Thermodynamics Solutions Manual

Author(s): Yunus A. Cengel, Michael A. Boles

Solution Manual for Thermodynamics: An Engineering Approach – 9th Edition in SI units

Solution Manual for Thermodynamics: An Engineering Approach – 5th, 6th, 7th, 8th and 9th Edition

8th Edition

Yunus A. Cengel, Michael A. Boles

McGraw-Hill, 2015

Chapter 2

ENERGY, ENERGY TRANSFER, AND

GENERAL ENERGY ANALYSIS

PROPRIETARY AND CONFIDENTIAL

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surface tension effects, the total energy of a system consists of the kinetic, potential, and internal energies.

2- 2C The internal energy of a system is made up of sensible, latent, chemical and nuclear energies. The sensible internal

energy is due to translational, rotational, and vibrational effects.

2-3C Thermal energy is the sensible and latent forms of internal energy, and it is referred to as heat in daily life.

2-4C The 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.

2- 5C Hydrogen is also a fuel, since it can be burned, but it is not an energy source since there are no hydrogen reserves in

the world. Hydrogen can be obtained from water by using another energy source, such as solar or nuclear energy, and then

the hydrogen obtained can be used as a fuel to power cars or generators. Therefore, it is more proper to view hydrogen is an

energy carrier than an energy source.

2- 6C In electric heaters, electrical energy is converted to sensible internal energy.

2- 7C The forms of energy involved are electrical energy and sensible internal energy. Electrical energy is converted to

sensible internal energy, which is transferred to the water as heat.

2-8E The total kinetic energy of an object is given is to be determined.

Analysis The total kinetic energy of the object is given by

 0 Btu

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KE
22
2 2
V

m

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2-12 Wind is blowing steadily at a certain velocity. The mechanical energy of air per unit mass and the power generation

potential are to be determined.

Assumptions The wind is blowing steadily at a constant

uniform velocity.

Properties The density of air is given to be  = 1 kg/m

3

.

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 V

2

/2 per unit mass, and 2/

2

Vm for

a given mass flow rate:

.0 050 kJ/kg

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  
V

e ke

35 , 340 kg/s

4

File Specification for 9th Edition.

2
3
2
   

 

 

D

m VA V

WmaxEmechemmech( 35 , 340 kg/s)(0 50 kJ/kg) 1770 kW

Therefore, 1770 kW of actual power can be generated by this wind turbine at the stated conditions.

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.

2-13 A water jet strikes the buckets located on the perimeter of a wheel at a specified velocity and flow rate. The power

generation potential of this system is to be determined.

Assumptions Water jet flows steadily at the specified speed and flow rate.

Analysis Kinetic energy is the only form of harvestable mechanical

energy the water jet possesses, and it can be converted to work entirely.

Therefore, the power potential of the water jet is its kinetic energy,

which is V

2

/2 per unit mass, and 2/

2

Vm for a given mass flow rate:

8 kJ/kg

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Therefore, 216 kW of power can be generated by this water jet at the

stated conditions.

Discussion An actual hydroelectric turbine (such as the Pelton wheel) can convert over 90% of this potential to actual

electric power.

Wind

10 m/s

60 m

Wind

turbine

Shaft

Nozzle

Vj

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2-14 Two sites with specified wind data are being considered for wind power generation. The site better suited for wind

power generation is to be determined.

Assumptions 1 The wind is blowing steadily at specified velocity during specified times. 2 The wind power generation is

negligible during other times.

Properties We take the density of air to be  = 1 kg/m

3

(it

does not affect the final answer).

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 V

2

/2 per unit mass, and 2/

2

Vm for a given mass flow

rate. Considering a unit flow area (A = 1 m

2

), the maximum wind

power and power generation becomes

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1 kJ/kg

2

7( m/s)

2
22
2 2
1

mech, 1 1  

  
V

e ke

.0 050 kJ/kg1000 m /s

1 kJ/kg

2

( 10 m/s)

2
22
2 2
2

mech, 2 2  

  
V

e ke

.1( 25 kg/m)(7 m/s)(1m )(0 kJ/kg) 0 kW

3 2

W 1 max, Emech, 1  1 em mech, 1 V 1 Ake 1  

 

.1( 25 kg/m)(10 m/s)(1m )(0 kJ/kg) 0 kW

3 2

W 2 max, Emech, 2  2 em mech, 2 V 2 Ake 2  

 

since 1 kW = 1 kJ/s. Then the maximum electric power generations per year become

.0( 2144 kW)(3000 h/yr) (per m flow area)

2

E 1 max, W t11 max,   643 kWh/yr

.0( 625 kW)(1500 h/yr) (per m flow area)

2

E 2 max, W 2 max, t 2   938 kWh/yr

Therefore, second site is a better one for wind generation.

Discussion Note the power generation of a wind turbine is proportional to the cube of the wind velocity, and thus the

average wind velocity is the primary consideration in wind power generation decisions.

Wind

V, m/s

Wind

turbine

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Energy Transfer by Heat and Work

2-17C The form of energy that crosses the boundary of a closed system because of a temperature difference is heat; all other

forms are work.

2- 18C (a) The car's radiator transfers heat from the hot engine cooling fluid to the cooler air. No work interaction occurs in

the radiator.

(b) The hot engine transfers heat to cooling fluid and ambient air while delivering work to the transmission.

(c) The warm tires transfer heat to the cooler air and to some degree to the cooler road while no work is produced.

No work is produced since there is no motion of the forces acting at the interface between the tire and road.

(d) There is minor amount of heat transfer between the tires and road. Presuming that the tires are hotter than the

road, the heat transfer is from the tires to the road. There is no work exchange associated with the road since it cannot move.

(e) Heat is being added to the atmospheric air by the hotter components of the car. Work is being done on the air as

it passes over and through the car.

2- 19C (a) From the perspective of the contents, heat must be removed in order to reduce and maintain the content's

temperature. Heat is also being added to the contents from the room air since the room air is hotter than the contents.

(b) Considering the system formed by the refrigerator box when the doors are closed, there are three interactions,

electrical work and two heat transfers. There is a transfer of heat from the room air to the refrigerator through its walls.

There is also a transfer of heat from the hot portions of the refrigerator (i., back of the compressor where condenser is

placed) system to the room air. Finally, electrical work is being added to the refrigerator through the refrigeration system.

(c) Heat is transferred through the walls of the room from the warm room air to the cold winter air. Electrical work

is being done on the room through the electrical wiring leading into the room.

2- 20C It is a work interaction.

2- 21C It is a work interaction since the electrons are crossing the system boundary, thus doing electrical work.

2- 22C It is a heat interaction since it is due to the temperature difference between the sun and the room.

2- 23C This is neither a heat nor a work interaction since no energy is crossing the system boundary. This is simply the

conversion of one form of internal energy (chemical energy) to another form (sensible energy).

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2-24 The power produced by an electrical motor is to be expressed in different units.

Analysis Using appropriate conversion factors, we obtain

(a)  N 5 m/s

 
J 1

N 1 m

1 W

J/s 1

W 5( W)

(b)

2 3

 5 kgm /s

 
 
N 1

1 kg m/s

J 1

N 1 m

1 W

J/s 15( W)

2

W

2-25E The power produced by a model aircraft engine is to be expressed in different units.

Analysis Using appropriate conversion factors, we obtain

(a) 7 ft/s

 

1 Btu/s

  1. 169 lbf ft/s
1055. 056 W

1 Btu/s

W ( 10 W)

(b) 0 hp

745 7. W

1 hp

W ( 10 W)

Mechanical Forms of Work

2- 26C The work done is the same, but the power is different.

2- 27E A construction crane lifting a concrete beam is considered. The amount of work is to be determined considering (a)

the beam and (b) the crane as the system.

Analysis (a) The work is done on the beam and it is determined from

185 Btu

144,000lbf ft


 
 
   
  1. 169 lbf ft

1 Btu( 144 , 000 lbf ft)

( 24 ft)32. 174 lbmft/s

lbf 13( 2000 lbm)( 32. 174 ft/s )2

2

W mg z

(b) Since the crane must produce the same amount of work as is required

to lift the beam, the work done by the crane is

W144,000lbf ft 185 Btu

24 ft

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2- 31E The engine of a car develops 450 hp at 3000 rpm. The torque transmitted through the shaft is to be determined.

Analysis The torque is determined from

 

 788 lbfft

 
 

1 hp

550 lbf ft/s

2 3000/60/s

450 hp

2
T

sh

n 

W

2-32E The work required to expand a soap bubble is to be determined.

Analysis Noting that there are two gas-liquid interfaces in a soap bubble, the surface tension work is determined from

 

12 10 Btu

 6
 
   
    

778 ft

1 Btu.0 001909 lbf ft .0( 001909 lbf ft)

2 ( ) .0(2 005 lbf/ft /3() 12 ft) /5( 12 ft)

2 2
1 2
2
1

W sdA  A A 

2- 33 A linear spring is elongated by 20 cm from its rest position. The work done is to be determined.

Analysis The spring work can be determined from

   (70kN/m)(0 0)m  1m1 kJ

2
1
( )
2
1222
1
2

Wspring xk 2 x

PROPRIETARY MATERIAL. © 2015 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation.

2-34 A ski lift is operating steadily at 10 km/h. The power required to operate and also to accelerate this ski lift from rest to

the operating speed are to be determined.

Assumptions 1 Air drag and friction are negligible. 2 The average mass of each loaded chair is 250 kg. 3 The mass of

chairs is small relative to the mass of people, and thus the contribution of returning empty chairs to the motion is disregarded

(this provides a safety factor).

Analysis The lift is 1000 m long and the chairs are spaced 20 m apart. Thus at any given time there are 1000/20 = 50 chairs

being lifted. Considering that the mass of each chair is 250 kg, the load of the lift at any given time is

Load = (50 chairs)(250 kg/chair) = 12,500 kg

Neglecting the work done on the system by the returning empty chairs, the work needed to raise this mass by 200 m is

  24,525 kJ

1000 kgm /s

1 kJ(12,500 kg)(9 m/s)(200 m)22

2
21 

Wgmgz z 

At 10 km/h, it will take

t   

distance

velocity

1 km

10 km/ h

0 h 360 s

to do this work. Thus the power needed is

 68 kW

360 s

24,525 kJ

t

W
W

gg

The velocity of the lift during steady operation, and the acceleration during start up are

2 m/s

3 km/h

1 m/s( 10 km/h) 

V

2

.0 556 m/ss 5

.2 778 m/s 0- 

t

V

a

During acceleration, the power needed is

  /(5s) 9 kW

1000 m /s

1 kJ/kg(12,500 kg) .2( 778 m/s) 02

1
( /)
2
1
22
2 2
1
2
2  

Wa Vm V t 

Assuming the power applied is constant, the acceleration will also be constant and the vertical distance traveled during

acceleration will be

(0 m/s)(5 s)(0) 1

2
1

1000 m

200 m

2

1

sin2

12222

h at  at  

and

  /(5s) 34 1. kW

1000 kgm/s

1 kJ/kg/ (12,500 kg)(9 m/s)(1 m)22

2
21  

Wgmgz z t

Thus,

WtotalWaWg 6  34 1. 43 kW

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The First Law of Thermodynamics

2- 37C Energy can be transferred to or from a control volume as heat, various forms of work, and by mass transport.

2- 38C Warmer. Because energy is added to the room air in the form of electrical work.

2- 39 Water is heated in a pan on top of a range while being stirred. The energy of the water at the end of the process is to be

determined.

Assumptions The pan is stationary and thus the changes in kinetic and potential energies are negligible.

Analysis We take the water in the pan as our system. This is a closed system since no mass enters or leaves. Applying the

energy balance on this system gives

35 kJ

   
    
  
2
2

in sh,in out 2 1

potenti etc. al, energies

Change in internal, kinetic,

system

by heat, work,and mass

Netenergy transfer

in out

30 kJ kJ5 kJ5 10 kJ

U
U
Q W Q U U U
E E E
 

Therefore, the final internal energy of the system is 35 kJ.

2- 40E Water is heated in a cylinder on top of a range. The change in the energy of the water during this process is to be

determined.

Assumptions The pan is stationary and thus the changes in kinetic and potential energies are negligible.

Analysis We take the water in the cylinder as the system. This is a closed system since no mass enters or leaves. Applying

the energy balance on this system gives

    52 Btu

  
    
  
2 1

in out out 2 1

potenti etc. al, energies

Change in internal, kinetic,

system

by heat, work,and mass

Netenergy transfer

in out

65 Btu 5 Btu 8 Btu

U U U
U
Q W Q U U U
E E E
 

Therefore, the energy content of the system increases by 52 Btu during this process.

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2- 41E The heat loss from a house is to be made up by heat gain from people, lights, appliances, and resistance heaters. For a

specified rate of heat loss, the required rated power of resistance heaters is to be determined.

Assumptions 1 The house is well-sealed, so no air enters or heaves the house. 2 All the lights and appliances are kept on. 3

The house temperature remains constant.

Analysis Taking the house as the system, the energy balance can be written as

/ 0

potential, etc. energies

Rateof changein internal, kinetic,

0 (steady)system

by heat, work, mass and

Rateofnet energy transfer

in out      

E E dE dt 

Ein Eout

 

where EoutQout 60 , 000 Btu/h and

EinEpeopleElightsEapplianceEheater 6000 Btu/hEheater

Substituting, the required power rating of the heaters becomes

15 kW

  

3412 Btu/h

1 kW

Eheater 60 , 000 6000 54 , 000 Btu/h

Discussion When the energy gain of the house equals the energy loss, the temperature of the house remains constant. But

when the energy supplied drops below the heat loss, the house temperature starts dropping.

2-42E A water pump increases water pressure. The power input is to be determined.

Analysis The power input is determined from

11 hp

 
 

0 Btu/s

1 hp

5 psia ft

1 Btu8( ft /s)( 70 15 )psia

( )
3
3

W V P 2 P 1

 

The water temperature at the inlet does not have any significant effect on the required power.

Water

15 psia

70 psia

HOUSE

Energy

  • Lights

  • People

  • Appliance

  • Heaters

Qout

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2-45 The classrooms and faculty offices of a university campus are not occupied an average of 4 hours a day, but the lights

are kept on. The amounts of electricity and money the campus will save per year if the lights are turned off during

unoccupied periods are to be determined.

Analysis The total electric power consumed by the lights in the classrooms and faculty offices is

264 264 528 kW

(Power consumed per lamp) (No. of lamps)=(400 6 110 W)=264,000 264 kW

(Power consumed per lamp) (No. of lamps)=(200 12 110 W)=264,000 264 kW

lighting, total lighting, classroom lighting, offices

lighting, offices

lighting, classroom

    
    
    
E E E
E
E
  

Noting that the campus is open 240 days a year, the total number of unoccupied work hours per year is

Unoccupied hours = (4 hours/day)(240 days/year) = 960 h/yr

Then the amount of electrical energy consumed per year during unoccupied work period and its cost are

  $55,757/yr

  

Cost savings (Energy savings)(Unit cost of energy) (506,880 kWh/yr)($0/kWh)

Energy savings (Elighting, total)(Unoccupied hours) (528 kW)(960 h/yr) 506,880 kWh

Discussion Note that simple conservation measures can result in significant energy and cost savings.

2- 46 An industrial facility is to replace its 40-W standard fluorescent lamps by their 35-W high efficiency counterparts. The

amount of energy and money that will be saved a year as well as the simple payback period are to be determined.

Analysis The reduction in the total electric power consumed by the lighting as a result of switching to the high efficiency

fluorescent is

Wattage reduction = (Wattage reduction per lamp)(Number of lamps)

= (40 - 34 W/lamp)(700 lamps)

= 4200 W

Then using the relations given earlier, the energy and cost savings associated with the replacement of the high efficiency

fluorescent lamps are determined to be

Energy Savings = (Total wattage reduction)(Ballast factor)(Operating hours)

= (4 kW)(1)(2800 h/year)

= 12,936 kWh/year

Cost Savings = (Energy savings)(Unit electricity cost)

= (12,936 kWh/year)($0. 105 /kWh)

= $1358/year

The implementation cost of this measure is simply the extra cost of the energy efficient

fluorescent bulbs relative to standard ones, and is determined to be

Implementation Cost = (Cost difference of lamps)(Number of lamps)

= [($2-$1)/lamp](700 lamps)

= $

This gives a simple payback period of

(3 months)

$ 1358 /year

$ 343

Annual cost savings

Implementation cost

Simple payback period =  0 year

Discussion Note that if all the lamps were burned out today and are replaced by high-efficiency lamps instead of the

conventional ones, the savings from electricity cost would pay for the cost differential in about 4 months. The electricity

saved will also help the environment by reducing the amount of CO 2 , CO, NOx, etc. associated with the generation of

electricity in a power plant.

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2- 47 A room contains a light bulb, a TV set, a refrigerator, and an iron. The rate of increase of the energy content of the

room when all of these electric devices are on is to be determined.

Assumptions 1 The room is well sealed, and heat loss from the room is negligible. 2 All the appliances are kept on.

Analysis Taking the room as the system, the rate form of the energy balance can be written as

 
 

potential, etc. energies

Rateof change in internal, kinetic,

system

by heat, work,and mass

Rateof netenergy transfer

EinEout  dE /dt  dEroom/dtEin

since no energy is leaving the room in any form, and thus Eout 0. Also,

1650 W
40 110 300 1200 W

in lights TV refrig iron

   
E E E E E

Substituting, the rate of increase in the energy content of the room becomes

dEroom/dtEin 1650 W

Discussion Note that some appliances such as refrigerators and irons operate intermittently, switching on and off as

controlled by a thermostat. Therefore, the rate of energy transfer to the room, in general, will be less.

2- 48E A fan accelerates air to a specified velocity in a square duct. The minimum electric power that must be supplied to the

fan motor is to be determined.

Assumptions 1 The fan operates steadily. 2 There are no conversion losses.

Properties The density of air is given to be  = 0 lbm/ft

3

.

Analysis A fan motor converts electrical energy to mechanical shaft energy, and the fan transmits the mechanical energy of

the shaft (shaft power) to mechanical energy of air (kinetic energy). For a control volume that encloses the fan-motor unit,

the energy balance can be written as

/ 0

potential, etc. energies

Rateof change in internal, kinetic,

0 (steady)system

by heat, work,and mass

Rateof netenergy transfer

  
    

E E dE dt

in out  Ein Eout

 

2

ke

2

outelect,in air out air

V

W m m

where

.0( 075 lbm/ft)(3 3 ft )(22 ft/s) 14. 85 lbm/s

3 2

mairVA  

Substituting, the minimum power input required is determined to be

  151 W

  .0 1435 Btu/s 25,037 ft /s

1 Btu/lbm

2

( 22 ft/s)1(4 lbm/s)2 22

2 2

outin air

V

W m

since 1 Btu = 1 kJ and 1 kJ/s = 1000 W.

Discussion The conservation of energy principle requires the energy to be conserved as it is converted from one form to

another, and it does not allow any energy to be created or destroyed during a process. In reality, the power required will be

considerably higher because of the losses associated with the conversion of electrical-to-mechanical shaft and mechanical

shaft-to-kinetic energy of air.

ROOM

Electricity

  • Lights

  • TV

  • Refrig

  • Iron

PROPRIETARY MATERIAL. © 2015 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation.

2-51 An inclined escalator is to move a certain number of people upstairs at a constant velocity. The minimum power

required to drive this escalator is to be determined.

Assumptions 1 Air drag and friction are negligible. 2 The average mass of each person is 75 kg. 3 The escalator operates

steadily, with no acceleration or breaking. 4 The mass of escalator itself is negligible.

Analysis At design conditions, the total mass moved by the escalator at any given time is

Mass = (50 persons)(75 kg/person) = 375 0 kg

The vertical component of escalator velocity is

VvertVsin 45  6( m/s)sin45

Under stated assumptions, the power supplied is used to increase the potential energy of people. Taking the people on

elevator as the closed system, the energy balance in the rate form can be written as

/ 0

potential, etc. energies

Rateof changein internal, kinetic,

system

by heat, work, mass and

Rateofnet energy transfer

in out   

E E dE dt 

t

E

E dE dt 

 

sysin sys/

in mgVvert t

mg z

t

PE
W 

 

That is, under stated assumptions, the power input to the escalator must be equal to the rate of increase of the potential

energy of people. Substituting, the required power input becomes

 15 kW

   12 5. kJ/s 1000 m /s

1 kJ/kg(3750 kg)(9 m/s )(0 m/s)sin22

2

Win mgVvert

When the escalator velocity is doubled to V = 1 m/s, the power needed to drive the escalator becomes

 31 kW

   25 0. kJ/s 1000 m /s

1 kJ/kg(3750 kg)(9 m/s)(1 m/s)sin22

2

Win mgVvert

Discussion Note that the power needed to drive an escalator is proportional to the escalator velocity.

PROPRIETARY MATERIAL. © 2015 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation.

2- 52 A car cruising at a constant speed to accelerate to a specified speed within a specified time. The additional power

needed to achieve this acceleration is to be determined.

Assumptions 1 The additional air drag, friction, and rolling resistance are not considered. 2 The road is a level road.

Analysis We consider the entire car as the system, except that let’s assume the power is supplied to the engine externally for

simplicity (rather that internally by the combustion of a fuel and the associated energy conversion processes). The energy

balance for the entire mass of the car can be written in the rate form as

/ 0

potential, etc. energies

Rateof changein internal, kinetic,

system

by heat, work, mass and

Rateofnet energy transfer

in out   

E E dE dt 

t

E

E dE dt 

 

sysin sys/

t

Vm V

t

KE
W
( 2/)
2
1
2
2

in

since we are considering the change in the energy content of the car due to

a change in its kinetic energy (acceleration). Substituting, the required

additional power input to achieve the indicated acceleration becomes

 77 kW

 77 8. kJ/s 1000 m /s

1 kJ/kg

2(5s)

(110/3 m/s) -(70/3 m/s)(1400 kg)2 22

2 2 2
1
2
2

in t

V V

W m

since 1 m/s = 3 km/h. If the total mass of the car were 700 kg only, the power needed would be

38 kW

22
2 2 2
1
2
2

in 1000 m /s

1 kJ/kg

2(5s)

(110/3 m/s) -(70/3 m/s)(700 kg)2 t

V V

W m

Discussion Note that the power needed to accelerate a car is inversely proportional to the acceleration time. Therefore, the

short acceleration times are indicative of powerful engines.

Energy Conversion Efficiencies

2-53C 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,

elect,in

pump

elect,in

mech,fluid

elect,in

mech,out mech,inpump-motor pumpmotorW

W
W
E
W
E E
 

   

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.

2-54C T he turbine efficiency, generator efficiency, and combined turbine-generator efficiency are defined as follows:

Mechanical energy extracted from the fluid | |

Mechanical energy output

mech,fluid

shaft,outturbineE

W

  

shaft,in

elect,outgeneratorMechanical power input

Electrical power output

W
W

  

| mech,fluid|

elect,out

mech,in mech,out

elect,outturbinegen- turbinegeneratorE

W
E E
W
 

   

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