Homework #8 Solutions

#1 4-63 The internal energy change of hydrogen gas during a heating process is to bedetermined using an empirical specific heat relation, constant specific heat at averagetemperature, and constant specific heat at room temperature.Analysis (a) Using the empirical relation for )(Tc p from Table A-2c and relating it to

)(Tcv ,( ) 32)( dTcTbTRaRcTc uup +++-=-=v

where a = 29.11, b = -0.1916¥10-2, c = 0.4003¥10-5, and d = -0.8704¥10-9. Then,

( )[ ]( )

kJ/kg 6194==D

=D

=

-¥--¥+

-¥---=

-+-+++--=

+++-==D

--

-

ÚÚ

kg/kmol 2.016

kJ/kmol 12,487

kJ/kmol 12,487

)200800)(108704.0()200800)(104003.0(

)200800)(101961.0()200800)(314.811.29(

)()()()(

)(

44941335

31

22221

41

424

131

323

121

222

112

2

1

322

1

M

uu

TTdTTcTTbTTRa

dTdTcTbTRadTTcu

u

uv

(b) Using a constant cp value from Table A-2b at the average temperature of 500 K,

kJ/kg 6233=-⋅=-=D

⋅==

200)KK)(800kJ/kg (10.389)(

KkJ/kg 10.389

12avg,

K 500@avg,

TTcu

cc

v

vv

(c) Using a constant cp value from Table A-2a at room temperature,

kJ/kg 6110=-⋅=-=D

⋅==

200)KK)(800kJ/kg (10.183)(

KkJ/kg 10.183

12avg,

K 300@avg,

TTcu

cc

v

vv

#2 4-120 The compression work from P1 to P2 using a polytropic process is to becompared for neon and air.Assumptions The process is quasi-equilibrium.Properties The gas constants for neon and air R = 0.4119 and 0.287 kJ/kg⋅K,respectively (Table A-2a).Analysis For a polytropic process,

Constant=nPvThe boundary work during a polytropic process of an ideal gas is

˙˙

˚

˘

ÍÍ

Î

È-˜̃

¯

ˆÁÁË

Ê

-=

˙˙

˚

˘

ÍÍ

Î

È-˜̃

¯

ˆÁÁË

Ê

-===

---ÚÚ 1

11

1Constant

/)1(

1

211

1

2112

1

2

1

out,

nnnn

b P

P

n

RT

n

PdPdw

vvvvvv

The negative of this expression gives the compression work during a polytropic process.Inspection of this equation reveals that the gas with the smallest gas constant (i.e., largestmolecular weight) requires the least work for compression. In this problem, air willrequire the least amount of work.

#3 5-10 Air is expanded and is accelerated as it is heated by a hair dryer of constantdiameter. The percent increase in the velocity of air as it flows through the drier is to bedetermined.Assumptions Flow through the nozzle is steady.Properties The density of air isgiven to be 1.20 kg/m3 at the inlet,and 1.05 kg/m3 at the exit.Analysis There is only one inlet and oneexit, and thus & & &m m m1 2= = . Then,

)of increaseand (or, 1.14 kg/m1.05

kg/m1.203

3

2

1

1

2

2211

21

14% ===

=

=

r

r

rr

V

V

AVAV

mm &&

Therefore, the air velocity increases 14% as it flows through the hair drier.

#4 5-18 Refrigerant-134a flows through a pipe. Heat is supplied to R-134a. The volumeflow rates of air at the inlet and exit, the mass flow rate, and the velocity at the exit are tobe determined.

Properties The specific volumes of R-134a at the inlet and exit are (Table A-13)

/kgm 1142.0C20

kPa 200 31

1

1 =˛˝¸

°=

=v

T

P/kgm 1374.0

C40

kPa 180 32

1

1 =˛˝¸

°=

=v

T

P

Analysis (a) (b) The volume flow rate at the inlet and the mass flow rate are

kg/s 2.696

/sm 0.3079 3

====

====

m/s) 5(4

m) 28.0(

/kgm 1142.0

1

4

11

m/s) 5(4

m) 28.0(

42

31

2

11

1

2

1

2

11

pp

pp

VD

VAm

VD

VA

c

c

vv

V

&

&

(c) Noting that mass flow rate is constant, the volume flow rate and the velocity at theexit of the pipe are determined from

m/s 6.02

/sm 0.3705 3

===

===

4

m) 28.0(

s/m 3705.0

/kg)m 74kg/s)(0.13 696.2(

2

32

2

322

pcAV

m

VvV&

&&

V2 V1

R-134a200 kPa

20°C5 m/s

Q

180 kPa40°C

5. 5-31 Air is accelerated in a nozzle from 30 m/s to 180 m/s. The mass flow rate, the exittemperature, and the exit area of the nozzle are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air isan ideal gas with constant specific heats. 3 Potential energy changes are negligible. 4 Thedevice is adiabatic and thus heat transfer is negligible. 5 There are no work interactions.Properties The gas constant of air is 0.287kPa.m3/kg.K (Table A-1). The specific heatof air at the anticipated average temperatureof 450 K is cp = 1.02 kJ/kg.°C (Table A-2).Analysis (a) There is only one inlet and oneexit, and thus & & &m m m1 2= = . Using the idealgas relation, the specific volume and themass flow rate of air are determined to be

/kgm 0.4525kPa 300

)K 473)(K/kgmkPa 0.287( 33

1

11 =

⋅⋅==

P

RTv

kg/s 0.5304=== )m/s 30)(m 0.008(/kgm 0.4525

11 2311

1

VAmv

&

(b) We take nozzle as the system, which is a control volume since mass crosses theboundary. The energy balance for this steady-flow system can be expressed in the rateform as

outin

energies etc. potential, kinetic, internal,in change of Rate

(steady) 0system

mass and work,heat,by nsferenergy tranet of Rate

outin 0

EE

EEE

&&

44 344 21&

43421&&

=

=D=- ä

( )

20

20

0)peW (since /2)+()2/(2

12

212,

21

22

12

222

211

VVTTc

VVhh

QVhmVhm

avep-

+-=æÆæ-

+-=

@D@@=+ &&&&

Substituting, ˜˜¯

ˆÁÁË

Ê-+-⋅=

22

22

2/sm 1000

kJ/kg 1

2

)m/s 30()m/s 180()C200)(KkJ/kg 1.02(0 oT

It yields T2 = 184.6°C(c) The specific volume of air at the nozzle exit is

/kgm 1.313kPa 100

)K 273184.6)(K/kgmkPa 0.287( 33

2

22 =

+⋅⋅==

P

RTv

( )m/s 180/kgm 1.313

1kg/s 0.5304

12322

2

AVAm =æÆæ=v

& _ A2 = 0.00387 m2 = 38.7 cm2

AIR P2 = 100kPaV2 = 180m/s

P1 = 300kPaT1 =200°CV1 = 30m/sA1 = 80cm2