 The system operates every day of the year for 10 hours in each mode. Properties kg/m3. 3 Reservoir 2 Pumpturbine 50 m Lake 4 1 We take the density of water to be ρ = 1000 Analysis We choose points 1 and 2 at the free surfaces of the lake and the reservoir, respectively, and take the surface of the lake as the reference level. Both points are open to the atmosphere (P1 = P2 = Patm) and the velocities at both locations are negligible (V1 = V2 = 0). Then the energy equation in terms of heads for steady incompressible flow through a control volume between these two points that includes the pump (or the turbine) and the pipes reduces to Pump mode: P1 V2 P V2 + α 1 1 + z1 + h pump, u = 2 + α 2 2 + z 2 + hturbine, e + h L → ρg ρg 2g 2g hpump, u = z 2 + hL = 50 + 4 = 54 m Turbine mode: (switch points 1 and 2 so that 1 is on inlet side) → hturbine, e = z1 − hL = 50 − 4 = 46 m The pump and turbine power corresponding to these heads are W pump, elect = W pump, u η pump-motor = ρVghpump, u η pump-motor (1000 kg/m )(2 m 3 /s)(9.81 m/s 2 )(54 m) ⎛⎜ 1N ⎜ 1 kg ⋅ m/s 2 0.75 ⎝ ⎞⎛ 1 kW ⎞ ⎟⎜ = 1413 kW ⎟⎝ 1000 N ⋅ m/s ⎟⎠ ⎠ ⎛ 1N = 0.75(1000 kg/m 3 )(2 m 3 /s)(9.81 m/s 2 )(46 m)⎜⎜ 2 ⎝ 1 kg ⋅ m/s ⎞⎛ 1 kW ⎞ ⎟⎜ = 677 kW ⎟⎝ 1000 N ⋅ m/s ⎟⎠ ⎠ = 3 W turbine = η turbine-gen m ghturbine, e = η turbine-gen ρVghturbine, e Then the power cost of the pump, the revenue generated by the turbine, and the net income (revenue minus cost) per year become Cost = W pump, elect Δt × Unit price = (1413 kW)(365 × 10 h/year)($0.03/kWh) = $154,724/year Revenue = W turbine Δt × Unit price = (677 kW)(365 × 10 h/year)($0.08/kWh) = $197,684/year Net income = Revenue – Cost = 197,684 – 154,724 = $42,960/year ≅ $43,000/year Discussion It appears that this pump-turbine system has a potential annual income of about $43,000. A decision on such a system will depend on the initial cost of the system, its life, the operating and maintenance costs, the interest rate, and the length of the contract period, among other things. 5-60 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-94E Solution Water is pumped from a lake to a nearby pool by a pump with specified power and efficiency. The head loss of the piping system and the mechanical power used to overcome it are to be determined. Assumptions 1 The flow is steady and incompressible. 2 The elevation difference between the lake and the free surface of the pool is constant. 3 All the losses in the pump are accounted for by the pump efficiency and thus hL represents the losses in piping. Properties We take the density of water to be ρ = 62.4 lbm/ft3. Analysis The useful pumping power and the corresponding useful pumping head are W pump, u = η pump W pump = (0.73)(12 hp) = 8.76 hp h pump, u = = W pump, u m g = W pump, u ρVg ⎛ 32.2 lbm ⋅ ft/s 2 ⎜ 1 lbf (62.4 lbm/ft 3 )(1.2 ft 3 /s)(32.2 ft/s 2 ) ⎜⎝ 8.76 hp ⎞⎛ 550 lbf ⋅ ft/s ⎞ ⎟⎜⎜ ⎟⎟ = 64.3 ft ⎟ 1 hp ⎠ ⎠⎝ We choose points 1 and 2 at the free surfaces of the lake and the pool, respectively. Both points are open to the atmosphere (P1 = P2 = Patm) and the velocities at both locations are negligible (V1 = V2 = 0). Then the energy equation for steady incompressible flow through a control volume between these two points that includes the pump and the pipes reduces to P1 V2 P V2 + α 1 1 + z1 + hpump, u = 2 + α 2 2 + z 2 + h turbine, e + h L ρg ρg 2g 2g → Pool Substituting, the head loss is determined to be h L = h pump, u − ( z 2 − z1 ) = 64.3 − 35 = 29.3 ft h L = h pump, u + z1 − z 2 2 Pump 35 ft Lake Then the power used to overcome it becomes 1 E mech loss, piping = ρVgh L 1 hp 1 lbf ⎞⎛ ⎛ ⎞ = (62.4 lbm/ft 3 )(1.2 ft 3 /s)(32.2 ft/s 2 )(29.3 ft )⎜ ⎟⎜ ⎟ 2 ⎝ 32.2 lbm ⋅ ft/s ⎠⎝ 550 lbf ⋅ ft/s ⎠ = 4.0 hp Discussion Note that the pump must raise the water an additional height of 29.3 ft to overcome the frictional losses in pipes, which requires an additional useful pumping power of about 4 hp. 5-61 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-95 Solution Underground water is pumped to a pool at a given elevation. The maximum flow rate and the pressures at the inlet and outlet of the pump are to be determined. Assumptions 1 The flow is steady and incompressible. 2 The elevation difference between the inlet and the outlet of the pump is negligible. 3 We assume the frictional effects in piping to be negligible since the maximum flow rate is to be determined, E mech loss, pipping = 0. 4 The effect of the kinetic energy correction factors is negligible, α = 1. Properties We take the density of water to be 1 kg/L = 1000 kg/m3. Analysis (a) The pump-motor draws 5-kW of power, and is 78% efficient. Then the useful mechanical (shaft) power it delivers to the fluid is W pump, u = η pump-motorW electric = (0.78)(5 kW) = 3.9 kW We take point 1 at the free surface of underground water, which is also taken as the reference level (z1 = 0), and point 2 at the free surface of the pool. Also, both 1 and 2 are open to the atmosphere (P1 = P2 = Patm), the velocities are negligible at both points (V1 ≅ V2 ≅ 0), and frictional losses in piping are disregarded. Then the energy equation for steady incompressible flow through a control volume between these two points that includes the pump and the pipes reduces to ⎞ ⎛P ⎞ ⎛P V2 V2 m ⎜ 1 + α 1 1 + gz1 ⎟⎟ + W pump = m ⎜⎜ 2 + α 2 2 + gz 2 ⎟⎟ + W turbine + E mech,loss 2 ⎜ρ 2 2 ⎠ ⎝ ρ ⎠ ⎝ Pool and In the absence of a turbine, E = E + E mech, loss W pump, u = W pump − E mech loss, pump . mech loss, pump mech loss, piping Thus, W pump, u = m gz 2 . 30 m Then the mass and volume flow rates of water become m = W pump, u V = m gz 2 ρ = = ⎛ 1000 m /s ⎞ ⎟ = 13.25 kg/s ⎜ ⎟ 1 kJ (9.81 m/s )(30 m) ⎜⎝ ⎠ 3.9 kJ/s 13.25 kg/s 1000 kg/m 3 2 1 2 2 Pump = 0.01325 m 3 /s ≅ 0.0133 m 3 /s (b) We take points 3 and 4 at the inlet and the exit of the pump, respectively, where the flow velocities are V V V V 0.01325 m 3 /s 0.01325 m 3 /s = = = = = = = 6.748 m/s 3 . 443 m/s , V3 = V 4 A3 πD32 / 4 π (0.07 m) 2 / 4 A4 πD42 / 4 π (0.05 m) 2 / 4 We take the pump as the control volume. Noting that z3 = z4, the energy equation for this control volume reduces to ⎛P ⎞ ⎛P ⎞ V2 V2 ρα (V32 − V 42 ) W pump, u m ⎜ 3 + α 3 3 + gz 3 ⎟ + W pump = m ⎜⎜ 4 + α 4 4 + gz 4 ⎟⎟ + W turbine + E mech loss, pump → P4 − P3 = + ⎜ ρ ⎟ 2 2 2 V ⎝ ρ ⎠ ⎝ ⎠ Substituting, ⎞ (1000 kg/m 3 )[(3.443 m/s) 2 − (6.748 m/s) 2 ] ⎛⎜ 1 kN ⎟ + 3.9 kJ/s ⎛⎜ 1 kN ⋅ m ⎞⎟ P4 − P3 = 2 3 ⎟ ⎜ 2 ⎝ 1000 kg ⋅ m/s ⎠ 0.01325 m /s ⎝ 1 kJ ⎠ = (−16.8 + 294.3) kN/m 2 = 277.5 kPa ≅ 278 kPa In an actual system, the flow rate of water will be less because of friction in the pipes. Also, the effect of Discussion flow velocities on the pressure change across the pump is negligible in this case (under 2%) and can be ignored. 5-62 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-96 Solution Underground water is pumped to a pool at a given elevation. For a given head loss, the flow rate and the pressures at the inlet and outlet of the pump are to be determined. Assumptions 1 The flow is steady and incompressible. 2 The elevation difference between the inlet and the outlet of the pump is negligible. 3 The effect of the kinetic energy correction factors is negligible, α = 1. Properties We take the density of water to be 1 kg/L = 1000 kg/m3. Analysis (a) The pump-motor draws 5-kW of power, and is 78% efficient. Then the useful mechanical (shaft) power it delivers to the fluid is W pump, u = η pump-motorW electric = (0.78)(5 kW) = 3.9 kW We take point 1 at the free surface of underground water, which is also taken as the reference level (z1 = 0), and point 2 at the free surface of the pool. Also, both 1 and 2 are open to the atmosphere (P1 = P2 = Patm), and the velocities are negligible at both points (V1 ≅ V2 ≅ 0). Then the energy equation for steady incompressible flow through a control volume between these two points that includes the pump and the pipes reduces to ⎞ ⎛P ⎞ ⎛P V2 V2 m ⎜ 1 + α 1 1 + gz1 ⎟⎟ + W pump = m ⎜⎜ 2 + α 2 2 + gz 2 ⎟⎟ + W turbine + E mech,loss ⎜ρ 2 2 ⎠ ⎝ ρ ⎠ ⎝ In the absence of a turbine, E mech, loss = E mech loss, pump + E mech loss, piping and W pump, u = W pump − E mech loss, pump and thus W pump, u = m gz 2 + E mech loss, piping Noting that E mech, loss = m gh L , the mass and volume flow rates of water become m = = V = m ρ W pump, u gz 2 + gh L = g ( z 2 + hL ) 30 m ⎛ 1000 m /s ⎞ ⎟ = 11.69 kg/s ⎜ ⎟ 1 kJ (9.81 m/s )(30 + 4 m) ⎜⎝ ⎠ 2 2 1 2 11.69 kg/s 1000 kg/m 3 Pool W pump, u 3.9 kJ/s = 2 = 0.01169 m 3 / s ≅ 0.0117 m 3 /s Pump (b) We take points 3 and 4 at the inlet and the exit of the pump, respectively, where the flow velocities are 0.01169 m 3 /s 0.01169 m 3 /s V V V V V V3 = , = = = 3 . 038 m/s = = = = 5.954 m/s 4 A4 πD42 / 4 π (0.05 m) 2 / 4 A3 πD32 / 4 π (0.07 m) 2 / 4 We take the pump as the control volume. Noting that z3 = z4, the energy equation for this control volume reduces to ⎛P ⎞ ⎞ ⎛P V2 V2 ρα (V32 − V 42 ) W pump, u m ⎜ 3 + α 3 3 + gz 3 ⎟ + W pump = m ⎜ 4 + α 4 4 + gz 4 ⎟ + W turbine + E mech loss, pump → P4 − P3 = + ⎜ ρ ⎟ ⎜ ρ ⎟  2 2 2 V ⎠ ⎝ ⎝ ⎠ Substituting, ⎞ (1000 kg/m 3 )[(3.038 m/s) 2 − (5.954 m/s) 2 ] ⎛⎜ 1 kN 3.9 kJ/s ⎛ 1 kN ⋅ m ⎞ ⎟+ P4 − P3 = ⎜ ⎟ 2 3 ⎟ ⎜ 2 ⎝ 1000 kg ⋅ m/s ⎠ 0.01169 m /s ⎝ 1 kJ ⎠ = (−13.1 + 333.6) kN/m 2 = 320.5 kPa ≅ 321 kPa Discussion Note that frictional losses in the pipes causes the flow rate of water to decrease. Also, the effect of flow velocities on the pressure change across the pump is negligible in this case (about 1%) and can be ignored. 5-63 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-97 Solution A fireboat is fighting fires by drawing sea water and discharging it through a nozzle. The head loss of the system and the elevation of the nozzle are given. The shaft power input to the pump and the water discharge velocity are to be determined. Assumptions α = 1. Properties 1 The flow is steady and incompressible. 2 The effect of the kinetic energy correction factors is negligible, The density of sea water is given to be ρ =1030 kg/m3. 2 Analysis We take point 1 at the free surface of the sea and point 2 at the nozzle exit. Noting that P1 = P2 = Patm and V1 ≅ 0 (point 1 is at the free surface; not at the pipe inlet), the energy equation for the control volume between 1 and 2 that includes the pump and the piping system reduces to P1 V2 P V2 + α 1 1 + z1 + hpump, u = 2 + α 2 2 + z 2 + h turbine, e + h L 2g 2g ρg ρg 3m 1 Fireboat → hpump, u = z 2 − z1 + α 2 V 22 + hL 2g where the water discharge velocity is V2 = 0.1 m3 /s V V = = = 50.9296 m/s ≅ 50.9 m/s A2 π D22 / 4 π (0.05 m) 2 / 4 Substituting, the useful pump head and the corresponding useful pump power are determined to be hpump, u = (3 m) + (1) (50.9296 m/s) 2 + (3 m) = 138.203 m ≅ 138 m 2(9.81 m/s 2 ) ⎛ ⎞ ⎛ 1 kW ⎞ 1 kN Wpump,u = ρV ghpump, u = (1030 kg/m3 )(0.1 m3 /s)(9.81 m/s 2 )(138.203 m) ⎜ ⎟ 2 ⎟⎜ ⎝ 1000 kg ⋅ m/s ⎠ ⎝ 1 kN ⋅ m/s ⎠ = 139.644 kW Then the required shaft power input to the pump becomes Wpump, shaft = Discussion Wpump, u η pump = 139.644 kW = 199.49 kW ≅ 199 kW 0.70 Note that the pump power is used primarily to increase the kinetic energy of water. 5-64 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations Review Problems 5-98 Solution The air in a hospital room is to be replaced every 20 minutes. The minimum diameter of the duct is to be determined if the air velocity is not to exceed a certain value. Assumptions 1 The volume occupied by the furniture etc in the room is negligible. 2 The incoming conditioned air does not mix with the air in the room. The volume of the room is Analysis V = (6 m)(5 m)(4 m) = 120 m3 To empty this air in 20 min, the volume flow rate must be Hospital Room 6×5×4 m 10 bulbs 120 m 3 V = = = 0.10 m 3 /s Δt 20 × 60 s V If the mean velocity is 5 m/s, the diameter of the duct is V = AV = πD 2 4 V → D= 4V = πV 4(0.10 m 3 /s) = 0.16 m π (5 m/s) Therefore, the diameter of the duct must be at least 0.16 m to ensure that the air in the room is exchanged completely within 20 min while the mean velocity does not exceed 5 m/s. Discussion This problem shows that engineering systems are sized to satisfy certain constraints imposed by certain considerations. 5-65 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-99 Solution The rate of accumulation of water in a pool and the rate of discharge are given. The rate supply of water to the pool is to be determined. Assumptions 1 Water is supplied and discharged steadily. 2 The rate of evaporation of water is negligible. 3 No water is supplied or removed through other means. Analysis The conservation of mass principle applied to the pool requires that the rate of increase in the amount of water in the pool be equal to the difference between the rate of supply of water and the rate of discharge. That is, = m i − m e dmpool dt → m i = dmpool dt + m e → dVpool  Vi = + Ve dt since the density of water is constant and thus the conservation of mass is equivalent to conservation of volume. The rate of discharge of water is Ve = AeV e = (πD 2 /4)Ve = [π (0.05 m) 2 /4](5 m/s) = 0.00982 m 3 /s The rate of accumulation of water in the pool is equal to the cross-section of the pool times the rate at which the water level rises, dV pool dt = Across-sectionV level = (3 m × 4 m)(0.015 m/min) = 0.18 m 3 /min = 0.00300 m 3 /s Substituting, the rate at which water is supplied to the pool is determined to be dVpool  Vi = + Ve = 0.003 + 0.00982 = 0.01282 m3 /s ≅ 0.0128 m 3 /s dt Therefore, water is supplied at a rate of 0.01282 m3/s = 12.82 L/s. Discussion This is a very simple application of the conservation of mass equations. 5-100 Solution A fluid is flowing in a circular pipe. A relation is to be obtained for the average fluid velocity in terms of V(r), R, and r. Analysis Choosing a circular ring of area dA = 2πrdr as our differential area, the mass flow rate through a crosssectional area can be expressed as ∫ ∫ m = ρV (r )dA = ρV (r )2π r dr dr R 0 A R r Setting this equal to and solving for Vavg, Vavg = Discussion 2 R2 ∫ V ( r ) r dr R 0 If V were a function of both r and θ, we would also need to integrate with respect to θ. 5-66 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-101 Solution Air is accelerated in a nozzle. The density of air at the nozzle exit is to be determined. Assumptions Flow through the nozzle is steady. Properties The density of air is given to be 3.80 kg/m3 at the inlet. Analysis m 1 = m 2 1 = m 2 = m  . Then, There is only one inlet and one exit, and thus m ρ1 A1V1 = ρ 2 A2V2 A V 120 m/s ρ 2 = 1 1 ρ1 = 2 (3.80 kg/m 3 ) = 2.40 kg/m 3 A2 V2 380 m/s Discussion nozzle. 1 AIR 2 Note that the density of air decreases considerably despite a decrease in the cross-sectional area of the 5-67 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-102 Solution A water tank open to the atmosphere is initially filled with water. The tank discharges to the atmosphere through a long pipe connected to a valve. The initial discharge velocity from the tank and the time required to empty the tank are to be determined. Assumptions 1 The flow is incompressible. 2 The draining pipe is horizontal. 3 The tank is considered to be empty when the water level drops to the center of the valve. (a) Substituting the known quantities, the discharge velocity can be expressed as Analysis V = 2 gz 2 gz = = 0.1481gz 1.5 + fL / D 1.5 + 0.015(80 m)/(0.10 m) Then the initial discharge velocity becomes V1 = 0.1481gz1 = 0.1481(9.81 m/s 2 )(2 m) = 1.705 m/s z D0 D where z is the water height relative to the center of the orifice at that time. (b) The flow rate of water from the tank can be obtained by multiplying the discharge velocity by the pipe cross-sectional area, V = ApipeV2 = πD 2 0.1481gz 4 Then the amount of water that flows through the pipe during a differential time interval dt is πD 2 dV = Vdt = 0.1481gz dt (1) 4 which, from conservation of mass, must be equal to the decrease in the volume of water in the tank, dV = Atan k (−dz ) = − πD02 dz (2) 4 where dz is the change in the water level in the tank during dt. (Note that dz is a negative quantity since the positive direction of z is upwards. Therefore, we used –dz to get a positive quantity for the amount of water discharged). Setting Eqs. (1) and (2) equal to each other and rearranging, πD 2 4 0.1481gz dt = − πD02 4 dz → dt = − D02 D dz 2 0.1481gz =− D02 D 2 z − 12 dz 0.1481g The last relation can be integrated easily since the variables are separated. Letting tf be the discharge time and integrating it from t = 0 when z = z1 to t = tf when z = 0 (completely drained tank) gives ∫ tf dt = − t =0 0.1481g ∫ D02 D 2 0 −1 / 2 z = z1 z dz → t f = - 1 D02 D 2 0 z2 0.1481g 1 2 z1 = 2 D02 D 2 0.1481g 1 z12 Simplifying and substituting the values given, the draining time is determined to be tf = Discussion 2 D02 D2 z1 2(8 m) 2 = 0.1212 g (0.1 m) 2 2m 0.1481(9.81 m/s 2 ) = 15,018 s = 4.17 h The draining time can be shortened considerably by installing a pump in the pipe. 5-68 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-103 Solution Water discharges from the orifice at the bottom of a pressurized tank. The time it will take for half of the water in the tank to be discharged and the water level after 10 s are to be determined. Assumptions 1 The flow is incompressible, and the frictional effects are negligible. 2 The tank air pressure above the water level is maintained constant. Properties We take the density of water to be 1000 kg/m3. Analysis We take point 1 at the free surface of the tank, and point 2 at the exit of orifice. We take the positive direction of z to be upwards with reference level at the orifice (z2 = 0). Fluid at point 2 is open to the atmosphere (and thus P2 = Patm) and the velocity at the free surface is very low (V1 ≅ 0). Then, P1 V12 P V2 + + z1 = 2 + 2 + z2 ρg 2 g ρg 2 g → P1 P V2 + z1 = atm + 2 ρg ρg 2 g → V2 = 2 gz1 + 2 P1,gage / ρ or, V2 = 2 gz + 2 P1,gage / ρ where z is the water height in the tank at any time t. Water surface moves down as the tank drains, and the value of z changes from H initially to 0 when the tank is emptied completely. We denote the diameter of the orifice by D, and the diameter of the tank by Do. The flow rate of water from the tank is obtained by multiplying the discharge velocity by the orifice cross-sectional area, πD 2 V = AorificeV 2 = 2 gz + 2 P1,gage / ρ 4 Then the amount of water that flows through the orifice during a differential time interval dt is πD 2 (1) dV = Vdt = 2 gz + 2 P1,gage / ρ dt 4 which, from conservation of mass, must be equal to the decrease in the volume of water in the tank, πD 2 dV = Atank (−dz ) = − 0 dz (2) 4 where dz is the change in the water level in the tank during dt. (Note that dz is a negative quantity since the positive direction of z is upwards. Therefore, we used –dz to get a positive quantity for the amount of water discharged). Setting Eqs. (1) and (2) equal to each other and rearranging, πD 2 D2 1 πD 2 → dz 2 gz + 2 P1,gage / ρ dt = − 0 dz dt = − 02 + 2 2 gz P 4 4 D 1, gage / ρ The last relation can be integrated since the variables are separated. Letting tf be the discharge time and integrating it from t = 0 when z = z0 to t = t when z = z gives 1 2 2 z 0 2 P1, gage 2 z 2 P1, gage D0 Air, 450 kPa + − + = 2t g g ρg 2 ρg 2 D where 2 P1,gage ρg 2 = ⎛ 1000 kg ⋅ m/s 2 ⎜ 1 kN (1000 kg/m 3 )(9.81 m/s 2 ) 2 ⎜⎝ 2(450 − 100) kN/m 2 ⎞ ⎟ = 7.274 s 2 ⎟ ⎠ Water tank D0 z0= 3 m The time for half of the water in the tank to be discharged (z = z0 /2) is 2(3 m) 9.81 m/s 2 + 7.274 s 2 − 2(1.5 m) 9.81 m/s 2 + 7.274 s 2 = (0.1 m) 2 (2 m) 2 t → t = 22.0 s D=10 cm 2 (10 s) → z =2.31 m 9.81 m/s 2 9.81 m/s 2 (2 m) 2 Discussion Note that the discharging time is inversely proportional to the square of the orifice diameter. Therefore, the discharging time can be reduced to one-fourth by doubling the diameter of the orifice. (b) Water level after 10s is 2(3 m) + 7.274 s 2 − 2z + 7.274 s 2 = (0.1 m) 2 5-69 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-104 Solution Air flows through a pipe that consists of two sections at a specified rate. The differential height of a water manometer placed between the two pipe sections is to be determined. Assumptions 1The flow through the pipe is steady, incompressible, and irrotational with negligible friction (so that the Bernoulli equation is applicable). 2 The losses in the reducing section are negligible. 3 The pressure difference across an air column is negligible because of the low density of air, and thus the air column in the manometer can be ignored. The density of air is given to be ρair = 1.20 kg/m3. We take the density of water to be ρw = 1000 kg/m3. Properties Analysis We take points 1 and 2 along the centerline of the pipe over the two tubes of the manometer. Noting that z1 = z2 (or, the elevation effects are negligible for gases), the Bernoulli equation between points 1 and 2 gives P V2 P1 V12 + + z1 = 2 + 2 + z2 ρg 2 g ρg 2 g → P1 − P2 = ρair (V22 − V12 ) 2 (1) We let the differential height of the water manometer be h. Then the pressure difference P2 – P1 can also be expressed as P1 − P2 = ρ w gh (2) Combining Eqs. (1) and (2) and solving for h, ρ air (V 22 − V12 ) 2 = ρ w gh → ρ (V 2 − V12 ) V 2 − V12 h = air 2 = 2 2 gρ w 2 gρ w / ρ air 18 cm Air 170 L/s 2 1 10 cm Calculating the velocities and substituting, V1 = V V2 = V h= A1 A2 = = V = V = πD12 / 4 πD22 / 4 0.170 m 3 /s π (0.18 m) 2 / 4 0.170 m 3 /s π (0.1 m) 2 / 4 = 6.681 m/s h Water = 21.65 m/s (21.65 m/s) 2 − (6.681 m/s) 2 = 0.0260 m = 2.60 cm 2(9.81 m/s 2 )(1000 /1.20) Therefore, the differential height of the water column will be 2.6 cm. Discussion Note that the differential height of the manometer is inversely proportional to the density of the manometer fluid. Therefore, heavy fluids such as mercury are used when measuring large pressure differences. 5-70 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-105 [Also solved using EES on enclosed DVD] Solution Air flows through a horizontal duct of variable cross-section. For a given differential height of a water manometer placed between the two pipe sections, the downstream velocity of air is to be determined, and an error analysis is to be conducted. Assumptions 1 The flow through the duct is steady, incompressible, and irrotational with negligible friction (so that the Bernoulli equation is applicable). 2 The losses in this section of the duct are negligible. 3 The pressure difference across an air column is negligible because of the low density of air, and thus the air column in the manometer can be ignored. The gas constant of air is R = 0.287 kPa⋅m3/kg⋅K. We take the density of water to be ρw = 1000 kg/m3. Properties Analysis We take points 1 and 2 along the centerline of the duct over the two tubes of the manometer. Noting that z1 = z2 (or, the elevation effects are negligible for gases) and V1 ≅ 0, the Bernoulli equation between points 1 and 2 gives P1 V12 P V2 + + z1 = 2 + 2 + z 2 ρg 2 g ρg 2 g where P1 − P2 = ρ w gh and ρ air = → P1 P V2 = 2 + 2 ρg ρg 2 g → V2 = P 100 kPa = = 1.17 kg/m 3 RT (0.287 kPa ⋅ m 3 /kg ⋅ K)(298 K) 2( P1 − P2 ) ρ air (1) V1<<V2 Air V2 2 1 Substituting into (1), the downstream velocity of air V2 is determined to be V2 = 2 ρ w gh ρ air = 2(1000 kg/m 3 )(9.81 m/s 2 )(0.08 m) 1.17 kg/m 3 = 36.6 m/s (2) h Water Therefore, the velocity of air increases from a low level in the first section to 36.6 m/s in the second section. Error Analysis We observe from Eq. (2) that the velocity is proportional to the square root of the differential height of the manometer fluid. That is, V 2 = k h . dh Taking the differential: dV 2 = 12 k h dV 2 1 dh 1 dV2 dh ±2 mm Dividing by V2: = 2k → = = = ± 0.013 V2 2h 2 × 80 mm V2 h k h Therefore, the uncertainty in the velocity corresponding to an uncertainty of 2 mm in the differential height of water is 1.3%, which corresponds to 0.013×(36.6 m/s) = 0.5m/s. Then the discharge velocity can be expressed as V2 = 36.6 ± 0.5 m/s The error analysis does not include the effects of friction in the duct; the error due to frictional losses is Discussion most likely more severe than the error calculated here. 5-71 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-106 Solution A tap is opened on the wall of a very large tank that contains air. The maximum flow rate of air through the tap is to be determined, and the effect of a larger diameter lead section is to be assessed. Assumptions Flow through the tap is steady, incompressible, and irrotational with negligible friction (so that the flow rate is maximum, and the Bernoulli equation is applicable). Properties The gas constant of air is R = 0.287 kPa⋅m3/kg⋅K. Analysis The density of air in the tank is ρ air = P 102 kPa = = 1.21 kg/m 3 RT (0.287 kPa ⋅ m 3 /kg ⋅ K)(293 K) We take point 1 in the tank, and point 2 at the exit of the tap along the same horizontal line. Noting that z1 = z2 (or, the elevation effects are negligible for gases) and V1 ≅ 0, the Bernoulli equation between points 1 and 2 gives P1 V12 P V2 + + z1 = 2 + 2 + z 2 ρg 2 g ρg 2 g → P1 P V2 = 2 + 2 ρg ρg 2 g → V2 = 2( P1 − P2 ) ρ air Substituting, the discharge velocity and the flow rate becomes V2 = 2( P1 − P2 ) ρ air V = AV 2 = πD 22 4 = V2 = 2(102 − 100) kN/m ⎛ 1000 kg ⋅ m/s ⎜ ⎜ 1 kN 1.21 kg/m 3 ⎝ 2 π (0.02 m) 2 4 2 ⎞ ⎟ = 57.5 m/s ⎟ ⎠ (57.5 m/s) = 0.0181 m 3 /s 100 kPa 20°C 2 cm 1 2 2 cm Air 102 kPa 4 cm This is the maximum flow rate since it is determined by assuming frictionless flow. The actual flow rate will be less. Adding a 2-m long larger diameter lead section will have no effect on the flow rate since the flow is frictionless (by using the Bernoulli equation, it can be shown that the velocity in this section increases, but the pressure decreases, and there is a smaller pressure difference to drive the flow through the tab, with zero net effect on the discharge rate). Discussion If the pressure in the tank were 300 kPa, the flow is no longer incompressible, and thus the problem in that case should be analyzed using compressible flow theory. 5-72 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-107 Solution Water is flowing through a venturi meter with known diameters and measured pressures. The flow rate of water is to be determined for the case of frictionless flow. Assumptions 1 The flow through the venturi is steady, incompressible, and irrotational with negligible friction (so that the Bernoulli equation is applicable). 2 The flow is horizontal so that elevation along the centerline is constant. 3 The pressure is uniform at a given cross-section of the venturi meter (or the elevation effects on pressure measurement are negligible). We take the density of water to be ρ = 1000 kg/m3. Properties Analysis We take point 1 at the main flow section and point 2 at the throat along the centerline of the venturi meter. Noting that z1 = z2, the application of the Bernoulli equation between points 1 and 2 gives P1 V12 P V2 + + z1 = 2 + 2 + z 2 ρg 2 g ρg 2 g → P1 − P2 = ρ V 22 − V12 2 (1) The flow is assumed to be incompressible and thus the density is constant. Then the conservation of mass relation for this single stream steady flow device can be expressed as V1 = V2 = V → A1V1 = A2V 2 = V → V1 = Substituting into Eq. (1), (V / A2 ) 2 − (V / A1 ) 2 ρV 2 P1 − P2 = ρ = 2 2 A22 ⎛ ⎜1 − ⎜ ⎝ V and A1 A22 A12 V2 = ⎞ ⎟ ⎟ ⎠ Solving for V gives the desired relation for the flow rate, V = A2 2( P1 − P2 ) V A2 (2) 380 kPa 1 150 kPa 4 cm 7 cm 2 ρ[1 − ( A2 / A1 ) 2 ] (3) The flow rate for the given case can be determined by substituting the given values into this relation to be V = πD22 4 2( P1 − P2 ) ρ[1 − ( D2 / D1 ) ] 4 = π (0.04 m) 2 4 ⎛ 1000 kg ⋅ m/s 2 ⎜ 1 kN (1000 kg/m 3 )[1 - (4/7) 4 ] ⎜⎝ 2(380 − 150) kN/m 2 ⎞ ⎟ = 0.0285 m 3 /s ⎟ ⎠ Discussion Venturi meters are commonly used as flow meters to measure the flow rate of gases and liquids by simply measuring the pressure difference P1 - P2 by a manometer or pressure transducers. The actual flow rate will be less than the value obtained from Eq. (3) because of the friction losses along the wall surfaces in actual flow. But this difference can be as little as 1% in a well-designed venturi meter. The effects of deviation from the idealized Bernoulli flow can be accounted for by expressing Eq. (3) as V = C d A2 2( P1 − P2 ) ρ [1 − ( A2 / A1 ) 2 ] where Cd is the venturi discharge coefficient whose value is less than 1 (it is as large as 0.99 for well-designed venturi meters in certain ranges of flow). For Re > 105, the value of venturi discharge coefficient is usually greater than 0.96. 5-73 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-108 Solution Water flows through the enlargement section of a horizontal pipe at a specified rate. For a given head loss, the pressure change across the enlargement section is to be determined. Assumptions 1 The flow through the pipe is steady and incompressible. 2 The pipe is horizontal. 3 The kinetic energy correction factors are given to be α1 = α2 = α = 1.05. Properties We take the density of water to be ρ = 1000 kg/m3. Analysis We take points 1 and 2 at the inlet and exit of the enlargement section along the centerline of the pipe. Noting that z1 = z2 , the energy equation for a control volume between these two points (in terms of heads) simplifies to P1 V2 P V2 + α 1 1 + z1 + hpump, u = 2 + α 2 2 + z 2 + h turbine, e + h L 2g 2g ρg ρg where the inlet and exit velocities are 0.025 m 3 /s V V V1 = = = = 8.84 m/s A1 πD12 / 4 π (0.06 m) 2 / 4 V2 = V A2 = V πD 22 / 4 = 0.025 m 3 /s π (0.11 m) 2 / 4 → P2 − P1 = ρ Water 0.025 m3/s = 2.63 m/s α (V12 − V 22 ) 2 − ρgh L 6 cm 1 2 11 cm Substituting, the change in static pressure across the enlargement section is determined to be ⎞⎛ ⎛ 1.05[(8.84 m/s) 2 − (2.63 m/s) 2 ] 1N P2 − P1 = (1000 kg/m 3 )⎜ − (9.81 m/s 2 )(0.45 m) ⎟⎜ ⎟⎜ 1 kg ⋅ m/s 2 ⎜ 2 ⎠⎝ ⎝ = 33.0 kPa ⎞⎛ 1 kPa ⎞ ⎟⎜ ⎟⎝ 1000 N/m 2 ⎟⎠ ⎠ Therefore, the water pressure increases by 33 kPa across the enlargement section. Discussion Note that the pressure increases despite the head loss in the enlargement section. This is due to dynamic pressure being converted to static pressure. But the total pressure (static + dynamic) decreases by 0.45 m (or 4.41 kPa) as a result of frictional effects. 5-109 Solution A water tank open to the atmosphere is initially filled with water. A sharp-edged orifice at the bottom drains to the atmosphere through a long pipe with a specified head loss. The initial discharge velocity is to be determined. Assumptions 1 The flow is incompressible. 2 The draining pipe is horizontal. 3 There are no pumps or turbines in the system. 4 The effect of the kinetic energy correction factor is negligible, α = 1. Analysis We take point 1 at the free surface of the tank, and point 2 at the exit of the pipe. We take the reference level at the centerline of the orifice (z2 = 0. Noting that the fluid at both points is open to the atmosphere (and thus P1 = P2 = Patm) and that the fluid velocity at the free surface is very low (V1 ≅ 0), the energy equation between these two points (in terms of heads) simplifies to P1 V2 P V2 + α 1 1 + z1 + hpump, u = 2 + α 2 2 + z 2 + h turbine, e + h L 2g 2g ρg ρg → z1 = α 2 1 Water 3m 80 m V 22 + hL 2g 10 cm 2 where α2 = 1 and the head loss is given to be hL = 1.5 m. Solving for V2 and substituting, the discharge velocity of water is determined to be V2 = 2 g ( z1 − hL ) = 2(9.81 m/s 2 )(3 − 1.5) m = 5.42 m/s Discussion Note that this is the discharge velocity at the beginning, and the velocity will decrease as the water level in the tank drops. The head loss in that case will change since it depends on velocity. 5-74 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-110 Solution The previous problem is reconsidered. The effect of the tank height on the initial discharge velocity of water from the completely filled tank as the tank height varies from 2 to 20 m in increments of 2 m at constant heat loss is to be investigated. The EES Equations window is printed below, followed by the tabulated and plotted results. Analysis g=9.81 "m/s2" rho=1000 "kg/m3" h_L=1.5 "m" D=0.10 "m" V_initial=SQRT(2*g*(z1-h_L)) "m/s" 16 14 Tank height, z1, m 2 3 4 5 6 7 8 9 10 11 12 Discussion Head Loss, hL, m 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 Initial velocity Vinitial, m/s 3.13 5.42 7.00 8.29 9.40 10.39 11.29 12.13 12.91 13.65 14.35 Vinitial, m/s 12 10 8 6 4 2 2 4 6 8 10 12 z1, m The dependence of V on height is not linear, but rather V changes as the square root of z1. 5-111 Solution A water tank open to the atmosphere is initially filled with water. A sharp-edged orifice at the bottom drains to the atmosphere through a long pipe equipped with a pump with a specified head loss. The required pump head to assure a certain velocity is to be determined. Assumptions 1 The flow is incompressible. 2 The draining pipe is horizontal. 3 The effect of the kinetic energy correction factor is negligible, α = 1. Analysis We take point 1 at the free surface of the tank, and point 2 at the exit of the pipe. We take the reference level at the centerline of the orifice (z2 = 0), and take the positive direction of z to be upwards. Noting that the fluid at both points is open to the atmosphere (and thus P1 = P2 = Patm) and that the fluid velocity at the free surface is very low (V1 ≅ 0), the energy equation between these two points (in terms of heads) simplifies to P1 V2 P V2 + α 1 1 + z1 + hpump, u = 2 + α 2 2 + z 2 + h turbine, e + h L 2g 2g ρg ρg → where α2 = 1 and the head loss is given to be hL = 1.5 m. Solving for hpump, u and substituting, the required useful pump head is determined to be hpump, u = V22 − z1 + hL = 2g (5 m/s) 2 2 2(9.81 m/s ) − (2 m) + (1.5 m) = 0.880 m Discussion Note that this is the required useful pump head at the beginning, and it will need to be increased as the water level in the tank drops to make up for the lost elevation head to maintain the constant discharge velocity. z1 + hpump, u = α 2 V 22 + hL 2g 1 Water 2m Pump 100 m 10 cm 5-75 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2 Chapter 5 Mass, Bernoulli, and Energy Equations 5-112E Solution A hose is connected to the bottom of a water tank open to the atmosphere. The hose is equipped with a pump and a nozzle at the end. The maximum height to which the water stream could rise is to be determined. Assumptions 1 The flow is incompressible with negligible friction. 2 The friction between the water and air is negligible. 3 We take the head loss to be zero (hL = 0) to determine the maximum rise of water jet. Properties 2 We take the density of water to be 62.4 lbm/ft3. Analysis We take point 1 at the free surface of the tank, and point 2 at the top of the water trajectory where V2 = 0. We take the reference level at the bottom of the tank. Noting that the fluid at both points is open to the atmosphere (and thus P1 = P2 = Patm) and that the fluid velocity at the free surface is very low (V1 ≅ 0), the energy equation for a control volume between these two points (in terms of heads) simplifies to 1 h 120 ft P1 V2 P V2 + α 1 1 + z1 + hpump, u = 2 + α 2 2 + z 2 + h turbine, e + h L → z1 + h pump, u = z 2 2g 2g ρg ρg where the useful pump head is ΔPpump ⎛ 144 lbf/ft 2 ⎞⎛ 32.2 lbm ⋅ ft/s 2 ⎞ 10 psi ⎟ = 23.1 ft ⎟⎜ ⎜ hpump, u = = ⎟ 1 lbf ρg (62.4 lbm/ft 3 )(32.2 ft/s 2 ) ⎜⎝ 1 psi ⎟⎠⎜⎝ ⎠ Substituting, the maximum height rise of water jet from the ground level is determined to be z 2 = z1 + hpump, u = 120 + 23.2 = 143.2 ft ≅ 143 ft Discussion The actual rise of water will be less because of the frictional effects between the water and the hose walls and between the water jet and air. 5-113 Solution A wind tunnel draws atmospheric air by a large fan. For a given air velocity, the pressure in the tunnel is to be determined. Assumptions 1The flow through the pipe is steady, incompressible, and irrotational with negligible friction (so that the Bernoulli equation is applicable). 2 Air is and ideal gas. Properties The gas constant of air is R = 0.287 kPa⋅m3/kg⋅K. Analysis We take point 1 in atmospheric air before it enters the wind tunnel (and thus P1 = Patm and V1 ≅ 0), and point 2 in the wind tunnel. Noting that z1 = z2 (or, the elevation effects are negligible for gases), the Bernoulli equation between points 1 and 2 gives P V2 P1 V12 + + z1 = 2 + 2 + z2 ρg 2 g ρg 2 g → P2 = P1 − ρV22 2 (1) Wind tunnel 1 where 20°C 101.3 kPa P 101.3 kPa ρ= = = 1.205 kg/m 3 RT (0.287 kPa ⋅ m 3 /kg ⋅ K)(293 K) 80 m/s 2 Substituting, the pressure in the wind tunnel is determined to be P2 = (101.3 kPa ) − (1.205 kg/m 3 ) (80 m/s) 2 2 ⎞⎛ 1 kPa ⎞ ⎛ 1N ⎟ ⎜⎜ = 97.4 kPa 2 ⎟⎜ 2⎟ ⎝ 1 kg ⋅ m/s ⎠⎝ 1000 N/m ⎠ Discussion Note that the velocity in a wind tunnel increases at the expense of pressure. In reality, the pressure will be even lower because of losses. 5-76 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations Design and Essay Problems 5-114 to 5-118 Solution Students’ essays and designs should be unique and will differ from each other. 5-119 Solution We are to evaluate a proposed modification to a wind turbine. Analysis Using the mass and the Bernoulli equations, it can be shown that this is a bad idea – the velocity at the exit of nozzle is equal to the wind velocity. (The velocity at nozzle inlet is much lower). Sample calculation using EES using a wind velocity of 10 m/s: V0=10 "m/s" rho=1.2 "kg/m3" g=9.81 "m/s2" A1=2 "m2" A2=1 "m2" A1*V1=A2*V2 P1/rho+V1^2/2=V2^2/2 m=rho*A1*V1 m*V0^2/2=m*V2^2/2 V0 =10 m/s V2 =10 m/s V1 =5 m/s Results: V1 = 5 m/s, V2 = 10 m/s, m = 12 kg/s (mass flow rate). Discussion Students’ approaches may be different, but they should come to the same conclusion – this does not help. 5-77 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-120 Solution We are to plot average pressure and average velocity versus axial location for flow through a convergingdiverging duct, and we are to discuss the results. Assumptions Analysis 1 The flow is steady, incompressible, and two-dimensional. The FlowLab template was run at V = 2 m/s. The velocity and pressure plots are shown below. Velocity distribution down the duct Pressure distribution down the duct The CFD results agree with our intuition – namely, the average velocity increases as the duct converges (cross-sectional area decreases), reaches a maximum at the throat, and then decreases as the duct diverges (cross-sectional area increases). According to the Bernoulli principle, we expect pressure to behave in the opposite manner, namely, P should decrease in the converging part of the duct, reach a minimum at the throat, and then increase in the diverging part of the duct. This also agrees with the CFD results. Discussion The CFD calculations, like the real flow, include irreversibilities, and therefore the Bernoulli approximation is not expected to yield exact quantitative results. This is tested in the following problem. However, qualitatively, we expect the trends to be predicted well. 5-78 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5 Mass, Bernoulli, and Energy Equations 5-121 Solution We are to calculate and plot average velocity and average pressure vs. axial location for convergingdiverging duct flow at a given inlet velocity, both analytically and using CFD, and we are to compare the results. Assumptions 1 The flow is steady, incompressible, and two-dimensional. 2 For the analytical analysis, we neglect irreversibilities such as friction. 3 In the CFD analysis, the width (into the page) is unity, i.e., 1 m. 4 The duct is horizontal – elevation differences do not play a role in this analysis. Analysis (a) We apply conservation of mass from the inlet to the outlet: ρVinlet Ainlet = ρVoutlet Aoutlet → Voutlet = Vinlet Ainlet m ⎛ 0.043 m 2 ⎞ m =2 ⎜ ⎟ = 3.44 Aoutlet s ⎝ 0.025 m 2 ⎠ s where density has dropped out because of the incompressible flow approximation. To estimate the pressure at the inlet, we neglect irreversibilities and apply the Bernoulli equation along a streamline from the inlet to the outlet, Pinlet + 1 1 ρVinlet 2 + ρ gzinlet = Poutlet + ρVoutlet 2 + ρ gzoutlet 2 2 → 1 Pinlet = Poutlet + ρ (Voutlet 2 − Vinlet 2 ) 2 2 2 ⎞ 1 1⎛ kg ⎞ ⎛ ⎛ m ⎞ ⎛ m ⎞ ⎞⎛ N N ρ (Voutlet 2 − Vinlet 2 ) = 0 + ⎜ 998.2 3 ⎟ ⎜ ⎜ 3.44 ⎟ − ⎜ 2 ⎟ ⎟ ⎜ = 3909.75 2 2 ⎟ ⎜ ⎟ 2 2⎝ m ⎠⎝⎝ s ⎠ ⎝ s ⎠ ⎠ ⎝ kg ⋅ m/s ⎠ m and when we substitute the outlet velocity from conservation of mass we get Pinlet = Poutlet + Ainlet 1 and P = Pinlet + ρ (Vinlet 2 − V 2 ) . We tabulate some of these analytical results below. A 2 where all pressures are gage pressures. In like manner, we calculate the average velocity and average pressure at other axial locations using V = Vinlet (b) We run FlowLab for the same inlet velocity and plot V and P as a functions of axial location x. On the same plot, we show the analytical predictions. The agreement is excellent for the velocity, since mass must be conserved whether irreversibilities are accounted for or not. The pressure agreement is reasonable, but friction leads to higher pressure drop in the duct; thus the CFD results (which account for irreversibilities) yield a higher required pressure to force the flow through the duct, as expected. …. etc. … table shortened for brevity ….. Discussion The Bernoulli equation does a reasonable job for this duct flow, but is not exact since the real flow has irreversibilites such as friction along the walls. We must keep in mind that the Bernoulli equation is only an approximation. KJ 5-79 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.