Thermo 1 (MEP 261)
Thermodynamics An Engineering Approach
Yunus A. Cengel & Michael A. Boles
7th Edition, McGraw-Hill Companies,
ISBN-978-0-07-352932-5, 2008
Sheet 3:Chapter 3
3-26) Complete the table for H2O:
T, °C p, kPa h, kJ/kg v, m3/kg Phase description x
50 4.16
200 Saturated vapor
250 400
110 600
Solution
a) T1 = 50°C, v1 = 4.16 m3/kg.
From Table A-4, Saturated water,
Temperature table
vf = 0.0001012 m3/kg, vg = 12.026 m3/kg, hf = 209.34 kJ/kg, hg =
2591.3 kJ/kg,
As vf (=0.0001012 m3/kg) < v1 (= 4.16 m3/kg)< vg (=
m3/kg).
12.026
Hence, State 1 is saturated Liquid-vapor
mixture.
p1 = psat @ T = 50°C = 12.352 kPa.
x1 = (v1 - vf)/( vg - vf) = (4.16 - 0.0001012)/( 12.026 -
0.0001012)= 0.3459
h1 = hf +x1 ( hg - hf) = 209.34 +0.3459 (2591.3 - 209.34) =
1033.26 kJ/kg.
b) p2 = 200 kPa, State 2 is saturated vapor. Hence, x2 = 1.
T2 = Tsat @ p = 200 kPa = 120.21°C. (from Table A-5)
v2 = vg @ p = 200 kPa = 0.88578 m3/kg. (from Table
A-5)
h2 = hg @ p = 200 kPa = 2706.3 kJ/kg. (from Table A-
5)
c) T3 = 250°C, p3 = 400 kPa.
From Table A-5, Saturated water, Pressure
table, Tsat @ p = 400 kPa = 143.61°C.
As T3 (= 400 °C) > Tsat @ p = 400 kPa = 143.61°C,
Hence, State 3 is superheated steam.
From Table A-6, for superheated steam,
For T3 = 250°C, p3 = 400 kPa (0.4 MPa), v3 =
0.59520 m3/kg, h3 = 2964.5 kJ/kg.
d) T4 = 110°C, p4 = 600 kPa.
From Table A-5, Saturated water, Pressure
table, Tsat @ p = 600 kPa = 158.83°C.
As T4 (= 110 °C) < Tsat @ p = 600 kPa = 158.83°C,
Hence, State 4 is compressed (subcooled)
liquid.
As there is no data for compressed liquid
water for p = 600 kPa (= 0.6 MPa) in Table
A-7. Hence, by utilizing the approximation:
v4 = vf @ T = 110 °C = 0.001052 m3/kg. (from Table
A-4)
h4 = hf @ T = 110 °C = 461.42 kJ/kg. (from Table A-
4)
Point T, °C p, kPa h, kJ/kg v, m3/kg Phase description x
1 50 12.352 1033.26 4.16 Saturated Liquid- 0.3459
vapor mixture
2 120.21 200 2706.3 0.88578 Saturated vapor 1.0
3 250 400 2964.5 0.5952 Superheated steam -
4 110 600 461.42 0.001052 Compressed -
(Subcooled liquid)
T psat = 12.352 kPa
State 1
T = 50 °C
1
v=
4.16 m3/kg
vf = vg = v
0.0001012 m3/kg 12.026 m3/kg
T
State 2
p = 200 kPa
T2 =Tsat = 120.21 °C
vf = v = vg =
0.0007499 m3/kg 0.88578 m3/kg
State 3 T
p = 400 kPa
T3 = 250°C 3
Tsat = 143.61 °C v=
0.5952 m3/kg
vf = vg = v
0.001084 m3/kg 0.46242 m3/kg
State 4 T
p = 600 kPa
4 Tsat = 158.83 °C
T = 110 °C
3-30) Complete the table for H2O:
3
T, °C p, kPa h, kJ/kg v, m /kg x Phase description
200 0.7
140 1800
950 0.0
80 500
800 3162.2
Solution
a) p1 = 200 kPa, x = 0.7.
From Table A-5, Saturated water, Pressure
table
T1 = Tsat @ p = 200 kPa = 120.21°C.
vf = 0.001061 m3/kg, vg = 0.88578 m3/kg, hf = 504.71 kJ/kg, hg =
2706.3 kJ/kg,
As 0 < x1 (= 0.7)< 1.
Hence, State 1 is saturated Liquid-vapor
mixture.
v1 = vf + x1 ( vg - vf) = 0.001061 + 0.7 (0.88578 - 0.001061) = 0.62
m3/kg.
h1 = hf +x1 ( hg - hf) = 504.71 +0.7 (2706.3 - 504.71) = 2045.8 kJ/kg.
b) T2 = 140 °C, h2 = 1800 kJ/kg.
From Table A-4, Saturated water, Temperature
table,
vf = 0.001080 m3/kg, vg = 0.50850 m3/kg, hf = 589.16 kJ/kg, hg =
2733.5 kJ/kg,
As hf (=589.16 kJ/kg ) < h2 (=1800 kJ/kg )< hg (=
2733.5 kJ/kg ).
Hence, State 2 is saturated Liquid-vapor
mixture.
p2 = psat @ T = 140 °C = 361.53 kPa. (from Table A-
4)
x2 = (h2- hf)/(hg - hf) = (1800- 589.16)/( 2733.5 - 589.16) = 0.565
v2 = vf + x2 ( vg - vf) = 0.001080 + 0.565 (0.50850 - 0.001080) =
0.288 m3/kg.
c) x3 = 0, p3 = 950 kPa.
As x3 = 0, hence, State 3 is saturated liquid.
From Table A-5, Saturated water, Pressure
table,
T3 = Tsat @ p = 950 kPa = 177.66 °C.
v3 = vf = 0.001124 m3/kg, h3 = hf = 752.74 kJ/kg,
d) T4 = 80°C, p4 = 500 kPa.
From Table A-5, Saturated water, Pressure table,
Tsat @ p = 500 kPa = 151.83°C.
As T4 (= 80 °C) < Tsat @ p = 500 kPa = 151.83°C, Hence,
State 4 is compressed (subcooled) liquid.
As there is no data for compressed liquid water for p = 500
kPa (= 0.5 MPa) in Table A-7. Hence, by utilizing the
approximation:
v4 = vf @ T = 80 °C = 0.001029 m3/kg. (from Table
A-4)
h4 = hf @ T = 80 °C = 461.42 kJ/kg. (from Table A-4)
e) P5 = 800kPa, h5 = 3162.2 kJ/kg.
From Table A-5, Saturated water, Pressure table, hf @ p =
500 kPa = 720.87 kJ/kg, hg @ p = 500 kPa = 2770.8 kJ/kg, Tsat
@ p = 800 kPa = 170.41°C.
As h5 (= 3162.2 kJ/kg) > hg @ p = 500 kPa = 2770.8 kJ/kg,
Hence, State 5 is superheated steam.
From Table A-6, for superheated steam,
For p5 = 800 kPa =0.8 MPa, h5 = 3162.2 kJ/kg, T5 = 350 °C,
v5 = 0.35442 m3/kg.
Point T, °C p, kPa h, kJ/kg v, m3/kg Phase description x
1 120.21 200 2045.8 0.62 Saturated Liquid- 0.7
vapor mixture
2 140 361.53 1800 0.288 Saturated Liquid- 0.565
vapor mixture
3 177.66 950 752.74 0.001124 Saturated liquid 0.0
4 80 500 461.42 0.001029 Compressed liquid -
5 350 800 3162.2 0.35442 Superheated steam -
T
State 1 p = 200 kPa
T = 120.21 °C
1
g
f
v=
0.62 m3/kg
vf = vg = v
0.001061 m3/kg 0.88578 m3/kg
T
State 2
p = 361.53 kPa
T = 140°C g
2
f
v=
0.288 m3/kg
vf = vg = v
0.001080 m3/kg 0.50850 m3/kg
T
State 3
p = 950 kPa
T = 177.66°C
v = vf = v
0.001124 m3/kg
T
State 4
p = 500 kPa
4 Tsat = 151.83°C
T = 80 °C
State 5 T
p = 800 kPa
T5 = 350°C
Tsat = 170.41°C v=
0.35442 m3/kg
vf = vg = v
0.001115 m3/kg 0.24035 m3/kg
3-31) Complete the following table for refrigerant
134a:
Point T, °C p, kPa u, kJ/kg v, m3/kg x Phase description
1 -8 320
2 30 0.015
3 180 Saturated vapor
4 80 600
Solution
For refrigerant 134a:
State 1:T = 8 °C, p = 320 kPa , From Table A 12 (Saturated
refrigerant 134a Pressure table),
For p = 320 kPa, Tsat = 2.46 °C,
As T < Tsat , Hence, State 1 is compressed (subcooled) liquid,
Hence, conditions are nearly the same as those at Saturated liquid
corresponding to T = 8 °C.
Hence, From Table A 11, corresponding to T = 8°C
v = vf = 0.0007571 m3/kg & u = uf = 41.03 kJ/kg
T
p = 320 kPa
Tsat = 2.46 °C
T = -8
State 2: T = 30 °C, v = 0.015 m3/kg, From Table A 11 (Saturated
refrigerant 134a Temperature table)
For T = 30 °C, psat = 770.64 kPa, vf = 0.0008421 m3/kg, vg =
0.026622 m3/kg , uf = 92.93 kJ/kg, ug = 246.14 kJ/kg.
As vf (= 0.0008421 m3/kg) < v (= 0.015 m3/kg) < vg (= 0.026622
m3/kg)
Hence, state 2 is saturated liquid-vapor mixture
Hence,
. 0.0008421
0.026622 0.0008421
Hence, x = 0.5492,
u = uf + x ( v (ug uf) = 92.93 kJ/kg +0.5492 (246.14 -92.93) kJ/kg=
177.07 kJ/kg
psat = 770.64 kPa
T = 30 °C
v = 0.015 m3/kg
vf = vg =
0.0008421 m3/kg 0.026622 m3/kg
State 3: p = 180 kPa , Satureated vapor, x = 1;
From Table A-12 (Saturated refrigerant 134a Pressure table),
For p = 180 kPa, T = Tsat = 12.73 °C, v = vg = 0.11041 m3/kg , u =
ug = 222.99 kJ/kg.
T
p = 180 kPa
T=Tsat = 12.73 °C
v = vg = v
0.11041 m3/kg
State 4: T = 80 °C, p = 600 kPa ,
From Table A-12 (Saturated refrigerant 134a Pressure table),
Tsat at p = 600 kPa (=21.55 °C),
As T (= 80 °C) > Tsat at p = 600 kPa (=21.55 °C), hence state 4 is
superheated vapor.
Hence, From Table A 13, Superheated refrigerant 134a
For p = 600 kPa (=0.6 MPa), T = 80 °C , v = 0.044710 m3/kg & u
= 292.73 kJ/kg
T
p = 600 kPa
T = 80°C
Tsat = 21.55 °C
v=
0.044710
m3/kg
vf = vg = v
0.0008199 m3/kg 0.034295 m3/kg
Point T, °C p, kPa u, kJ/kg v, m3/kg x Phase description
1 -8 320 41.03 0.0007571 - Compressed
liquid
2 30 770.64 177.07 0.015 0.5492 Saturated liquid-
vapor mixture
3 12.73 180 222.99 0.11041 1.0 Saturated vapor
4 80 600 292.73 0.044710 - Superheated vapor
3 32: Complete the table for refrigerant 134a:
T, °C p, kPa u, kJ/kg v, m3/kg Phase description x
20 95
-12 Saturated liquid
400 300
8 600
Solution
For refrigerant 134a:
T = 20 °C, u = 95 kJ/kg, From Table A 11,
For T = 20 °C, psat = 572.07 kPa, uf = 78.86 kJ/kg, ug = 241.02 kJ/kg,
vf = 0.0008161 m3/kg, vg = 0.035969 m3/kg,
As uf < u < ug , Hence, State is saturated liquid vapor mixture,
95 78.86
.
241.02 78.86
As
Hence,
0.0008161
.
0.035969 0.0008161
Hence,v = 0.004315 m3/kg,
T psat = 572.07 kPa
T = 20 °C
v=
0.004315 m3/kg
vf = vg = v
0.0008161 m3/kg 0.035969 m3/kg
b)
T = 12 °C, Saturated liquid, From Table A 11,
For T = 12 °C, psat = 185.37 kPa, u = uf = 35.78 kJ/kg, v = vf =
0.0007499 m3/kg& x = 0.
T
psat = 185.37 kPa
T = -12 °C
v = vf =
0.0007499 m3/kg
c)
p = 400 kPa , u = 300 kJ/kg, From Table A 12,Data for p = 400 kPa;
p, kPa T , °C v , m3/kg u, kJ/kg
400 80 0.068747 294.53
T v 300
90 0.071023 303.32
Hence,
300 294.53
90 80 303.32 294.53
Hence, T = 86.22 °C
0.068747 300 294.53
0.071023 0.068747 303.32 294.53
v = 0.07016 m3/kg,
From Table A-12,
For p = 400 kPa, Tsat = 8.91 °C, Tsat = 185.37 kPa, vf = 0.0007907
m3/kg& vg = 0.051201 m3/kg.
T
p = 400 kPa
T = 86.22°C
Tsat = 8.91 °C
v=
0.07016 m3/kg
vf = vg = v
0.0007907 m3/kg 0.0512012 m3/kg
d)
p = 600 kPa, From Table A 12,
For p = 600 kPa, Tsat = 21.55 °C.
As T (= 8 °C) < Tsat at p = 600 kPa (=21.55 °C), hence state is sucbcooled
(compressed) liquid.
Hence, conditions are nearly the same as those at Saturated liquid
corresponding to T = 8 °C.
Hence, From Table A 11, corresponding to T = 8°C
v = vf = 0.0007887 m3/kg & u = uf = 62.39 kJ/kg
T
p = 400 kPa
Tsat = 21.55 °C
T = 8 °C
T, °C p, kPa u, kJ/kg v, m3/kg Phase description x
20 572.07 95 0.004315 Saturated liquid-vapor 0.09953
mixture
-12 185.37 35.78 0.0007499 Saturated liquid 0
86.22 400 300 0.07016 Superheated vapor -
8 600 62.39 0.0007887 Sucbcooled -
(compressed) liquid
3 34: Complete the table for H2O:
T, °C p, kPa h, kJ/kg v, m3/kg x Phase description
140 0.05
550 Saturated liquid
125 750
500 0.140
Solution
a) T = 140 °C, v = 0.05 m3/kg.
From Table A-4, Saturated water,
Temperature table, psat at T = 140 °C (=361.53 kPa
For T = 140°C.
vf = 0.001080 m3/kg, vg = 0.50850 m3/kg, hf = 589.16 kJ/kg, hg =
2733.5 kJ/kg,
As vf (=0.001080 m3/kg) < v (=0.05 m3/kg)< vg(=0.50850
m3/kg).
Hence, State 1 is saturated Liquid-vapor
mixture.
x1 =( v1 - vf)/( vg - vf) = (0.05 - 0.001080) / (0.50850 - 0.001080) =
0.0964.
h1 = hf +x1 ( hg - hf) = 589.16 +0.0964 (2733.5 - 589.16) = 795.87
kJ/kg.
b) P2 = 550 kPa, Saturated liquid.
From Table A-5, Saturated water, Pressure
table, Tsat @ p = 550 kPa = 155.46°C.
v = vf = 0.001097 m3/kg, h = hf = 655.77 kJ/kg.
c) T3 = 125°C, p3 = 750 kPa.
From Table A-5, Saturated water, Pressure table,
Tsat @ p = 750 kPa = 167.75 °C.
As T3 (= 125 °C) < Tsat @ p = 750 kPa = 167.75°C, Hence,
State 4 is compressed (subcooled) liquid.
As there is no data for compressed liquid water for p = 750
kPa (= 0.75 MPa) in Table A-7. Hence, by utilizing the
approximation:
v3 = vf @ T = 125 °C = 0.001065 m3/kg. (from Table
A-4)
h3 = hf @ T = 125 °C = 525.07 kJ/kg. (from Table A-4)
3
d) T4 = 500 °C, v4 =0.140 m /kg.
From Table A-6, Superheated steam, T4 = 500 °C , v4 = 0.140
m3/kg , h4 = 3462.8 kJ/kg , p4 = 2.5 MPa.
T, °C p, kPa h, kJ/kg v, m3/kg x Phase description
140 361.53 795.87 0.05 0.0964 Saturated liquid-
vapor mixture
155.46 550 655.77 0.001097 0.0 Saturated liquid
125 750 525.07 0.001065 - Compressed
(Subcooled) liquid
500 2500 3462.8 0.140 - Superheated
steam
T
State 1
p = 361.53 kPa
T =140 °C
1
g
f
v=
0.05 m3/kg
vf = vg = v
0.001080 m3/kg 0.50850 m3/kg
T
State 2
p = 550 kPa
T = 155.46°C
v = vf = 0.001097 m3/kg v
T
State 3
p = 750 kPa
3 Tsat = 167.75 °C
T = 125 °C
v = 0.001065 m3/kg
State 4 T
p = 2.5 MPa
T4 = 500°C
v=
0.140 m3/kg
v
3-35) Complete the following table for H2O
T, °C p, kPa u, kJ/kg v, m3/kg x Phase description
400 1450
220 Saturated vapor
190 2500
4000 3040
Solution
a) p = 400 kPa, u = 1450 kJ /kg.
From Table A-5, Saturated water,
Teperature table, Tsat at p = 400 kPa (=143.61°C).
vf = 0.001084 m3/kg, vg = 0.46242 m3/kg, uf = 604.22 kJ/kg, ug =
2553.1 kJ/kg,
As uf (=604.22 kJ/kg) < u (=1450 kJ /kg)< ug(=2553.1
kJ/kg).
Hence, State 1 is saturated Liquid-vapor
mixture.
x1 =( u1 - uf)/( ug - uf) = (1450- 604.22) / (2553.1 - 604.22) = 0.44.
v1 = vf +x1 ( vg - vf) = 0.001084 +0.44 (0.46242 - 0.001084) = 0.2013
m3/kg.
b) T2 = 220 °C, Saturated vapor.
From Table A-4, Saturated water,
Tempererature table, psat @ T = 220 °C = 2319.6
kPa.
v = vg = 0.078405 m3/kg, u = ug= 2602.3 kJ/kg.
c) T3 = 190°C, p3 = 2500 kPa.
From Table A-5, Saturated water, Pressure table,
Tsat @ p = 2500 kPa = 223.95 °C.
As T3 (=190 °C) < Tsat @ p = 2500 kPa = 223.95 °C, Hence,
State 3 is compressed (subcooled) liquid.
As there is no data for compressed liquid water for p =
2500 kPa (= 2.5 MPa) in Table A-7. Hence, by utilizing the
approximation:
v3 = vf @ T = 190 °C = 0.001141 m3/kg. (from Table
A-4)
u3 = uf @ T = 190 °C = 806.00 kJ/kg. (from Table A-4)
d) p4 = 4000 kPa, u4 = 3040 kJ/kg.
From Table A-5, Saturated water, Pressure table,
uf = 1082.4 kJ/kg, ug = 2601.7 kJ/kg,
As u4 = 3040 kJ/kg > ug = 2601.7 kJ/kg, hence state 4 is superheated
vapor.
From Table A-6, Superheated water,
P4 = 4000 kPa (4.0 MPa),
T , °C v , m3/kg u , kJ/kg
450 0.08004 3011.0
T4 v4 3040
500 0.08644 3100.3
Hence, (T4-450)/(500-450)=(3040-3011.0)/(3100.3-
3011.0)
i.e., T4 = 466.24 °C
Similarly, (v4-0.08004)/( 0.08644-0.08004)=(3040-
3011.0)/( 3100.3- 3011.0)
i.e., v4 = 0.082118 m3/kg
T, °C p, kPa u, kJ/kg v, m3/kg x Phase description
143.61 400 1450 0.2013 0.44 Saturated liquid-
vapor mixture
220 2319.6 2602.3 0.078405 1.0 Saturated vapor
190 2500 806.00 0.001141 - compressed
(subcooled) liquid
466.24 4000 3040 0.082118 - Superheated
vapor
T
State 1
p = 400 kPa
T =143.61 °C
1
g
f
v=
0.2013 m3/kg
vf = vg = v
0.001084 m3/kg 0.46242 m3/kg
T
State 2
p = 2319.6 kPa
T = 220°C 2
g
v = vg = 0.078405 m3/kg v
T
State 3
p = 2500 kPa
3 Tsat = 223.95 °C
T = 190 °C
v = 0.001065 m3/kg
State 4 T
p = 4000 kPa
T4 = 466.24 °C
4
v4 =
0.082118 m3/kg
v
Solution
Solution
Solution
Solution
