PROBLEM 3.117

As shown in Fig. P3.117, 20 ft3 of air at T1 = 600

oR, 100 lbf/in.

2 undergoes a polytropic

expansion to a final pressure of 51.4 lbf/in.2 The process follows pV

1.2 = constant. The work is

W = 194.34 Btu. Assuming ideal gas behavior for the air, and neglecting kinetic and potential

energy effects, determine

(a) the mass of air, in lb, and the final temperature, in oR.

(b) the heat transfer, in Btu.

KNOWN: Air undergoes a polytropic process in a piston-cylinder assembly. The work is

known.

FIND: Determine the mass of air, the final temperature, and the heat transfer.

SCHEMATIC AND GIVEN DATA:

ENGINEERING MODEL: 1. The air is a closed system.

2. Volume change is the only work mode. 3. The process is

polytropic, with pV1.2

= constant and W = 194.34 Btu.

4. Kinetic and potential energy effects can be neglected.

ANALYSIS: (a) The mass is determined using the ideal gas equation of state.

m =

=

= 9.00 lb

To get the final temperature, we use the polytropic process, pV1.2

= constant, to evaluate V2 as

follows.

V2 =

=

(20 ft

3) = 34.83 ft

3

Now

Air

W = 194.34 Btu

Q T1 = 600

oR

p1 = 100 lbf/in.2

V1 = 20 ft3

P2 = 51.4 lbf/in.2

pV1.2

= constant

p

v

100

51.4 2

1 T1

T2

pv1.2 = constant

.

.

PROBLEM 3.117 (CONTINUED)

T2 =

=

= 537

oR

Alternative solution for T2

The work for the polytropic process can be evaluated using W =

. For the process pV

1.2 =

constant, and incorporating the ideal gas equation of state, we get

W =

=

Solving for T2 and inserting values

T2 =

+ T1 =

+ (600

oR) = 537

oR

(b) Applying the energy balance; ΔKE + ΔPE + ΔU = Q – W. With ΔU = m(u2 – u1), we get

Q = m(u2 – u1) + W

From Table A-22E: u(600oR) = 102.34 Btu/lb and u(537

oR) = 91.53 Btu/lb. Thus,

Q = (9.00 lb)(91.53 – 102.34)Btu/lb + (194.34 Btu) = 97.05 Btu (in)

1

Corrected December, 2011

4.59 Refrigerant 134a enters an air conditioner compressor at 4 bar, 20°C, and is

compressed at steady state to 12 bar, 80°C. The volumetric flow rate of the refrigerant

entering is 4 m3/min. The power input to the compressor is 60 kJ per kg of refrigerant

flowing. Neglecting kinetic and potential energy effects, determine the heat transfer rate,

in kW.

KNOWN: Refrigerant 134a with known inlet and exit conditions flows through a

compressor operating at steady state.

FIND: Determine the heat transfer rate, in kW.

SCHEMATIC AND GIVEN DATA:

ENGINEERING MODEL:

(1) The control volume shown in the accompanying schematic operates at steady state.

(2) Potential and kinetic energy changes from inlet to exit can be neglected.

ANALYSIS:

To determine the heat transfer rate, begin with the steady state mass and energy balances.

Simplify based on assumptions and solve for cvQ .

Refrigerant 134a

p1 = 4 bar

T1 = 20oC

A1V1 = 4 m3/min

2

1

p2 = 12 bar

T2 = 80oC

Compressorm

W

cv

= –60 kJ/kg

cvQ = ?

21

2

2

2

121cvcv

21

2

VV0 zzghhmWQ

mmm

2

(1)

To obtain m , in kg/s, fix state 1 by referencing Table A-11 at 4 bar. T1 > Tsat and

therefore a superheated condition exists at state 1. From Table A-12 at T1 and p1: v1 =

0.05397 m3/kg.

From Table A-12: h1 = 262.96 kJ/kg and h2 = 310.24 kJ/kg. To obtain cvQ , in kW,

substitute into Eq. (1).

1. The negative sign indicates that there is energy rejected from the system by heat

transfer.

s

kg1.235

60s

min 1

kg

m0.05397

min

m4

AV3

3

1

1 v

m

12

cv

12cvcv hhm

WmhhmWQ

1# kW 71.15

s

kJ1

kW 1

kg

kJ262.96310.24

kg

kJ60

s

kg1.235cv

Q

PROBLEM 6.45

Steam undergoes an adiabatic expansion in a piston-cylinder assembly from 100 bar, 360oC to 1

bar, 160oC. What is work in kJ per kg of steam for the process? Calculate the amount of entropy

produced, in kJ/K per kg of steam. What is the maximum theoretical work that could be obtained

from the given initial state to the same final pressure. Show both processes on a properly-labeled

sketch of the T-s diagram.

KNOWN: Steam undergoes an adiabatic expansion in a piston-cylinder assembly. Data are

given at the initial and final states.

FIND: Determine the work and entropy produced, each per unit mass of steam. Find the

maximum theoretical work that could be obtained from the given initial state to the same final

pressure. Show both processes on a T-s diagram.

SCHEMATIC AND GIVEN DATA:

ANALYSIS: The work is determined by reducing the energy balance, as follows.

ΔKE +ΔPE + ΔU = Q – W → W/m = (u1 – u2)

From Table A-3: p1 = 100 bar and T1 = 360oC; u1 = 2729.1 kJ/kg and s1 = 6.0060 kJ/kg∙K

From Table A-3: p2 = 1 bar and T2 = 160oC; u2 = 2597.8 kJ/kg and s2 = 7.6597 kJ/kg∙K

W/m = (2729.1 - 2597.8) = 131.3 kJ/kg (out, as expected)

Now, the entropy production can be evaluated using the entropy balance, as follows.

ΔS =

+ σ → σ/m = (s2 – s1) = (7.6597 – 6.0060) kJ/kg∙K = 1.6537 kJ/kg∙K

To find the maximum theoretical work, we note that since (s2 – s1) = σ/m ≥ 0; s2 ≥ s1

Steam

p1 = 100 bar

T1 = 360oC

p2 1 bar

T2 = 160oC

ENGINEERING MODEL: (1) The

steam is a closed system. (2) Q = 0.

(3) ΔKE +ΔPE = 0.

PROBLEM 6.45 (CONTINUED)

Graphically

From Table A-3: p2 = 1 bar, s2s = 6.0060 kJ/kg∙K gives

x2s =

=

= 0.7765

and

u2s = uf@1 bar + x2s(ug@1 bar – uf@1 bar) = 417.36 + (0.7765)(2506.1 – 417.36) = 2039.3 kJ/kg

Finally, the maximum theoretical work is

(W/m)max = (u1 – u2s) = (2729.1 – 2039.3) = 689.8 kJ/kg

T

s

100 bar

360oC

1 bar

160oC . .

.

(2)

(2s)

(1) Note that since s2 ≥ s1, state 2s is the limiting

case when s2s = s1 (reversible expansion with

no entropy production) . It is not possible to

access states to the left of state 2s adiabatically

from state 1.

Expansion to state 2s give the biggest

difference in u, and hence the maximum

theoretical work.

Not accessible

adiabatically

PROBLEM 6.140

m3/s and

1

8.43 Compare the results of Problem 8.40 with those for the same cycle whose processes of the

working fluid are not internally reversible in the turbines and pumps. Assume that both turbine

stages and both pumps have an isentropic efficiency of 83%.

KNOWN: A regenerative vapor power cycle with one open feedwater heater operates with

steam as the working fluid. Operational data are provided.

FIND: Determine (a) the cycle thermal efficiency, (b) the mass flow rate into the first turbine

stage, in kg/s, and (c) the rate of entropy production in the open feedwater heater, in kW/K.

Compare results with those of Problem 8.40.

SCHEMATIC AND GIVEN DATA:

tW

3

Condenser

outQ

Pump

2

p2W

4

7

1Steam

Generator

inQ

p4 = p3 = 6 kPa

x4 = 0 (saturated liquid)

p1 = 12 MPa

T1 = 560oC

p3 = 6 kPa

p7 = p1 = 12 MPa

MW 330cycle W

Open

Feedwater

Heater

Pump

1

p1W

56

2

(1)

(1)

(1–y)

(1–y)

(1–y)

(y)

p6 = p5 = p2 = 1 MPa

x6 = 0 (saturated liquid)

Turbine

p2 = 1 MPa

ht = 83%

hp2 = 83% hp1 = 83%

s

T

65s

2s

1

p = 6 kPa

p = 12 MPa

3s4

p = 1 MPa

7s

2

3

7

5

2

ENGINEERING MODEL:

1. Each component of the cycle is analyzed as a control volume at steady state. The control

volumes are shown on the accompanying sketch by dashed lines.

2. All processes of the working fluid are internally reversible except for processes in the

turbines and pumps and mixing in the open feedwater heater.

3. The turbines, pumps, and open feedwater heater operate adiabatically.

4. Kinetic and potential energy effects are negligible.

5. Saturated liquid exits the open feedwater heater, and saturated liquid exits the condenser.

ANALYSIS: First fix each principal state.

State 1: p1 = 12 MPa (120 bar), T1 = 560oC → h1 = 3506.2 kJ/kg, s1 = 6.6840 kJ/kg∙K

State 2s: p2s = p2 = 1 MPa (10 bar), s2s = s1 = 6.6840 kJ/kg∙K → h2s = 2823.3 kJ/kg

State 2: p2 = 1 MPa (10 bar), h2 = 2939.4 kJ/kg (see below) → s2 = 6.9174 kJ/kg∙K

kg

kJ)3.28232.3506)(83.0(

kg

kJ2.3506)( 21t12

21

21t

s

s

hhhhhh

hhhh = 2939.4 kJ/kg

State 3s: p3s = p3 = 6 kPa (0.06 bar), s3s = s2 = 6.9174 kJ/kg∙K → x3s = 0.8191, h3s = 2130.4

kJ/kg

State 3: p3 = 6 kPa (0.06 bar), h3 = 2267.9 kJ/kg (see below) → x3 = 0.8760, s3 = 7.3620

kJ/kg∙K

kg

kJ)4.21304.2939)(83.0(

kg

kJ4.2939)( 32t23

32

32t

s

s

hhhhhh

hhhh = 2267.9 kJ/kg

State 4: p4 = 6 kPa (0.06 bar), saturated liquid → h4 = 151.53 kJ/kg, v4 = 0.0010064 m3/kg, s4 =

0.5210 kJ/kg∙K

State 5: p5 = p2 = 1 MPa (10 bar), h5 = 152.74 kJ/kg (see below) → s5 ≈ 0.5249 kJ/kg∙K

(assuming the saturated liquid state corresponding to h5 = hf in Table A-2 and interpolating

for s5 = sf)

p1

45445

45

454p1

)()(

hh

pphh

hh

pp

vv

mN 1000

kJ 1

kPa 1m

N1000

83.0

kPa )61000)(kg

m0010064.0(

kg

kJ53.151

2

3

5

h = 152.74 kJ/kg

State 6: p6 = 1 MPa (10 bar), saturated liquid → h6 = 762.81 kJ/kg, s6 = 2.1387 kJ/kg∙K,

3

v6 = 0.0011273 m3/kg

State 7: p7 = p1 = 12 MPa (120 bar), h7 = 777.75 kJ/kg (see below)

p2

67667

67

676p2

)()(

hh

pphh

hh

pp

vv

mN 1000

kJ 1

kPa 1m

N1000

83.0

kPa )100012000)(kg

m0011273.0(

kg

kJ81.762

2

3

7

h = 777.75 kJ/kg

(a) Applying energy and mass balances to the control volume enclosing the open feedwater

heater, the fraction of flow, y, extracted at location 2 is

kJ/kg )74.1524.2939(

kJ/kg )74.15281.762(

52

56

hh

hhy = 0.2189

For the control volume surrounding the turbine stages

))(1()( 32211

t hhyhhm

W

kg

kJ)9.22674.2939)(2189.01(

kg

kJ)4.29392.3506(

1

t m

W

= 1091.3 kJ/kg

For the pumps

))(1()( 45671

phhyhh

m

W

kg

kJ)53.15174.152)(2189.01(

kg

kJ)81.76275.777(

1

p

m

W

= 15.89 kJ/kg

For the working fluid passing through the steam generator

kg

kJ)75.7772.3506(71

1

in hhm

Q

= 2728.5 kJ/kg

Thus, the thermal efficiency is

4

kJ/kg 5.2728

kJ/kg )89.153.1091(

/

//

1in

1p1t

mQ

mWmW

h = 0.394 (39.4%)

(b) The net power developed is

)//( 1p1t1cycle mWmWmW

Thus,

)//( 1p1t

cycle1

mWmW

Wm

MW 1

s

kJ1000

kg

kJ)89.153.1091(

MW 3301

m = 306.9 kg/s

(c) The rate of entropy production in the open feedwater heater is determined using the steady-

state form of the entropy rate balance:

cv0

e

e

ei

i

i

j j

jsmsm

T

Q

Since the feedwater heater is adiabatic, the heat transfer term drops. Thus,

552266cv smsmsmsmsm i

i

ie

e

e

])1([ 5261cv syyssm

kJ/s 1

kW 1

Kkg

kJ)]5249.0)(2189.01()9174.6)(2189.0(1387.2[

s

kg9.306cv

= 65.82 kW/K

Compared to the ideal cycle in problem 8.40, the presence of internal irreversibilities in the

turbine stages and the pumps results in lower cycle thermal efficiency, higher required mass

flow rate of steam entering the first-stage turbine, and greater rate of entropy production in

the open feedwater heater.