ejercicios tema 7

PROPRIETARY MATERIAL. 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and

educators for course preparation. If you are a student using this Manual, you are using it without permission.

7-154

7-203 Refrigerant-134a is expanded adiabatically in a capillary tube. The rate of entropy generation is to be

determined.

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis The rate of entropy generation within the expansion device during this process can be determined

by applying the rate form of the entropy balance on the system. Noting that the system is adiabatic and thus

there is no heat transfer, the entropy balance for this steady-flow system can be expressed as

,

12gen

12gen

gen2211

entropy of change of Rate

(steady) 0system

generation entropy of Rate

gen

mass andheat by ansferentropy trnet of Rate

outin

)(

0

sss

ssmS

Ssmsm

SSSS

'

It may be easily shown with an energy balance that the enthalpy remains constant during the throttling

process. The properties of the refrigerant at the inlet and exit states are (Tables A-11 through A-13)

KkJ/kg 44193.0

KkJ/kg 50.123

0

C50

1

1

1

1 q shxT

KkJ/kg 48038.0)79406.0)(4223.0(14504.0

4223.038.207

92.3550.123

K kJ/kg 50.123

C12

22

2

2

12

2 q

fgf

fg

f

sxss

h

hhx

hh

T

Substituting,

kW/K0.00769 KkJ/kg )44193.0038kg/s)(0.48 2.0()( 12gen ssmS

R-134a

50qC sat. liq.

12qC Capillary tube



7-166

7-214 The pressure in a hot water tank rises to 2 MPa, and the tank explodes. The explosion energy of the

water is to be determined, and expressed in terms of its TNT equivalence.

Assumptions 1 The expansion process during explosion is isentropic. 2 Kinetic and potential energy

changes are negligible. 3 Heat transfer with the surroundings during explosion is negligible.

Properties The explosion energy of TNT is 3250 kJ/kg. From the steam tables (Tables A-4 through 6)

kJ/kg 811.832.20881889.040.4171889.0

0562.6

3028.14467.2

KkJ/kg 6.0562

kJ/kg 2088.2

,3028.1

,40.417kPa 100

KkJ/kg 2.4467

kJ/kg 906.12

/kgm 0.001177

liquid sat.

MPa 2

22

22

12

2

MPa 2@1

MPa 2@1

3MPa 2@1

1

fgf

fg

f

fg

fg

f

f

f

f

f

uxuu

s

ssx

s

u

s

u

ss

P

ss

uu

vP

v

Analysis We idealize the water tank as a closed system that undergoes a reversible adiabatic process with

negligible changes in kinetic and potential energies. The work done during this idealized process represents

the explosive energy of the tank, and is determined from the closed system energy balance to be

21outb,exp 12outb,energies etc. potential,

kinetic, internal,in Change

system

mass and work,heat,by nsferenergy traNet

outin

)(

uumWE

uumUW

EEE

' '

where

kg 99.67/kgm 0.001177

m 0.0803

3

1

v

Vm

Substituting, kJ 6410kJ/kg811.83906.12kg 67.99exp E which is equivalent to

TNTkg1.972 kJ/kg 3250

kJ 6410TNTm

Water

Tank

2 MPa



7-174

7-221 Two rigid tanks that contain water at different states are connected by a valve. The valve is opened

and steam flows from tank A to tank B until the pressure in tank A drops to a specified value. Tank B loses

heat to the surroundings. The final temperature in each tank and the entropy generated during this process

are to be determined.

Assumptions 1 Tank A is insulated, and thus heat transfer is negligible. 2 The water that remains in tank A

undergoes a reversible adiabatic process. 3 The thermal energy stored in the tanks themselves is negligible.

4 The system is stationary and thus kinetic and potential energy changes are negligible. 5 There are no

work interactions.

Analysis (a) The steam in tank A undergoes a reversible, adiabatic process, and thus s2 = s1. From the

steam tables (Tables A-4 through A-6),

Tank A:

kJ/kg 2125.9kJ/kg 1982.17895.011.561 /kgm 0.47850001073.060582.07895.0001073.07895.0

3200.5

6717.18717.5

mixture sat.

kPa 300

KkJ/kg 5.87171191.58.07765.1

kJ/kg 2163.39.19488.022.604

/kgm 0.37015001084.046242.08.0001084.0

8.0

kPa 400

,2,2

3,2,2

,2

,2

kPa 300@sat,2

12

1

1,1

1,1

31,1

1

1

q

fgAfA

fgAfA

fg

fA

A

A

fgfA

fgfA

fgfA

uxuu

x

s

ssx

TT

ss

P

sxss

uxuu

x

x

P

vvv

vvv

C133.52

Tank B:

KkJ/kg 7.7100

kJ/kg 2731.4

/kgm 1.1989

C250

kPa 200

,1

,1

3,1

1

1

q B

B

B

s

uT

Pv

The initial and the final masses in tank A are

and

kg 0.4180/kgm 0.47850

m 0.2

kg 0.5403/kgm 0.37015

m 0.2

3

3

,2,2

3

3

,1,1

A

AA

A

AA

m

m

v

V

v

V

Thus, 0.5403 - 0.4180 = 0.1223 kg of mass flows into tank B. Then,

kg 3.12231223.031223.0,1,2 BB mm The final specific volume of steam in tank B is determined from

/kgm 1.1519kg 3.1223

/kgm 1.1989kg 3 33

,2

11

,2,2

B

B

B

BB

m

m

m

vVv

We take the entire contents of both tanks as the system, which is a closed system. The energy balance for

this stationary closed system can be expressed as

A

steam

V = 0.2 m3

P = 400 kPa

x = 0.8

B

steam

m = 3 kg

T = 250qC P = 200 kPa

u600 kJ



7-175

BA

BA

umumumumQ

WUUUQ

EEE

11221122out

out

energies etc. potential, kinetic, internal,in Change

system


outin

0)=PE=KE (since )()( '' ' '

Substituting,

^ ` ^ `

kJ/kg 2522.0

4.273131223.33.21635403.09.2125418.0600

,2

,2 B Bu u Thus,

KkJ/kg 7.2274kJ/kg 2522.0

/kgm 1.1519

,2

,2

,2

3,2 q

B BBB s Tuv C113.2 (b) The total entropy generation during this process is determined by applying the entropy balance on an

extended system that includes both tanks and their immediate surroundings so that the boundary

temperature of the extended system is the temperature of the surroundings at all times. It gives

,

BAgensurrb,

out

entropyin Change

system

generationEntropy

gen

mass andheat by ansferentropy trNet

outin

SSST

Q

SSSS

'' '

Rearranging and substituting, the total entropy generated during this process is determined to be ^ ` ^ `

kJ/K0.916 ''

K 273

kJ 6007100.732274.71223.38717.55403.08717.5418.0

surrb,

out11221122

surrb,

outgen

T

Qsmsmsmsm

T

QSSS

BABA



7-196

7-235 An evacuated bottle is surrounded by atmospheric air. A valve is opened, and air is allowed to fill

the bottle. The amount of heat transfer through the wall of the bottle when thermal and mechanical

equilibrium is established and the amount of entropy generated are to be determined.

Assumptions 1 This is an unsteady process since the conditions within the device are changing during the

process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains

constant. 2 Air is an ideal gas. 3 Kinetic and potential energies are negligible. 4 There are no work

interactions involved. 5 The direction of heat transfer is to the air in the bottle (will be verified).

Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1).

Analysis We take the bottle as the system, which is a control volume since mass crosses the boundary.

Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and

internal energy u, respectively, the mass and energy balances for this uniform-flow system can be

expressed as

Mass balance: )0 (since initialout2systemoutin o' mmmmmmm i Energy balance:

)0peke (since initialout22in

energies etc. potential, kinetic, internal,in Change

system


outin

## # '

EEWumhmQ

EEE

ii

Combining the two balances: ihumQ 22in where

kJ/kg 206.91

kJ/kg 290.16K 290

kg 0.0060K 290K/kgmkPa 0.287

m 0.005kPa 100

2

17-A Table2

3

3

2

22

o

u

hTT

RT

Pm

ii

V

Substituting,

Qin = (0.0060 kg)(206.91 - 290.16) kJ/kg = - 0.5 kJ o Qout = 0.5 kJ Note that the negative sign for heat transfer indicates that the assumed direction is wrong. Therefore, we

reverse the direction.

The entropy generated during this process is determined by applying the entropy balance on an

extended system that includes the bottle and its immediate surroundings so that the boundary temperature

of the extended system is the temperature of the surroundings at all times. The entropy balance for it can be

expressed as

,

220

1122tankgeninb,

out

entropyin Change

system

generationEntropy

gen

mass andheat by ansferentropy trNet

outin

smsmsmSST

Qsm

SSSS

ii ' '

Therefore, the total entropy generated during this process is

kJ/K0.0017 K 290

kJ 0.5

surr

out

outb,

out022

outb,

out22gen

T

Q

T

Qssm

T

QsmsmS iii

5 L

Evacuated

10 kPa

17qC



7-200

7-238 The turbocharger of an internal combustion engine consisting of a turbine driven by hot exhaust

gases and a compressor driven by the turbine is considered. The air temperature at the compressor exit and

the isentropic efficiency of the compressor are to be determined.

Assumptions 1 Steady operating conditions exist. 2

Kinetic and potential energy changes are negligible. 3

Exhaust gases have air properties and air is an ideal gas

with constant specific heats.

Properties The specific heat of exhaust gases at the

average temperature of 425C is cp = 1.075 kJ/kg.K and

properties of air at an anticipated average temperature of

100C are cp = 1.011 kJ/kg.K and k =1.397 (Table A-2).

Analysis (a) The turbine power output is determined from

kW 075.1C400)-C)(450kJ/kg. 5kg/s)(1.07 02.0(

)( 21exhT qq TTcmW p For a mechanical efficiency of 95% between the turbine and the compressor,

kW 021.1kW) 075.1)(95.0(TC WW m K Then, the air temperature at the compressor exit becomes

C126.1q qq

2

2

12airC

C70)-C)(kJ/kg. 1kg/s)(1.01 018.0(kW 021.1

)(

T

T

TTcmW p

(b) The air temperature at the compressor exit for the case of isentropic process is

C106K 379kPa 95

kPa 135K) 27370(

1)/1.397-(1.397/)1(

1

212 q

kks

P

PTT

The isentropic efficiency of the compressor is determined to be

0.642 701.126 7010612 12C TT TT sK

Compressor Turbine

Exh. gas

450qC 0.02 kg/s

Air, 70qC 95 kPa

0.018 kg/s 400qC

135 kPa



7-202

7-240 A cryogenic turbine in a natural gas liquefaction plant produces 350 kW of power. The efficiency of

the turbine is to be determined.

Assumptions 1 The turbine operates steadily. 2 The properties of methane is used for natural gas.

Properties The density of natural gas is given to be 423.8 kg/m3.

Analysis The maximum possible power that can be obtained from this

turbine for the given inlet and exit pressures can be determined from

kW 2.480kPa)3004000(kg/m 8.423

kg/s) 55()(

3outinmax PPmW U

Given the actual power, the efficiency of this cryogenic turbine becomes

72.9%0.729 kW 2.480

kW 350

maxW

W

K

This efficiency is also known as hydraulic efficiency since the cryogenic

turbine handles natural gas in liquid state as the hydraulic turbine handles

liquid water.

Cryogenic

turbine

LNG, 40 bar

-160qC, 55 kg/s

3 bar

7-203.pdf (p.1)7-203.pdf (p.2)7-214.pdf (p.3)7-214.pdf (p.4)7-221.pdf (p.5)7-221.pdf (p.6-7)7-235.pdf (p.8)7-235.pdf (p.9)7-238.pdf (p.10)7-238.pdf (p.11)7-240.pdf (p.12)7-240.pdf (p.13)

ejercicios tema 7

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