# 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

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#### Related posts:

#### 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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###### 2

Solution manuals for 9th edition and 9th SI Edition are sold separately.

###### KE

###### 22

###### 2 2

###### V

#### m

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

#### 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

#### WmaxEmechemmech( 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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###### V

#### e ke

Payment for second product (Solution manual for 9th SI edition).

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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

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

#### 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

If you have any questions, contact us here.

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

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

#### 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).

#### 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 kgm /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

- 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

######

######

######

######

######

######

######

######

######

######

######

######

######

######

######

######

######

- 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

#### W144,000lbf ft 185 Btu

#### 24 ft

#### 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 lbfft

######

######

######

######

######

######

######

######

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 1m1 kJ

###### 2

###### 1

###### ( )

###### 2

###### 1222

###### 1

###### 2

#### Wspring xk 2 x

#### 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

######

######

######

######

######

######

######

######

#### Wgmgz 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

######

######

######

######

######

######

######

#### Wa 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

######

######

######

######

######

######

######

######

#### Wgmgz z t

#### Thus,

#### WtotalWaWg 6 34 1. 43 kW

#### 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.

#### 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 EoutQout 60 , 000 Btu/h and

#### EinEpeopleElightsEapplianceEheater 6000 Btu/hEheater

#### Substituting, the required power rating of the heaters becomes

15 kW

######

######

######

######

######

3412 Btu/h

1 kW

#### Eheater 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

#### 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 (Elighting, 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.

#### 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

#### EinEout dE /dt dEroom/dtEin

#### since no energy is leaving the room in any form, and thus Eout 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/dtEin 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

### mairVA

#### 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

#### 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

#### VvertVsin 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.

#### 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