Solution Manual Thermodynamics 8th Edition

  0 Btu 25 , 037 ft /s. ( 50 )ft/s( 10 lbm)2. (60m).1( 25 kg/m)( 10 m/s)4. 1 kW( 120 kg/s)(1 kJ/kg).

Yunus A. Cengel, Michael A. Boles

Wmax Emech emmech. .0 0245 kJ/kg1000 m /s. .0 050 kJ/kg1000 m /s. .1( 25 kg/m)(7 m/s)(1m )(0 kJ/kg) 0 kW. .1( 25 kg/m)(10 m/s)(1m )(0 kJ/kg) 0 kW. .0( 2144 kW)(3000 h/yr) (per m flow area).

ENERGY, ENERGY TRANSFER, AND

Chapter 2

GENERAL ENERGY ANALYSIS

Solutions Manual for

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Thermodynamics: An Engineering Approach

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mairVA  

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

8th Edition

E 1 max, W t11 max,   643 kWh/yr. .0( 625 kW)(1500 h/yr) (per m flow area). E 2 max, W 2 max, t 2   938 kWh/yr.

PROPRIETARY AND CONFIDENTIAL

1 Btu( 144 , 000 lbf ft). lbf 13( 2000 lbm)( 32. 1 Btu.0 001909 lbf ft .0( 001909 lbf ft). 2 ( ) .0(2 005 lbf/ft /3() 12 ft) /5( 12 ft). 1 kJ(12,500 kg)(9 m/s)(200 m)22. 1 m/s( 10 km/h) .

.2 778 m/s 0- . 1 kJ/kg(12,500 kg) .2( 778 m/s) 02. 1 kJ/kg/ (12,500 kg)(9 m/s)(1 m)22. in sh,in out 2 1.

Change in internal, kinetic,.

by heat, work,and mass. Netenergy transfer. 30 kJ kJ5 kJ5 10 kJ. Change in internal, kinetic,. by heat, work,and mass. Netenergy transfer. 65 Btu 5 Btu 8 Btu. potential, etc. Rateof changein internal, kinetic,. 0 (steady)system.

by heat, work, mass and. Rateofnet energy transfer. in out      .

professors and instructors for use in preparing for the classes using the affiliated textbook. No

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

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

(Power consumed per lamp) (No.

without the prior written permission of McGraw-Hill Education.
(b) Considering the system formed by the refrigerator box when the doors are closed, there are three interactions,
Analysis We take the water in the pan as our system. This is a closed system since no mass enters or leaves. Applying the
(b)
2- 38C Warmer. Because energy is added to the room air in the form of electrical work.
in out  Ein Eout

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.   $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.

Annual cost savings.

Implementation cost.

Analysis The reduction in the total electric power consumed by the lighting as a result of switching to the high efficiency
The First Law of Thermodynamics
a
2-3C Thermal energy is the sensible and latent forms of internal energy, and it is referred to as heat in daily life.
/2 per unit mass, and 2/
The water temperature at the inlet does not have any significant effect on the required power.
2- 48E A fan accelerates air to a specified velocity in a square duct. The minimum electric power that must be supplied to the
2-8E The total kinetic energy of an object is given is to be determined.
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
WmaxEmechemmech( 35 , 340 kg/s)(0 50 kJ/kg) 1770 kW
Assumptions The pan is stationary and thus the changes in kinetic and potential energies are negligible.
The combined pump-motor efficiency cannot be greater than either of the pump or motor efficiency since both pump and
Energy Savings = (Total wattage reduction)(Ballast factor)(Operating hours)
Analysis The power input is determined from
EinEpeopleElightsEapplianceEheater 6000 Btu/hEheater
a change in its kinetic energy (acceleration). Substituting, the required
Discussion Note that the power needed to drive an escalator is proportional to the escalator velocity.
 
Analysis (a) The work is done on the beam and it is determined from
directly and completely. The forms of mechanical energy of a fluid stream are kinetic, potential, and flow energies.
(c) The warm tires transfer heat to the cooler air and to some degree to the cooler road while no work is produced.
W
power and power generation becomes
e ke
WtotalWaWg 6  34 1. 43 kW
Analysis At design conditions, the total mass moved by the escalator at any given time is
Analysis We consider the entire car as the system, except that let’s assume the power is supplied to the engine externally for
Noting that the campus is open 240 days a year, the total number of unoccupied work hours per year is
energy the water jet possesses, and it can be converted to work entirely.
where
2-51 An inclined escalator is to move a certain number of people upstairs at a constant velocity. The minimum power
The implementation cost of this measure is simply the extra cost of the energy efficient
stated conditions.
kinetic energy, which is V
Cost Savings = (Energy savings)(Unit electricity cost)
energy carrier than an energy source.
specified rate of heat loss, the required rated power of resistance heaters is to be determined.

potential, etc.

Rateof change in internal, kinetic,.

by heat, work,and mass.

Rateof netenergy transfer.

in lights TV refrig iron.

potential, etc.

W 5( W)

Rateof change in internal, kinetic,.

0 (steady)system.

by heat, work,and mass.

Rateof netenergy transfer.

outelect,in air out air.

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

  .0 1435 Btu/s 25,037 ft /s.

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

potential, etc.

Rateof changein internal, kinetic,.

by heat, work, mass and.

Rateofnet energy transfer.

in out   .

1 kJ/kg(3750 kg)(9 m/s )(0 m/s)sin22. 1 kJ/kg(3750 kg)(9 m/s)(1 m/s)sin22. potential, etc. Rateof changein internal, kinetic,.

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

by heat, work, mass and. Rateofnet energy transfer. in out   . (110/3 m/s) -(70/3 m/s)(1400 kg)2 22. (110/3 m/s) -(70/3 m/s)(700 kg)2 t. mech,out mech,inpump-motor pumpmotorW.

Mechanical energy extracted from the fluid | |. Mechanical energy output. elect,outgeneratorMechanical power input.

Electricity

Electrical power output.

Vm for

mech,in mech,out.

= 12,936 kWh/year
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
is being done on the room through the electrical wiring leading into the room.
Substituting, the rate of increase in the energy content of the room becomes
simplicity (rather that internally by the combustion of a fuel and the associated energy conversion processes). The energy
At 10 km/h, it will take

elect,outturbinegen- turbinegeneratorE.

Then the net force that acts on the rock is

F net F up F down 200  29. 37  170 N 6.

From the Newton's second law, the acceleration of the rock becomes

 56 m/s 2 

2-12 Wind is blowing steadily at a certain velocity. The mechanical energy of air per unit mass and the power generation
(it
V, m/s
= (12,936 kWh/year)($0. 105 /kWh)
 31 kW
Vm for a given mass flow
temperature. Heat is also being added to the contents from the room air since the room air is hotter than the contents.

1 kg m/s3 kg

   (70kN/m)(0 0)m  1m1 kJ

m

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

a

Stone

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

4- 13 2 Problem 4- 13 1 is reconsidered. The effect of the environment temperature on the final pressure and the heattransfer as the environment temperature varies from 0°C to 50°C is to be investigated. The final results are to be plottedagainst the environment temperature.Analysis The problem is solved using EES, and the solution is given below.

"Knowns"Vol_A=0 [m^3]P_A[1]=400 [kPa]x_A[1]=0.T_B[1]=250 [C]P_B[1]=200 [kPa]Vol_B=0 [m^3]T_final=25 [C] "T_final = T_surroundings. To do the parametric studyor to solve the problem when Q_out = 0, place this statement in {}."{Q_out=0 [kJ]} "To determine the surroundings temperature thatmakes Q_out = 0, remove the {} and resolve the problem."

"Solution""Conservation of Energy for the combined tanks:"E_in-E_out=DELTAEE_in=E_out=Q_outDELTAE=m_A(u_A[2]-u_A[1])+m_B(u_B[2]-u_B[1])m_A=Vol_A/v_A[1]m_B=Vol_B/v_B[1]Fluid$='Steam_IAPWS'u_A[1]=INTENERGY(Fluid$,P=P_A[1], x=x_A[1])v_A[1]=volume(Fluid$,P=P_A[1], x=x_A[1])T_A[1]=temperature(Fluid$,P=P_A[1], x=x_A[1])u_B[1]=INTENERGY(Fluid$,P=P_B[1],T=T_B[1])v_B[1]=volume(Fluid$,P=P_B[1],T=T_B[1])"At the final state the steam has uniform properties through out the entire system."u_B[2]=u_finalu_A[2]=u_finalm_final=m_A+m_BVol_final=Vol_A+Vol_Bv_final=Vol_final/m_finalu_final=INTENERGY(Fluid$,T=T_final, v=v_final)P_final=pressure(Fluid$,T=T_final, v=v_final)

Pfinal [kPa] Qout [kJ] Tfinal [C] 0 2300 0 0 2274 5. 1 2247 11. 1 2218 16. 2 2187 22. 3 2153 27. 5 2116 33. 6 2075 38. 9 2030 44. 12 1978 50

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6-6- 27E An OTEC power plant operates between the temperature limits of 86F and 41F. The cooling water experiences atemperature rise of 6F in the condenser. The amount of power that can be generated by this OTEC plans is to bedetermined.Assumptions1 Steady operating conditions exist. 2 Water is anincompressible substance with constant properties.Properties The density and specific heat of water are taken  = 64.lbm/ft 3 and c = 1 Btu/lbm.F, respectively.Analysis The mass flow rate of the cooling water is

=113,790 lbm/min= 1897 lbm/s7

1 ft( 64 0. lbm/ft)(13,300 gal/min)

33water water 

2- 26C The work done is the same, but the power is different.
Assumptions 1 The wind is blowing steadily at specified velocity during specified times. 2 The wind power generation is
negligible during other times.
The velocity of the lift during steady operation, and the acceleration during start up are
potential are to be determined.
Analysis Kinetic energy is the only form of harvestable mechanical

The rate of heat rejection to the cooling water isQ out m water ( TC out T in)( 1897 lbm/s)(1 Btu/lbm.F)(6F)=11,380 Btu/sNoting that the thermal efficiency of this plant is 2%, the power generation is determined to be

292 Btu/s= 308 kW ( 11 , 380 Btu/s)

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

in out

2- 52 A car cruising at a constant speed to accelerate to a specified speed within a specified time. The additional power
Energy
Analysis Kinetic energy is the only form of mechanical energy
Discussion When the energy gain of the house equals the energy loss, the temperature of the house remains constant. But
2-42E A water pump increases water pressure. The power input is to be determined.
shaft-to-kinetic energy of air.
electric power.
2-13 A water jet strikes the buckets located on the perimeter of a wheel at a specified velocity and flow rate. The power
2- 6C In electric heaters, electrical energy is converted to sensible internal energy.
does not affect the final answer).
room when all of these electric devices are on is to be determined.
the shaft (shaft power) to mechanical energy of air (kinetic energy). For a control volume that encloses the fan-motor unit,
electricity in a power plant.
conventional ones, the savings from electricity cost would pay for the cost differential in about 4 months. The electricity
(b) 0 hp
generation will change strongly with the wind conditions.
2-24 The power produced by an electrical motor is to be expressed in different units.
the world. Hydrogen can be obtained from water by using another energy source, such as solar or nuclear energy, and then

since 1 kW = 0 Btu/s.

Refrigerators and Heat Pumps

6-28C The difference between the two devices is one of purpose. The purpose of a refrigerator is to remove heat from acold medium whereas the purpose of a heat pump is to supply heat to a warm medium.

6- 29C The difference between the two devices is one of purpose. The purpose of a refrigerator is to remove heat from arefrigerated space whereas the purpose of an air-conditioner is remove heat from a living space.

6-30C No. Because the refrigerator consumes work to accomplish this task.

6-31C No. Because the heat pump consumes work to accomplish this task.

6-32C The coefficient of performance of a refrigerator represents the amount of heat removed from the refrigerated spacefor each unit of work supplied. It can be greater than unity.

6-33C The coefficient of performance of a heat pump represents the amount of heat supplied to the heated space for eachunit of work supplied. It can be greater than unity.

6-34C No. The heat pump captures energy from a cold medium and carries it to a warm medium. It does not create it.

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7- 1337- 176 A piston-cylinder device contains air that undergoes a reversible thermodynamic cycle composed of three processes.The work and heat transfer for each process are to be determined.Assumptions1 All processes are reversible. 2 Kinetic and potentialenergy changes are negligible. 3 Air is an ideal gas with variable specificheats.Properties The gas constant of air is R = 0 kPa 3 /kg (Table A-1).Analysis Using variable specific heats, the properties can be determinedusing the air table as follows

 

h at  at  

283 kJ/kg1 .3 696150 kPa

400 kPa

2-4C The mechanical energy is the form of energy that can be converted to mechanical work completely and directly by a

.1 70203 kJ/kg

214 kJ/kg300 K

3

322

33

1 2

02

01

1 21 2

(c) Heat is transferred through the walls of the room from the warm room air to the cold winter air. Electrical work
HOUSE
n 
 
= (40 - 34 W/lamp)(700 lamps)
t   
Assumptions The wind is blowing steadily at a constant
elevator as the closed system, the energy balance in the rate form can be written as

uPP

Analysis Kinetic energy is the only form of mechanical
  /(5s) 9 kW
Discussion Note the power generation of a wind turbine is proportional to the cube of the wind velocity, and thus the

s s

u uT T

r r

r r

The mass of the air and the volumes at the various states are

3

3

3

33

3

3

2

22

3

3

1

11

.0 3967 m400 kPa

(1 kg)(0 kPam /kgK)(396 K)

m 8.150 kPa

(1 kg)(0 kPa m/kgK)(300 K)

.1 394 kg(0 kPam/kgK)(300 K)

(400 kPa)(0 )

Eheater 60 , 000 6000 54 , 000 Btu/h
conversion of one form of internal energy (chemical energy) to another form (sensible energy).
The vertical component of escalator velocity is
Wgmgz z 
Neglecting the work done on the system by the returning empty chairs, the work needed to raise this mass by 200 m is
uniform velocity.
2- 33 A linear spring is elongated by 20 cm from its rest position. The work done is to be determined.
dEroom/dtEin 1650 W
2- 21C It is a work interaction since the electrons are crossing the system boundary, thus doing electrical work.

mRT

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

mRT

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

m

2-17C The form of energy that crosses the boundary of a closed system because of a temperature difference is heat; all other
the wind possesses, and it can be converted to work entirely.
Assumptions 1 The room is well sealed, and heat loss from the room is negligible. 2 All the appliances are kept on.

Process 1-2: Isothermal expansion ( T 2 = T 1 )

.0 3924 kJ/kg400 kPa

150 kPaln (1 kg)(0 kJ/kg)ln1

2  21   P

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

S mR

Q in,1 2  T 1  S  21 ( 300 K)(0 kJ/K) 117 kJ

W out,1 2  Q in,1 2  117 kJ

Process 2-3: Isentropic (reversible-adiabatic) compression ( s 2 = s 1 )W in,2 3  ( um 3  u 2 )(1 kg)(283.71-214) kJ/kg 97 kJ

Q 2-3 = 0 kJProcess 3-1: Constant pressure compression process ( P 1 = P 3 )

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

Q out,3 1  W in,3 1  ( um 1  u 3 ) 37 0. kJ-(1 kg)(214.07-283) kJ/kg 135 kJ

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

T = const. 2

EinEout  dE /dt  dEroom/dtEin

s = const.

P = const.

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8- 38

8-51 Air is compressed steadily by an 8-kW compressor from a specified state to another specified state. Theincrease in the exergy of air and the rate of exergy destruction are to be determined.Assumptions1 Air is an ideal gas with variable specific heats. 2 Kinetic and potential energy changes are negligible.Properties The gas constant of air is R = 0 kJ/kg (Table A-1). From the air table (Table A-17)

T hsT hs

1 1

2 2

24 ft
Qout
Therefore, the final internal energy of the system is 35 kJ.
2-32E The work required to expand a soap bubble is to be determined.
additional power input to achieve the indicated acceleration becomes
2-1C The sum of all forms of the energy a system possesses is called total energy. In the absence of magnetic, electrical and
W m
2- 23C This is neither a heat nor a work interaction since no energy is crossing the system boundary. This is simply the

K kJ /kg kJ /kgKK kJ /kgkJ /kgK

1

o

2

o

Wattage reduction = (Wattage reduction per lamp)(Number of lamps)
Wind
controlled by a thermostat. Therefore, the rate of energy transfer to the room, in general, will be less.
2- 22C It is a heat interaction since it is due to the temperature difference between the sun and the room.

Analysis The increase in exergy is the difference between the exit andinlet flow exergies,

Substituting, the required power rating of the heaters becomes
Analysis The total kinetic energy of the object is given by

Increasein exergy

2 1 0 2 1

2 1 0 2 1

2 1 0 0

h h T s s

h h ke pe T s s   

Assumptions 1 The additional air drag, friction, and rolling resistance are not considered. 2 The road is a level road.
Simple payback period =  0 year

 

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

where

.0 09356 kJ/kg K

100 kPa

600 kPa.2( 0887 .1 66802 )kJ/kg -K (0 kJ/kg K)ln

( ) ln 1

o 21

o2 1 2

Mechanical Forms of Work
determined.
(d) There is minor amount of heat transfer between the tires and road. Presuming that the tires are hotter than the
Analysis We take the water in the cylinder as the system. This is a closed system since no mass enters or leaves. Applying
placed) system to the room air. Finally, electrical work is being added to the refrigerator through the refrigeration system.

s s s s R

Substituting,

 

 178 kJ/kg

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

(441 290)kJ/kg-(290K)( .0 09356 kJ/kgK)

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

Then the reversible power input is

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

( b ) The rate of exergy destruction (or irreversibility) is determined from its definition,

X destroyed W in W rev,in 8  .6 25  1 kW

Discussion Note that 1 kW of power input is wasted during this compression process.

100 kPa 17 C

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

600 kPa 167 C

8 kW

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9- 919-1 12 An expression for the thermal efficiency of an ideal Brayton cycle with an ideal regenerator is to be developed.Assumptions1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at roomtemperature. 3 Kinetic and potential energy changes are negligible.Analysis The expressions for the isentropic compression and expansion processes are

T 2  1 rT pk  /)1( k

k k

rp

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

/)1(4 3

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

Energy Transfer by Heat and Work
 
/ 0
unoccupied periods are to be determined.
Wa Vm V t 
(a) 7 ft/s
2-14 Two sites with specified wind data are being considered for wind power generation. The site better suited for wind
No work is produced since there is no motion of the forces acting at the interface between the tire and road.

For an ideal regenerator,

6 2

5 4T T

fluorescent bulbs relative to standard ones, and is determined to be
surface tension effects, the total energy of a system consists of the kinetic, potential, and internal energies.
 0 Btu

The thermal efficiency of the cycle is

k kp

k kp

k kp

rT

motor efficiencies are less than 1, and the product of two numbers that are less than one is less than either of the numbers.

r

rT

  
E E dE dt 
Assumptions 1 Air drag and friction are negligible. 2 The average mass of each person is 75 kg. 3 The escalator operates
15 psia
Analysis The torque is determined from
W144,000lbf ft 185 Btu
2- 18C (a) The car's radiator transfers heat from the hot engine cooling fluid to the cooler air. No work interaction occurs in
= $1358/year
mechanical device such as a propeller. It differs from thermal energy in that thermal energy cannot be converted to work
since we are considering the change in the energy content of the car due to
 

q

q

/)1(3

1

/)1(

/)1(

3

1

4 3

2 13

1

5 3

6 13

13 5

6 1in

outth

Water
power generation is to be determined.
Then the amount of electrical energy consumed per year during unoccupied work period and its cost are
38 kW
energy is due to translational, rotational, and vibrational effects.
.0 050 kJ/kg
the energy balance can be written as
considerably higher because of the losses associated with the conversion of electrical-to-mechanical shaft and mechanical
= $
energy of people. Substituting, the required power input becomes



VvertVsin 45  6( m/s)sin45
another, and it does not allow any energy to be created or destroyed during a process. In reality, the power required will be
  151 W
Energy Conversion Efficiencies
70 psia
2- 37C Energy can be transferred to or from a control volume as heat, various forms of work, and by mass transport.
it passes over and through the car.
Therefore, the power potential of the water jet is its kinetic energy,
35 , 340 kg/s
saved will also help the environment by reducing the amount of CO 2 , CO, NOx, etc. associated with the generation of
Therefore, the energy content of the system increases by 52 Btu during this process.
 15 kW
the hydrogen obtained can be used as a fuel to power cars or generators. Therefore, it is more proper to view hydrogen is an

s

Under stated assumptions, the power supplied is used to increase the potential energy of people. Taking the people on
to lift the beam, the work done by the crane is
Therefore, 1770 kW of actual power can be generated by this wind turbine at the stated conditions.
When the escalator velocity is doubled to V = 1 m/s, the power needed to drive the escalator becomes

q in 3

5

fan motor is to be determined.

q out

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11 -

0 0 0 0 0 0 1

9

12

15

18

21

24

27

30

 comp

turbine

in,net

[kW]

turb=1.

turb=0.

turb=0.

0 0 0 0 0 0 1

8

12

16

20

 comp

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

Refrig

[kW]

turb=0.

turb=1.

turb=0.

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

16 -49 Problem 16- 48 is reconsidered. The effect of the final temperature on the rate of heat supplied for the twocases is to be studied.Analysis The problem is solved using EES, and the solution is given below.

"Given"T1=298 [K]T2=3000 [K]"P=1 [atm]"m_dot=0 [kg/min]T0=298 [K]

"The equilibrium constant for these two reactions at 3000 K are determined from Table A-28"K_p1=exp(-3)K_p2=exp(-2)

"Properties"MM_H2O=molarmass(H2O)

"Analysis""(a)""Actual reaction: H2O = N_H2O H2O + N_H2 H2 + N_O2 O2 + N_OH OH"2=2N_H2O+2N_H2+N_OH "H balance"1=N_H2O+2N_O2+N_OH "O balance"N_total=N_H2O+N_H2+N_O2+N_OH"Stoichiometric reaction 1: H2O = H2 + 1/2 O2""Stoichiometric coefficients for reaction 1"nu_H2O_1=nu_H2_1=nu_O2_1=1/"Stoichiometric reaction 2: H2O = 1/2 H2 + OH""Stoichiometric coefficients for reaction 2"nu_H2O_2=nu_H2_2=1/nu_OH_2="K_p relations are"K_p1=(N_H2^nu_H2_1*N_O2^nu_O2_1)/N_H2O^nu_H2O_1(P/N_total)^(nu_H2_1+nu_O2_1-nu_H2O_1)K_p2=(N_H2^nu_H2_2N_OH^nu_OH_2)/N_H2O^nu_H2O_2(P/N_total)^(nu_H2_2+nu_OH_2-nu_H2O_2)

"Enthalpy of formation data from Table A-26"h_f_OH="Enthalpies of products"h_H2O_R=enthalpy(H2O, T=T1)h_H2O_P=enthalpy(H2O, T=T2)h_H2=enthalpy(H2, T=T2)h_O2=enthalpy(O2, T=T2)h_OH=98763 "at T2 from the ideal gas tables in the text""Standard state enthalpies"h_o_OH=9188 "at T0 from the ideal gas tables in the text"

"Heat transfer"H_P=N_H2Oh_H2O_P+N_H2h_H2+N_O2h_O2+N_OH(h_f_OH+h_OH-h_o_OH)H_R=N_H2O_Rh_H2O_RN_H2O_R=Q_in_a=H_P-H_RQ_dot_in_a=(m_dot/MM_H2O)*Q_in_a"(b)"Q_in_b=N_H2O_R(h_H2O_P-h_H2O_R)Q_dot_in_b=(m_dot/MM_H2O)*Q_in_b

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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

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18 -64E A wind turbine is to generate power with two specified wind speeds. The amount of electricity that can be producedby the turbine and the blade tip speed are to be determined.Assumptions Wind flows steadily at the specified speeds.Properties The density of air is given to be  = 0 lbm/ft 3.Analysis The blade span area is

8 kJ/kg

The wind power potential at the wind speed of 16 ft/s is

174 0. kW 23,730 lbmft /s

1 kW(0 lbm/ft )(26,880 ft )(16 ft/s)2

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

32

3 3 2 3available, 11 

W sdA  A A 
Analysis The spring work can be determined from
Analysis Taking the house as the system, the energy balance can be written as
Load = (50 chairs)(250 kg/chair) = 12,500 kg
0 h 360 s
= 4200 W
), the maximum wind

The wind power potential at the wind speed of 24 ft/s is

587 2. kW 23,730 lbm ft /s

1 kW(0 lbm/ft )(26,880 ft )(24 ft/s)2

which is V
There is also a transfer of heat from the hot portions of the refrigerator (i., back of the compressor where condenser is
= [($2-$1)/lamp](700 lamps)

32

3 3 2 3available, 22 

Therefore, second site is a better one for wind generation.
Forms of Energy
entirely. Therefore, the power potential of the wind is its
short acceleration times are indicative of powerful engines.
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
2- 46 An industrial facility is to replace its 40-W standard fluorescent lamps by their 35-W high efficiency counterparts. The
= (4 kW)(1)(2800 h/year)

The overall wind turbine efficiency at a wind speed of 16 ft/s is

energy balance on this system gives

The overall wind turbine efficiency at a wind speed of 24 ft/s is

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

The electric power generated at a wind speed of 16 ft/s is

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

The electric power generated at a wind speed of 24 ft/s is

acceleration will be

The amount of electricity produced at a wind speed of 16 ft/s is

W electric,1 W electric,1Operating hours 1 ( 48. 54 kW)(3000 h)145,630 kWh

The amount of electricity produced at a wind speed of 24 ft/s is

W electric,2 W electric,2Operating hours 2 ( 191 1. kW)(4000 h) 764 ,560 kWh

The total amount of electricity produced isW electric,total W electric,1 W electric,2 145 , 560  764 , 560  910,000 kWh

Noting that the tip of blade travels a distance of  D per revolution, the tip velocity of the turbine blade for a rotational speedof n  becomes

 99mph

Analysis Using appropriate conversion factors, we obtain
(0 m/s)(5 s)(0) 1
the radiator.
Analysis Taking the room as the system, the rate form of the energy balance can be written as
2 m/s
Discussion Note that some appliances such as refrigerators and irons operate intermittently, switching on and off as
Thus,
m
Then using the relations given earlier, the energy and cost savings associated with the replacement of the high efficiency
Assumptions 1 The fan operates steadily. 2 There are no conversion losses.
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

1/s

1 mi/h60 s

1 min

W V P 2 P 1

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