# Thermodynamics Solutions Manual

## GENERAL ENERGY ANALYSIS

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It have solution for all chapters of textbook (chapters 1 to 17).

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1 kJ/kg

7( m/s)

mech, 1 1  

#### e ke

.0 050 kJ/kg1000 m /s

1 kJ/kg

( 10 m/s)

mech, 2 2  

#### e ke

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

### 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

### 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

N 1 m

1 W

J/s 1

 5 kgm /s

1 kg m/s

J 1

N 1 m

1 W

J/s 15( W)

#### (a) 7 ft/s

###### 

1 Btu/s

1. 169 lbf ft/s

1 Btu/s

1 hp

#### 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

W mg z

 788 lbfft

1 hp

550 lbf ft/s

2 3000/60/s

450 hp

sh

####  

12 10 Btu

###### 

778 ft

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

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

####   24,525 kJ

1000 kgm /s

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

distance

velocity

1 km

10 km/ h

 68 kW

360 s

24,525 kJ

t

gg

#### 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

####   /(5s) 9 kW

1000 m /s

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

1000 m

200 m

2

sin2

####   /(5s) 34 1. kW

1000 kgm/s

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

#### energy balance on this system gives

35 kJ

###### 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

#### 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

#### / 0

potential, etc. energies

Rateof changein internal, kinetic,

by heat, work, mass and

Rateofnet energy transfer

in out      

Ein Eout

15 kW

3412 Btu/h

1 kW

#### Analysis The power input is determined from

11 hp

######  

0 Btu/s

1 hp

5 psia ft

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

• Lights

• People

• Appliance

• Heaters

#### 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

#### 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

#### (3 months)

\$ 1358 /year

###### \$ 343

Annual cost savings

Implementation cost

#### 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

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

###### 40 110 300 1200 W

in lights TV refrig iron

#### the energy balance can be written as

###### / 0

potential, etc. energies

Rateof change in internal, kinetic,

by heat, work,and mass

Rateof netenergy transfer

E E dE dt

####  

ke

###### 2

outelect,in air out air

#### where

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

### mairVA  

####   151 W

###### 

  .0 1435 Btu/s 25,037 ft /s

1 Btu/lbm

2

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

outin air

• Lights

• TV

• Refrig

• Iron

#### 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   

t

E dE dt 

sysin sys/

in mgVvert t

mg z

t

####  15 kW

###### 

   12 5. kJ/s 1000 m /s

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

Win mgVvert

####  31 kW

###### 

   25 0. kJ/s 1000 m /s

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

Win mgVvert

#### 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   

t

E dE dt 

sysin sys/

t

Vm V

t

in

#### 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

in t

#### 38 kW

###### 2

in 1000 m /s

1 kJ/kg

2(5s)

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

#### 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

#### 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

####   

shaft,in

elect,outgeneratorMechanical power input

Electrical power output

####   

| mech,fluid|

elect,out

mech,in mech,out

elect,outturbinegen- turbinegeneratorE