Chapter 3 Problems

3.7

The following table lists temperatures and specific volumes of water vapor at two

pressures:

P=1.0 MPa P=1.5 MPa

T(C) v(m3/kg) T(C) v(m

3/kg)

200 0.2060 200 0.1325

240 0.2275 240 0.1483

280 0.2480 280

Data encountered in solving problems often do not fall exactly on the grid of values

provided by property tables, and linear interpolation between adjacent table entries

becomes necessary. Using the data provided here, estimate:

a) The specific volume at T=240C, P=1.25MPa, (in m3/kg).

b) The temperature at P=1.5MPa, v=0.1555m3/kg, (in C).

c) The specific volume at T=220C, P=1.4MPa, (in m3/kg).

a) At a temperature of 240C, the specified pressure of 1.25 MPa falls between the table

values of 1.0 and 1.5 MPa. To determine the specific volume corresponding to 1.25

MPa, we find the slope of a straight line joining the adjacent table states, as follows:

Similar triangles:

3

0.1483 0.2275 0.1483 0.250.1483 0.2275 0.1483

1.5 1.25 1.5 1.0 0.50

0.1879 ( )

vslope v

mv a

kg

b) At a pressure of 1.5MPa, the given specific volume of 0.1555m3/kg falls between the

table values of 240 and 280C. To determine the temperature corresponding to the given

specific volume, we find the slope of a straight line joining the adjacent table states, as

follows:

240 280 240

.1555 .1483 .1627 .1483

.1555 .1483240 40

.1627 .1483

260 ( )

Tslope

T

C b

c) In this case, the specified pressure falls between the table values of 1.0 and 1.5MPa

and the specified temperature falls between the table values of 200 and 240C. Thus,

double interpolation is required.

At 220C, the specific volume at each pressure is simply the average over the interval:

At 1.0 MPa, 220C; 3.2060 .2275

0.216752

mv

kg

At 1.5 MPa, 220C; 3.1325 .1483

0.14042

mv

kg

Thus with the same approach as in (a)

3

0.1404 0.21675 0.1404 0.10.1404 0.21675 0.1404

1.5 1.4 1.5 1.0 0.5

0.15567 ( )

vv

mc

kg

3.8

The following data lists the temperature and specific volume of NH3 at two pressures:

2 2

3 3

2

3

2

50 Lbf/in 60 Lbf/in

( F) (ft /Lb) (ft /Lb)

100 6.836 100 5.659

120 7.110 120 5.891

140 7.380 140 6.120

a) ? at 120 F, 54Lbf/in

60 545.891 (7.110 5.891)

60 50

6.622 ft /Lb

b) ? 60Lbf/in and 5.982 f

P P

T v T v

v T P

v

v

T at P v 3

2

2

t /Lb

5.982 5.891120 (20)

6.12 5.891

127.9F

c) ? at 110F, 58Lbf/in

Double interpoloation

At 110 F, the specified volume at each pressure is simply the average over the interval:

lbfat 50 , 1

in

T

T

v T P

3

3

2

3

7.110 6.83610F; 6.973ft /lb

2

lbf 5.891 5.659at 60 , 110 F; 5.775ft /lb

in 2

With the same approach as in (a)

- 5.775 6.973 5.775 25.775 [6.973 5.775] 6.015ft /lb

60 - 58 60 50 10

v

v

vv

3.10

For H2O, determine the specified property at the indicated state. Locate the state on a

sketch of the T-v diagram.

a) P=300 kPa, v=0.5 m3/kg. Find T, in C.

3 3 3Table A-3, 1.0732 /10 / , 0.6058 /

Since , the state is in the two-phase,

liquid-vapor region, as shown on the T-v diagram.

Therefore,

3 133.6

f g

f g

sat

v m kg v m kg

v v v

T T bar C T

b) P=28 MPa, T=200C. Find v, in m3/kg.

State is in liquid region. From Table A-5,

@200C

250 bar: 31.1344 *10v

300 bar: 31.1302 *10v

So, at 200 bar 3

31.1319 *10m

vkg

c) P=1 MPa, T=405C. Find v, in m3/kg.

From superheat Table A-4

at 10 bar, interpolation gives

v = 0.309m3/kg.

d) T=100C, x=60%. Find v, in m3/kg.

Two phase equation:

1 f gv x v xv

With data from Table A-2 at 100C,

3 30.4 1.0435 *10 0.6 1.673 1.0042 /v m kg

3.13

3.18

3

2 2

Determine the qualityof a two phase liquid vapor mixture of:

Lbf fta) @ 10 with 15

in Lbm

(1 )

15 (1 )0.01659 (38.42)

0.39

Btub) 134 @ 60 F with u=50.5

Lbm

30.39 101.27

(1

f g

f g

H O P v

v x v xv

x x

x

R a T

u and u

u x

2

3

)

0.284

Lbf Btuc) @ 80 with h=350

in Lbm

91.22 623.32

(1 )

0.486

ftd) Pr @ 20F with 1

Lbm

(1 )

0.246

f g

f g

f g

f g

u xu

x

Ammonia P

h and h

h x h xh

x

opane T v

v x v xv

x

3.21

As shown in Fig. P3.21, a closed, rigid cylinder contains different volumes of saturated

liquid water and saturated water vapor at a temperature of 150C. Determine the quality

of the mixture, expressed as a percent.

Fig P3.21

Analysis:

, / . Thus, ,

/

/ /

vap vap liq

vap liq

vap liq g f

vap g

vap g liq f

m V Vx m V v m m

m m v v

V vx

V v V v

30vapV A and 20liqV A , where area A is in the

same units as the vertical measure shown. Then

30 / 1

30 / 20 / 201

30

g

g f g

f

A vx

A v A v v

v

Since ratios appear in the last expression, the

quantities can be in any consistent units.

Using vf and vg from Table A-2 at 150C, 3 3

3

1.0905 10 /

0.3928 /

f

g

v m kg

v m kg

3

10.0041 0.41%

20 0.39281

30 1.0905 10

x

3.24

Water is contained in a closed tank initially saturated vapor at 200 C is cooled to 100 C.

Determine the initial and the final pressures each in bar. Sketch the T-v and the P-v

diagram.

3

1

1

1

1 2

2

2

0.1274

15.54Bar

1

mass & volume does not change (closed system)

Since , the state is in the two-phase,

100 1.014Bar

f g

sat

mv

kg

P

x

v v

v v v

P P at C

3.25

3.29

Ammonia contained in a piston cylinder arrangement initially at

1 0 FT , saturated vapor

undergoes an isothermal process during which its volume (a) doubles, (b) reduces to half.

For each case find the final state giving quality or pressure as appropriate. Sketch the

process on a T-v and p-v diagram.

1

3

1

3

2 1

2 2

2

3

2 1

2 2

0 F

ft9.11

Lb

ft2 1.822

Lb

Lbf15.6 interpolate table 15E

in

0 F

1 ft4.555

2 Lb

Since at 0 F State 2 is two phase

4.555 0.024190.499 0.5

9.11 0.02419

g

f g

T

v v

v v

P

T

v v

T v v v

x

3.32

1 2

3

Two Lb of water vapor in a piston-cylinder is compressed at constant 250 from a volume

of 6.88 ft to saturated vapor state. Determine the temperatue at initial and final state and the work fo

LbfP

in

3

1

1 1 2

1

3

2 2 2

2

33

2

2 1

r this

process in Btu.

Solution:

6.88 ft3.44

2

250 , sup ( 4 )

1000

ft250 1.845

401

ft(2 )(1.845 ) 3.69 ft

= P (

250(3

g

vLb

LbfKnowing v and P The state is erheated A E

in

T F

Lbfv v at P

in Lb

T F

V LbLb

W dV P V V

W144

.69 6.88) 147.6778

Work done on the system (compression)

Btu

3.40

Determine the values of the specified properties at each of the following conditions.

a) For Refrigerant 134a at P=140 lbf/in2 and h=100 Btu/lb, determine T in F and v in

ft3/lb.

b) For ammonia at T = 0F and v =15 ft3/lb, determine P in lbf/in

2 and h in Btu/lb.

c) For Refrigerant 22 at T=30F and v =1.2 ft3/lb, determine P in lbf/in

2 and h in Btu/lb.

a) Refrigerant 134a; p=140lbf/in2, h=100 Btu/lb

from Table A-11E; hf

3.52

1

2 1

2

2

2 2 2

:

Btu113.83 (from table A-12E)

Lbm

( )

1.2 4.32 0.5( 113.83)

Btu107.59

Lbm

Knowing & u 80

Solution

u

Q W m u u

u

u

P T F

3.56

As shown in Fig. P3.56, 0.1 kg of propane is contained within a piston-cylinder assembly

at a constant pressure of 0.2 MPa. Energy transfer by heat occurs slowly to the propane,

and the volume of the propane increase from 0.0277m3 to 0.0307m

3. Friction between

the piston and cylinder is negligible. The local atmospheric pressure and acceleration of

gravity are 100 kPa and 9.81 m/s2, respectively. The propane experiences no significant

kinetic and potential energy effects. For the propane, determine (a) the initial and final

temperature, in C, (b) the work, in kJ, and (c) the heat transfer, in kJ.

Schematic & Given Data:

Engr. Model:

1. The propane within the piston-cylinder

assembly is the close system.

2. Friction between the piston and cylinder

is negligible and the expansion occurs

slowly at a constant pressure of 0.2 MPa.

3. Volume change is the only work mode.

4. For the system there are no significant

kinetic and potential energy effects.

Analysis:

a) To find T1 and T2 requires two property values to fix each state. Since pressure is

constant, it is one of the properties. The specific volume provides the other: 3 3

11

0.02770.277

0.1

V m mv

m kg kg,

3 3

22

0.03070.307

0.1

V m mv

m kg kg. Thus, from Table A-

18 at 0.2MPa=2bar, T1=30C, T2=60C.

b) Since volume change is the only work mode and pressure is constant, 2

12 2 1

1

6 23

123

10 / 10.2MPa 0.0307 0.0277 0.6

1 10

W pdV p V V

N m kJm kJ W

MPa N m

c) An energy balance reduces to give, U KE PE 12 12Q W or

12 12 2 1Q W m u u

With data from Table A-18 at 0.2MPa=2bar

12

12

0.6 0.1 524.3 476.3

5.4

kJQ kJ kg

kg

Q

3.77

A system consisting of 1kg H2O undergoes a power cycle composed of the following

processes:

Process 1-2: Constant-pressure heating at 10 bar from saturated vapor.

Process 2-3: Constant-volume cooling to P3= 5bar, T3= 160C.

Process 3-4: Isothermal compression with Q34 = -815.8kJ.

Process 4-1: Constant-volume heating.

Sketch the cycle on T-v and p-v diagrams. Neglecting kinetic and potential energy

effects, determine the thermal efficiency.

Analysis: The thermal efficiency of a power cycle is = Wcycle/Qin where Wcycle=W12+W23+W34+W41.

And Qin= 12 41Q Q

Process 1-2: State 1: P1=10 bar and x = 1. Therefore v1 = 0.1944 and u1=2583.6

State 2 is fixed by P2=10 bar, but need another property

From Table A-4 at P3= 5 bar, T3= 160C by interpolation between T=151.86 and T=1803

3 30.3835 2575.2m KJ

v and ukg Kg

3

2 3 2 20.3835 P =10 bar, therefore 3231.8m KJ

v v and ukg Kg

2 5 2 3

12 2 1 3

1

10 / 11 10 0.3835 0.1944 189.1

1 10

N m m kJW pdV mp v v kg bar kJ

bar kg N m

The first law for closed system is: Q U W

12 2 1 12 1 3231.8 2583.6 189.1 837.3Q m u u W kg kJ kJ

Process 2-3: 23 0W

23 3 2 23Q m u u w0

1 2575.2 3231.8 656.6kg kJ

Process 3-4: 34 815.8Q kJ (given).

34 34 34 34 34 4 3Q W U W Q U Q m u u

State 4 is fixed by 4 160T C , 4 1v v

34

4 3

4

0.1944 1.102 /100.6317

0.3071 1.1021/10

f

g

v vx

v v

4 4 4

34

(1 ) (1 0.6317)674.86 0.6317 2568.4 1871 /

815.8 1 1871 2575.2 111.6

f gu x u x u kJ kg

w kJ

Process 4-1: 41 0W , and

41Q U W0

1 41 1 2583.6 1871 712.6kJ

m u u kg kJkg

The net work is then

12 23 34 41

189.1 0 111.6 0 77.5

cycle

cycle

W W W W W

W kJ

To obtain Qin,

12 41 837.3 712.6 1549.9inQ Q Q kJ

Then

77.50.05 (5%)

1349.9

As a check, note that for every cycle Qcycle= Wcycle.

12 23 34 41

837.3 656.6 815.8 712.6 77.5

cycleQ Q Q Q Q

kJ

Which agrees with Wcycle calculated using the work quantities.

3.85

A gallon of milk at 68F is placed in a refrigerator. If energy is removed from the milk

by heat transfer at a constant rate of 0.08 Btu/s, how long would it take, in minutes, for

the milk to cool to 40F? The specific heat and density of the milk are 0.94 Btu/lb*R

and 64 lb/ft3, respectively.

Schematic & Given Data:

Engr. Model:1. The milk is the closed system. 2. For the system, 0W and kinetic and

potential energy play no role. 3. The milk is modeled as incompressible with constant

specific heat, C.

Analysis: An energy rate balance reduces as follows, dv dKE

dt dt

dPE

dtQ W or

2 1 .dv

Q U Q dt m u u Q tdt

In this expression, 3

3

0.1336864 1 8.56

1

lb ftm V gal lb

ft gal. Also, with Eq. 3.20a

2 1 2 1u u c T T . Collecting results

2 1

8.56 0.94 281min

47 min0.08 / 60

Btulb R

mc T T lb Rt

Btu s sQ

3.98

Five lbmol of carbon dioxide (CO2), initially at 320 lbf/in2, 660R, is compressed at

constant pressure in a piston-cylinder assembly. For the gas, W=-2000Btu. Determine

the final temperature, in R.

Engr. Model: 1. The CO2 is a closed system. 2. Pressure remains constant during the

compression process.

Analysis: The final state of the gas is fixed by pressure, 320 lbf/in2, and the specific

volume 2v , which can be evaluated from the work.

2

2 1 2 1 2 1

1

WW pdV p V V np v v v v

np(*)

1v can be found from the compressibility chart. From Table A-1E, pc=72.9 atm,

Tc=548R. Then

1

320 /14.7 6600.30 1.2

72.9 548R R

atm Rp T

atm R

Using these, Fig A-1 gives ' 1 4.0Rv , Then

2' 3

1 1 2 2

1545 5481

4.0 21.95 /72.9 14.7 / 144

cR

c

ft lbfR

RT ftlbmol Rv v ft lbmol

p lbf in in

Returning to Eq. (*)

3

2 1 2

3

778( 2000 )

121.95

5 320 14.8 /

15.21 /

ft lbfBtu

W ft Btuv v

np lbmol lbmol lbf ft

ft lbmol

So, at state 2, 2 1 0.30R Rp p and

'

2 2 /R c cv v p RT

3 2

'

2

15.21 / 72.9 14.7 144 /2.77

1545 548R

ft lbmol lbf ftV

ft lbfR

lbmol R

Returning to Fig. A-1, 2 0.95RT . Thus

2 2 0.95 548 521R CT T T R R

3.128

Air is confined to one side of a rigid container divided by a partition, as shown in Fig.

P3.128. The other side is initially evacuated. The air is initially at p1=5 bar, T1=500 K,

and V1=0.2m3. When the partition is removed, the air expands to fill the entire chamber.

Measurements show that V2=2V1 and p2=p1/4. Assuming the air behaves as an ideal gas,

determine (a) the final temperature, in K, and (b) the heat transfer, kJ.

P3.123

Engr. Model:

1. The closed system is the region within

the container, ignoring the partition.

2. The air is modeled as an ideal gas.

3. There are no overall changes in kinetic or

potential energy.

4. W=0.

Analysis:

a) Using the ideal gas model equation of state, 1 1 1PV mRT , 2 2 2PV mRT . Thus,

2 22 1

1 1

1500 2 250

4

PVT T K K

PV

b) An energy balance reduces to Q W U KE PE , or

2 1Q m u u

Where 5 2 2

1 1

1

5 10 / 0.20.7

8314500

28.97

N m mPVm kg

RT N mK

kg K

So, with data from the air Table A-22

0.7 178.28 359.49 126.8kJ

Q kg kJkg

3.135

Air is compressed in a piston cylinder assembly from 1 2

10Lbf

pin

, 1 500T R to a final

volume of V2 = 1 ft3 in a process described by 1.25pv const . The mass of the air is

0.5 Lb. Assuming ideal gas behavior, and ignoring kinetic and potential energy effects,

determine the work and heat transfer, each in Btu. Solve the problem each of two ways:

(a) Using a constant specified heat evaluated at 500 R. (b) Using data from Table A-22E. Compare the results and discuss

3

2

3

1

1

1.25 1.25

1 1 2 2

2 2

02 22

2 2 1 1

2 1

ft.Lbf1545( )

Lbm.R 53.328.97

12

0.5

(53.3)(500)18.52

(10)(144)

161.5

(161.5)(144)(2)872

(53.3)

1(0.5) 51

1 778

U

air

air

RR

M

ftv

Lbm

RT ftv

p Lbm

p v p v

Lbfp

in

p vT R

R

p v p vW Btu

n

Q W U U

2 1 2 1

1

2

sing the ideal gas model:

Btu( ) 0.173(872 500) 64.356

Lbm

51 0.5(64.356) 18.82

Using the Air Table A22E

Btuu =85.2

Lbm

Btuu =149.61

Lbm

51 0.5(149.61 85.2) 18.80

avgvu u C T T

Q Btu

Q Btu

3.138

Two-tenths kmol of nitrogen (N2) in a piston-cylinder assembly undergoes two processes

in series as follows:

Process 1-2: Constant pressure at 5 bar from V1=1.33m3 to V2=1m

3.

Process 2-3: Constant volume to p3=4bar.

Assuming ideal gas behavior and neglecting kinetic and potential energy effects,

determine the work and heat transfer for each process, in kJ.

Schematic & Given Data:

Engr. Model:

1) The N2 contained in the piston-cylinder assembly if

the closed system. 2) Volume change is the only work

mode. 3) The N2 is modeled as an ideal gas. 4) Kinetic

and potential energy effects play no role.

Analysis: Process 1-2 occurs at constant pressure. Thus 2

12 2 11

W pdV p V V . That is:

5 23 3

12 3

10 / 15 1 1.33 165

1 10

N m kJW bar m m kJ

bar N m

Process 2-3 occurs at constant volume. That is W23=0.

Reducing an energy balance, V KE PE Q W . Thus Q W n u . This

requires T1,T2,T3. To find T1 was the ideal gas equation of state:

5 3

21 1

1

5 10 1.33

4008314

0.2

Nm

PV mT K

N mnRkmol

Kmol K

For process 1-2, p=constant; so the ideal gas equation of state gives 1 1pV nRT ,

2 2pV nRT , or

3

22 1 3

1

1400 301

1.33

V mT T K

V m

For process 2-3, V=constant; so 2 2p V nRT , 3 3p V nRT or

33 2

2

4301 241

5

P barT T K K

P bar

For process 1-2, with 1u and 2u from Table A-23

2 112 12 165 0.2 6250 8314 578kJ

Q W n u u kJ kmol kJkmol

For process 2-3 3 223 23 0.2 5000 6250 250kJ

Q W n u u kmol kJkmol

3.140

Air is contained in a piston cylinder assembly undergoes the power cycle shown.

Assuming ideal gas behavior, evaluate the thermal efficiency of the cycle.

3.140 continued:

Analysis:

2

2 212 1 1

11 1

5

12 3

12 12 12

12

12

2

23 3 21

5

23 3

2 22

ln ln

1*10 5ln 804.7

10 1

Since no change in temperature (isothermal) 0

804.7

1*10(1 5) 400

10

1742.2 from air t

v vw pdv c Pv

v v

KJw

Kg

q w u

u

KJq

Kg

w pdv p v v

KJw

Kg

P vT K

R2

3 33 3

23 23 3 2

31

12 23

able A22 1432.5

348.4 from air table A22 248.9

1583.6

0 .

404.7 404.7

1988.3

0.204 (20.4%)

net net

in

net

in

KJu

Kg

P v KJT K u

R Kg

KJq w u u

Kg

w Volume is const

KJ KJw and q

Kg Kg

KJq q q

Kg

w

q