Problem set 1

The University of Western Australia

Energy (ENSC2002)
Adapted from YA engel, MA Boles, Thermodynamics: an engineering approach, 7th ed.,
2008.
1. Complete this table for water.
T, C
50
250
110

P, kPa

, m3/kg

Phase description

s, kJ/kgK

7.72
400
500
350

Saturated vapour

2. Complete this table for R134a.

T, C
-10
-14
44

P, kPa

h, kJ/kg

600

180

Phase
description

0.6
500
1200

300.61
1.0

3. Examine the pressure-enthalpy (P-h) and temperature-entropy (T-s) diagrams of

R134a.
For the P-h diagram, identify the subcooled, two phase, and superheated regions.
Identify the isotherms (constant temperature lines), isentropes (constant entropy
lines), iso-density lines, and iso-quality lines.
The P-h diagram is very useful for the analysis of chiller thermodynamic cycle.
For the T-s diagram, identify the subcooled, two phase, and superheated regions.
Identify the isenthalps (constant enthalpy lines), isobars (constant pressure lines),
iso-density lines, and iso-quality lines.
The T-s diagram is very useful for the analysis of both chiller and power plant
thermodynamic cycles.

Prof. Hui Tong Chua

362

4. Examine temperature-entropy (T-s) diagram of water. Identify the subcooled, two

phase, and superheated regions. Identify the isenthalps (constant enthalpy lines),
isobars (constant pressure lines), iso-density lines, and iso-quality lines.
This is the cornerstone for the analysis of steam turbine power plant.
5. Plot the state points in question 1 on the T-s diagram of water.
6. Plot the state points in question 2 on the P-h diagram of R134a.

Prof. Hui Tong Chua

363

Prof. Hui Tong Chua

364

Prof. Hui Tong Chua

365

Prof. Hui Tong Chua

366

Problem set 1 solutions

T C

p kPa

v m3/kg

s kJ/kg-K

Phase

50

12.35

7.72

5.435

Two-phase

143.61

400

0.4624

6.8955

Saturated vapor

250

500

0.4744

7.2724

Superheated

110

350

0.001052

1.4188

Subcooled

1.

a. From the Properties of Saturated Water Properties by Temperature Table at

50 C, = 0.001007 m3/kg and = 12.03 m3/kg. Hence from the problem
statement < < and therefore the system is two-phase with = =
12.35 kPa.
=

(7.72 0.001007) m3 /kg

= 0.642
(12.03 0.001007) m3 /kg

= (1 0.642) 0.7038 kJkg K + 0.642 8.0748 kJkg K

= 5.435 kJkg K
b. From the Properties of Saturated Water Properties by Pressure Table at 400
kPa:
= 143.61 C,

= = 0.4624 m3 /kg,

= = 6.8955 kJkg K

c. From the Properties of Saturated Water Properties by Temperature Table at

250 C, = 3.976 MPa > 500 kPa, therefore the steam is superheated. From
the Superheated Water Properties Table at 250 C, 0.50 MPa:
= 0.4744m3 /kg,

= 7.2724 kJkg K

d. From the Properties of Saturated Water Properties by Temperature Table at

110 C, = 0.14338 MPa < 350 kPa, therefore the system is a subcooled

367

liquid. Away from the critical point, liquids are nearly incompressible and as a
consequence of this the properties of a subcooled liquid are a function of
temperature but only very weakly a function of pressure. Hence we can
approximate the properties of the subcooled liquid by looking at the properties of
the saturated liquid at the same temperature.

3
110 C,350 kPa ,110
C = 0.001052 m kg

110 C,350 kPa ,110

C = 1.4188 kJkg K

368

2.

T C

p kPa

h kJ/kg

Phase

-15.11

600

180

Subcooled

-10

201.1

310.28

0.6

Two-phase

-14

500

181.45

Subcooled

46.31

1200

300.61

0.22

Two-phase

1.0

Under constrained

44

a. From the Saturated R134a Properties by Pressure Table at 600 kPa, =

229.68 kJ/kg > 180 kJ/kg. Therefore the system is subcooled and x = 0.
Using linear interpolation from the saturated liquid enthalpy values:
= 15.59 C +

(180 179.37) kJkg

(12.71 + 15.59) C
(183.13 179.37) kJkg

= 15.11 C
b. From the Saturated R134a Properties by Temperature Table at -10 C:
= 0.18524 MPa +

(10 + 12) C
(0.21693 0.18524) MPa
(8 + 12) C

= 0.2011 MPa
= 184.07 kJkg +

(10 + 12) C
(189.34 184.07) kJkg
(8 + 12) C

= 186.70 kJ/kg
= 391.46 kJkg +

(10 + 12) C
(393.87 391.46) kJkg
(8 + 12) C

= 392.66 kJ/kg
With x = 0.6,
= (1 0.6) 186.70 kJkg + 0.6 392.66 = 310.28 kJ/kg
c. From the Saturated R134a Properties by Pressure Table at 500 kPa, =
15.73 C > 14 C. Therefore the system is subcooled and x = 0. From the
Saturated R134a Properties by Temperature Table at -14 C

369

 ,14 = 178.83 kJkg +

(14 + 16) C
(184.07 178.83) kJkg
(12 + 16) C

= 181.45 kJ/kg
d. From the Saturated R134a Properties by Pressure Table at 1.2 MPa: =
265.95 kJkg < 300.61 kJkg < 422.04 kJkg = . Therefore the system
is two-phase and = = 46.31 C.
=

(300.61 265.95) kJ/kg

= 0.22
(422.04 265.95) kJ/kg

e. The system is under constrained. It could be a saturated vapor such that

= = 1.1301 Mpa and = = 421.11 kJ/kg. Alternately it could be
superheated with < 1.301 MPa . For instance if = 1.00 MPa then from the
Superheated R134a Properties Table:
= 419.86 kJkg +

(44 40) C
(430.88 419.86) kJkg
(50 40) C

= 424.27 kJkg

3. If you have questions about this problem, please discuss it with your group and if
you still arent sure about the lines and regions on the R134a p-h and T-s
diagrams ask your facilitator for help.
4. If you have questions about this problem, please discuss it with your group and if
you still arent sure about the lines and regions on the water T-s diagrams ask
your facilitator for help.

370

Problem 5 Solution.
371

Problem 6 Solution.

372