CHAPTER 5 BERNOULLI AND ENERGY EQUATIONS

Chapter 5 Mass, Bernoulli, and Energy Equations5-13SolutionA smoking lounge that can accommodate 40 smokers is considered. The required minimum flow rate of airthat needs to be supplied to the lounge and the diameter of the duct are to be determined.AssumptionsInfiltration of air into the smoking lounge is negligible.PropertiesThe minimum fresh air requirements for a smoking lounge is given to be 30 L/s per person.AnalysisThe required minimum flow rate of air that needs to be supplied to the lounge is determined directly from V air V air per person ( No. of persons) (30 L/s person)(40 persons) 1200 L/s 1.2 m 3 /sSmoking LoungeThe volume flow rate of fresh air can be expressed asV VA V ( D 2 / 4 )40 smokers30 L/s personSolving for the diameter D and substituting,D 4V V4(1.2 m 3 /s) 0.437 m (8 m/s)Therefore, the diameter of the fresh air duct should be at least 43.7 cm if the velocity of air is not to exceed 8 m/s.DiscussionFresh air requirements in buildings must be taken seriously to avoid health problems.5-14SolutionThe 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.The volume of the building and the required minimum volume flow rate of fresh air areAnalysisV room (2.7 m)(200 m 2 ) 540 m 3V V room ACH (540 m 3 )(0.35/h ) 189 m 3 / h 189,000 L/h 3150 L/minThe volume flow rate of fresh air can be expressed asV VA V ( D 2 / 4 )HouseSolving for the diameter D and substituting,D 4V V0.35 ACH200 m24(189 / 3600 m 3 /s) 0.116 m (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.DiscussionFresh air requirements in buildings must be taken seriously to avoid health problems.5-7PROPRIETARY MATERIAL. 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, orposted on a website, in whole or part.

Chapter 5 Mass, Bernoulli, and Energy Equations5-15SolutionAir is accelerated in a nozzle. The mass flow rate and the exit area of the nozzle are to be determined.AssumptionsFlow through the nozzle is steady.PropertiesThe 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 bem 1 A1V1 (2.21 kg/m 3 )(0.006 m 2 )(20 m/s) 0.2652 kg/s 0.265 kg/s 1 m 2 m .(b) There is only one inlet and one exit, and thus mV1 20 m/sA1 60 cm2AIRV2 150 m/sThen the exit area of the nozzle is determined to bem 2 A2V2 A2 0.2652 kg/sm 0.00232 m 2 23.2 cm 2 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.Discussion5-16SolutionAssumptionsAnalysisAir flows in a varying crosss section pipe. The speed at a specified section is to be determined.Flow through the pipe is steady.Applying conservation of mass for the cv shown, t cv .d .V . n .dA 0, .V1. A1 2 . A2 .V2 0csPabsRT, 1 P1( abs )RT1, d 2 2 P2( abs ) 1 P1( abs ), 2 V1 2 . A2 .V2, 1. A1P2( abs )RT2A2 d 4 A1 D 2 D 422110 1 V1 . .30 2.44 m/s150 3 5-8PROPRIETARY MATERIAL. 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, orposted on a website, in whole or part.

Chapter 5 Mass, Bernoulli, and Energy Equations5-17SolutionAir is expanded and is accelerated as it is heated by a hair dryer of constant diameter. The percent increasein the velocity of air as it flows through the drier is to be determined.AssumptionsFlow through the nozzle is steady.PropertiesThe 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 mm 1 m 2 1 AV1 2 AV2V2V2 1 1.20 kg/m 1.14V1 2 1.05 kg/m3V13(or, an increase of 14%)Therefore, the air velocity increases 14% as it flows through the hair drier.DiscussionIt makes sense that the velocity increases since the density decreases, but the mass flow rate is constant.Mechanical Energy and Efficiency5-18CSolutionAnalysisWe are to define and discuss turbine, generator, and turbine-generator efficiency.Turbine efficiency, generator efficiency, and combined turbine-generator efficiency are defined as follows:W shaft,outMechanical energy output turbine Mechanical energy extracted from the fluid E mech,fluid generator Electrical power output W elect,out Mechanical power input W shaft,in turbine-gen turbine generaor DiscussionW elect,out E E mech,inmech,out W elect,out E mech,fluid Most turbines are connected directly to a generator, so the combined efficiency is a useful parameter.5-9PROPRIETARY MATERIAL. 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, orposted on a website, in whole or part.

Chapter 5 Mass, Bernoulli, and Energy Equations5-19CSolutionWe are to define and discuss mechanical efficiency.AnalysisMechanical efficiency is defined as the ratio of the mechanical energy output to the mechanical energyinput. A mechanical efficiency of 100% for a hydraulic turbine means that the entire mechanical energy of the fluid isconverted to mechanical (shaft) work.DiscussionNo real fluid machine is 100% efficient, due to frictional losses, etc. – the second law of thermodynamics.5-20CSolutionWe are to define and discuss pump-motor efficiency.AnalysisThe combined pump-motor efficiency of a pump/motor system is defined as the ratio of the increase in themechanical energy of the fluid to the electrical power consumption of the motor,W pumpE mech,out E mech,in E mech,fluid pump -motor pump motor W W W elect,inelect,inelect,inThe combined pump-motor efficiency cannot be greater than either of the pump or motor efficiency since both pump andmotor efficiencies are less than 1, and the product of two numbers that are less than one is less than either of the numbers.DiscussionSince many pumps are supplied with an integrated motor, pump-motor efficiency is a useful parameter.5-21CSolutionWe are to discuss mechanical energy and how it differs from thermal energy.AnalysisMechanical energy is the form of energy that can be converted to mechanical work completely anddirectly by a mechanical device such as a propeller. It differs from thermal energy in that thermal energy cannot beconverted to work directly and completely. The forms of mechanical energy of a fluid stream are kinetic, potential, andflow energies.DiscussionIt would be nice if we could convert thermal energy completely into work. However, this would violate thesecond law of thermodynamics.5-10PROPRIETARY MATERIAL. 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, orposted on a website, in whole or part.

Chapter 5 Mass, Bernoulli, and Energy Equations5-22SolutionWind is blowing steadily at a certain velocity. The mechanical energy of air per unit mass, the powergeneration 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 isindependent of the wind speed.PropertiesThe density of air is given to be 1.25 kg/m3.AnalysisKinetic energy is the only form of mechanical energy the wind possesses, and it can be converted to workentirely. Therefore, the power potential of the wind is its kinetic energy, which is V2/2 per unit mass, and m V 2 / 2 for agiven mass flow rate:emech ke V 2 (8 m/s) 2 1 kJ/kg 22 1000 m 2 /s 2m VA V D 24 0.032 kJ/kg (1.25 kg/m 3 )(8 m/s) (50 m) 24 19,635 kg/sW max E mech m emech (19,635 kg/s)(0.032 kJ/kg) 628 kWThe actual electric power generation is determined by multiplying the power generation potential by the efficiency,W elect wind turbineW max (0.30)(628 kW) 188 kWTherefore, 283 kW of actual power can be generated by this wind turbine at the stated conditions.WindWindturbine8 m/s50 mDiscussionThe power generation of a wind turbine is proportional to the cube of the wind velocity, and thus the powergeneration will change strongly with the wind conditions.5-11PROPRIETARY MATERIAL. 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, orposted on a website, in whole or part.

Chapter 5 Mass, Bernoulli, and Energy Equations5-23SolutionThe previous problem is reconsidered. The effect of wind velocity and the blade span diameter on windpower 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 to80 m in increments of 20 m is to be investigated.AnalysisThe 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 0.30rho 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, m20406080V, m/s5101520510152051015205101520m, ct, 5-12PROPRIETARY MATERIAL. 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, orposted on a website, in whole or part.

Chapter 5 Mass, Bernoulli, and Energy EquationsDiscussionWind turbine power output is obviously nonlinear with respect to both velocity and diameter.5-13PROPRIETARY MATERIAL. 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, orposted on a website, in whole or part.

Chapter 5 Mass, Bernoulli, and Energy Equations5-24ESolutionA differential thermocouple indicates that the temperature of water rises a certain amount as it flowsthrough 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 ofwater is completely due to frictional heating. 2 Water is an incompressible substance.PropertiesWe take the density of water to be 62.4 lbm/ft3 and its specific heat to be c 1.0 Btu/lbm F.AnalysisThe 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 V (62.4 lbm/ft 3 )(1.5 ft 3/s) 93.6 lbm/s T 0.048 FE mech, loss U m c T1 hp (93.6 lbm/s)(1.0 Btu/lbm F)(0.048 F) 6.36 hp 0.7068 Btu/s The mechanical efficiency of the pump is determined from the generaldefinition of mechanical efficiency, pump 1 E mech,loss6.36 hp 1 0.724 or 72.4% 23 hpWmech, inPump23 hpDiscussionNote 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 tofrictional heating is usually negligible in heat transfer analysis.5-14PROPRIETARY MATERIAL. 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, orposted on a website, in whole or part.

Chapter 5 Mass, Bernoulli, and Energy Equations5-25SolutionA hydraulic turbine-generator is generating electricity from the water of a large reservoir. The combinedturbine-generator efficiency and the turbine efficiency are to be determined.1 The elevation of the reservoir remains constant. 2 The mechanical energy of water at the turbine exit isAssumptionsnegligible.AnalysisWe take the free surface of the reservoir to be point 1 and the turbine exit to be point 2. We also take theturbine exit as the reference level (z2 0), and thus the potential energy at points 1 and 2 are pe1 gz1 and pe2 0. The flowenergy P/ at both points is zero since both 1 and 2 are open to the atmosphere (P1 P2 Patm). Further, the kinetic energyat both points is zero (ke1 ke2 0) since the water at point 1 is essentially motionless, and the kinetic energy of water atturbine 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 )(110 m) 1.079 kJ/kg 1000 m 2 /s 2 1Then the rate at which the mechanical energy of the fluid is suppliedto the turbine become E mech,fluid m (e mech,in e mech,out ) m ( pe1 0) m pe1 (900 kg/s)(1.079 kJ/kg)750 kW110 m 971.2 kWThe combined turbine-generator and the turbine efficiency aredetermined from their definitions, turbine-gen turbine W elect,out750 kW 0.772 or 77.2% E mech,fluid 971.2 kWGeneratorTurbine2W shaft,out800 kW 0.824 or 82.4% E mech,fluid 971.2 kWTherefore, the reservoir supplies 971.2 kW of mechanical energy to the turbine, which converts 800 kW of it to shaft workthat drives the generator, which generates 750 kW of electric power.DiscussionThis problem can also be solved by taking point 1 to be at the turbine inlet, and using flow energy instead ofpotential energy. It would give the same result since the flow energy at the turbine inlet is equal to the potential energy atthe free surface of the reservoir.5-15PROPRIETARY MATERIAL. 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, orposted on a website, in whole or part.

Chapter 5 Mass, Bernoulli, and Energy Equations5-26SolutionA river is flowing at a specified velocity, flow rate, and elevation. The total mechanical energy of the riverwater 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 averagevelocity. 3 The mechanical energy of water at the turbine exit is negligible.PropertiesWe take the density of water to be 1000 kg/m3.AnalysisNoting that the sum of the flow energy and the potential energy is constant for a given fluid body, we cantake the elevation of the entire river water to be the elevation of the free surface, and ignore the flow energy. Then the totalmechanical energy of the river water per unit mass becomesemech pe ke gh V22 (4 m/s) 2 (9.81 m/s 2 )(70 m) 2 0.695 kJ/kgRiver 1 kJ/kg 1000 m 2 /s 2 4 m/s70 mThe power generation potential of the river water is obtained by multiplying the totalmechanical energy by the mass flow rate,m V (1000 kg/m 3 )(500 m 3 /s) 500,000 kg/sW max E mech m e mech (500,000 kg/s)(0.695 kg/s) 347,350 kW 347 MWTherefore, 347 MW of power can be generated from this river as it discharges into the lake if its power potential can berecovered completely.DiscussionNote that the kinetic energy of water is negligible compared to the potential energy, and it can be ignored inthe analysis. Also, the power output of an actual turbine will be less than 347 MW because of losses and inefficiencies.5-16PROPRIETARY MATERIAL. 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, orposted on a website, in whole or part.

Chapter 5 Mass, Bernoulli, and Energy Equations5-27SolutionWater is pumped from a lake to a storage tank at a specified rate. The overall efficiency of the pump-motorunit 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. 3The changes in kinetic energy are negligible. 4 The elevation difference across the pump is negligible.PropertiesWe take the density of water to be 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 andpe2 gz2. The flow energy at both points is zero since both 1 and 2 are open to the atmosphere (P1 P2 Patm). Further, thekinetic energy at both points is zero (ke1 ke2 0) since the water at both locations is essentially stationary. The mass flowrate of water and its potential energy at point 2 arem V (1000 kg/m 3 )(0.070 m 3/s) 70 kg/s 1 kJ/kgpe1 gz1 (9.81 m/s 2 )(18 m) 1000 m 2 /s 2 0.177 kJ/kg Then the rate of increase of the mechanical energy of water becomes E mech,fluid m (e mech,out e mech,in ) m ( pe2 0) m pe2 (70 kg/s)(0.177 kJ/kg) 12.4 kWThe overall efficiency of the combined pump-motor unit is determined from its definition, pump-motor E mech,fluid 12.4 kW 0.606 or 60.6%20.4 kWW elect,in(b) Now we consider the pump. The change in the mechanical energy ofwater as it flows through the pump consists of the change in the flowenergy only since the elevation difference across the pump and thechange in the kinetic energy are negligible. Also, this change must beequal to the useful mechanical energy supplied by the pump, which is12.4 kW:P P1 E mech,fluid m (e mech,out e mech,in ) m 2 V P 2Storagetank18 mPump1Solving for P and substituting, P E mech,fluid12.4 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 inorder to raise its elevation by 18 m.DiscussionNote that only two-thirds of the electric energy consumed by the pump-motor is converted to themechanical energy of water; the remaining one-third is wasted because of the inefficiencies of the pump and the motor.5-17PROPRIETARY MATERIAL. 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use.Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, orposted on a website, in whole or part.

Chapter 5 Mass, Bernoulli, and Energy EquationsBernoulli Equation5-28CSolutionWe are to define stagnation pressure and discuss how it can be meas

Fluid Mechanics: Fundamentals and Applications Third Edition Yunus A. Çengel & John M. Cimbala McGraw-Hill, 2013 CHAPTER 5 BERNOULLI AND ENERGY EQUATIONS PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and pr