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THERMODYNAMICS 2

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THERMODYNAMICS 2
RANKINE CYCLE (ACTUAL)
1. A 300 MW steam power plant operates on the rankine cycle with turbine and pump efficiencies of
85%. The turbine inlet conditions are 10.0 MPa and 460 °C, while the condenser pressure is 25
KPa. Calculate the (a) cycle thermal efficiency, (b) steam flow rate, and (c) condenser coolingwater flow rate.
Given:
Pnet = 300 MW
P1 = 10 MPa
T1 = 460 °C
πœ‚T = 85%
πœ‚P = 85%
P3 = 25 KPa
tH20 inlet = 15 °C
tH20 outlet = 30 °C
Required:
(a) πœ‚TH’
(b) ṁs
(c) ṁcw
Solution:
οƒ˜
State 1
At P1 = 10 MPa and T1 = 460 °C
tsat = 311.06 °C
tsat < T1 ; superheated vapor
οƒ˜ State 2
𝐾𝐽
S1 = S2 = 6.5966 π‘˜π‘”βˆ™πΎ
𝐾𝐽
P2 = P3 = 0.025 π‘˜π‘”βˆ™πΎ
𝐾𝐽
𝐾𝐽
Sg = 7.8314 π‘˜π‘”βˆ™πΎ
𝐾𝐽
SF = 0.8931 π‘˜π‘”βˆ™πΎ
h1 = 3373.7 π‘˜π‘”
𝐾𝐽
s1 = 6.5966 π‘˜π‘”βˆ™πΎ
For x2,
Sβ‚‚− Sf
x2 = 𝑆𝑓𝑔
x2 =
𝐾𝐽
(6.5966−0.8931 )
π‘˜π‘”βˆ™πΎ
𝐾𝐽
6.9383
π‘˜π‘”βˆ™πΎ
x2 = 0.8220
x2 = 82.20 %
Sf < S2 < Sg; mixture
For h2,
h2 = hf2 + x2 (hfg2)
𝐾𝐽
𝐾𝐽
h2 = 271.93 π‘˜π‘” + (0.8220) (2346.3 π‘˜π‘”)
𝐾𝐽
h2 = 2200.5886 π‘˜π‘”
1
οƒ˜ State 3
P2 = P3 = 0.025 MPa
𝐾𝐽
hf2 = h3 = 271.93 π‘˜π‘”
𝜐f2 = 𝜐3 =1.0199 × 10-3 π‘š³
π‘˜π‘”
οƒ˜ State 4
P4 = P1 = 10 MPa
Wp = 𝜐3 (P4 - P3)
π‘š³
Wp = 1.0199 × 10-3 π‘˜π‘” (10 MPa – 0.025 MPA)
π‘š³
1000 πΎπ‘ƒπ‘Ž
/
/ ×
Wp = 1.0300 × 10-3 / (9.975 MPa
×
1 π‘€π‘ƒπ‘Ž
/
π‘˜π‘”
1000
𝑁
/
π‘š²
1 πΎπ‘ƒπ‘Ž
/
1 𝐽/
1 𝐾𝐽
× 1 π‘βˆ™π‘š
× 1000 𝐽 )
/
/
/
𝐾𝐽
Wp = 10.7135 π‘˜π‘”
For WT,
For WT
h4 = h3 + Wp
𝐾𝐽
𝐾𝐽
h4 = 271.93 π‘˜π‘” + 10.7135 π‘˜π‘”
WT = h1 - h2
𝐾𝐽
𝐾𝐽
WT = 3373.7 π‘˜π‘” - 2200.5886 π‘˜π‘”
𝐾𝐽
𝐾𝐽
WT = 1173.1114 π‘˜π‘”
h4 = 242.1035 π‘˜π‘”
For h4’,
For h2’,
β„Žβ‚„−β„Žβ‚ƒ
πœ‚P = β„Žβ‚„′−β„Žβ‚ƒ
h4 ’ =
h4 ’ =
β„Žβ‚„−β„Žβ‚ƒ
ηβ‚š
β„Žβ‚−β„Žβ‚‚′
πœ‚T = β„Žβ‚−β„Žβ‚‚
+ h3
𝐾𝐽
(242.1035−271.93)
π‘˜π‘”
0.85
𝐾𝐽
+ 271.93 π‘˜π‘”
𝐾𝐽
h4’ = 283.8988 π‘˜π‘”
h2’ = - [πœ‚T(h1-h2)] + h1
h2’ = h1 - [πœ‚T(h1-h2)]
𝐾𝐽
𝐾𝐽
𝐾𝐽
h2’ = 3373.7 π‘˜π‘” – (0.85)( 3373.7 π‘˜π‘” - 2200.5886 π‘˜π‘”)
𝐾𝐽
h2’ = 2376.5553 π‘˜π‘”
For Wp’,
Wp’ = h4’ – h3
For WT’,
WT’ = h1 - h2’
𝐾𝐽
𝐾𝐽
Wp’ = 283.8988 π‘˜π‘” - 271.93 π‘˜π‘”
𝐾𝐽
𝐾𝐽
WT’ = 3373.7 π‘˜π‘” - 2376.5553 π‘˜π‘”
𝐾𝐽
Wp’ = 11.9688 π‘˜π‘”
𝐾𝐽
WT’ = 997.1447 π‘˜π‘”
For Wnet’,
Wnet’ = WT’ - Wp’
𝐾𝐽
𝐾𝐽
Wnet’ = 997.1447 π‘˜π‘” - 11.9688 π‘˜π‘”
𝐾𝐽
Wnet’ = 985.1759 π‘˜π‘”
For QA’,
QA’ = h1 – h4’
𝐾𝐽
𝐾𝐽
QA’ = 3373.7 π‘˜π‘” - 283.8988 π‘˜π‘”
𝐾𝐽
QA’ = 3089.8012 π‘˜π‘”
2
(a) For πœ‚TH’,
π‘Šπ‘›π‘’π‘‘′
πœ‚TH’ = 𝑄𝐴′
πœ‚TH’ =
πœ‚TH’ =
π‘Šπ‘‡′−π‘Šπ‘′
𝑄𝐴′
𝐾𝐽
𝐾𝐽
− 11.9688
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
3089.8012
π‘˜π‘”
997.1447
πœ‚TH’ = 0.32
πœ‚TH’ = 32%
(b) For ṁs,
𝑃𝑛𝑒𝑑
ṁs = π‘Šπ‘›π‘’π‘‘
ṁs =
𝐾𝐽
𝑠
𝐾𝐽
985.1759
π‘˜π‘”
300 000
𝐾𝑔
ṁs = 304.51 𝑠
(c) For ṁcw,
tH20 inlet = 15 °C
𝐾𝐽
hf = 62.99 π‘˜π‘”
tH20 outlet = 30 °C
𝐾𝐽
hf = 125.79 π‘˜π‘”
ṁs (β„Žβ‚‚′− β„Žβ‚ƒ)
ṁcw = (β„Žπ‘œπ‘’π‘‘−β„Žπ‘–π‘›)
ṁcw =
304.51 𝐾𝑔(2376.5553 𝐾𝐽− 271.93 𝐾𝐽)
𝑠
(125.79
π‘˜π‘”
𝐾𝐽
𝐾𝐽
− 62.99 )
π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
𝐾𝑔
ṁcw = 10205.0868 𝑠
TS DIAGRAM
P1 = P4 = 10 MPa
T1 = 460 °C
P2 = P3 = 0.025 MPa
3
2. A steam power plant with a capacity of 150 MW runs on the rankine cycle, featuring turbine and
pump efficiencies of 82%. The steam enters the turbine at 8.0 MPa and 550 °C and is condensed
at 25 KPa. Calculate the (a) cycle’s thermal efficiency, (b) the steam rate, and (c) the cooling
water flow rate in the condenser, assuming the cooling water enters at 20 °C and leaving at 35 °C.
Given:
P = 150 MW
P1 = 8.0 MPa
T1 = 550 °C
πœ‚T = 82%
πœ‚P = 82%
P3 = 25 KPa
tH20 inlet = 20 °C
tH20 outlet = 35 °C
Required:
(a)
πœ‚TH’
(b)
ṁs
(c)
ṁcw
Solution:
οƒ˜ State 1
At P1 = 8.0 MPa and T1 = 550 °C
tsat = 295.06 °C
tsat < T1 ; superheated vapor
𝐾𝐽
h1 = 3521.0 π‘˜π‘”
𝐾𝐽
s1 = 6.8778 π‘˜π‘”βˆ™πΎ
οƒ˜ State 2
For x2,
𝐾𝐽
Sβ‚‚− Sf
S1 = S2 = 6.8778 π‘˜π‘”βˆ™πΎ
x2 = 𝑆𝑓𝑔
𝐾𝐽
P2 = P3 = 0.025 π‘˜π‘”βˆ™πΎ
x2 =
𝐾𝐽
Sg = 7.8314 π‘˜π‘”βˆ™πΎ
(6.8778−0.8931 )
6.9383
𝐾𝐽
π‘˜π‘”βˆ™πΎ
𝐾𝐽
π‘˜π‘”βˆ™πΎ
x2 = 0.8626
x2 = 86.26 %
𝐾𝐽
SF = 0.8931 π‘˜π‘”βˆ™πΎ
Sf < S2 < Sg; mixture
For h2,
h2 = hf2 + x2 (hfg2)
𝐾𝐽
οƒ˜ State 3
P2 = P3 = 0.025 MPa
𝐾𝐽
h2 = 271.93 π‘˜π‘” + (0.8626) (2346.3 π‘˜π‘”)
𝐾𝐽
h2 = 2295.8484 π‘˜π‘”
𝐾𝐽
hf2 = h3 = 271.93 π‘˜π‘”
𝜐f2 = 𝜐3 =1.0199 × 10-3 π‘š³
π‘˜π‘”
4
οƒ˜ State 4
P4 = P1 = 8.0 MPa
Wp = 𝜐3 (P4 - P3)
π‘š³
Wp = 1.0199 × 10-3 π‘˜π‘” (8.0 MPa – 0.025 MPA)
π‘š³
Wp = 1.0199 × 10-3 π‘˜π‘” ( 7.975 MPa ×
1000
/ πΎπ‘ƒπ‘Ž
1000
𝑁
1𝐽
1 𝐾𝐽
π‘š²
× 1 πΎπ‘ƒπ‘Ž
× 1 π‘βˆ™π‘š × 1000 𝐽 )
1 π‘€π‘ƒπ‘Ž
𝐾𝐽
Wp = 8.1337 π‘˜π‘”
WT = h1 - h2
𝐾𝐽
𝐾𝐽
WT = 3521.0 π‘˜π‘” - 2295.8484 π‘˜π‘”
𝐾𝐽
WT = 1225.1516 π‘˜π‘”
h4 = h3 + Wp
𝐾𝐽
𝐾𝐽
h4 = 271.93 π‘˜π‘” + 8.1337 π‘˜π‘”
𝐾𝐽
h4 = 280.0637 π‘˜π‘”
For h4’,
β„Žβ‚„−β„Žβ‚ƒ
πœ‚P = β„Žβ‚„′−β„Žβ‚ƒ
h4 ’ =
h4 ’ =
β„Žβ‚„−β„Žβ‚ƒ
ηβ‚š
+ h3
(280.0637 − 271.93)
𝐾𝐽
π‘˜π‘”
0.82
𝐾𝐽
+ 271.93 π‘˜π‘”
𝐾𝐽
h4’ = 281.8491 π‘˜π‘”
For h2’,
β„Žβ‚−β„Žβ‚‚′
πœ‚T = β„Žβ‚−β„Žβ‚‚
h2’ = - [πœ‚T (h1-h2)] + h1
h2’ = h1 - [πœ‚T (h1-h2)]
𝐾𝐽
𝐾𝐽
𝐾𝐽
h2’ = 3521.0 π‘˜π‘” – (0.82) (3521.0 π‘˜π‘” - 2295.8484π‘˜π‘”)
h2’ = 2516.3757
𝐾𝐽
π‘˜π‘”
For WT’,
WT’ = h1 - h2’
For Wp’,
Wp’ = h4’ – h3
𝐾𝐽
𝐾𝐽
WT’ =3521.0 π‘˜π‘” - 2516.3757 π‘˜π‘”
𝐾𝐽
WT’ = 1004.6243 π‘˜π‘”
𝐾𝐽
𝐾𝐽
Wp’ = 281.8491 π‘˜π‘” - 271.93 π‘˜π‘”
𝐾𝐽
Wp’ = 9.9191 π‘˜π‘”
5
For Wnet’,
Wnet’ = WT’ - Wp’
Wnet’ = 1004.6243
𝐾𝐽
π‘˜π‘”
𝐾𝐽
- 9.9191
𝐾𝐽
π‘˜π‘”
Wnet’ = 994.7052 π‘˜π‘”
(a) For πœ‚TH’,
πœ‚TH’ = 𝑄𝐴′
πœ‚TH’ =
𝐾𝐽
𝐾𝐽
QA’ = 3521.0 π‘˜π‘” - 281.8491 π‘˜π‘”
𝐾𝐽
QA’ = 3239.1509 π‘˜π‘”
(b) For ṁs,
π‘Šπ‘›π‘’π‘‘′
πœ‚TH’ =
For QA’,
QA’ = h1 – h4’
𝑃𝑛𝑒𝑑
ṁs = π‘Šπ‘›π‘’π‘‘
π‘Šπ‘‡′−π‘Šπ‘′
ṁs =
𝑄𝐴′
𝐾𝐽
𝐾𝐽
− 9.9191
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
3239.1509
π‘˜π‘”
1004.6243
𝐾𝐽
𝑠
𝐾𝐽
994.7052
π‘˜π‘”
150 000
𝐾𝑔
ṁs = 150.7984 𝑠
πœ‚TH’ = 0.3071
πœ‚TH’ = 30.71 %
(c) For ṁcw,
tH20 inlet = 20 °C
𝐾𝐽
hf = 83.96 π‘˜π‘”
tH20 outlet = 35 °C
𝐾𝐽
hf = 146.68 π‘˜π‘”
ṁs (β„Žβ‚‚′− β„Žβ‚ƒ)
ṁcw = (β„Žπ‘œπ‘’π‘‘−β„Žπ‘–π‘›)
ṁcw =
𝐾𝑔
𝐾𝐽
𝐾𝐽
(2516.3757
− 271.93 )
𝑠
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
𝐾𝐽
(146.68
− 83.96 )
π‘˜π‘”
π‘˜π‘”
150.7984
𝐾𝑔
ṁcw = 5396.3460 𝑠
T-S DIAGRAM
P1 = P4 = 8 MPa
T1 = 550 °C
P2 = P3 = 0.025 MPa
6
3. A steam power plant operates 200 Mw on the rankine cycle with rubine and pump effieciencies
of 80%. Steam enters the turbine at 6.8 MPa and 480 °C and is condensed at 15 KPa.
Determine the (a) cycle’s thermal efficiency, (b) the steam rate, and (c) the cooling water flow
rate in the condenser if the cooling water enters at 25 °C and leaves at 40 °C.
Given:
Pnet = 200 MW
P1 = 6.8 MPa
T1 = 480 °C
P3 = 15 KPa
tH20 inlet = 25 °C
tH20 outlet = 40 °C
πœ‚T = 80%
πœ‚P = 80%
P1 = 6.8 MPa
T1 = 480 °C
πœ‚T = 80%
πœ‚P = 80%
= 25 °C
= 40 °C
P2 = P3 = 15 KPa = 0.015 MPa
Required:
(a) πœ‚TH’
(b) ṁs
(c) ṁcw
Solution:
οƒ˜ State 1
At P1 = 6.8 MPa and T1 = 480 °C,
tsat = 283.93 °C
tsat < T1 ; superheated vapor
𝐾𝐽
h1 = 3364.1 π‘˜π‘”
𝐾𝐽
s1 = 6.7495 π‘˜π‘”βˆ™πΎ
7
οƒ˜ State 2
𝐾𝐽
S1 = S2 = 6.7495 π‘˜π‘”βˆ™πΎ
𝐾𝐽
P2 = P3 = 0.015 π‘˜π‘”βˆ™πΎ
𝐾𝐽
Sg = 8.0085 π‘˜π‘”βˆ™πΎ
𝐾𝐽
SF = 0.7549 π‘˜π‘”βˆ™πΎ
Sf < S2 < Sg; mixture
For x2,
Sβ‚‚− Sf
x2 = 𝑆𝑓𝑔
x2 =
(6.7495 −0.7549 )
7.2536
𝐾𝐽
π‘˜π‘”βˆ™πΎ
𝐾𝐽
π‘˜π‘”βˆ™πΎ
x2 = 0.8264
x2 = 82.64 %
For h2,
h2 = hf2 + x2 (hfg2)
𝐾𝐽
𝐾𝐽
h2 = 225.94 π‘˜π‘” + (0.82.64) (2373.1 π‘˜π‘”)
𝐾𝐽
h2 = 2187.0698 π‘˜π‘”
οƒ˜ State 3
P2 = P3 = 0.015 MPa
𝐾𝐽
hf2 = h3 = 225.94 π‘˜π‘”
𝜐f2 = 𝜐3 =1.0141 × 10-3 π‘š³
π‘˜π‘”
οƒ˜ State 4
P4 = P1 = 6.8 MPa
Wp = 𝜐3 (P4 - P3)
Wp = 1.0141 × 10-3
π‘š³
π‘˜π‘”
(6.8 MPa – 0.015 MPA)
π‘š³
1000 πΎπ‘ƒπ‘Ž
/
/
/ ×
Wp = 1.0141 × 10-3
( 6.785 MPa
×
1 π‘€π‘ƒπ‘Ž
/
π‘˜π‘”
𝑁
π‘š²
1000
/ /
/ /
1 πΎπ‘ƒπ‘Ž
1 𝐽/
1 𝐾𝐽
× 1 π‘βˆ™π‘š × 1000 𝐽 )
/
/
𝐾𝐽
Wp = 6.8807 π‘˜π‘”
WT = h1 - h2
𝐾𝐽
𝐾𝐽
WT = 3364.1 π‘˜π‘” - 2187.0698 π‘˜π‘”
𝐾𝐽
WT = 1177.0302 π‘˜π‘”
8
h4 = h3 + Wp
𝐾𝐽
𝐾𝐽
h4 = 225.94 π‘˜π‘” + 6.8807 π‘˜π‘”
𝐾𝐽
h4 = 232.8207 π‘˜π‘”
For h4’,
β„Žβ‚„−β„Žβ‚ƒ
πœ‚P = β„Žβ‚„′−β„Žβ‚ƒ
h4 ’ =
h4 ’ =
β„Žβ‚„−β„Žβ‚ƒ
ηβ‚š
+ h3
(232.8207 − 225.94)
𝐾𝐽
π‘˜π‘”
0.82
𝐾𝐽
+ 225.94 π‘˜π‘”
𝐾𝐽
h4’ =234.5409 π‘˜π‘”
For h2’,
β„Žβ‚−β„Žβ‚‚′
πœ‚T = β„Žβ‚−β„Žβ‚‚
h2’ = - [πœ‚T (h1-h2)] + h1
h2’ = h1 - [πœ‚T (h1-h2)]
𝐾𝐽
𝐾𝐽
𝐾𝐽
h2’ = 3364.1 π‘˜π‘” – (0.80) (3364.1 π‘˜π‘” - 2187.0698 π‘˜π‘”)
𝐾𝐽
h2’ = 2422.4758 π‘˜π‘”
For WT’,
WT’ = h1 - h2’
For Wp’,
Wp’ = h4’ – h3
𝐾𝐽
𝐾𝐽
WT’ =3364.1 π‘˜π‘” - 2422.4758 π‘˜π‘”
𝐾𝐽
𝐾𝐽
𝐾𝐽
Wp’ = 234.5409 π‘˜π‘” - 225.94 π‘˜π‘”
𝐾𝐽
WT’ = 941.6242 π‘˜π‘”
Wp’ = 8.6009 π‘˜π‘”
For QA’,
QA’ = h1 – h4’
For Wnet’,
Wnet’ = WT’ - Wp’
𝐾𝐽
𝐾𝐽
Wnet’ = 941.6242 π‘˜π‘” - 8.6009 π‘˜π‘”
𝐾𝐽
Wnet’ = 933.0.233 π‘˜π‘”
𝐾𝐽
𝐾𝐽
QA’ = 3364.1 π‘˜π‘” - 234.5409 π‘˜π‘”
𝐾𝐽
QA’ = 3129.5591 π‘˜π‘”
(a) For πœ‚TH’,
π‘Šπ‘›π‘’π‘‘′
πœ‚TH’ = 𝑄𝐴′
πœ‚TH’ =
πœ‚TH’ =
π‘Šπ‘‡′−π‘Šπ‘′
𝑄𝐴′
𝐾𝐽
𝐾𝐽
− 8.6009
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
3129.5591
π‘˜π‘”
941.6242
πœ‚TH’ = 0.2981
πœ‚TH’ = 29.81 %
9
(b) For ṁs,
𝑃𝑛𝑒𝑑
ṁs = π‘Šπ‘›π‘’π‘‘
ṁs =
𝐾𝐽
𝑠
𝐾𝐽
933.0.233
π‘˜π‘”
200 000
𝐾𝑔
ṁs = 214.3569 𝑠
(c) For ṁcw,
tH20 inlet = 25 °C
𝐾𝐽
hf = 104.89 π‘˜π‘”
tH20 outlet = 40 °C
𝐾𝐽
hf = 167.57 π‘˜π‘”
ṁcw =
ṁcw =
ṁs (β„Žβ‚‚′− β„Žβ‚ƒ)
(β„Žπ‘œπ‘’π‘‘−β„Žπ‘–π‘›)
𝐾𝑔
𝐾𝐽
𝐾𝐽
(2422.4758
− 225.94
)
𝑠
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
𝐾𝐽
(167.57
− 104.89
)
π‘˜π‘”
π‘˜π‘”
214.3569
𝐾𝑔
ṁcw =7511.8476 𝑠
T-S DIAGRAM
P1 = P4 = 6.8 MPa
T1 = 480 °C
P2 = P3 = 0.015 MPa
10
RANKINE CYCLE (IDEAL)
4. In a steam power plant operating on the ideal Rankine Cycle, steam enters the turbine at
10 MPa and 500 °C and is condensed in the condenser at a pressure of 50 KPa. Fine the
thermal efficiency of this cycle.
Given:
P1 = 10 MPa
T1 = 500 °C
H20 in
H20 out
P3 = 50 KPa = 0.050 MPa
Required: πœ‚
Solution:
οƒ˜ State 1
At P1 = 10 MPa and T1 = 500 °C
tsat = 311.06 °C
tsat < T1 ; superheated vapor
𝐾𝐽
h1 = 3373.7 π‘˜π‘”
𝐾𝐽
s1 = 6.5966 π‘˜π‘”βˆ™πΎ
οƒ˜ State 2
𝐾𝐽
S1 = S2 = 6.5966 π‘˜π‘”βˆ™πΎ
P2 = P3 = 0.050 MPa
𝐾𝐽
Sg = 7.5939 π‘˜π‘”βˆ™πΎ
𝐾𝐽
SF = 1.0910 π‘˜π‘”βˆ™πΎ
Sf < S2 < Sg; mixture
11
For x2,
For h2,
Sβ‚‚− Sf
x2 = 𝑆𝑓𝑔
x2 =
h2 = hf2 + x2 (hfg2)
𝐾𝐽
(6.5966−1.0910)
π‘˜π‘”βˆ™πΎ
𝐾𝐽
6.5029
π‘˜π‘”βˆ™πΎ
𝐾𝐽
𝐾𝐽
h2 = 340.49 π‘˜π‘” + (0.8466) (2305.4 π‘˜π‘”)
𝐾𝐽
h2 = 2292.2416 π‘˜π‘”
x2 = 0.8466
x2 = 84.66 %
οƒ˜ State 3
P2 = P3 = 0.050 MPa
𝐾𝐽
hf2 = h3 = 340.49 π‘˜π‘”
𝜐f2 = 𝜐3 =1.0300 × 10-3 π‘š³
π‘˜π‘”
οƒ˜ State 4
P4 = P1 = 10 MPa
Wp = 𝜐3 (P4 - P3)
π‘š³
Wp = 1.0300 × 10-3 π‘˜π‘” (10 MPa – 0.050 MPA)
π‘š³
1000 πΎπ‘ƒπ‘Ž
Wp = 1.0300 × 10-3
( 9.95 MPa ×
×
π‘˜π‘”
Wp = 10.2485
1 π‘€π‘ƒπ‘Ž
1000
𝑁
π‘š²
1 πΎπ‘ƒπ‘Ž
1𝐽
1 𝐾𝐽
× 1 π‘βˆ™π‘š × 1000 𝐽 )
𝐾𝐽
π‘˜π‘”
WT = h 1 - h 2
𝐾𝐽
𝐾𝐽
WT = 3373.7 π‘˜π‘” - 2292.2416 π‘˜π‘”
𝐾𝐽
WT = 1081.4584 π‘˜π‘”
h4 = h3 + Wp
𝐾𝐽
𝐾𝐽
h4 = 340.49 π‘˜π‘” + 10.2485 π‘˜π‘”
h4 = 350. 7385
𝐾𝐽
π‘˜π‘”
Q A = h1 - h4
𝐾𝐽
𝐾𝐽
QA = 3373.7 π‘˜π‘” - 350. 7385 π‘˜π‘”
𝐾𝐽
QA = 3022.9615 π‘˜π‘”
12
For πœ‚,
πœ‚=
π‘Šπ‘›π‘’π‘‘
πœ‚=
π‘Šπ‘‡−π‘Šπ‘
πœ‚=
𝑄𝐴
𝑄𝐴
𝐾𝐽
𝐾𝐽
− 10.2485
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
3022.9615
π‘˜π‘”
1081.4584
πœ‚ = 0.3544
πœ‚ = 35.44 %
T-S DIAGRAM
T
P4 = P1 = 10 MPa
T1 = 500 °C
50 KPa
P2 = P3 = 0.050 MPa
0.050 MPa
S
s3 = s4
s1 = s2
13
5. For an ideal Rankine cycle in a steam power plant, steam enters the turbine at 6 MPa
and 450 °C and is condensed in the condenser at a pressure of 10 KPa. Calculate the
net work and its thermal efficiency.
Given:
P1 = 6 MPa
T1 = 450 °C
H20 in
H20 out
P3 = 10 KPa = 0.010 MPa
Required: Wnet and πœ‚
Solution:
οƒ˜ State 1
At P1 = 6 MPa and T1 = 450 °C
tsat = 275.64 °C
tsat < T1 ; superheated vapor
𝐾𝐽
h1 = 3301.8 π‘˜π‘”
𝐾𝐽
s1 = 6.7193 π‘˜π‘”βˆ™πΎ
οƒ˜ State 2
𝐾𝐽
S1 = S2 = 6.7193 π‘˜π‘”βˆ™πΎ
P2 = P3 = 0.010 MPa
𝐾𝐽
Sg = 8.1502 π‘˜π‘”βˆ™πΎ
𝐾𝐽
SF = 0.6493 π‘˜π‘”βˆ™πΎ
Sf < S2 < Sg; mixture
14
For x2,
For h2,
Sβ‚‚− Sf
x2 = 𝑆𝑓𝑔
x2 =
h2 = hf2 + x2 (hfg2)
(6.7193−0.6493)
𝐾𝐽
𝐾𝐽
π‘˜π‘”βˆ™πΎ
7.5009
𝐾𝐽
h2 = 191.83 π‘˜π‘” + (0.8092) (2392.8 π‘˜π‘”)
𝐾𝐽
π‘˜π‘”βˆ™πΎ
𝐾𝐽
h2 = 2128.0838 π‘˜π‘”
x2 = 0.8092
x2 = 80.92 %
οƒ˜ State 3
P2 = P3 = 0.010 MPa
𝐾𝐽
hf2 = h3 = 191.83
π‘˜π‘”
𝜐f2 = 𝜐3 =1.0102 × 10-3 π‘š³
π‘˜π‘”
οƒ˜ State 4
P4 = P1 = 6 MPa
Wp = 𝜐3 (P4 - P3)
π‘š³
Wp = 1.0102 × 10-3 π‘˜π‘” (6 MPa – 0.010 MPA)
π‘š³
1000 πΎπ‘ƒπ‘Ž
Wp = 1.0300 × 10-3
(5.99 MPa ×
×
π‘˜π‘”
1 π‘€π‘ƒπ‘Ž
1000
𝑁
π‘š²
1 πΎπ‘ƒπ‘Ž
1𝐽
1 𝐾𝐽
× 1 π‘βˆ™π‘š × 1000 𝐽 )
𝐾𝐽
Wp = 6.0511 π‘˜π‘”
WT = h 1 - h 2
𝐾𝐽
𝐾𝐽
WT = 3301.8 π‘˜π‘” - 2128.0838 π‘˜π‘”
𝐾𝐽
WT = 1173.7162 π‘˜π‘”
h4 = h3 + Wp
𝐾𝐽
𝐾𝐽
h4 = 191.83 π‘˜π‘” + 6.0511 π‘˜π‘”
𝐾𝐽
h4 = 197.8811 π‘˜π‘”
Q A = h1 - h4
𝐾𝐽
𝐾𝐽
QA = 3301.8 π‘˜π‘” - 197.8811 π‘˜π‘”
𝐾𝐽
QA = 3103.9189 π‘˜π‘”
16
For Wnet,
Wnet = WT - WP
𝐾𝐽
𝐾𝐽
Wnet = 1173.7162 π‘˜π‘” - 6.0511 π‘˜π‘”
𝐾𝐽
Wnet = 1167.6651 π‘˜π‘”
For πœ‚,
πœ‚=
πœ‚=
πœ‚=
π‘Šπ‘›π‘’π‘‘
𝑄𝐴
π‘Šπ‘‡−π‘Šπ‘
𝑄𝐴
𝐾𝐽
𝐾𝐽
− 6.0511
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
3103.9189
π‘˜π‘”
1173.7162
πœ‚ = 0.3762
πœ‚ = 37.62 %
T-S DIAGRAM
T
P4 = P1 = 6 MPa
T1 = 450 °C
10 KPa
P2 = P3 = 0.010 MPa
0.010 MPa
S
s3 = s4
s1 = s2
17
SINGLE-STAGE REHEATING CYCLE (ACTUAL)
1. Steam enters the initial turbine stage at 7.0 MPa and 400°C, expanding down to 0.6 MPa. It is
then reheated to 440 °C before entering the second turbine stage, where it expands further to
the condenser pressure of 0.02 MPa. With an isentropic efficiency of 85% for both the turbines
and pump. Determine (a) WTotal’, (b) WP’, and (c) πœ‚TH’.
Given:
Required:
(a) WTotal’
(b) WP’
(c) πœ‚TH’
Solution:
οƒ˜ State 1
At P1 = 7.0 MPa and T1 = 400 °C
tsat = 285.88 °C
tsat < T1 ; superheated vapor
οƒ˜ State 2
𝐾𝐽
S1 = S2 = 6.4478 π‘˜π‘”βˆ™πΎ
P2 = 0.6 MPa
𝐾𝐽
h1 = 3158.1 π‘˜π‘”
𝐾𝐽
s1 = 6.4478 π‘˜π‘”βˆ™πΎ
S (π‘˜π‘”βˆ™πΎ)
𝐾𝐽
h (π‘˜π‘”)
𝐾𝐽
6.4127
S2 = 6.4478
6.4507
2616.7
h2
2631.1
By interpolation,
𝐾𝐽
π‘˜π‘”βˆ™πΎ
𝐾𝐽
(6.4507−6.4127)
π‘˜π‘”βˆ™πΎ
(6.4478−6.4127)
h2 = 2630.0011
𝐾𝐽
=
β„Žβ‚‚ −2616.7 π‘˜π‘”
𝐾𝐽
π‘˜π‘”
(2631.1−2616.7)
𝐾𝐽
π‘˜π‘”
18
οƒ˜ State 2’
β„Žβ‚−β„Žβ‚‚′
πœ‚T = β„Žβ‚−β„Žβ‚‚
0.85 =
3158.1
3158.1
𝐾𝐽
−β„Žβ‚‚′
π‘˜π‘”
𝐾𝐽
𝐾𝐽
− 2630.0011
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
h2’ = 2709.2159 π‘˜π‘”
οƒ˜ State 3
P2 = P3 = 0.6 MPa and T3 = 440 °C
tsat = 158.85 °C
tsat < T1 ; superheated vapor
𝐾𝐽
h3 = 3354.7 π‘˜π‘”
𝐾𝐽
s3 = 7.8297π‘˜π‘”βˆ™πΎ
For x4,
Sβ‚„− Sfβ‚„
X4 = 𝑆𝑓𝑔₄
οƒ˜ State 4
𝐾𝐽
S3 = S4 = 7.8297 π‘˜π‘”βˆ™πΎ
P2 = 0.02 MPa
𝐾𝐽
Sf = 0.8320 π‘˜π‘”βˆ™πΎ
𝐾𝐽
π‘˜π‘”βˆ™πΎ
𝐾𝐽
7.0766 π‘˜π‘”βˆ™πΎ
X4 = 0.9889
X4 = 98.89 %
𝐾𝐽
Sg = 7.9085 π‘˜π‘”βˆ™πΎ
S g > S4
For h4,
(7.8297−0.8320 )
X4 =
οƒ˜ State 4’
h4 = hf4 + x4 (hfg4)
𝐾𝐽
𝐾𝐽
h4 = 251.40 π‘˜π‘” + (0.9889) (2358.3 π‘˜π‘”)
β„Žβ‚ƒ−β„Žβ‚„′
πœ‚T = β„Žβ‚ƒ−β„Žβ‚„
𝐾𝐽
h4 = 2583.5229 π‘˜π‘”
0.85 =
3354.7
3354.7
𝐾𝐽
−β„Žβ‚„′
π‘˜π‘”
𝐾𝐽
𝐾𝐽
− 2583.5229
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
h4’ = 2699.1995 π‘˜π‘”
οƒ˜ State 5
P4 = P5 = 0.02 MPa
𝐾𝐽
hf4 = h5 = 251.40 π‘˜π‘”
π‘š³
𝜐f4 = 𝜐5 =1.0172 × 10-3 π‘˜π‘”
οƒ˜ State 6
P1 = P6 = 7.0 MPa
Wp = 𝜐5 (P6 – P5)
π‘š³
Wp = 1.0172 × 10-3 π‘˜π‘” (7.0 MPa – 0.02 MPa)
π‘š³
Wp = 1.0172 × 10-3 π‘˜π‘” (6.98 MPa ×
1000 πΎπ‘ƒπ‘Ž
×
1 π‘€π‘ƒπ‘Ž
1000
𝑁
π‘š²
1 πΎπ‘ƒπ‘Ž
1𝐽
1 𝐾𝐽
× 1 π‘βˆ™π‘š × 1000 𝐽 )
𝐾𝐽
Wp = 7.1001 π‘˜π‘”
οƒ˜ State 6’
h6= h5 + Wp
β„Žβ‚†−β„Žβ‚…
𝐾𝐽
𝐾𝐽
h6 = 251.40 π‘˜π‘” + 7.1001π‘˜π‘”
𝐾𝐽
h6 = 258.5001 π‘˜π‘”
πœ‚T = β„Žβ‚†′−β„Žβ‚…
0.85 =
𝐾𝐽
𝐾𝐽
− 251.40
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
β„Žβ‚†′− 251.40
π‘˜π‘”
258.5001
𝐾𝐽
h6’ = 259.7531 π‘˜π‘”
19
(a) For WTotal’,
WTotal’= WT1’ – WT2’
WTotal’ = (β„Žβ‚ − β„Žβ‚‚′) + (β„Žβ‚ƒ − β„Žβ‚„′)
𝐾𝐽
𝐾𝐽
𝐾𝐽
𝐾𝐽
WTotal’ = (3158.1 π‘˜π‘” − 2709.2159π‘˜π‘”) + (3354.7 π‘˜π‘” − 2699.1995π‘˜π‘”)
𝐾𝐽
WTotal’ = 1104.3846 π‘˜π‘”
(b) For WP’,
WP’ = h6’ - h5
𝐾𝐽
𝐾𝐽
WP’ = 259.7531 π‘˜π‘” - 251.40 π‘˜π‘”
𝐾𝐽
WP’ = 8.3531 π‘˜π‘”
(c) For πœ‚TH’,
πœ‚TH’ =
πœ‚TH’ =
πœ‚TH’ =
πœ‚TH’ =
π‘Šπ‘›π‘’π‘‘′
𝑄𝐴′
π‘Šπ‘‡π‘œπ‘‘π‘Žπ‘™′−π‘Šπ‘′
𝑄𝐴′
[(β„Žβ‚−β„Žβ‚‚′) + (β„Žβ‚ƒ−β„Žβ‚„′)] − [h₆’ − hβ‚…]
(β„Žβ‚−β„Žβ‚†′) +(β„Žβ‚ƒ−β„Žβ‚‚′)
[ (3158.1
𝐾𝐽
𝐾𝐽
𝐾𝐽
𝐾𝐽
𝐾𝐽
𝐾𝐽
− 2709.2159 )+ (3354.7
− 2699.1995 )]+[259.7531
− 251.40 ]
π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
𝐾𝐽
𝐾𝐽
𝐾𝐽
(3158.1 −259.7531 )+(3354.7
−2709.2159 )
π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
πœ‚TH’ = 0.3093
πœ‚TH’ = 30.93 %
TS DIAGRAM
P1 = P6 = 7.0 MPa
T1 = 400°C
P2 = P3 = 0.6 MPa
T3 = 440°C
P4 = P5 = 0.02 MPa
s1 = s2
s3 = s4
x4
20
2. Steam is supplied to the first turbine at 5.5 MPa and 420°C and expands to 0.4 MPa. It is then
reheated to 460°C and enters the second turbine stage, expanding to a condenser pressure of
0.010 MPa. Both the turbines and pump have the efficiency of 80%. Calculate the (a) h 2’, (b) h4’,
(c) h6’, and (d) πœ‚TH’. Show the TS diagram.
Given:
Required:
(a) h2’
(b) h4’
(c) h6’
(d) πœ‚TH’
Solution:
οƒ˜ State 1
At P1 = 5.5 MPa and T1 = 420 °C
tsat = 270.02 °C
tsat < T1 ; superheated vapor
οƒ˜ State 2
𝐾𝐽
S1 = S2 = 6.6642 π‘˜π‘”βˆ™πΎ
P2 = 0.4 MPa
𝐾𝐽
h1 = 3236.2 π‘˜π‘”
𝐾𝐽
s1 = 6.6642 π‘˜π‘”βˆ™πΎ
S (π‘˜π‘”βˆ™πΎ)
𝐾𝐽
h (π‘˜π‘”)
𝐾𝐽
6.6319
S2 = 6.6642
6.6981
2634.5
h2
2659.5
By interpolation,
𝐾𝐽
π‘˜π‘”βˆ™πΎ
𝐾𝐽
(6.6981−6.6319)
π‘˜π‘”βˆ™πΎ
( 6.6642−6.6319)
𝐾𝐽
=
β„Žβ‚‚ −2634.5 π‘˜π‘”
𝐾𝐽
π‘˜π‘”
(2659.5−2634.5 )
𝐾𝐽
h2 = 2646.6979 π‘˜π‘”
21
οƒ˜ State 3
P2 = P3 = 0.4 MPa and T3 = 460 °C
tsat = 143.63 °C
tsat < T1 ; superheated vapor
h3 = 3399.7 π‘˜π‘”
οƒ˜ State 4
For x4,
𝐾𝐽
𝐾𝐽
s3 = 8.0781 π‘˜π‘”βˆ™πΎ
𝐾𝐽
S3 = S4 = 8.0781 π‘˜π‘”βˆ™πΎ
Sβ‚„− Sfβ‚„
X4 = 𝑆𝑓𝑔₄
P4 = 0.010 MPa
X4 =
𝐾𝐽
Sf4 = 0.6493 π‘˜π‘”βˆ™πΎ
(8.0781−06493 )
𝐾𝐽
π‘˜π‘”βˆ™πΎ
𝐾𝐽
7.5009 π‘˜π‘”βˆ™πΎ
X4 = 0.9904
X4 = 99.04 %
𝐾𝐽
Sg4 = 8.1502 π‘˜π‘”βˆ™πΎ
Sg4 > S4; mixture
For h4,
οƒ˜ State 5
P4 = P5 = 0.010 MPa
h4 = hf4 + x4 (hfg4)
𝐾𝐽
𝐾𝐽
𝐾𝐽
h4 = 191.83 π‘˜π‘” + (0.9904 (2392.8 π‘˜π‘”)
hf4 = h5 = 191.83 π‘˜π‘”
𝐾𝐽
π‘š³
h4 = 2561.6591 π‘˜π‘”
𝜐f4 = 𝜐5 =1.0102 × 10-3 π‘˜π‘”
οƒ˜ State 6
P1 = P6 = 5.5 MPa
Wp = 𝜐5 (P6 – P5)
π‘š³
Wp = 1.0102 × 10-3 π‘˜π‘” (5.5 MPa – 0.010 MPA)
π‘š³
1000 πΎπ‘ƒπ‘Ž
Wp = 1.0102 × 10-3
(5.49 MPa ×
×
π‘˜π‘”
1 π‘€π‘ƒπ‘Ž
1000
𝑁
π‘š²
1 πΎπ‘ƒπ‘Ž
1𝐽
1 𝐾𝐽
× 1 π‘βˆ™π‘š × 1000 𝐽 )
𝐾𝐽
Wp = 5.5460 π‘˜π‘”
h6= h5 + Wp
𝐾𝐽
𝐾𝐽
h6 = 191.83 π‘˜π‘” + 5.5460 π‘˜π‘”
𝐾𝐽
h6 = 197.376 π‘˜π‘”
(a) For h2’,
(b) For h4’,
β„Žβ‚−β„Žβ‚‚′
β„Žβ‚„−β„Žβ‚ƒ
πœ‚T = β„Žβ‚−β„Žβ‚‚
0.80 =
πœ‚P = β„Žβ‚„′−β„Žβ‚ƒ
3236.2
3236.2
𝐾𝐽
−β„Žβ‚‚′
π‘˜π‘”
h4 ’ =
𝐾𝐽
𝐾𝐽
− 2646.6979
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
h2’ = 2764.5983 π‘˜π‘”
h4 ’ =
β„Žβ‚„−β„Žβ‚ƒ
ηβ‚š
+ h3
2561.6591
𝐾𝐽
𝐾𝐽
− 3399.7
π‘˜π‘”
π‘˜π‘”
0.80
𝐾𝐽
h4’ = 2729.2673 π‘˜π‘”
22
(b) For h6’,
β„Žβ‚†−β„Žβ‚…
πœ‚T = β„Žβ‚†′−β„Žβ‚…
0.80 =
𝐾𝐽
𝐾𝐽
−191.83
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
β„Žβ‚†′− 191.83
π‘˜π‘”
197.376
𝐾𝐽
h6’ = 204.3085 π‘˜π‘”
(c) For πœ‚TH’,
πœ‚TH’ =
πœ‚TH’ =
πœ‚TH’ =
πœ‚TH’ =
π‘Šπ‘›π‘’π‘‘′
𝑄𝐴′
π‘Šπ‘‡π‘œπ‘‘π‘Žπ‘™′−π‘Šπ‘′
𝑄𝐴′
[(β„Žβ‚−β„Žβ‚‚′) + (β„Žβ‚ƒ−β„Žβ‚„′)] − [h₆’ − hβ‚…]
(β„Žβ‚−β„Žβ‚†′) +(β„Žβ‚ƒ−β„Žβ‚‚′)
[ (3236.2
𝐾𝐽
𝐾𝐽
𝐾𝐽
𝐾𝐽
𝐾𝐽
𝐾𝐽 𝐾𝐽
− 2646.6979 )+ (3399.7
− 2729.2673 )]+[204.3085
− 191.83
]
π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
π‘˜π‘” π‘˜π‘”
𝐾𝐽 𝐾𝐽
𝐾𝐽
𝐾𝐽
𝐾𝐽
(3236.2
−204.3085 )+(3399.7
− 2646.6979 )
π‘˜π‘” π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
πœ‚TH’ = 0.3537
πœ‚TH’ = 35.3 %
TS DIAGRAM
P1 = P6 = 5.5 MPa
T1 = 420°C
P2 = P3 = 0.4 MPa
T3 = 460°C
P4 = P5 = 0.010 MPa
s1 = s2
s3 = s4
x4
23
3. The steam enters the first stage turbine at 6.5 MPa amd 440°C, undergoing expansion to 0.56
MPa. It is then reheated to 480°C before entering the second stage turbine where it further
expands to a condenser pressure of 0.011 MPa. Both turbine and pump has 90% efficiency.
Determine Wnet’, QA’, and πœ‚TH’.
Given:
P₁ = 6.5 MPa
T1 = 440 °C
P2 = 0.56 MPa
πœ‚T = 90%
2’
P4 = 0.011 MPa
T3 = 480 °C
4’
πœ‚P = 90%
H20 in
H20 out
6’
Required:
(a) Wnet’
(b) QA’
(c) πœ‚TH’
Solution:
οƒ˜ State 1
At P1 = 6.5 MPa and T1 = 440 °C
tsat = 280.91 °C
tsat < T1 ; superheated vapor
𝐾𝐽
h1 = 3269.7 π‘˜π‘”
s1 = 6.6401
𝐾𝐽
π‘˜π‘”βˆ™πΎ
οƒ˜ State 2
𝐾𝐽
S1 = S2 = 6.6401 π‘˜π‘”βˆ™πΎ
P2 =0.56 MPa
S (π‘˜π‘”βˆ™πΎ)
𝐾𝐽
h (π‘˜π‘”)
𝐾𝐽
6.6300
S2 = 6.6401
6.6610
2690.0
h2
2702.6
24
By interpolation,
οƒ˜ State 2’
𝐾𝐽
𝐾𝐽
(6.6401 − 6.6300)
β„Žβ‚‚ − 2690.0
π‘˜π‘” βˆ™ 𝐾
π‘˜π‘”
=
𝐾𝐽
𝐾𝐽
(6.6610 − 6.6300)
(2702.6 − 2690.0)
π‘˜π‘” βˆ™ 𝐾
π‘˜π‘”
β„Žβ‚−β„Žβ‚‚′
πœ‚T = β„Žβ‚−β„Žβ‚‚
0.90 =
𝐾𝐽
−β„Žβ‚‚′
π‘˜π‘”
3269.7
3269.7
𝐾𝐽
𝐾𝐽
− 2694.1052
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
h2’ = 2751.6647 π‘˜π‘”
𝐾𝐽
h2 = 2694.1052 π‘˜π‘”
οƒ˜ State 3
P2 = P3 = 0.56 MPa and T3 = 480 °C
tsat = 156.17 °C
tsat < T1 ; superheated vapor
𝐾𝐽
h3 = 3440.3 π‘˜π‘”
𝐾𝐽
s3 = 7.9782π‘˜π‘”βˆ™πΎ
οƒ˜ State 4
For x4,
S3 = S4 = 7.9782
𝐾𝐽
Sβ‚„− Sfβ‚„
X4 = 𝑆𝑓𝑔₄
π‘˜π‘”βˆ™πΎ
P4 = 0.011MPa
𝐾𝐽
Sf4 = 0.6738 π‘˜π‘”βˆ™πΎ
X4 =
(7.9782−0.6738 )
𝐾𝐽
7.4430 π‘˜π‘”βˆ™πΎ
𝐾𝐽
Sg4 = 8.1168 π‘˜π‘”βˆ™πΎ
𝐾𝐽
π‘˜π‘”βˆ™πΎ
X4 = 0.9814
Sg4 > S4; mixture
X4 = 98.14 %
For h4,
h4 = hf4 + x4 (hfg4)
οƒ˜ State 4’
β„Žβ‚ƒ−β„Žβ‚„′
𝐾𝐽
πœ‚T = β„Žβ‚ƒ−β„Žβ‚„
𝐾𝐽
h4 = 199.67 π‘˜π‘” + (0.9814) (2388.3 π‘˜π‘”)
0.85 =
𝐾𝐽
h4 = 2543.5476 π‘˜π‘”
3440.3
𝐾𝐽
−β„Žβ‚„′
π‘˜π‘”
3440.3 − 2543.5476
𝐾𝐽
π‘˜π‘”
𝐾𝐽
h4’ = 2633.2228 π‘˜π‘”
οƒ˜ State 5
P4 = P5 = 0.011 MPa
𝐾𝐽
hf4 = h5 = 199.67 π‘˜π‘”
π‘š³
𝜐f4 = 𝜐5 =1.0111 × 10-3 π‘˜π‘”
οƒ˜ State 6
P1 = P6 = 6.5 MPa
Wp = 𝜐5 (P6 – P5)
π‘š³
Wp = 1.0111 × 10-3 π‘˜π‘” (6.5 MPa – 0.011 MPA)
π‘š³
1000 πΎπ‘ƒπ‘Ž
Wp = 1.0111 × 10-3
(6.489 MPa ×
×
π‘˜π‘”
1 π‘€π‘ƒπ‘Ž
1000
𝑁
π‘š²
1 πΎπ‘ƒπ‘Ž
1𝐽
1 𝐾𝐽
× 1 π‘βˆ™π‘š × 1000 𝐽 )
𝐾𝐽
Wp = 6.5610 π‘˜π‘”
25
h6= h5 + Wp
𝐾𝐽
𝐾𝐽
h6 = 199.67 π‘˜π‘” + 6.5610 π‘˜π‘”
οƒ˜ State 6’
β„Žβ‚†−β„Žβ‚…
πœ‚T = β„Žβ‚†′−β„Žβ‚…
𝐾𝐽
h6 = 206.231 π‘˜π‘”
0.85 =
𝐾𝐽
𝐾𝐽
− 199.67
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
β„Žβ‚†′− 199.67
π‘˜π‘”
206.231
𝐾𝐽
h6’ = 206.96 π‘˜π‘”
(a) For Wnet’,
Wnet’= WTotal’ – WP’
Wnet’= [(β„Žβ‚ − β„Žβ‚‚′) + (β„Žβ‚ƒ − β„Žβ‚„′)] – (β„Žβ‚†′ − β„Žβ‚…)
𝐾𝐽
𝐾𝐽
𝐾𝐽
𝐾𝐽
𝐾𝐽
𝐾𝐽
Wnet’= [(3269.7 π‘˜π‘” − 2751.6647 π‘˜π‘”) + (3440.3 π‘˜π‘” − 2633.2228 π‘˜π‘”)] – (206.96 π‘˜π‘” - 199.67π‘˜π‘”)
𝐾𝐽
Wnet’= 1317.8225 π‘˜π‘”
(b) For QA’,
QA’ = QAB’ + QAR’
QA’ = (β„Žβ‚ − β„Žβ‚†′) + (β„Žβ‚ƒ − β„Žβ‚‚′)]
𝐾𝐽
𝐾𝐽
𝐾𝐽
𝐾𝐽
QA’ = (3269.7 π‘˜π‘” − 206.96 π‘˜π‘”) + (3440.3π‘˜π‘” – 2451.6647 π‘˜π‘”)
𝐾𝐽
QA’ = 3751.3753π‘˜π‘”
(c) For πœ‚TH’,
πœ‚TH’ =
πœ‚TH’ =
π‘Šπ‘›π‘’π‘‘′
𝑄𝐴′
𝐾𝐽
π‘˜π‘”
𝐾𝐽
3751.3753
π‘˜π‘”
1317.8225
πœ‚TH’ = 0.3513
πœ‚TH’ = 35.13 %
TS DIAGRAM
P1 = P6 = 6.5 MPa
T1 = 440°C
P2 = P3 = 0.56 MPa
T3 = 480°C
P4 = P5 = 0.011 MPa
s1 = s2
s3 = s4
x4
26
SINGLE-STAGE REHEATING CYCLE (IDEAL)
4. A reheat turbine with one stage of reheat receives steam at 8.0 MPa, 480°C and expands to 0.7
MPa. It is then reheated to 440°C before entering the second stages turbine, where it expands to the
condenser pressure of 0.008 MPa. Determine thermal efficiency.
Given:
P₁ = 8.0 MPa
T1 = 480 °C
P2 = 0.7 MPa
P4 = 0.008 MPa
T3 = 440 °C
H20 in
H20 out
Required: πœ‚TH
Solution:
οƒ˜ State 1
At P1 = 8.0 MPa and T1 = 480 °C
tsat = 295.06 °C
tsat < T1 ; superheated vapor
𝐾𝐽
h1 = 3348.4 π‘˜π‘”
𝐾𝐽
s1 = 6.6586 π‘˜π‘”βˆ™πΎ
S (π‘˜π‘”βˆ™πΎ)
𝐾𝐽
h (π‘˜π‘”)
6.6516
S2 = 6.6586
6.6803
2739.0
h2
2751.4
οƒ˜ State 2
𝐾𝐽
S1 = S2 = 6.6586 π‘˜π‘”βˆ™πΎ
P2 =0.7 MPa
𝐾𝐽
By interpolation,
𝐾𝐽
𝐾𝐽
β„Žβ‚‚ − 2739.0
π‘˜π‘” βˆ™ 𝐾
π‘˜π‘”
=
𝐾𝐽
𝐾𝐽
(6.6803 − 6.6516)
(2751.4 − 2739)
π‘˜π‘” βˆ™ 𝐾
π‘˜π‘”
(6.6586 − 6.6516)
𝐾𝐽
h2 = 2742.0244 π‘˜π‘”
27
οƒ˜ State 3
P2 = P3 = 0.7 MPa and T3 = 440 °C
tsat = 164.97 °C
tsat < T1 ; superheated vapor
𝐾𝐽
h3 = 3353.3 π‘˜π‘”
𝐾𝐽
s3 = 7.7571π‘˜π‘”βˆ™πΎ
οƒ˜ State 4
For x4,
𝐾𝐽
S3 = S4 = 7.7571 π‘˜π‘”βˆ™πΎ
Sβ‚„− Sfβ‚„
X4 = 𝑆𝑓𝑔₄
P4 = 0.008 MPa
𝐾𝐽
Sf4 = 0.5926 π‘˜π‘”βˆ™πΎ
X4 =
(7.7571−0.5926)
𝐾𝐽
Sg4 = 8.2287π‘˜π‘”βˆ™πΎ
𝐾𝐽
π‘˜π‘”βˆ™πΎ
𝐾𝐽
7.6361 π‘˜π‘”βˆ™πΎ
X4 = 0.9382
Sg > S4; mixture
X4 = 93.82 %
For h4,
h4 = hf4 + x4 (hfg4)
𝐾𝐽
𝐾𝐽
h4 = 173.88 π‘˜π‘” + (0.9382) (2403.1 π‘˜π‘”)
𝐾𝐽
h4 = 2428.4684 π‘˜π‘”
οƒ˜ State 5
P4 = P5 = 0.008 MPa
𝐾𝐽
hf4 = h5 = 173.88 π‘˜π‘”
π‘š³
𝜐f4 = 𝜐5 =1.0084 × 10-3 π‘˜π‘”
οƒ˜ State 6
P1 = P6 = 6.5 MPa
Wp = 𝜐5 (P6 – P5)
π‘š³
Wp = 1.0084 × 10-3 π‘˜π‘” (8.0 MPa – 0.008 MPA)
π‘š³
1000 πΎπ‘ƒπ‘Ž
Wp = 1.0111 × 10-3
(7.992 MPa ×
×
π‘˜π‘”
Wp = 8.0591
1 π‘€π‘ƒπ‘Ž
1000
𝑁
π‘š²
1 πΎπ‘ƒπ‘Ž
1𝐽
1 𝐾𝐽
× 1 π‘βˆ™π‘š × 1000 𝐽 )
𝐾𝐽
π‘˜π‘”
h6= h5 + Wp
𝐾𝐽
𝐾𝐽
h6 = 173.88 π‘˜π‘” + 5.0591 π‘˜π‘”
𝐾𝐽
h6 = 181.9391 π‘˜π‘”
28
For πœ‚TH,
πœ‚TH =
πœ‚TH =
πœ‚TH =
πœ‚TH =
π‘Šπ‘›π‘’π‘‘
𝑄𝐴
π‘Šπ‘‡π‘œπ‘‘π‘Žπ‘™−π‘Šπ‘
𝑄𝐴
[(β„Žβ‚−β„Žβ‚‚) + (β„Žβ‚ƒ−β„Žβ‚„)] − [h₆− hβ‚…]
(β„Žβ‚−β„Žβ‚†) +(β„Žβ‚ƒ−β„Žβ‚‚)
[ (3348.4
𝐾𝐽
𝐾𝐽
𝐾𝐽 𝐾𝐽
𝐾𝐽
𝐾𝐽
𝐾𝐽
− 2742.0244 )+ (3353.3
− 2428.4684 )]+[181.9391
− 173.88 ]
π‘˜π‘”
π‘˜π‘”
π‘˜π‘” π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
𝐾𝐽 𝐾𝐽
𝐾𝐽
𝐾𝐽
𝐾𝐽
(3348.4
−181.9391 )+(3353.3
− 2742.0244 )
π‘˜π‘” π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
πœ‚TH = 0.4032
πœ‚TH = 40.32 %
TS DIAGRAM
P1 = P6 = 8 MPa
T1 = 480°C
P2 = P3 = 0.7 MPa
T3 = 440°C
P4 = P5 = 0.08 MPa
s1 = s2
s3 = s4
x4
29
5. Steam initially enters the first-stage turbine at 6.0 MPa and 480°C, undergoes expansion to 0.5
MPa, is subsequently reheated to 460°C prior to entering the second stage turbine and finally
expands to the condenser pressure at 0.01 MPa. Determine the thermal efficiency.
Given:
P₁ = 6.0 MPa
T1 = 480°C
P2 = 0.5 MPa
P4 = 0.01 MPa
T3 = 460 °C
H20 in
H20 out
Required: πœ‚TH
Solution:
οƒ˜ State 1
At P1 = 8.0 MPa and T1 = 480 °C
tsat = 275.64°C
tsat < T1 ; superheated vapor
𝐾𝐽
h1 = 3227.8 π‘˜π‘”
𝐾𝐽
s1 = 6.6148 π‘˜π‘”βˆ™πΎ
οƒ˜ State 2
𝐾𝐽
S1 = S2 = 6.6148 π‘˜π‘”βˆ™πΎ
P2 =0.5 MPa
S (π‘˜π‘”βˆ™πΎ)
𝐾𝐽
h (π‘˜π‘”)
𝐾𝐽
6.6000
S2 = 6.6148
6.6328
2658.9
h2
2671.7
30
By interpolation,
KJ
KJ
hβ‚‚ -2658.9
kgβˆ™K
kg
=
KJ
KJ
(6.6328-6.6000)
(2671.7-2658.9)
kgβˆ™K
kg
(6.6148-6.6000)
𝐾𝐽
h2 = 2664.6756 π‘˜π‘”
οƒ˜ State 3
P2 = P3 = 0.5 MPa and T3 = 460 °C
tsat = 151.86°C
tsat < T1 ; superheated vapor
𝐾𝐽
h3 = 3398.4 π‘˜π‘”
𝐾𝐽
s3 = 7.9738π‘˜π‘”βˆ™πΎ
οƒ˜ State 4
For x4,
𝐾𝐽
S3 = S4 = 7.9738 π‘˜π‘”βˆ™πΎ
Sβ‚„− Sfβ‚„
X4 = 𝑆𝑓𝑔₄
P4 = 0.01 MPa
𝐾𝐽
Sf4 = 0.6493 π‘˜π‘”βˆ™πΎ
X4 =
(7.9738−0.6493)
𝐾𝐽
7.5009 π‘˜π‘”βˆ™πΎ
𝐾𝐽
Sg4 = 8.1502π‘˜π‘”βˆ™πΎ
𝐾𝐽
π‘˜π‘”βˆ™πΎ
X4 = 0.9765
Sg > S4; mixture
X4 = 97.65%
For h4,
h4 = hf4 + x4 (hfg4)
𝐾𝐽
𝐾𝐽
h4 = 191.83π‘˜π‘” + (0.9765) (2392.8 π‘˜π‘”)
𝐾𝐽
h4 = 2528.3992 π‘˜π‘”
οƒ˜ State 5
P4 = P5 = 0.01 MPa
𝐾𝐽
hf4 = h5 = 191.83 π‘˜π‘”
π‘š³
𝜐f4 = 𝜐5 =1.0102 × 10-3 π‘˜π‘”
οƒ˜ State 6
P1 = P6 = 6.0 MPa
Wp = 𝜐5 (P6 – P5)
π‘š³
Wp = 1.0102× 10-3 π‘˜π‘” (6.0 MPa – 0.01 MPA)
π‘š³
Wp = 1.0102 × 10-3 π‘˜π‘” (5.99 MPa ×
1000 πΎπ‘ƒπ‘Ž
×
1 π‘€π‘ƒπ‘Ž
1000
𝑁
π‘š²
1 πΎπ‘ƒπ‘Ž
1𝐽
1 𝐾𝐽
× 1 π‘βˆ™π‘š × 1000 𝐽 )
𝐾𝐽
Wp = 6.0511 π‘˜π‘”
31
h6= h5 + Wp
h6 = 191.83
𝐾𝐽
π‘˜π‘”
+ 6.0511
𝐾𝐽
π‘˜π‘”
𝐾𝐽
h6 = 197.8811 π‘˜π‘”
For πœ‚TH,
πœ‚TH =
πœ‚TH =
πœ‚TH =
πœ‚TH =
π‘Šπ‘›π‘’π‘‘
𝑄𝐴
π‘Šπ‘‡π‘œπ‘‘π‘Žπ‘™−π‘Šπ‘
𝑄𝐴
[(β„Žβ‚−β„Žβ‚‚) + (β„Žβ‚ƒ−β„Žβ‚„)] − [h₆− hβ‚…]
(β„Žβ‚−β„Žβ‚†) +(β„Žβ‚ƒ−β„Žβ‚‚)
[ (3327.8
𝐾𝐽
𝐾𝐽
𝐾𝐽 𝐾𝐽
𝐾𝐽
𝐾𝐽
− 2664.6756 )+ (3398.4
− 2528.3992)]+[197.8811
− 191.83 ]
π‘˜π‘”
π‘˜π‘”
π‘˜π‘” π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
𝐾𝐽 𝐾𝐽
𝐾𝐽
𝐾𝐽
𝐾𝐽
(3327.8
−197.8811 )+(3398.4
− 2664.6756 )
π‘˜π‘” π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
πœ‚TH = 0.3792
πœ‚TH = 37.92 %
TS DIAGRAM
P1 = P6 = 6 MPa
T1 = 480°C
P2 = P3 = 0.5 MPa
T3 = 460°C
P4 = P5 = 0.01 MPa
s1 = s2
s3 = s4
x4
32
TWO-STAGES REHEATING CYCLE
1. A two-stage reheating cycle is designed with steam initially expanding from 20 MPa at 520 °C.
The two reheaters operate at pressures of 5 MPa and 2 MPa, and the steam exits each reheater
at 520 °C. Its reheater enters the turbine and exhausts at 0.0075 MPa. Both the turbine and
pump have 85% efficiency. Determine the quality, network, actual network, and actual work
pump.
Given:
P1 = P8 = 20 MPa
P2 = P3 = 5 MPa
P4 = P5 = 2 MPa
P6 = P7 = 0.0075 MPa
T1 = T3 = T5 = 520 °C
πœ‚T = πœ‚P = 85%
Required:
(a) x6
(b) Wnet
(c) Wnet’
(d) WP’
Solution:
οƒ˜ State 1
At P1 = 20 MPa and T1 = 520 °C
tsat = 365.81 °C
tsat < T1 ; superheated vapor
𝐾𝐽
h1 = 3302.2 π‘˜π‘”
𝐾𝐽
s1 = 6.2218 π‘˜π‘”βˆ™πΎ
οƒ˜ State 2
𝐾𝐽
S1 = S2 = 6.2218 π‘˜π‘”βˆ™πΎ
P2 = 5 MPa
31
𝐾𝐽
𝐾𝐽
S (π‘˜π‘”βˆ™πΎ)
h (π‘˜π‘”)
6.2084
S2 = 6.2218
6.2621
2924.5
h2
2955.6
οƒ˜ State 2’
β„Žβ‚−β„Žβ‚‚′
πœ‚T = β„Žβ‚−β„Žβ‚‚
0.85 =
3302.2
3302.2
𝐾𝐽
−β„Žβ‚‚′
π‘˜π‘”
𝐾𝐽
𝐾𝐽
− 2932.2605
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
h2’ =2987.7514 π‘˜π‘”
By interpolation,
𝐾𝐽
𝐾𝐽
β„Žβ‚‚ − 2924.5
π‘˜π‘” βˆ™ 𝐾
π‘˜π‘”
=
𝐾𝐽
𝐾𝐽
(6.2621 − 6.2084)
(2955.6 − 2924.5)
π‘˜π‘” βˆ™ 𝐾
π‘˜π‘”
(6.2218 − 6.2084)
𝐾𝐽
h2 = 2932.2605 π‘˜π‘”
οƒ˜ State 3
P2 = P3 = 5 MPa and T3 = 520 °C
tsat = 263.99 °C
tsat < T1 ; superheated vapor
𝐾𝐽
h3 = 3480.5 π‘˜π‘”
𝐾𝐽
s3 = 7.0355π‘˜π‘”βˆ™πΎ
οƒ˜ State 4
𝐾𝐽
S3 = S4 = 7.0355 π‘˜π‘”βˆ™πΎ
οƒ˜ State 4’
β„Žβ‚ƒ−β„Žβ‚„′
πœ‚T = β„Žβ‚ƒ−β„Žβ‚„
P4 = 2 MPa
S (π‘˜π‘”βˆ™πΎ)
𝐾𝐽
h (π‘˜π‘”)
𝐾𝐽
7.0265
S4 = 7.0355
7.0606
3181.4
h4
3203.6
0.85 =
3480.5
3480.5
𝐾𝐽
−β„Žβ‚„′
π‘˜π‘”
𝐾𝐽
𝐾𝐽
− 3187.5592
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
h4’ =3231.5003 π‘˜π‘”
By interpolation,
𝐾𝐽
𝐾𝐽
β„Žβ‚‚ − 3181.4
π‘˜π‘” βˆ™ 𝐾
π‘˜π‘”
=
𝐾𝐽
𝐾𝐽
(7.0606 − 7.0265)
(3203.6 − 3181.4)
π‘˜π‘” βˆ™ 𝐾
π‘˜π‘”
(7.0355 − 7.0265)
𝐾𝐽
h4 = 3187.5592 π‘˜π‘”
οƒ˜ State 5
P4 = P5 = 2 MPa and T3 = 520 °C
tsat = 212.42 °C
tsat < T1 ; superheated vapor
𝐾𝐽
h5 = 3511.8 π‘˜π‘”
𝐾𝐽
s5 = 7.4882π‘˜π‘”βˆ™πΎ
32
οƒ˜ State 6
𝐾𝐽
S5 = S6 = 7.4882 π‘˜π‘”βˆ™πΎ
h6 = hf6 + x6(hfg6)
P6 = 0.0075 MPa
𝐾𝐽
Sf6 = 0.5764 π‘˜π‘”βˆ™πΎ
𝐾𝐽
𝐾𝐽
h6 = 168.79π‘˜π‘” + (0.9006) (2406.0π‘˜π‘”)
𝐾𝐽
h6 = 2335.6336 π‘˜π‘”
𝐾𝐽
Sg6 = 8.2515 π‘˜π‘”βˆ™πΎ
Sg > S4; mixture
For x6,
οƒ˜ State 6’
S₆− Sf₆
X6 = 𝑆𝑓𝑔₆
X6 =
β„Žβ‚…−β„Žβ‚†′
πœ‚T = β„Žβ‚…−β„Žβ‚†
𝐾𝐽
(7.4882−0.5764 )
π‘˜π‘”βˆ™πΎ
𝐾𝐽
7.6750 π‘˜π‘”βˆ™πΎ
0.85 =
X6 = 0.9006
X6 = 90.06 %
3511.8
3511.8
𝐾𝐽
−β„Žβ‚†′
π‘˜π‘”
𝐾𝐽
𝐾𝐽
− 2335.6336
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
h6’ = 2512.0586 π‘˜π‘”
οƒ˜ State 7
P6 = P7 = 0.0075 MPa
𝐾𝐽
hf6 = h7 = 168.79 π‘˜π‘”
π‘š³
𝜐f6 = 𝜐7 =1.0079 × 10-3 π‘˜π‘”
οƒ˜ State 8
P1 = P8 = 20 MPa
Wp = 𝜐5 (P8 – P7)
π‘š³
Wp = 1.0079 × 10-3 π‘˜π‘” (20 MPa – 0.02 MPa)
π‘š³
1000 πΎπ‘ƒπ‘Ž
Wp = 1.0079 × 10-3
(19.9925 MPa ×
×
π‘˜π‘”
1 π‘€π‘ƒπ‘Ž
1000
𝑁
π‘š²
1 πΎπ‘ƒπ‘Ž
1𝐽
1 𝐾𝐽
× 1 π‘βˆ™π‘š × 1000 𝐽 )
𝐾𝐽
Wp = 20.1504 π‘˜π‘”
h8= h7 + Wp
𝐾𝐽
𝐾𝐽
h8 = 168.79 π‘˜π‘” + 20.1504π‘˜π‘”
𝐾𝐽
h8 = 188.9404 π‘˜π‘”
οƒ˜ State 8’
β„Žβ‚ˆ−β„Žβ‚‡
πœ‚T = β„Žβ‚ˆ′−β„Žβ‚‡
0.85 =
𝐾𝐽
𝐾𝐽
− 168.79
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
β„Žβ‚ˆ′− 168.79
π‘˜π‘”
188.9404
𝐾𝐽
h8’ = 192.4964 π‘˜π‘”
WP’ = h8’ – h7
𝐾𝐽
𝐾𝐽
WP’ = 192.4964 π‘˜π‘” – 168.79 π‘˜π‘”
𝐾𝐽
WP’ = 23.7064 π‘˜π‘”
33
For Wnet,
Wnet = WTtotal - WP
WTtotal = WT1 + WT2 + WT3
WTtotal = (h1 – h2) + (h3 – h4) + (h5 – h6)
𝐾𝐽
WTtotal = [(3302.2 – 2932.2605) + (3480.5– 3187.5592) + (3511.8 – 2335.6336)] π‘˜π‘”
𝐾𝐽
WTtotal = 1839.0467 π‘˜π‘”
Wnet = WTtotal - WP
𝐾𝐽
𝐾𝐽
Wnet = 1839.0467 π‘˜π‘” - 20.1504 π‘˜π‘”
𝐾𝐽
Wnet = 1818.8963 π‘˜π‘”
For Wnet’,
Wnet’ = WTtotal’ - WP’
WTtotal’ = (h1 – h2’) + (h3 – h4’) + (h5 – h6’)
𝐾𝐽
WTtotal’ = [(3302.2 – 2987.7514) + (3480.5 – 3231.5003) + (3511.8 – 2512.0586)] π‘˜π‘”
𝐾𝐽
WTtotal’ = 1563.1897 π‘˜π‘”
Wnet’ = WTtotal’ - WP’
𝐾𝐽
𝐾𝐽
Wnet’ = 1563.1897 π‘˜π‘” – 23.7064 π‘˜π‘”
𝐾𝐽
Wnet’ = 1539.4833 π‘˜π‘”
TS DIAGRAM
P1 = P8
T1
T3
P2 = P3
20 MPa
5 MPa
P4 = P5
2 MPa
0.0075 MPa
P6 = P7
34
2. A supercritical reheat Rankine cycle with two stages of reheat operates as follows: steam
enters the high-pressure turbine at 22 MPa and 550 °C. It expands to 3.80 MPa and is then
reheated to 450 °C. The steam reenters the turbine, expands to 1.10 MPa, and is reheated
again to 400 °C. After this, it reenters the turbine once more and expands until it exhausts at
0.0080 MPa. Both the turbine and pump have an efficiency of 80%. Determine the work of the
turbine, the net work, the actual work of the pump, and the actual work of the turbine.
Given:
P1 = P8 = 22 MPa
P2 = P3 = 3.80 MPa
P4 = P5 = 1.10 MPa
P6 = P7 = 0.0080 MPa
T1 = 550 °C
T3 = 450 °C
T5 = 400 °C
πœ‚T = πœ‚P = 80%
P1 = 22 MPa
T1 = 550 °C
P2 = 3.80 MPa
2’
Reheater 1
πœ‚T = 80%
T3 = 450 °C
6’
P6 = 0.0080 MPa
Reheater 2
4’
πœ‚P = 80%
P5 = 1.10 MPa
T5 = 400 °C
H20 in
H20 out
8’
Required:
(a) WTtotal
(b) Wnet
(c) WP’
(d) WTtotal’
Solution:
οƒ˜ State 1
At P1 = 22 MPa and T1 = 550 °C
tsat = 373.80 °C
tsat < T1 ; superheated vapor
𝐾𝐽
h1 = 3370.6 π‘˜π‘”
𝐾𝐽
s1 = 6.2691 π‘˜π‘”βˆ™πΎ
35
οƒ˜ State 2
οƒ˜ State 2’
𝐾𝐽
S1 = S2 = 6.2691 π‘˜π‘”βˆ™πΎ
β„Žβ‚−β„Žβ‚‚′
πœ‚T = β„Žβ‚−β„Žβ‚‚
P2 = 3.80 MPa
0.80 =
S (π‘˜π‘”βˆ™πΎ)
𝐾𝐽
h (π‘˜π‘”)
𝐾𝐽
6.2650
S2 = 6.2691
6.2925
2894.9
h2
2910.1
3370.6
3370.6
𝐾𝐽
−β„Žβ‚‚′
π‘˜π‘”
𝐾𝐽
𝐾𝐽
− 2897.1662
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
h2’ = 2991.8530 π‘˜π‘”
By interpolation,
𝐾𝐽
𝐾𝐽
β„Žβ‚‚ − 2894.9
π‘˜π‘” βˆ™ 𝐾
π‘˜π‘”
=
𝐾𝐽
𝐾𝐽
(6.2925 − 6.2650)
(2910.1 − 2894.9)
π‘˜π‘” βˆ™ 𝐾
π‘˜π‘”
(6.2691 − 6.2650)
𝐾𝐽
h2 =2897.1662 π‘˜π‘”
οƒ˜ State 3
P2 = P3 = 3.80 MPa and T3 = 450 °C
tsat = 247.38 °C
tsat < T1 ; superheated vapor
𝐾𝐽
h3 = 3333.0 π‘˜π‘”
𝐾𝐽
s3 = 6.9629 π‘˜π‘”βˆ™πΎ
οƒ˜ State 4
οƒ˜ State 4’
𝐾𝐽
S3 = S4 = 6.9629 π‘˜π‘”βˆ™πΎ
β„Žβ‚ƒ−β„Žβ‚„′
πœ‚T = β„Žβ‚ƒ−β„Žβ‚„
P4 = 1.10 MPa
0.80 =
S (π‘˜π‘”βˆ™πΎ)
𝐾𝐽
h (π‘˜π‘”)
𝐾𝐽
6.9584
S4 = 6.9629
6.9984
2983.2
h4
3005.1
3333.0
3333.0
𝐾𝐽
−β„Žβ‚„′
π‘˜π‘”
𝐾𝐽
𝐾𝐽
− 2985.6638
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
h4’ =3055.1310 π‘˜π‘”
By interpolation,
𝐾𝐽
𝐾𝐽
β„Žβ‚‚ − 2983.2
π‘˜π‘” βˆ™ 𝐾
π‘˜π‘”
=
𝐾𝐽
𝐾𝐽
(6.9984 − 6.9584)
(3005.1 − 2983.2)
π‘˜π‘” βˆ™ 𝐾
π‘˜π‘”
(6.9629 − 6.9584)
𝐾𝐽
h4 = 2985.6638 π‘˜π‘”
36
οƒ˜ State 5
P4 = P5 = 1.10 MPa and T3 = 400 °C
tsat = 184.09 °C
tsat < T1 ; superheated vapor
𝐾𝐽
h5 = 3262.3 π‘˜π‘”
𝐾𝐽
s5 = 7.4193π‘˜π‘”βˆ™πΎ
οƒ˜ State 6
𝐾𝐽
h6 = hf6 + x6(hfg6)
S5 = S6 = 7.4193 π‘˜π‘”βˆ™πΎ
𝐾𝐽
𝐾𝐽
h6 = 173.88 π‘˜π‘” + (0.89.40) (2403.1π‘˜π‘”)
P6 = 0.0080 MPa
𝐾𝐽
Sf6 = 0.5926 π‘˜π‘”βˆ™πΎ
𝐾𝐽
h6 = 2322.2514π‘˜π‘”
𝐾𝐽
Sg6 = 8.2287 π‘˜π‘”βˆ™πΎ
Sg > S4; mixture
For x6,
οƒ˜ State 6’
β„Žβ‚…−β„Žβ‚†′
πœ‚T = β„Žβ‚…−β„Žβ‚†
S₆− Sf₆
X6 = 𝑆𝑓𝑔₆
X6 =
(7.4193−0.5926 )
𝐾𝐽
π‘˜π‘”βˆ™πΎ
0.80 =
𝐾𝐽
7.6361 π‘˜π‘”βˆ™πΎ
3262.3
3262.3
𝐾𝐽
−β„Žβ‚†′
π‘˜π‘”
𝐾𝐽
𝐾𝐽
− 2322.2514
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
h6’ = 2510.2611 π‘˜π‘”
X6 = 0.8940
X6 = 89.40 %
οƒ˜ State 7
P6 = P7 = 0.0080 MPa
𝐾𝐽
hf6 = h7 = 173.88 π‘˜π‘”
π‘š³
𝜐f6 = 𝜐7 =1.0084 × 10-3 π‘˜π‘”
οƒ˜ State 8
P1 = P8 = 22 MPa
Wp = 𝜐5 (P8 – P7)
π‘š³
Wp = 1.0084 × 10-3 π‘˜π‘” (22 MPa – 0.0080 MPa)
π‘š³
Wp = 1.0084 × 10-3 π‘˜π‘” (21.992 MPa ×
1000 πΎπ‘ƒπ‘Ž
×
1 π‘€π‘ƒπ‘Ž
1000
𝑁
π‘š²
1 πΎπ‘ƒπ‘Ž
1𝐽
1 𝐾𝐽
× 1 π‘βˆ™π‘š × 1000 𝐽 )
𝐾𝐽
Wp = 22.1767 π‘˜π‘”
h8= h7 + Wp
37
𝐾𝐽
𝐾𝐽
h8 = 173.88 π‘˜π‘” + 22.1767 π‘˜π‘”
𝐾𝐽
h8 = 196.0567 π‘˜π‘”
οƒ˜ State 8’
β„Žβ‚ˆ−β„Žβ‚‡
πœ‚T = β„Žβ‚ˆ′−β„Žβ‚‡
0.80 =
𝐾𝐽
𝐾𝐽
− 173.88
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
β„Žβ‚ˆ′− 173.88
π‘˜π‘”
196.0567
𝐾𝐽
h8’ = 201.6009 π‘˜π‘”
WP’ = h8’ – h7
WP’ = 201.6009
WP’ = 27.7209
𝐾𝐽
π‘˜π‘”
𝐾𝐽
– 173.88
𝐾𝐽
π‘˜π‘”
π‘˜π‘”
For Wnet,
Wnet = WTtotal - WP
WTtotal = WT1 + WT2 + WT3
WTtotal = (h1 – h2) + (h3 – h4) + (h5 – h6)
𝐾𝐽
WTtotal = [(3370.6 - 2897.1662) + (3333.0 – 2985.6638) + (3262.3 – 2322.2514)] π‘˜π‘”
𝐾𝐽
WTtotal = 1760.8186 π‘˜π‘”
Wnet = WTtotal - WP
𝐾𝐽
𝐾𝐽
Wnet = 1760.8186 π‘˜π‘” – 22.1767 π‘˜π‘”
𝐾𝐽
Wnet = 1738.6419 π‘˜π‘”
For Wnet’,
Wnet’ = WTtotal’ - WP’
WTtotal’ = (h1 – h2’) + (h3 – h4’) + (h5 – h6’)
𝐾𝐽
WTtotal’ = [(3370.6 – 2991.8530) + (3333.0 – 3055.1310) + (3262.3 –n2510.2611)] π‘˜π‘”
𝐾𝐽
WTtotal’ = 1408.6549 π‘˜π‘”
Wnet’ = WTtotal’ - WP’
𝐾𝐽
𝐾𝐽
Wnet’ = 1408.6549 π‘˜π‘” – 27.7209 π‘˜π‘”
𝐾𝐽
Wnet’ = 1380.934 π‘˜π‘”
38
TS DIAGRAM
P1 = P8
T1
T3
P2 = P3
22 MPa
3.80 MPa
P4 = P5
1.10 MPa
0.0080 MPa
P6 = P7
39
3. A reheat cycle features two stages of reheat. Steam enters the high pressure turbine at 19.5
MPa at 600 °C. It expands to 4 MPa where it is reheated to 470 °C before expanding further
to 1.20 MPa and being reheated to 420 °C. Finally the steam reenters the turbine and exhaust
at 0.010 MPa. The turbine and pump have the same efficiency of 90%. Determine the ideal
thermal efficiency and the actual thermal efficiency and the actual thermal efficiency.
Given:
P1 = P8 = 19.5 MPa
P2 = P3 = 4 MPa
P4 = P5 = 1.20 MPa
P6 = P7 = 0.0010 MPa
T1 = 600 °C
T3 = 470 °C
T5 = 420 °C
πœ‚T = πœ‚P = 80%
P1 = 19.5 MPa
T1 = 600 °C
P2 = 4 MPa
2’
πœ‚T = 80%
Reheater 1
T3 = 470 °C
6’
P6 = 0.0010 MPa
Reheater 2
4’
πœ‚P = 80%
8’
P5 = 1.20 MPa
T5 = 420 °C
H20 in
H20 out
Required:
(a) πœ‚TH
(b) πœ‚TH’
Solution:
οƒ˜ State 1
At P1 = 19.5 MPa and T1 = 550 °C
tsat = 600 °C
tsat < T1 ; superheated vapor
𝐾𝐽
h1 = 3542.1 π‘˜π‘”
𝐾𝐽
s1 = 6.5206 π‘˜π‘”βˆ™πΎ
40
οƒ˜ State 2
οƒ˜ State 2’
𝐾𝐽
S1 = S2 = 6.5206 π‘˜π‘”βˆ™πΎ
β„Žβ‚−β„Žβ‚‚′
πœ‚T = β„Žβ‚−β„Žβ‚‚
P2 = 4 MPa
S (π‘˜π‘”βˆ™πΎ)
𝐾𝐽
h (π‘˜π‘”)
𝐾𝐽
6.4991
S2 = 6.5206
6.5413
3041.6
h2
3067.3
0.90 =
3542.1
3542.1
𝐾𝐽
−β„Žβ‚‚′
π‘˜π‘”
𝐾𝐽
𝐾𝐽
− 3054.6936
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
h2’ = 3103.4342 π‘˜π‘”
By interpolation,
𝐾𝐽
𝐾𝐽
β„Žβ‚‚ − 3041.6
π‘˜π‘” βˆ™ 𝐾
π‘˜π‘”
=
𝐾𝐽
𝐾𝐽
(6.5413 − 6.4991)
(3067.3 − 3041.6)
π‘˜π‘” βˆ™ 𝐾
π‘˜π‘”
(6.5206 − 6.4991)
𝐾𝐽
h2 =3054.6936 π‘˜π‘”
οƒ˜ State 3
P2 = P3 = 4 MPa and T3 = 470 °C
tsat = 250.40 °C
tsat < T1 ; superheated vapor
οƒ˜ State 4
𝐾𝐽
h3 = 3376.4 π‘˜π‘”
𝐾𝐽
s3 = 6.9992π‘˜π‘”βˆ™πΎ
οƒ˜ State 4’
𝐾𝐽
S3 = S4 = 6.9992 π‘˜π‘”βˆ™πΎ
β„Žβ‚ƒ−β„Žβ‚„′
πœ‚T = β„Žβ‚ƒ−β„Žβ‚„
P4 = 1.20 MPa
S (π‘˜π‘”βˆ™πΎ)
𝐾𝐽
h (π‘˜π‘”)
𝐾𝐽
6.9984
S4 = 6.9992
7.0373
3005.1
h4
3826.9
0.90 =
3376.4
3376.4
𝐾𝐽
−β„Žβ‚„′
π‘˜π‘”
𝐾𝐽
𝐾𝐽
− 3305.5483
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
h4’ = 3042.6335 π‘˜π‘”
By interpolation,
𝐾𝐽
𝐾𝐽
β„Žβ‚‚ − 3005.1
π‘˜π‘” βˆ™ 𝐾
π‘˜π‘”
=
𝐾𝐽
𝐾𝐽
(7.0373 − 6.9984)
(3826.9 − 3005.1)
π‘˜π‘” βˆ™ 𝐾
π‘˜π‘”
(6.9992 − 6.9984)
𝐾𝐽
h4 = 3005.5483 π‘˜π‘”
οƒ˜ State 5
P4 = P5 = 1.20 MPa and T3 = 420 °C
tsat = 187.99 °C
tsat < T1 ; superheated vapor
𝐾𝐽
h5 = 3303.6 π‘˜π‘”
𝐾𝐽
s5 = 7.4401π‘˜π‘”βˆ™πΎ
41
οƒ˜ State 6
h6 = hf6 + x6(hfg6)
𝐾𝐽
S5 = S6 = 7.4401 π‘˜π‘”βˆ™πΎ
h6 = 191.83
P6 = 0.0010 MPa
𝐾𝐽
π‘˜π‘”
𝐾𝐽
+ (0.9053) (2392.8 )
π‘˜π‘”
𝐾𝐽
𝐾𝐽
h6 = 2358.0318 π‘˜π‘”
Sf6 = 0.6493 π‘˜π‘”βˆ™πΎ
𝐾𝐽
Sg6 = 8.1502 π‘˜π‘”βˆ™πΎ
Sg > S4; mixture
For x6,
οƒ˜ State 6’
β„Žβ‚…−β„Žβ‚†′
πœ‚T = β„Žβ‚…−β„Žβ‚†
S₆− Sf₆
X6 = 𝑆𝑓𝑔₆
X6 =
(7.4401−0.6493 )
𝐾𝐽
π‘˜π‘”βˆ™πΎ
3303.6
0.90 =
𝐾𝐽
7.5009 π‘˜π‘”βˆ™πΎ
3303.6
𝐾𝐽
−β„Žβ‚†′
π‘˜π‘”
𝐾𝐽
𝐾𝐽
− 2358.0318
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
h6’ = 2454.5886 π‘˜π‘”
X6 = 0.9053
X6 = 90.53 %
οƒ˜ State 7
P6 = P7 = 0.0010 MPa
𝐾𝐽
hf6 = h7 = 191.83 π‘˜π‘”
π‘š³
𝜐f6 = 𝜐7 =1.0102 × 10-3 π‘˜π‘”
οƒ˜ State 8
P1 = P8 = 19.5 MPa
Wp = 𝜐5 (P8 – P7)
π‘š³
Wp = 1.0102 × 10-3 π‘˜π‘” (19.5 MPa – 0.0010 MPa)
π‘š³
1000 πΎπ‘ƒπ‘Ž
Wp = 1.0102 × 10-3
(19.49 MPa ×
×
π‘˜π‘”
1 π‘€π‘ƒπ‘Ž
1000
𝑁
π‘š²
1 πΎπ‘ƒπ‘Ž
1𝐽
1 𝐾𝐽
× 1 π‘βˆ™π‘š × 1000 𝐽 )
𝐾𝐽
Wp = 19.6888 π‘˜π‘”
h8= h7 + Wp
𝐾𝐽
𝐾𝐽
h8 = 191.83 π‘˜π‘” + 19.6888 π‘˜π‘”
𝐾𝐽
h8 = 211.5188 π‘˜π‘”
οƒ˜ State 8’
β„Žβ‚ˆ−β„Žβ‚‡
πœ‚T = β„Žβ‚ˆ′−β„Žβ‚‡
0.90 =
𝐾𝐽
𝐾𝐽
− 191.83
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
β„Žβ‚ˆ′− 191.83
π‘˜π‘”
211.5188
𝐾𝐽
h8’ = 213.7064 π‘˜π‘”
42
For Wnet,
Wnet = WTtotal - WP
WTtotal = WT1 + WT2 + WT3
WTtotal = (h1 – h2) + (h3 – h4) + (h5 – h6)
𝐾𝐽
WTtotal = [(3542.1 – 3054.6936) + (3376.4 – 3005.5483) + (3303.6 – 2358.0318)] π‘˜π‘”
𝐾𝐽
WTtotal = 1803.8263 π‘˜π‘”
Wnet = WTtotal - WP
𝐾𝐽
𝐾𝐽
Wnet = 1803.8263π‘˜π‘” – 19.6888 π‘˜π‘”
𝐾𝐽
Wnet = 1784.1375 π‘˜π‘”
QA = QAB + QAR1 + QAR2
QA = (h1 – h8) + (h3 – h2) + (h5 – h4)
𝐾𝐽
QA = [(3542.1 – 211.5188) + (3376.4 – 6054.6936) + (3303.6 – 3005.5483)] π‘˜π‘”
𝐾𝐽
QA = 3950.3393 π‘˜π‘”
For πœ‚TH,
π‘Šπ‘›π‘’π‘‘
πœ‚TH = 𝑄𝐴
πœ‚TH =
𝐾𝐽
π‘˜π‘”
𝐾𝐽
3950.3393
π‘˜π‘”
1784.1375
πœ‚TH = 0.4516
πœ‚TH = 45.16%
For πœ‚TH’,
π‘Šπ‘›π‘’π‘‘′
πœ‚TH’ = 𝑄𝐴′
π‘Šπ‘‡π‘‘π‘œπ‘‘π‘Žπ‘™′−π‘Šπ‘ƒ′
πœ‚TH’ = 𝑄𝐴𝐡′ + 𝑄𝐴𝑅1′+𝑄𝐴𝑅2′
πœ‚TH’ =
πœ‚TH’ =
[(β„Žβ‚−β„Žβ‚‚′)+(β„Žβ‚ƒ−β„Žβ‚„′)+(β„Žβ‚…−β„Žβ‚†′)]−[β„Žβ‚ˆ′−β„Žβ‚‡]
(β„Ž1 −β„Ž8′ )+(β„Ž3 −β„Ž2′ )+(β„Žβ‚…−β„Žβ‚„′)
KJ
kg
KJ
[(3542.1-213.7064)+(3376.4-3103.4342)+(3303.6-3042.6335)]kg
[(3542.1-3103.4342)+(3376.4-3042.6335)+(3303.6-2454.5886)]
πœ‚TH’ = 0.4141
πœ‚TH’ = 41.41%
43
TS DIAGRAM
P1 = P8
T1
T3
P2 = P3
19.5 MPa
4 MPa
P4 = P5
1.20 MPa
0.0010 MPa
P6 = P7
44
REGENERATIVE OPEN FEED WATER HEATER CYCLE
1. A steam power plant running on a regenerative Rankine cycle with one open feed water heater.
Steam enters the turbine at 8 MPa and 450 °C and is condensed in the condenser at a pressure
of 0.02 MPa. Some steam exits the turbine at 1.5 MPa and enters the open feed water heater.
Both the turbine and pump have an efficiency of 85%. Determine the fraction of steam extracted
from the turbine and the ideal and actual thermal efficiency of the cycle.
Given:
Required:
(a) ṁ
(b) πœ‚TH
(c) πœ‚TH’
Solution:
οƒ˜ State 1
P1 = 10 MPa
T1 = 500 °C
tsat = 311.06 °C
tsat < T1; superheated vapor
οƒ˜ State 2
𝐾𝐽
S1 = S2 = 6.5966 π‘˜π‘”βˆ™πΎ
P2 = 2 MPa
𝐾𝐽
h1 = 3373.7 π‘˜π‘”
𝐾𝐽
s1 = 6.5966 π‘˜π‘”βˆ™πΎ
οƒ˜ State 2’
β„Žβ‚−β„Žβ‚‚′
πœ‚T = β„Žβ‚−β„Žβ‚‚
0.80 =
3373.7
3373.7
𝐾𝐽
−β„Žβ‚‚′
π‘˜π‘”
𝐾𝐽
𝐾𝐽
− 2929.5840
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
h2’ = 3018.4072 π‘˜π‘”
45
S (π‘˜π‘”βˆ™πΎ)
𝐾𝐽
h (π‘˜π‘”)
𝐾𝐽
6.5931
S2 = 6.5966
6.6388
2927.7
h2
2952.3
By interpolation,
𝐾𝐽
𝐾𝐽
β„Žβ‚‚ − 2927.7
π‘˜π‘” βˆ™ 𝐾
π‘˜π‘”
=
𝐾𝐽
𝐾𝐽
(6.6388 − 6.5931)
(2952.3 − 2927.7)
π‘˜π‘” βˆ™ 𝐾
π‘˜π‘”
𝐾𝐽
h2 = 2929.5840 π‘˜π‘”
(6.5966 − 6.5931)
For x6,
οƒ˜ State 3
s2 = s3 = 6.5966
𝐾𝐽
οƒ˜ State 3’
S₃− Sf₃
X3 = 𝑆𝑓𝑔₃
π‘˜π‘”βˆ™πΎ
P3 = 0.01 MPa
(6.5966−0.6493 )
sg3 = 8.1502 π‘˜π‘”βˆ™πΎ
X3 =
sg3 > s3; mixture
X3 = 0.7929
𝐾𝐽
β„Žβ‚−β„Žβ‚ƒ′
πœ‚T = β„Žβ‚−β„Žβ‚ƒ
𝐾𝐽
π‘˜π‘”βˆ™πΎ
0.80 =
𝐾𝐽
7.5009 π‘˜π‘”βˆ™πΎ
3373.7
3373.7
𝐾𝐽
−β„Žβ‚ƒ′
π‘˜π‘”
𝐾𝐽
𝐾𝐽
− 2089.0811
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
h3’ =2346.0049 π‘˜π‘”
X3 = 79.29 %
h3 = hf3 + x3 (hfg3)
𝐾𝐽
𝐾𝐽
h3 = 191.83 π‘˜π‘” + (0.7929) (2392.8π‘˜π‘”)
𝐾𝐽
h3 = 2089.0811 π‘˜π‘”
οƒ˜ State 4
P3 = P4 =0.01 MPa MPa
𝐾𝐽
hf3 = h4 = 191.83 π‘˜π‘”
π‘š³
𝜐f3 = 𝜐4 =1.0079 × 10-3 π‘˜π‘”
οƒ˜ State 5
P2 = P5 = 2 MPa
WP1 = 𝜐4 (P5 – P4)
π‘š³
WP1 = 1.0079 × 10-3 π‘˜π‘” (2 MPa – 0.01 MPa)
WP1
π‘š³
1000 πΎπ‘ƒπ‘Ž
= 1.0079 × 10-3
(1.99 MPa ×
×
π‘˜π‘”
1 π‘€π‘ƒπ‘Ž
1000
𝑁
π‘š²
1 πΎπ‘ƒπ‘Ž
1𝐽
1 𝐾𝐽
× 1 π‘βˆ™π‘š × 1000 𝐽 )
𝐾𝐽
WP1 = 2.0103 π‘˜π‘”
h5= h4 + Wp
𝐾𝐽
𝐾𝐽
h5 = 191.83 π‘˜π‘” + 2.0103 π‘˜π‘”
𝐾𝐽
h5 = 193.8403 π‘˜π‘”
46
οƒ˜ State 5’
β„Žβ‚…−β„Žβ‚„
πœ‚T = β„Žβ‚…′−β„Žβ‚„
0.80 =
𝐾𝐽
𝐾𝐽
−191.83
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
hβ‚…′− 191.83
π‘˜π‘”
193.8403
𝐾𝐽
h5’ = 194.3429 π‘˜π‘”
οƒ˜ State 6
P5 = P6 = 2 MPa
𝐾𝐽
hf5 = h6 = 908.79 π‘˜π‘”
π‘š³
𝜐f5 = 𝜐6 = 1.1767× 10-3 π‘˜π‘”
οƒ˜ State 7
P1 = P7 = 10 MPa
WP2 = 𝜐6 (P7 – P6)
π‘š³
WP2 = 1.1767 × 10-3 π‘˜π‘” (10 MPa – 2 MPa)
WP2
π‘š³
1000 πΎπ‘ƒπ‘Ž
= 1.1767 × 10-3
(8MPa ×
×
π‘˜π‘”
1 π‘€π‘ƒπ‘Ž
1000
𝑁
π‘š²
1 πΎπ‘ƒπ‘Ž
1𝐽
1 𝐾𝐽
× 1 π‘βˆ™π‘š × 1000 𝐽 )
𝐾𝐽
WP2 = 9.4136 π‘˜π‘”
h7= h6 + Wp
h7 = 908.79
𝐾𝐽
π‘˜π‘”
+ 9.4136
𝐾𝐽
π‘˜π‘”
𝐾𝐽
h7 = 918.2036 π‘˜π‘”
οƒ˜ State 7’
β„Žβ‚‡−β„Žβ‚†
πœ‚T = β„Žβ‚‡′−β„Žβ‚†
0.80 =
𝐾𝐽
𝐾𝐽
−908.79
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
h₇′− 908.79
π‘˜π‘”
918.2036
𝐾𝐽
h7’ = 920.557 π‘˜π‘”
For m,
(m)(h2)
h6
1 kg
β„Žβ‚†−β„Žβ‚…
m = β„Žβ‚‚−β„Žβ‚…
(1-m)(h5)
m=
𝐾𝐽
π‘˜π‘”
𝐾𝐽
(2929.5840−193.8403)
π‘˜π‘”
(908.79−193.8403)
m = 0.2613
47
ENERGY BALANCE @ OPEN FWH
Ein = Eout
(m)(h2) + (1-m)(h5) = h6
(m)(h2) + h5 – m(h5) = h6
m(h2 + h5) + h5 = h6
m(h2 + h5) = h6 - h5
m' =
m' =
β„Žβ‚†−β„Žβ‚…′
β„Žβ‚‚′−β„Žβ‚…′
𝐾𝐽
π‘˜π‘”
𝐾𝐽
(3018.4072−194.3429)
π‘˜π‘”
(908.79−194.3429)
m’ = 0.2530
ENERGY BALANCE @ TURBINE
1 kg
h1
WT = h1 – mh2 – (1-m)h3 + h2 – h2
WT = h1 – h2 - mh2 + h2 - (1-m)h3
WT = (h1 – h2) + (1-m) h2 – (1-m)h3
WT = (h1 – h2) + (1-m) h2 – (1-m)(h2 – h3)
WT = (h1 – h2) + (1-m) – (1-m)(h2 – h3)
WT
m
(1-m)
h2
h3
WT = (h1 – h2) + (1-m) – (1-m) (h2 – h3)
𝐾𝐽
𝐾𝐽
WT = (3373.7-2929.5840) π‘˜π‘” + (1 – 0.2613) (2929.5840 – 2089.0049) π‘˜π‘”
𝐾𝐽
WT = 1064.9955 π‘˜π‘”
WT’ = (h1 – h2’) + (1-m’) – (1-m) (h2’ – h3’)
𝐾𝐽
𝐾𝐽
WT’ = (3373.7 – 3018.4072) π‘˜π‘” + (1-0.2530) (3018.4072 – 2346.0049) π‘˜π‘”
𝐾𝐽
WT’ = 857.5773 π‘˜π‘”
For WPtotal,
WPtotal = WP1 + WP2
𝐾𝐽
WPtotal = (2.0103 + 9.4136) π‘˜π‘”
𝐾𝐽
WPtotal = 11.4239 π‘˜π‘”
WPtotal’ = WP1’ + WP2’
WPtotal’ = (h5’ – h4) + (h7’ – h6)
𝐾𝐽
WPtotal’ = [(194.3429 – 191.83) + (920.557 – 908.79)] π‘˜π‘”
𝐾𝐽
WPtotal’ = 14.2799 π‘˜π‘”
For πœ‚TH,
πœ‚TH =
πœ‚TH =
π‘Šπ‘›π‘’π‘‘
𝑄𝐴
π‘Šπ‘‘−π‘Šπ‘ƒπ‘‘π‘œπ‘‘π‘Žπ‘™
β„Žβ‚ −β„Žβ‚‡
48
πœ‚TH =
𝐾𝐽
π‘˜π‘”
𝐾𝐽
(3373.7−918.2036)
π‘˜π‘”
(1064.9955−11.4239)
πœ‚TH = 0.4291
πœ‚TH = 42.91%
For πœ‚TH’,
πœ‚TH’ =
πœ‚TH’ =
πœ‚TH’ =
π‘Šπ‘›π‘’π‘‘ ′
𝑄𝐴′
π‘Šπ‘‡ ′ −π‘Šπ‘ƒπ‘‘π‘œπ‘‘π‘Žπ‘™′
β„Ž1 −β„Ž7′
𝐾𝐽
(857.5773−14.2799)
π‘˜π‘”
𝐾𝐽
π‘˜π‘”
(3373.7−920.557)
πœ‚TH’ = 0.3438
πœ‚TH’ = 34.38%
TS DIAGRAM
T1
P1 = P7 = 10 MPa
P2 = P6 = 2 MPa
P3 = P4 = 0.01 MPa
x3
S4 = S5
S6 = S7
x3’
S1 = S2
49
2. In a steam power plant utilizing a regenerative Rankine cycle with one open feed water heater,
steam is admitted to the turbine at 16 MPa and 620 °C, and then condensed in the condenser at
0.04 MPa. A portion of the steam exits the turbine at 5 MPa and enters the open feed water
heater. Both the turbine and pump have efficiencies of 75%. Calculate the heat added to the
cycle, heat rejected and the actual of it.
Given:
P1 = 16 MPa
T1 = 620 °C
1 kg
πœ‚T = 75%
P2 = 5 MPa
P3 = 0.04 MPa
(1-m)
m
(1-m)
1 kg
πœ‚p = 75%
Required:
(a) QA, QA’
(b) QR, QR’
Solution:
οƒ˜ State 1
P1 = 16 MPa
T1 = 620 °C
tsat =347.44 °C ; tsat < T1; superheated vapor
οƒ˜ State 2
𝐾𝐽
S1 = S2 = 6.6999 π‘˜π‘”βˆ™πΎ
𝐾𝐽
s1 = 6.6999 π‘˜π‘”βˆ™πΎ
οƒ˜ State 2’
β„Žβ‚−β„Žβ‚‚′
πœ‚T = β„Žβ‚−β„Žβ‚‚
P2 = 5 MPa
𝐾𝐽
𝐾𝐽
h1 = 3626.5 π‘˜π‘”
𝐾𝐽
3626.5
𝐾𝐽
−β„Žβ‚‚′
π‘˜π‘”
S (π‘˜π‘”βˆ™πΎ)
h (π‘˜π‘”)
0.75 =
6.6820
S2 = 6.6999
6.7172
3220.2
h2
3244.4
h2’ = 3331.0047 π‘˜π‘”
3626.5
𝐾𝐽
𝐾𝐽
− 3232.5063
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
50
By interpolation,
𝐾𝐽
𝐾𝐽
β„Žβ‚‚ − 3220.2
π‘˜π‘” βˆ™ 𝐾
π‘˜π‘”
=
𝐾𝐽
𝐾𝐽
(6.7172 − 6.6820)
(3244.4 − 3220.2)
π‘˜π‘” βˆ™ 𝐾
π‘˜π‘”
𝐾𝐽
h2 = 3232.5063 π‘˜π‘”
(6.6999 − 6.6820)
οƒ˜ State 3
For x6,
𝐾𝐽
οƒ˜ State 3’
S₃− Sf₃
X3 = 𝑆𝑓𝑔₃
s2 = s3 = 6.6999 π‘˜π‘”βˆ™πΎ
P3 = 0.04 MPa
(6.699931.0259 )
sg3 = 7.6700 π‘˜π‘”βˆ™πΎ
X3 =
sg3 > s3; mixture
X3 = 0.8540
𝐾𝐽
β„Žβ‚−β„Žβ‚ƒ′
πœ‚T = β„Žβ‚−β„Žβ‚ƒ
𝐾𝐽
π‘˜π‘”βˆ™πΎ
0.80 =
𝐾𝐽
6.6441 π‘˜π‘”βˆ™πΎ
3626.5
3626.5
𝐾𝐽
−β„Žβ‚ƒ′
π‘˜π‘”
𝐾𝐽
𝐾𝐽
− 2298.1768
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
h3’ =2630.2576π‘˜π‘”
X3 = 85.40 %
h3 = hf3 + x3 (hfg3)
𝐾𝐽
𝐾𝐽
h3 = 317.58 π‘˜π‘” + (0.8540) (2319.2π‘˜π‘”)
𝐾𝐽
h3 = 2298.1768 π‘˜π‘”
οƒ˜ State 4
P3 = P4 =0.04MPa
𝐾𝐽
hf3 = h4 = 317.58 π‘˜π‘”
π‘š³
𝜐f3 = 𝜐4 =1.0265 × 10-3 π‘˜π‘”
οƒ˜ State 5
P2 = P5 = 5 MPa
WP1 = 𝜐4 (P5 – P4)
π‘š³
WP1 = 1.0265 × 10-3 π‘˜π‘” (5 MPa – 0.04 MPa)
π‘š³
WP1 = 1.0265 × 10-3 π‘˜π‘” (4.96 MPa ×
1000 πΎπ‘ƒπ‘Ž
×
1 π‘€π‘ƒπ‘Ž
1000
𝑁
π‘š²
1 πΎπ‘ƒπ‘Ž
1𝐽
1 𝐾𝐽
× 1 π‘βˆ™π‘š × 1000 𝐽 )
𝐾𝐽
WP1 = 5.0914π‘˜π‘”
h5= h4 + Wp
𝐾𝐽
𝐾𝐽
h5 = 317.58 π‘˜π‘” + 5.0914π‘˜π‘”
𝐾𝐽
h5 = 322.6714 π‘˜π‘”
οƒ˜ State 5’
β„Žβ‚…−β„Žβ‚„
πœ‚T = β„Žβ‚…′−β„Žβ‚„
0.75 =
𝐾𝐽
𝐾𝐽
−317.58
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
hβ‚…′− 317.58
π‘˜π‘”
322.6714
𝐾𝐽
h5’ = 324.3685 π‘˜π‘”
51
οƒ˜ State 6
P5 = P6 = 5 MPa
hf5 = h6 = 1154.23
𝐾𝐽
π‘˜π‘”
π‘š³
𝜐f5 = 𝜐6 = 1.2859 × 10-3 π‘˜π‘”
οƒ˜ State 7
P1 = P7 = 16 MPa
WP2 = 𝜐6 (P7 – P6)
π‘š³
WP2 = 1.2859 × 10-3 π‘˜π‘” (16 MPa – 5 MPa)
π‘š³
WP2 = 1.2859 × 10-3 π‘˜π‘” (11 MPa ×
1000 πΎπ‘ƒπ‘Ž
×
1 π‘€π‘ƒπ‘Ž
1000
𝑁
π‘š²
1 πΎπ‘ƒπ‘Ž
1𝐽
1 𝐾𝐽
× 1 π‘βˆ™π‘š × 1000 𝐽 )
𝐾𝐽
WP2 = 14.1449 π‘˜π‘”
h7= h6 + Wp
𝐾𝐽
𝐾𝐽
h7 = 1154.23 π‘˜π‘” + 14.1449 π‘˜π‘”
𝐾𝐽
h7 = 1168.3749 π‘˜π‘”
οƒ˜ State 7’
β„Žβ‚‡−β„Žβ‚†
πœ‚T = β„Žβ‚‡′−β„Žβ‚†
0.75 =
𝐾𝐽
𝐾𝐽
−1154.23
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
h₇′− 1154.23
π‘˜π‘”
1168.3749
𝐾𝐽
h7’ = 1173.0899 π‘˜π‘”
For m,
(m)(h2)
h6
1 kg
β„Žβ‚†−β„Žβ‚…
m = β„Žβ‚‚−β„Žβ‚…
(1-m)(h5)
m=
𝐾𝐽
π‘˜π‘”
𝐾𝐽
(3232.5063−322.6714)
π‘˜π‘”
(1154.23−322.6714)
m = 0.2858
ENERGY BALANCE @ OPEN FWH
Ein = Eout
(m)(h2) + (1-m)(h5) = h6
(m)(h2) + h5 – m(h5) = h6
m(h2 + h5) + h5 = h6
m(h2 + h5) = h6 - h5
β„Žβ‚†−β„Žβ‚…′
m' = β„Žβ‚‚′−β„Žβ‚…′
m' =
𝐾𝐽
π‘˜π‘”
𝐾𝐽
(3331.0047−324.3685)
π‘˜π‘”
(1154.23−324.3685)
m’ = 0.2760
52
(1)(h1)
ENERGY BALANCE @ BOILER
Q A + h7 = h1
QA = h1 - h7
𝐾𝐽
QA = (3626.5-1168.3749) π‘˜π‘”
QA
𝐾𝐽
QA = 3458.1251π‘˜π‘”
(1)(h7)
For QA’,
QA’ = h1 - h7’
𝐾𝐽
QA’ = (3626.5-1168.3749) π‘˜π‘”
𝐾𝐽
QA’ = 2458.1251π‘˜π‘”
For QR,
(1-m)(h3)
(1-m)(h4)
ENERGY BALANCE @ OPEN FWH
Ein = Eout
(1-m)(h3) = QR +(1-m)(h4)
QR = (1-m)(h3) – (1-m)(h4)
QR = (1-m) (h3-h4)
𝐾𝐽
QR = (1-0.2858) (2298.1768-317.58) π‘˜π‘”
𝐾𝐽
QR = 1414.5422 π‘˜π‘”
For QR’,
QR = (1-m’) (h3’-h4)
𝐾𝐽
QR = (1-0.2760) (2630.2576-317.58) π‘˜π‘”
𝐾𝐽
QR = 1674.3786 π‘˜π‘”
53
TS DIAGRAM
T1
P1 = P7 = 16 MPa
P2 = P6 = 5 MPa
P3 = P4 = 0.04 MPa
x3
S4 = S5
S6 = S7
x3’
S1 = S2
54
3. In a steam power plant running on regenerative Rankine cycle with a single open feed water
heater, steam is let into the turbine at 12 MPa and 670 °C before being condensed in the
condenser at 0.08 MPa. A fraction of the steam exits the turbine at 3.5 MPa and goes into the
open feed water heater. The turbine and pump have efficiencies of 90%. Find the actual fraction
of steam extracted from the turbine, work of the turbine and the actual work of the pump.
Given:
P1 = 12 MPa
T1 = 670 °C
1 kg
P2 = 3.5 MPa
πœ‚T = 90%
P3 = 0.08 MPa
(1-m)
m
(1-m)
1 kg
πœ‚p = 90%
Required:
(a) m'
(b) WT
(c) WPtotal’
Solution:
οƒ˜ State 1
P1 = 12 MPa
T1 = 670 °C
tsat =324.75 °C
tsat < T1; superheated vapor
οƒ˜ State 2
𝐾𝐽
S1 = S2 = 6.9971 π‘˜π‘”βˆ™πΎ
P2 = 3.5 MPa
𝐾𝐽
h1 = 3783.75 π‘˜π‘”
𝐾𝐽
s1 = 6.9971π‘˜π‘”βˆ™πΎ
S (π‘˜π‘”βˆ™πΎ)
𝐾𝐽
h (π‘˜π‘”)
𝐾𝐽
6.9734
S2 = 6.9971
7.0052
3314.4
h2
3337.2
55
οƒ˜ State 2’
By interpolation,
𝐾𝐽
𝐾𝐽
β„Žβ‚‚ − 3314.4
π‘˜π‘” βˆ™ 𝐾
π‘˜π‘”
=
𝐾𝐽
𝐾𝐽
(7.0052 − 6.9734)
(3337.2 − 3314.4)
π‘˜π‘” βˆ™ 𝐾
π‘˜π‘”
𝐾𝐽
h2 = 3331.3925 π‘˜π‘”
β„Žβ‚−β„Žβ‚‚′
πœ‚T = β„Žβ‚−β„Žβ‚‚
(6.9971 − 6.9734)
0.90 =
𝐾𝐽
3783.9
𝐾𝐽
𝐾𝐽
− 3331.3925
π‘˜π‘”
π‘˜π‘”
𝐾𝐽
οƒ˜ State 3’
S₃− Sf₃
X3 = 𝑆𝑓𝑔₃
s2 = s3 = 6.9971 π‘˜π‘”βˆ™πΎ
P3 = 0.08 MPa
X3 =
sg3 > s3; mixture
X3 = 0.9295
h3 = hf3 + x3 (hfg3)
𝐾𝐽
𝐾𝐽
h3 = 317.58 π‘˜π‘” + (0.8540) (2319.2π‘˜π‘”)
β„Žβ‚−β„Žβ‚ƒ′
πœ‚T = β„Žβ‚−β„Žβ‚ƒ
(6.9971−1.2329 )
sg3 = 7.64346 π‘˜π‘”βˆ™πΎ
𝐾𝐽
𝐾𝐽
−β„Žβ‚‚′
π‘˜π‘”
h2’ = 3376.6433 π‘˜π‘”
For x6,
οƒ˜ State 3
3783.9
𝐾𝐽
π‘˜π‘”βˆ™πΎ
0.90 =
𝐾𝐽
6.2017 π‘˜π‘”βˆ™πΎ
3783.9
3783.9
𝐾𝐽
−β„Žβ‚ƒ′
π‘˜π‘”
𝐾𝐽
𝐾𝐽
− 2505.4360
π‘˜π‘”
π‘˜π‘”
h3’ =2633.2824
𝐾𝐽
π‘˜π‘”
X3 = 92.95 %
𝐾𝐽
h3 = 2298.1768 π‘˜π‘”
οƒ˜ State 4
P3 = P4 =0.08 MPa
𝐾𝐽
hf3 = h4 = 391.66 π‘˜π‘”
π‘š³
𝜐f3 = 𝜐4 =1.0386 × 10-3 π‘˜π‘”
οƒ˜ State 5
P2 = P5 = 3.5 MPa
WP1 = 𝜐4 (P5 – P4)
π‘š³
WP1 = 1.0386 × 10-3 π‘˜π‘” (3.5 MPa – 0.08 MPa)
π‘š³
WP1 = 1.0386 × 10-3 π‘˜π‘” (3.42 MPa ×
1000 πΎπ‘ƒπ‘Ž
×
1 π‘€π‘ƒπ‘Ž
1000
𝑁
π‘š²
1 πΎπ‘ƒπ‘Ž
1𝐽
1 𝐾𝐽
× 1 π‘βˆ™π‘š × 1000 𝐽 )
𝐾𝐽
WP1 = 3.5520 π‘˜π‘”
h5= h4 + Wp1
𝐾𝐽
𝐾𝐽
h5 = 391.66 π‘˜π‘” + 3.5520π‘˜π‘”
𝐾𝐽
h5 = 395.212 π‘˜π‘”
οƒ˜ State 5’
β„Žβ‚…−β„Žβ‚„
πœ‚T = β„Žβ‚…′−β„Žβ‚„
0.90 =
𝐾𝐽
−391.66
π‘˜π‘”
𝐾𝐽
hβ‚…′− 391.66
π‘˜π‘”
95.232
𝐾𝐽
h5’ = 395.6067 π‘˜π‘”
56
For m,
(m)(h2)
h6
β„Žβ‚†−β„Žβ‚…
m = β„Žβ‚‚−β„Žβ‚…
(1-m)(h5)
1 kg
m=
𝐾𝐽
π‘˜π‘”
𝐾𝐽
(3331.3925−395.212)
π‘˜π‘”
(1049.75−395.212)
m = 0.2229
ENERGY BALANCE @ OPEN FWH
Ein = Eout
(m)(h2) + (1-m)(h5) = h6
(m)(h2) + h5 – m(h5) = h6
m(h2 + h5) + h5 = h6
m(h2 + h5) = h6 - h5
β„Žβ‚†−β„Žβ‚…′
m' = β„Žβ‚‚′−β„Žβ‚…′
m' =
𝐾𝐽
π‘˜π‘”
𝐾𝐽
(3376.6433−395.6067)
π‘˜π‘”
(1049.75−395.6067)
m’ = 0.2194
ENERGY BALANCE @ TURBINE
1 kg
h1
m
WT = h1 – mh2 – (1-m)h3 + h2 – h2
WT = h1 – h2 - mh2 + h2 - (1-m)h3
WT = (h1 – h2) + (1-m) h2 – (1-m)h3
WT = (h1 – h2) + (1-m) h2 – (1-m)(h2 – h3)
WT = (h1 – h2) + (1-m) – (1-m)(h2 – h3)
WT
(1-m)
h2
h3
WT = (h1 – h2) + (1-m) – (1-m) (h2 – h3)
𝐾𝐽
𝐾𝐽
WT = (3783.9-3331.3925) π‘˜π‘” + (1 – 0.2229) (3331.3925-2505.4360) π‘˜π‘”
𝐾𝐽
WT = 1049.3583 π‘˜π‘”
WT’ = (h1 – h2’) + (1-m’) – (1-m) (h2’ – h3’)
𝐾𝐽
𝐾𝐽
WT’ = (3783.9-3376.6433) π‘˜π‘” + (1-0.2194) (3376.6433-2633.2824) π‘˜π‘”
𝐾𝐽
WT’ = 987.5242π‘˜π‘”
For WPtotal,
WPtotal = WP1 + WP2
𝐾𝐽
WPtotal = (3.5520+10.4950) π‘˜π‘”
𝐾𝐽
WPtotal = 14.047 π‘˜π‘”
WPtotal’ = WP1’ + WP2’
WPtotal’ = (h5’ – h4) + (h7’ – h6)
𝐾𝐽
WPtotal’ = [(395.6067-391.66) + (1061.4111-1049.75)] π‘˜π‘”
𝐾𝐽
WPtotal’ = 15.6075 π‘˜π‘”
57
TS DIAGRAM
T1
P1 = P7 = 12 MPa
P2 = P6 = 3.5 MPa
P3 = P4 = 0.00 MPa
x3
S4 = S5
S6 = S7
x3’
S1 = S2
58
REGENERATIVE CLOSED FEED WATER HEATER CYCLE
1. Examine a steam power plant running on a regenerative Rankine cycle with a closed feed water
heater. The turbine inlet is kept at 10 MPa and 500 °C, while the condenser operates at 20 kPA.
Steam is withdrawn at 5 MPa to the closed feed water heater, then expanded to condenser
pressure before entering the condenser. Determine the turbine work, pump work, and the actual
thermal efficiency, assuming both the turbine and pump operate at 85% efficiency.
Given:
P1 = 10 MPa
T1 = 500 °C
P2 = 5 MPa
πœ‚T = 85%
P3 = 20 kPa
Closed
FWH
(1-m)
πœ‚P = 85%
Required:
(d) WT
(e) WPtotal
(f) πœ‚TH’
Solution:
οƒ˜ State 1
P1 = 10 MPa
T1 = 500 °C
tsat = 311.06 °C
tsat < T1; superheated vapor
οƒ˜ State 2
𝐾𝐽
S1 = S2 = 6.5966 π‘˜π‘”βˆ™πΎ
P2 = 5 MPa
𝐾𝐽
h1 = 3373.7 π‘˜π‘”
𝐾𝐽
s1 = 6.5966 π‘˜π‘”βˆ™πΎ
S (π‘˜π‘”βˆ™πΎ)
𝐾𝐽
h (π‘˜π‘”)
𝐾𝐽
6.5708
S2 = 6.5966
6.6089
3145.9
h2
3171.0
59
By interpolation,
𝐾𝐽
𝐾𝐽
β„Žβ‚‚ − 3145.9
π‘˜π‘” βˆ™ 𝐾
π‘˜π‘”
=
𝐾𝐽
𝐾𝐽
(6.6089 − 6.5708)
(3171.0 − 3145.9)
π‘˜π‘” βˆ™ 𝐾
π‘˜π‘”
𝐾𝐽
h2 = 3136.9583 π‘˜π‘”
(6.5966 − 6.5708)
οƒ˜ State 3
For x3,
𝐾𝐽
οƒ˜ State 3’
S₃− Sf₃
X3 = 𝑆𝑓𝑔₃
s2 = s3 = 6.5966 π‘˜π‘”βˆ™πΎ
P3 = 0.02 MPa
(6.5966−0.8320 )
sg3 = 7.9085 π‘˜π‘”βˆ™πΎ
X3 =
sg3 > s3; mixture
X3 = 0.8146
𝐾𝐽
β„Žβ‚−β„Žβ‚ƒ′
πœ‚T = β„Žβ‚−β„Žβ‚ƒ
𝐾𝐽
π‘˜π‘”βˆ™πΎ
0.85 =
𝐾𝐽
7.0766 π‘˜π‘”βˆ™πΎ
3373.7
𝐾𝐽
−β„Žβ‚ƒ′
π‘˜π‘”
𝐾𝐽
𝐾𝐽
− 2172.4712
π‘˜π‘”
π‘˜π‘”
3373.7
𝐾𝐽
h3’ =2352.6555 π‘˜π‘”
X3 = 81.46 %
h3 = hf3 + x3 (hfg3)
𝐾𝐽
𝐾𝐽
h3 = 251.40π‘˜π‘” + (0.8146) (2358.3π‘˜π‘”)
𝐾𝐽
h3 = 2172.4712 π‘˜π‘”
οƒ˜ State 4
P3 = P4 =0.02 MPa MPa
𝐾𝐽
hf3 = h4 = 251.40 π‘˜π‘”
𝜐f3 = 𝜐4 =1.0172 × 10-3
π‘š³
π‘˜π‘”
οƒ˜ State 5
P5 = P9 = P8 = P7 = P7 = 10 MPa
WP1 = 𝜐4 (P5 – P4)
π‘š³
WP1 = 1.0172 × 10-3 π‘˜π‘” (10 MPa – 0.02 MPa)
WP1
π‘š³
1000 πΎπ‘ƒπ‘Ž
= 1.0172 × 10-3
(9.98 MPa ×
×
π‘˜π‘”
1 π‘€π‘ƒπ‘Ž
1000
𝑁
π‘š²
1 πΎπ‘ƒπ‘Ž
1𝐽
1 𝐾𝐽
× 1 π‘βˆ™π‘š × 1000 𝐽 )
𝐾𝐽
WP1 = 10.1517 π‘˜π‘”
h5= h4 + Wp
𝐾𝐽
𝐾𝐽
h5 = 251.40 π‘˜π‘” + 10.1517 π‘˜π‘”
𝐾𝐽
h5 = 261.5517 π‘˜π‘”
οƒ˜ State 6
P2 = P6 = 5 MPa
π‘š³
𝜐f2 = 𝜐6 = 1.2859× 10-3 π‘˜π‘”
60
WP2 = 𝜐6 (P7 – P6)
π‘š³
WP2 = 1.2859 × 10-3 π‘˜π‘” (10 MPa – 5 MPa)
π‘š³
WP2 = 1.2859 × 10-3 π‘˜π‘” (5 MPa ×
1000 πΎπ‘ƒπ‘Ž
×
1 π‘€π‘ƒπ‘Ž
1000
𝑁
π‘š²
1 πΎπ‘ƒπ‘Ž
1𝐽
1 𝐾𝐽
× 1 π‘βˆ™π‘š × 1000 𝐽 )
𝐾𝐽
WP2 = 6.4295 π‘˜π‘”
οƒ˜ State 7
h7= h6 + WP2
𝐾𝐽
𝐾𝐽
h7 = 1154.23 π‘˜π‘” + 6.4259 π‘˜π‘”
𝐾𝐽
h7 = 1160.6595 π‘˜π‘”
Two fluids steam which are being mixed have the same enthalpy (h). Therefore, h7 = h8 = h9 =
1160.6595
𝐾𝐽
π‘˜π‘”
.
For m,
(m)(h2)
ENERGY BALANCE
H9
(1-m)
(1-m)(h5)
(m)(h6)
Ein = Eout
(m)(h2) + (1-m)(h5) = mh6 + (1-m)(h9)
(m)(h2) - mh6 = (1-m) (h9 - h5)
m(h2 – h6) = (h9 – h5) - m(h9 – h5)
m(h2 – h6) - m(h9 – h5) = h9 - h5
β„Žβ‚‰−β„Žβ‚…
m = (β„Žβ‚‚−β„Žβ‚†)+(β„Žβ‚‰−β„Žβ‚…)
m=
𝐾𝐽
π‘˜π‘”
(1160.6595−261.5517)
𝐾𝐽
𝐾𝐽
+(1160.6595−261.5517)
π‘˜π‘”
π‘˜π‘”
(3095.1803−1154.23)
m = 0.3166
For WT,
WT = (h1 – h2) + (1-m)(h2 – h3)
𝐾𝐽
𝐾𝐽
WT = (3373.7-3095.1803) π‘˜π‘” + (1 – 0.3166) (3095.1803-2172.4712) π‘˜π‘”
𝐾𝐽
WT = 909.0991 π‘˜π‘”
For WPtotal,
WPtotal = WP1 + WP2
𝐾𝐽
WPtotal = (10.1517+6.4295) π‘˜π‘”
𝐾𝐽
WPtotal = 16.5812 π‘˜π‘”
61
For πœ‚TH’,
πœ‚TH’ =
π‘Šπ‘›π‘’π‘‘ ′
𝑄𝐴′
=
π‘Šπ‘1
WP1’ = πœ‚π‘ =
π‘Šπ‘‡ ′ −π‘Šπ‘ƒπ‘‘π‘œπ‘‘π‘Žπ‘™′
𝑄𝐴′
𝐾𝐽
π‘˜π‘”
10.1517
π‘Šπ‘2
WP2’ = πœ‚π‘ =
0.85
𝐾𝐽
𝐾𝐽
π‘˜π‘”
6.4295
0.85
𝐾𝐽
WP1’ = 11.9432 π‘˜π‘”
WP2’ =7.5641 π‘˜π‘”
h5’ = h4 + Wp1’
Since h7’ = h8’ = h9’ in closed FWH,
h9’ = h7’ = h6 + Wp1’
𝐾𝐽
𝐾𝐽
h5’ = 251.40 π‘˜π‘” + 11.9432π‘˜π‘”
𝐾𝐽
𝐾𝐽
h9’ = 1154.23 π‘˜π‘” + 7.5641π‘˜π‘”
𝐾𝐽
h5’ = 263.3432 π‘˜π‘”
𝐾𝐽
h9’ = 1161.7941 π‘˜π‘”
β„Žβ‚‰′−β„Žβ‚…′
m’ = (β„Žβ‚‚′−β„Žβ‚†)+(β„Žβ‚‰′−β„Žβ‚…′)
m’ =
𝐾𝐽
π‘˜π‘”
(1161.7941−263.3432)
𝐾𝐽
𝐾𝐽
+(1161.7941−263.3432)
π‘˜π‘”
π‘˜π‘”
(3136.9583−1154.23)
m’ = 0.3118
For WT’,
WT’ = (h1 – h2’) + (1-m’)(h2’ – h3’)
𝐾𝐽
𝐾𝐽
WT’ = (3373.7-3136.9583) π‘˜π‘” + (1 – 0.3118) (3136.9483-2352.6555) π‘˜π‘”
𝐾𝐽
WT’ = 776.4989 π‘˜π‘”
WPtotal’ = WP1’ + WP2’
𝐾𝐽
WPtotal’ = (11.9432+7.7641) π‘˜π‘”
𝐾𝐽
WPtotal’ = 19.5073 π‘˜π‘”
QA’ = h1 - h7’
𝐾𝐽
QA’ = (3373.7-1161.7941) π‘˜π‘”
𝐾𝐽
QA’ = 2211.9059π‘˜π‘”
πœ‚TH’ =
π‘Šπ‘›π‘’π‘‘ ′
𝑄𝐴′
=
1. πœ‚TH’ =
π‘Šπ‘‡ ′ −π‘Šπ‘ƒπ‘‘π‘œπ‘‘π‘Žπ‘™′
𝑄𝐴′
𝐾𝐽
π‘˜π‘”
(776.4989−19.5073)
2211.9059
𝐾𝐽
π‘˜π‘”
πœ‚TH’ = 0.3422
πœ‚TH’ = 34.22%
62
TS DIAGRAM
T1
P1 = P7 = 10 MPa
P2 = P6 = 5 MPa
P3 = P4 = 0.020 MPa
x3
S1 = S2 = S3
x3’
63
2. Consider a steam power plant on a regenerative Rankine cycle with a closed feed water heater.
The turbine inlet conditions are set at 9 MPa and 480 °C, the condenser operates at 15 kPa.
Steam is extracted at 4 MPa to the closed feed water heater the expanded to condenser pressure
before entering the condenser. Determine the turbine work, pump work, and the actual thermal
efficiency, assuming both the turbine and pump have an efficiency of 88%.
Given:
P1 = 9 MPa
T1 = 480 °C
P2 = 4 MPa
πœ‚T = 88%
P3 = 15 kPa
Closed
FWH
(1-m)
πœ‚P = 88%
Required:
(a) WT
(b) WPtotal
(c) πœ‚TH’
Solution:
οƒ˜ State 1
P1 = 9 MPa
T1 = 480 °C
tsat = 303.40 °C
tsat < T1; superheated vapor
οƒ˜ State 2State 2
𝐾𝐽
S1 = S2 = 6.5906π‘˜π‘”βˆ™πΎ
P2 = 4 MPa
𝐾𝐽
h1 = 3335.0 π‘˜π‘”
𝐾𝐽
s1 = 6.5906 π‘˜π‘”βˆ™πΎ
S (π‘˜π‘”βˆ™πΎ)
𝐾𝐽
h (π‘˜π‘”)
𝐾𝐽
6.5821
S2 = 6.5906
6.6215
3092.5
h2
3117.2
64
By interpolation,
𝐾𝐽
𝐾𝐽
β„Žβ‚‚ − 3092.5
π‘˜π‘” βˆ™ 𝐾
π‘˜π‘”
=
𝐾𝐽
𝐾𝐽
(6.6215 − 6.5821)
(3117.2 − 3092.5)
π‘˜π‘” βˆ™ 𝐾
π‘˜π‘”
𝐾𝐽
h2 = 3126.2893 π‘˜π‘”
(6.5821 − 6.5906)
οƒ˜ State 3
For x3,
𝐾𝐽
οƒ˜ State 3’
S₃− Sf₃
X3 = 𝑆𝑓𝑔₃
s2 = s3 = 6.5906 π‘˜π‘”βˆ™πΎ
P3 = 0.015 MPa
β„Žβ‚−β„Žβ‚ƒ′
πœ‚T = β„Žβ‚−β„Žβ‚ƒ
(6.5906−0.7549 )
sg3 = 8.0085 π‘˜π‘”βˆ™πΎ
X3 =
sg3 > s3; mixture
X3 = 0.8045
𝐾𝐽
𝐾𝐽
π‘˜π‘”βˆ™πΎ
0.88 =
𝐾𝐽
7.2536 π‘˜π‘”βˆ™πΎ
3335.0
𝐾𝐽
−β„Žβ‚ƒ′
π‘˜π‘”
𝐾𝐽
𝐾𝐽
− 2135.0990
π‘˜π‘”
π‘˜π‘”
3335.0
𝐾𝐽
h3’ =2279.0871 π‘˜π‘”
X3 = 80.45 %
h3 = hf3 + x3 (hfg3)
𝐾𝐽
𝐾𝐽
h3 = 225.94π‘˜π‘” + (0.8045) (2373.1π‘˜π‘”)
𝐾𝐽
h3 = 2135.0990π‘˜π‘”
οƒ˜ State 4
P3 = P4 =0.015 MPa MPa
𝐾𝐽
hf3 = h4 = 225.94 π‘˜π‘”
𝜐f3 = 𝜐4 =1.0141 × 10-3
π‘š³
π‘˜π‘”
οƒ˜ State 5
P5 = P9 = P8 = P7 = P7 = 9 MPa
WP1 = 𝜐4 (P5 – P4)
π‘š³
WP1 = 1.0141 × 10-3 π‘˜π‘” (9 MPa – 0.015 MPa)
WP1
π‘š³
1000 πΎπ‘ƒπ‘Ž
= 1.0141 × 10-3
(8.985 MPa ×
×
π‘˜π‘”
1 π‘€π‘ƒπ‘Ž
1000
𝑁
π‘š²
1 πΎπ‘ƒπ‘Ž
1𝐽
1 𝐾𝐽
× 1 π‘βˆ™π‘š × 1000 𝐽 )
𝐾𝐽
WP1 = 9.1117 π‘˜π‘”
h5= h4 + Wp
𝐾𝐽
𝐾𝐽
h5 = 225.94 π‘˜π‘” + 9.1117 π‘˜π‘”
𝐾𝐽
h5 = 235.0517 π‘˜π‘”
οƒ˜ State 6
P2 = P6 = 4 MPa
π‘š³
𝜐f2 = 𝜐6 = 1.2522 × 10-3 π‘˜π‘”
65
WP2 = 𝜐6 (P7 – P6)
π‘š³
WP2 = 1.2522 × 10-3 π‘˜π‘” (9 MPa – 4 MPa)
π‘š³
WP2 = 1.2522 × 10-3 π‘˜π‘” (5 MPa ×
1000 πΎπ‘ƒπ‘Ž
×
1 π‘€π‘ƒπ‘Ž
1000
𝑁
π‘š²
1 πΎπ‘ƒπ‘Ž
1𝐽
1 𝐾𝐽
× 1 π‘βˆ™π‘š × 1000 𝐽 )
𝐾𝐽
WP2 = 6.261 π‘˜π‘”
οƒ˜ State 7
h7= h6 + WP2
𝐾𝐽
𝐾𝐽
h7 = 1087.31 π‘˜π‘” + 6.261 π‘˜π‘”
𝐾𝐽
h7 = 1093.571 π‘˜π‘”
Two fluids steam which are being mixed have the same enthalpy (h). Therefore, h7 = h8 = h9 =
1093.571
𝐾𝐽
π‘˜π‘”
.
For m,
(m)(h2)
ENERGY BALANCE
H9
(1-m)
(1-m)(h5)
(m)(h6)
Ein = Eout
(m)(h2) + (1-m)(h5) = mh6 + (1-m)(h9)
(m)(h2) - mh6 = (1-m) (h9 - h5)
m(h2 – h6) = (h9 – h5) - m(h9 – h5)
m(h2 – h6) - m(h9 – h5) = h9 - h5
β„Žβ‚‰−β„Žβ‚…
m = (β„Žβ‚‚−β„Žβ‚†)+(β„Žβ‚‰−β„Žβ‚…)
m=
𝐾𝐽
π‘˜π‘”
(1093.571−235.0517)
𝐾𝐽
𝐾𝐽
+(1093.571−235.0517)
π‘˜π‘”
π‘˜π‘”
(3097.8287−1087.31)
m = 0.2992
For WT,
WT = (h1 – h2) + (1-m)(h2 – h3)
𝐾𝐽
𝐾𝐽
WT = (3335.0-3097.8287) π‘˜π‘” + (1 – 0.2992) (3097.9523-2135.0990) π‘˜π‘”
𝐾𝐽
WT = 911.8523 π‘˜π‘”
For WPtotal,
WPtotal = WP1 + WP2
𝐾𝐽
WPtotal = (9.1117+6.261) π‘˜π‘”
𝐾𝐽
WPtotal = 15.3727 π‘˜π‘”
66
For πœ‚TH’,
πœ‚TH’ =
π‘Šπ‘›π‘’π‘‘ ′
𝑄𝐴′
=
π‘Šπ‘1
WP1’ = πœ‚π‘ =
π‘Šπ‘‡ ′ −π‘Šπ‘ƒπ‘‘π‘œπ‘‘π‘Žπ‘™′
𝑄𝐴′
𝐾𝐽
π‘˜π‘”
9.1117
π‘Šπ‘2
WP2’ = πœ‚π‘ =
0.88
𝐾𝐽
𝐾𝐽
π‘˜π‘”
6.261
0.88
𝐾𝐽
WP1’ = 10.3542 π‘˜π‘”
WP2’ =7.1148 π‘˜π‘”
h5’ = h4 + Wp1’
Since h7’ = h8’ = h9’ in closed FWH,
h9’ = h7’ = h6 + Wp1’
𝐾𝐽
𝐾𝐽
h5’ = 225.94 π‘˜π‘” + 10.3542π‘˜π‘”
𝐾𝐽
𝐾𝐽
h9’ = 1087.31 π‘˜π‘” + 7.1148π‘˜π‘”
𝐾𝐽
h5’ = 236.2942 π‘˜π‘”
𝐾𝐽
h9’ = 1094.4248 π‘˜π‘”
β„Žβ‚‰′−β„Žβ‚…′
m’ = (β„Žβ‚‚′−β„Žβ‚†)+(β„Žβ‚‰′−β„Žβ‚…′)
m’ =
𝐾𝐽
π‘˜π‘”
(1094.4248−236.2942)
𝐾𝐽
𝐾𝐽
+(1094.4248−236.2942)
π‘˜π‘”
π‘˜π‘”
(3126.2893−1087.31)
m’ = 0.2873
For WT’,
WT’ = (h1 – h2’) + (1-m’)(h2’ – h3’)
𝐾𝐽
𝐾𝐽
WT’ = (3335.0-3126.2893) π‘˜π‘” + (1 – 0.2873) (3126.2893-2279.0871) π‘˜π‘”
𝐾𝐽
WT’ = 812.5117 π‘˜π‘”
WPtotal’ = WP1’ + WP2’
𝐾𝐽
WPtotal’ = (10.3542+7.1148) π‘˜π‘”
𝐾𝐽
WPtotal’ = 17.469 π‘˜π‘”
QA’ = h1 - h7’
𝐾𝐽
QA’ = (3335.0-1094.4248) π‘˜π‘”
𝐾𝐽
QA’ = 2240.5752π‘˜π‘”
πœ‚TH’ =
πœ‚TH’ =
π‘Šπ‘›π‘’π‘‘ ′
𝑄𝐴′
=
π‘Šπ‘‡ ′ −π‘Šπ‘ƒπ‘‘π‘œπ‘‘π‘Žπ‘™′
𝑄𝐴′
𝐾𝐽
π‘˜π‘”
(812.5117−17.469)
2240.5752
𝐾𝐽
π‘˜π‘”
πœ‚TH’ = 0.3548
πœ‚TH’ = 35.48%
67
TS DIAGRAM
T1
P1 = P7 = 9 MPa
P2 = P6 = 4 MPa
P3 = P4 = 0.015 MPa
x3
S1 = S2 = S3
x3’
68
3. Investigate a steam power plant operating on a regenerative Rankine cycle with a closed feed
water heater. The turbine inlet conditions are set at 12 MPa and 540 °C, and the condenser
operates at 30 kPA. Steam is extracted at 7 MPa to the closed feed water heater, and then
expanded to condenser pressure before entering the condenser. Determine the fraction of steam
extracted, the ideal thermal efficiency, and the actual thermal efficiency, assuming both the
turbine and pump operate at 87% efficiency.
Given:
P1 = 12 MPa
T1 = 540 °C
P2 = 7 MPa
πœ‚T = 87%
P3 = 30 kPa
Closed
FWH
(1-m)
πœ‚P = 87%
Required:
(a) m
(b) πœ‚TH’
(c) πœ‚TH’
Solution:
οƒ˜ State 1
P1 = 12 MPa
T1 =540 °C
tsat = 324.75 °C
tsat < T1; superheated vapor
οƒ˜ State 2State 2
S1 = S2 = 6.6209
P2 = 4 MPa
𝐾𝐽
h1 = 3454.4 π‘˜π‘”
𝐾𝐽
s1 = 6.6209π‘˜π‘”βˆ™πΎ
S (π‘˜π‘”βˆ™πΎ)
𝐾𝐽
h (π‘˜π‘”)
𝐾𝐽
6.5821
S2 = 6.6209
6.6215
3092.5
h2
3117.2
69
οƒ˜ State 2’
By interpolation,
β„Žβ‚−β„Žβ‚ƒ′
πœ‚T = β„Žβ‚−β„Žβ‚ƒ
𝐾𝐽
𝐾𝐽
β„Žβ‚‚ − 3092.5
π‘˜π‘” βˆ™ 𝐾
π‘˜π‘”
=
𝐾𝐽
𝐾𝐽
(6.6215 − 6.5821)
(3117.2 − 3092.5)
π‘˜π‘” βˆ™ 𝐾
π‘˜π‘”
𝐾𝐽
h2 = 3116.3223π‘˜π‘”
(6.6201 − 6.5821)
οƒ˜ State 3
0.87=
P3 = 0.03 MPa
οƒ˜ State 3’
β„Žβ‚−β„Žβ‚ƒ′
πœ‚T = β„Žβ‚−β„Žβ‚ƒ
(6.6209−0.9439 )
sg3 = 7.7686 π‘˜π‘”βˆ™πΎ
X3 =
sg3 > s3; mixture
X3 = 0.8318
𝐾𝐽
𝐾𝐽
𝐾𝐽
− 3116.3223
π‘˜π‘”
π‘˜π‘”
3454.4
𝐾𝐽
S₃− Sf₃
X3 = 𝑆𝑓𝑔₃
s2 = s3 = 6209 π‘˜π‘”βˆ™πΎ
𝐾𝐽
−β„Žβ‚ƒ′
π‘˜π‘”
h3’ =3160.2730 π‘˜π‘”
For x3,
𝐾𝐽
3454.4
𝐾𝐽
π‘˜π‘”βˆ™πΎ
0.87 =
𝐾𝐽
6.8247 π‘˜π‘”βˆ™πΎ
3454.4
𝐾𝐽
−β„Žβ‚ƒ′
π‘˜π‘”
𝐾𝐽
𝐾𝐽
− 2232.3980
π‘˜π‘”
π‘˜π‘”
3454.4
𝐾𝐽
h3’ =2391.2583 π‘˜π‘”
X3 = 83.18 %
h3 = hf3 + x3 (hfg3)
For x3’,
𝐾𝐽
h3 = 289.23 + (0.8218) (2336.1π‘˜π‘”)
X3’ =
𝐾𝐽
h3 = 2232.3980π‘˜π‘”
X3’ =
οƒ˜ State 4
P3 = P4 =0.03 MPa MPa
h₃′− hf₃
β„Žπ‘“π‘”β‚ƒ
(2391.2583−289.23 )
𝐾𝐽
π‘˜π‘”
𝐾𝐽
2336.1π‘˜π‘”
X3’ = 0.8998
𝐾𝐽
hf3 = h4 = 289.23 π‘˜π‘”
X3’ = 89.98 %
π‘š³
𝜐f3 = 𝜐4 =1.0223 × 10-3 π‘˜π‘”
οƒ˜ State 5
P5 = P9 = P8 = P7 = P7 = 12 MPa
WP1 = 𝜐4 (P5 – P4)
π‘š³
WP1 = 1.0223 × 10-3 π‘˜π‘” (12 MPa – 0.03 MPa)
π‘š³
WP1 = 1.0223 × 10-3 π‘˜π‘” (11.97 MPa ×
WP1 = 12.2369
1000 πΎπ‘ƒπ‘Ž
×
1 π‘€π‘ƒπ‘Ž
1000
𝑁
π‘š²
1 πΎπ‘ƒπ‘Ž
1𝐽
1 𝐾𝐽
× 1 π‘βˆ™π‘š × 1000 𝐽 )
𝐾𝐽
π‘˜π‘”
h5= h4 + Wp
𝐾𝐽
𝐾𝐽
h5 = 289.23 π‘˜π‘” + 12.2369 π‘˜π‘”
𝐾𝐽
h5 = 301.4669 π‘˜π‘”
οƒ˜ State 6
P2 = P6 = 7 MPa
π‘š³
𝜐f2 = 𝜐6 = 1.3513 × 10-3 π‘˜π‘”
70
WP2 = 𝜐6 (P7 – P6)
π‘š³
WP2 = 1.3513 × 10-3 π‘˜π‘” (12 MPa – 7 MPa)
WP2
π‘š³
1000 πΎπ‘ƒπ‘Ž
= 1.3513 × 10-3
(5 MPa ×
×
π‘˜π‘”
1 π‘€π‘ƒπ‘Ž
1000
𝑁
π‘š²
1 πΎπ‘ƒπ‘Ž
1𝐽
1 𝐾𝐽
× 1 π‘βˆ™π‘š × 1000 𝐽 )
𝐾𝐽
WP2 = 6.7565 π‘˜π‘”
οƒ˜ State 7
h7= h6 + WP2
𝐾𝐽
𝐾𝐽
h7 = 1267.0 π‘˜π‘” + 6.7565 π‘˜π‘”
𝐾𝐽
h7 = 1273.7565 π‘˜π‘”
Two fluids steam which are being mixed have the same enthalpy (h). Therefore, h7 = h8 = h9 =
𝐾𝐽
1273.7565 π‘˜π‘”.
For m,
(m)(h2)
ENERGY BALANCE
H9
(1-m)
(1-m)(h5)
(m)(h6)
Ein = Eout
(m)(h2) + (1-m)(h5) = mh6 + (1-m)(h9)
(m)(h2) - mh6 = (1-m) (h9 - h5)
m(h2 – h6) = (h9 – h5) - m(h9 – h5)
m(h2 – h6) - m(h9 – h5) = h9 - h5
β„Žβ‚‰−β„Žβ‚…
m = (β„Žβ‚‚−β„Žβ‚†)+(β„Žβ‚‰−β„Žβ‚…)
m=
𝐾𝐽
π‘˜π‘”
(1273.7565−301.4669)
𝐾𝐽
𝐾𝐽
+(1273.7565−301.4669)
π‘˜π‘”
π‘˜π‘”
(3116.3223−1267.0)
m = 0.3446
For WT,
WT = (h1 – h2) + (1-m)(h2 – h3)
𝐾𝐽
𝐾𝐽
WT = (3454.6-3116.3223) π‘˜π‘” + (1 – 0.3446 (3116.3223-2232.3980) π‘˜π‘”
𝐾𝐽
WT = 917.6017 π‘˜π‘”
For WPtotal,
WPtotal = WP1 + WP2
𝐾𝐽
WPtotal = (12.2369+6.7565) π‘˜π‘”
𝐾𝐽
WPtotal = 18.9934 π‘˜π‘”
71
For πœ‚TH,
π‘Šπ‘›π‘’π‘‘
πœ‚TH =
π‘Šπ‘‡ −π‘Šπ‘ƒπ‘‘π‘œπ‘‘π‘Žπ‘™
=
𝑄𝐴
𝑄𝐴
(911.6017−18.9934)
πœ‚TH =
2180.6435
𝐾𝐽
π‘˜π‘”
𝐾𝐽
π‘˜π‘”
πœ‚TH = 0.4093
πœ‚TH = 40.93%
For πœ‚TH’,
πœ‚TH’ =
π‘Šπ‘›π‘’π‘‘ ′
𝑄𝐴′
π‘Šπ‘‡ ′ −π‘Šπ‘ƒπ‘‘π‘œπ‘‘π‘Žπ‘™′
=
π‘Šπ‘1
WP1’ = πœ‚π‘ =
𝑄𝐴′
𝐾𝐽
π‘˜π‘”
12.2369
π‘Šπ‘2
WP2’ = πœ‚π‘ =
0.87
𝐾𝐽
𝐾𝐽
π‘˜π‘”
6.7565
0.87
𝐾𝐽
WP2’ =7.7660 π‘˜π‘”
WP1’ = 14.0654 π‘˜π‘”
Since h7’ = h8’ = h9’ in closed FWH,
h9’ = h7’ = h6 + Wp1’
h5’ = h4 + Wp1’
𝐾𝐽
𝐾𝐽
h5’ = 225.94 π‘˜π‘” + 14.0654π‘˜π‘”
𝐾𝐽
𝐾𝐽
h9’ = 1267.0 π‘˜π‘” + 7.7660π‘˜π‘”
𝐾𝐽
h5’ = 240.0054 π‘˜π‘”
𝐾𝐽
h9’ = 1274.766 π‘˜π‘”
β„Žβ‚‰′−β„Žβ‚…′
m’ = (β„Žβ‚‚′−β„Žβ‚†)+(β„Žβ‚‰′−β„Žβ‚…′)
m’ =
𝐾𝐽
π‘˜π‘”
(1274.766−240.0054)
𝐾𝐽
𝐾𝐽
+(1274.766−240.0054)
π‘˜π‘”
π‘˜π‘”
(3160.2730−1267.0)
m’ = 0.3534
For WT’,
WT’ = (h1 – h2’) + (1-m’)(h2’ – h3’)
𝐾𝐽
𝐾𝐽
WT’ = (3454.4 − 3160.2730) π‘˜π‘” + (1 – 0.3534) (3160.2730-2391.2583) π‘˜π‘”
𝐾𝐽
WT’ = 791.3719 π‘˜π‘”
WPtotal’ = WP1’ + WP2’
𝐾𝐽
WPtotal’ = (14.0654+7.7660) π‘˜π‘”
𝐾𝐽
WPtotal’ = 21.8314 π‘˜π‘”
QA’ = h1 - h7’
𝐾𝐽
QA’ = (3454.4-1274.766) π‘˜π‘”
𝐾𝐽
QA’ = 2179.634π‘˜π‘”
72
πœ‚TH’ =
πœ‚TH’ =
π‘Šπ‘›π‘’π‘‘ ′
𝑄𝐴′
=
π‘Šπ‘‡ ′ −π‘Šπ‘ƒπ‘‘π‘œπ‘‘π‘Žπ‘™′
𝑄𝐴′
𝐾𝐽
π‘˜π‘”
(791.3719−21.8314)
𝐾𝐽
π‘˜π‘”
2179.634
πœ‚TH’ = 0.3531
πœ‚TH’ = 35.31%
TS DIAGRAM
T1
P1 = P7 = 12 MPa
P2 = P6 = 7 MPa
P3 = P4 = 0.03 MPa
x3
S1 = S2 = S3
x3’
73
REHEAT-REGENERATIVE RANKINE CYCLE
1. Examine a steam that enters a turbine at 20 MPa and 640 °C and is condensed in the condenser
at a pressure of 15 kPa. Some of the steam is extracted from the turbine at 5 MPa for a closed
feed water heater and the remaining steam is reheated at the same pressure to 640 °C. The
extracted steam is completely condensed in the heater and is pumped to 20 MPa before it mixes
with the feed water at the same pressure. Steam for an open feed water heater is extracted from
the low pressure turbine at a pressure of 0.6 MPa. Determine the fraction of steam extracted
from the turbine as well as the actual work done of the pump considering the turbine and pump
have efficiencies of 90%.
Given:
P1 = 20 MPa
T1 = 640 °C
P2 = 5 MPa
(1-m1)
P3 = 5 MPa
πœ‚T = 90%
P4 = 0.6 MPa
(1-m1-m2)
m2
(1-m1)
P6 = 0.015 MPa
m1
πœ‚p = 90%
Required:
a) m1, m2
b) WPtotal’
Solution:
οƒ˜ State 1
P1 = 20 MPa
T1 = 640 °C
tsat = 365.81 °C; tsat < T1; superheated vapor
οƒ˜ State 2State 2
𝐾𝐽
S1 = S2 = 6.6286π‘˜π‘”βˆ™πΎ
P2 = 5 MPa
𝐾𝐽
h1 = 3648.1 π‘˜π‘”
𝐾𝐽
s1 = 6.6286 π‘˜π‘”βˆ™πΎ
S (π‘˜π‘”βˆ™πΎ)
𝐾𝐽
h (π‘˜π‘”)
𝐾𝐽
6.6089
S2 = 6.6286
6.6459
3171.0
h2
3195.7
74
By interpolation,
𝐾𝐽
𝐾𝐽
β„Žβ‚‚ − 3171.0
π‘˜π‘” βˆ™ 𝐾
π‘˜π‘”
=
𝐾𝐽
𝐾𝐽
(3195.7 − 3171.0)
(6.6459 − 6.6089)
π‘˜π‘” βˆ™ 𝐾
π‘˜π‘”
𝐾𝐽
h2 = 3184.1511π‘˜π‘”
(6.6286 − 6.6089)
οƒ˜ State 3
P2 = P3 = 5 MPa
T3 = 640 °C
tsat = 263.99 °C; tsat < T1; superheated vapor
𝐾𝐽
h3 = 3759.6 π‘˜π‘”
𝐾𝐽
s3 = 7.3632 π‘˜π‘”βˆ™πΎ
οƒ˜ State 4
𝐾𝐽
S3 = S4 = 7.3632π‘˜π‘”βˆ™πΎ
P4 =0.6 MPa
𝐾𝐽
sg4 = 6.76.. π‘˜π‘”βˆ™πΎ
sg4 < s4; superheated vapor
S (π‘˜π‘”βˆ™πΎ)
𝐾𝐽
h (π‘˜π‘”)
𝐾𝐽
7.3357
S4 = 7.3632
7.3724
3040.8
h4
3061.6
𝐾𝐽
h4 = 3056.3858π‘˜π‘”
οƒ˜ State 5
𝐾𝐽
S3 = S4 = S5 = 7.3632π‘˜π‘”βˆ™πΎ
P5 =0.015 MPa
𝐾𝐽
sg5 = 8.0085 π‘˜π‘”βˆ™πΎ
h5 = hf5 + x5 (hfg5)
𝐾𝐽
𝐾𝐽
h5 = 225.94 π‘˜π‘”+ (0.9110) (2373.1π‘˜π‘”)
𝐾𝐽
h5 = 2387.8341π‘˜π‘”
sg4 > s4; mixture
S₃− Sf₃
X5 = 𝑆𝑓𝑔₃
X5 =
(7.3632−0.7549 )
𝐾𝐽
π‘˜π‘”βˆ™πΎ
𝐾𝐽
7.2536 π‘˜π‘”βˆ™πΎ
X5 = 0.9110
X5 = 91.10 %
οƒ˜ State 6
P5 = P6 = 0.015 MPa
π‘š³
𝜐f5 = 𝜐5 = 1.0141 × 10-3 π‘˜π‘”
𝐾𝐽
hf5 = h5 = 225.94π‘˜π‘”
75
οƒ˜ State 7
P4 = P7 = 0.6 MPa
WP1 = 𝜐6 (P7 – P6)
π‘š³
WP1 = 1.0.41 × 10-3 π‘˜π‘” (0.6 MPa – 0.015 MPa)
WP1
π‘š³
1000 πΎπ‘ƒπ‘Ž
= 1.0141 × 10-3
(0.585 MPa ×
×
π‘˜π‘”
1 π‘€π‘ƒπ‘Ž
1000
𝑁
π‘š²
1 πΎπ‘ƒπ‘Ž
1𝐽
1 𝐾𝐽
× 1 π‘βˆ™π‘š × 1000 𝐽 )
𝐾𝐽
WP1 = 0.5932 π‘˜π‘”
h7 = h6 + Wp1
𝐾𝐽
𝐾𝐽
h7 = 225.94 π‘˜π‘” + 0.5932π‘˜π‘”
𝐾𝐽
h7 = 226.5332 π‘˜π‘”
οƒ˜ State 8
P4 = P7 = P8 = 0.6MPa
π‘š³
𝜐f4 = 𝜐8 = 1.1006 × 10-3 π‘˜π‘”
𝐾𝐽
hf4 = h8 = 670.56 π‘˜π‘”
οƒ˜ State 9
P1 = P9 = 20 MPa
WP2 = 𝜐8 (P9 – P8)
π‘š³
WP2 = 1.1006× 10-3 π‘˜π‘” (20 MPa – 0.6 MPa)
π‘š³
WP2 = 1.1006 × 10-3 π‘˜π‘” (19.4 MPa ×
1000 πΎπ‘ƒπ‘Ž
×
1 π‘€π‘ƒπ‘Ž
1000
𝑁
π‘š²
1 πΎπ‘ƒπ‘Ž
1𝐽
1 𝐾𝐽
× 1 π‘βˆ™π‘š × 1000 𝐽 )
𝐾𝐽
WP2 = 21.3516 π‘˜π‘”
h9 = h8 + Wp2
𝐾𝐽
𝐾𝐽
h9 = 670.56 π‘˜π‘” + 21.3516π‘˜π‘”
𝐾𝐽
h9 = 391.9116 π‘˜π‘”
οƒ˜ State 10
P9 = P10 = 20MPa
h10 = h12
οƒ˜ State 11
P2 = P11 = 5MPa
π‘š³
𝜐f2 = 𝜐11 = 1.2859 × 10-3 π‘˜π‘”
𝐾𝐽
hf2 = h11 = 1154.23 π‘˜π‘”
76
οƒ˜ State 12
P10 = P12 = 20 MPa
WP3 = 𝜐11 (P12 – P11)
π‘š³
WP3 = 1.2859× 10-3 π‘˜π‘” (20 MPa – 5 MPa)
WP3
π‘š³
1000 πΎπ‘ƒπ‘Ž
= 1.2859 × 10-3
(15 MPa ×
×
π‘˜π‘”
1000
1 π‘€π‘ƒπ‘Ž
𝑁
π‘š²
1 πΎπ‘ƒπ‘Ž
1𝐽
1 𝐾𝐽
× 1 π‘βˆ™π‘š × 1000 𝐽 )
𝐾𝐽
WP3 = 19.2885 π‘˜π‘”
h12 = h11 + Wp3
𝐾𝐽
𝐾𝐽
h12 = 1154.23 π‘˜π‘” + 19.2885π‘˜π‘”
𝐾𝐽
h12 = 1173.5185 π‘˜π‘”
οƒ˜ State 13
P12 = P13 = 20MPa
𝐾𝐽
h10 = h12 = h10 =1173.5185π‘˜π‘”
For m1,
(1-m)
h10
m1 =
m1 =
Ein = Eout
(m1)(h2) + (1-m1)(h9) = m1h6 + (1-m1)(h10)
(m1)(h2) – h9 – m1h9 = m1h11 + h10 – m1h10
m1h2 – m1h9 - m1h11 + m1h10 = h10 – h9
m1[(h2 – h11) + (h10 – h9)] = h10 – h9
(m1)(h2)
h2
(1-m1)
h9
h11
m1
h₁₀ – h₉
(hβ‚‚ – h₁₁) + (h₁₀ – h₉)
(1173.5185−691.9116)
(3184.1511−1154.28)
𝐾𝐽
π‘˜π‘”
𝐾𝐽
𝐾𝐽
+(1173.5185−391.9116)
π‘˜π‘”
π‘˜π‘”
m1 = 0.1918
For m2,
h4
m2
(1-m1-m2)
h7
h8
Ein = Eout
(m2)(h4) + (1-m1-m2)(h7) = (1-m1)(h8)
(m2)(h4) – h7 – m1h7 – m2h7 = h8 – m1h8
m2(h4 – h7) + (1-m1)(h7) = (1-m1)(h8)
m2(h4 – h7) = (1-m1)(h8-h7)
(1-m1)
77
m2 =
m2 =
(1−m1)(h8−h7)
h4 – h7
𝐾𝐽
π‘˜π‘”
(1−0.1918)(670.56−226.5332)
(3056.3858−226.5332)
𝐾𝐽
π‘˜π‘”
m2 = 0.1268
QAR = (1-m1) (h3 – h2)
QAR = (1-0.1918)(3759.6-3184.1511)
QAR = 465.0778
𝐾𝐽
π‘˜π‘”
𝐾𝐽
π‘˜π‘”
QAB = h1 – h3
QAB = (3648.1-1173.5185)
QAB = 2474.5815
𝐾𝐽
π‘˜π‘”
𝐾𝐽
π‘˜π‘”
QA = QAB + QAR
QA = (2474.5815 + 465.0778)
QA = 2939.6593
𝐾𝐽
π‘˜π‘”
𝐾𝐽
π‘˜π‘”
For WPtotal’,
WPtotal’= Wp1’ + Wp2’ + Wp3’
WPtotal’=
WPtotal’=
π‘Šπ‘1
πœ‚β‚š
+
π‘Šπ‘2
𝐾𝐽
0.5932
π‘˜π‘”
0.90
πœ‚β‚š
+
WPtotal’= 45.8148
+
π‘Šπ‘3
πœ‚β‚š
𝐾𝐽
π‘˜π‘”
21.3516
0.90
+
19.2885
𝐾𝐽
π‘˜π‘”
0.90
𝐾𝐽
π‘˜π‘”
TS DIAGRAM
78
79
2. A steam enters a turbine at 10 MPa and 620 °C and is subsequently condensed in the condenser
at a pressure of 8 kPa. A portion of the steam is extracted from the turbine at 2.5 MPa for used
in a closed feed water heater while the remaining steam is reheated at the same pressure to 620
°C. The extracted steam is completely condensed in the heater and is pumped to 10 MPa before
mixing with the feed water at the same pressure. Additionally, steam is extracted from the low
pressure turbine at a pressure of 0.7 MPa for an open feed water heater. Determine the ideal
thermal efficiency as well as the actual work done of the pump, assuming that the turbine and
pump have efficiencies of 88%.
Given:
P1 = 10 MPa
T1 = 620 °C
P2 = 2.5 MPa
(1-m1)
P3 = 2.5 MPa
πœ‚T = 88%
P4 = 0.7 MPa
(1-m1-m2)
m2
(1-m1)
P6 = 0.08 MPa
m1
πœ‚p = 88%
Required:
a) πœ‚TH
b) WPtotal’
Solution:
οƒ˜ State 1
P1 =10 MPa
T1 = 620 °C
tsat = 311.06 °C; tsat < T1; superheated vapor
οƒ˜ State 2State 2
𝐾𝐽
S1 = S2 = 6.9587π‘˜π‘”βˆ™πΎ
P2 = 2.5 MPa
𝐾𝐽
h1 = 3674.6 π‘˜π‘”
𝐾𝐽
s1 = 6.9587π‘˜π‘”βˆ™πΎ
S (π‘˜π‘”βˆ™πΎ)
𝐾𝐽
h (π‘˜π‘”)
𝐾𝐽
6.9471
S2 = 6.9587
6.9813
3194.4
h2
3216.9
𝐾𝐽
h2 = 3202.0316π‘˜π‘”
80
οƒ˜ State 3
P2 = P3 = 2.5 MPa
T3 = 620 °C
tsat = 223.99 °C; tsat < T1; superheated vapor
𝐾𝐽
h3 = 3731.5 π‘˜π‘”
𝐾𝐽
s3 = 7.6473 π‘˜π‘”βˆ™πΎ
οƒ˜ State 4
𝐾𝐽
S3 = S4 = 7.6473π‘˜π‘”βˆ™πΎ
P4 =0.7 MPa
𝐾𝐽
sg4 = 6.7080 π‘˜π‘”βˆ™πΎ
sg4 < s4; superheated vapor
S (π‘˜π‘”βˆ™πΎ)
𝐾𝐽
h (π‘˜π‘”)
𝐾𝐽
7.6350
3268
S4 = 7.6473
h4
7.6968
3310.9
𝐾𝐽
h4 = 3277.0990 π‘˜π‘”
οƒ˜ State 5
𝐾𝐽
S3 = S4 = S5 = 7.64732π‘˜π‘”βˆ™πΎ
P5 =0.008 MPa
𝐾𝐽
sg5 = 8.2287 π‘˜π‘”βˆ™πΎ
h5 = hf5 + x5 (hfg5)
𝐾𝐽
𝐾𝐽
h5 = 173.88 π‘˜π‘”+ (0.9239) (2403.1π‘˜π‘”)
𝐾𝐽
h5 = 2387.8341π‘˜π‘”
sg4 > s4; mixture
S₃− Sf₃
X5 = 𝑆𝑓𝑔₃
X5 =
(7.6473−0.5926)
𝐾𝐽
π‘˜π‘”βˆ™πΎ
𝐾𝐽
7.6361 π‘˜π‘”βˆ™πΎ
X5 = 0.9239
X5 = 92.39 %
οƒ˜ State 6
P5 = P6 = 0.008 MPa
π‘š³
𝜐f5 = 𝜐5 = 1.0084 × 10-3 π‘˜π‘”
𝐾𝐽
hf5 = h5 = 173.88π‘˜π‘”
οƒ˜ State 7
P4 = P7 = 0.7 MPa
81
WP1 = 𝜐6 (P7 – P6)
π‘š³
WP1 = 1.0084 × 10-3 π‘˜π‘” (0.7 MPa – 0.008 MPa)
π‘š³
WP1 = 1.0084 × 10-3 π‘˜π‘” (0.692MPa ×
1000 πΎπ‘ƒπ‘Ž
×
1 π‘€π‘ƒπ‘Ž
1000
𝑁
π‘š²
1𝐽
1 𝐾𝐽
1 πΎπ‘ƒπ‘Ž
× 1 π‘βˆ™π‘š × 1000 𝐽 )
𝑁
π‘š²
1𝐽
𝐾𝐽
WP1 0.6978 π‘˜π‘”
h7 = h6 + Wp1
𝐾𝐽
𝐾𝐽
h7 = 173.88 π‘˜π‘” + 6978π‘˜π‘”
𝐾𝐽
h7 =174.5778 π‘˜π‘”
οƒ˜ State 8
P4 = P7 = P8 = 0.7MPa
π‘š³
𝜐f4 = 𝜐8 = 1.1080 × 10-3 π‘˜π‘”
𝐾𝐽
hf4 = h8 = 697.22 π‘˜π‘”
οƒ˜ State 9
P1 = P9 = 10 MPa
WP2 = 𝜐8 (P9 – P8)
π‘š³
WP2 = 1.1080× 10-3 π‘˜π‘” (10 MPa – 0.7 MPa)
WP2
π‘š³
1000 πΎπ‘ƒπ‘Ž
= 1.1080× 10-3
(9.3MPa ×
×
π‘˜π‘”
1 π‘€π‘ƒπ‘Ž
1000
1 πΎπ‘ƒπ‘Ž
1 𝐾𝐽
× 1 π‘βˆ™π‘š × 1000 𝐽 )
𝐾𝐽
WP2 = 10.3053 π‘˜π‘”
h9 = h8 + Wp2
𝐾𝐽
𝐾𝐽
h9 = 697.22 π‘˜π‘” + 10.3053π‘˜π‘”
𝐾𝐽
h9 = 707.5253 π‘˜π‘”
οƒ˜ State 10
P9 = P10 = 10MPa
h10 = h12
οƒ˜ State 11
P2 = P11 = 2.5MPa
π‘š³
𝜐f2 = 𝜐11 = 1.1973 × 10-3 π‘˜π‘”
𝐾𝐽
hf2 = h11 = 962.11 π‘˜π‘”
οƒ˜ State 12
P10 = P12 = 10 MPa
WP3 = 𝜐11 (P12 – P11)
82
π‘š³
WP3 = 1.1973× 10-3 π‘˜π‘” (10 MPa – 2.5 MPa)
WP3
π‘š³
1000 πΎπ‘ƒπ‘Ž
= 1.1973 × 10-3
(7.5 MPa ×
×
π‘˜π‘”
1 π‘€π‘ƒπ‘Ž
1000
𝑁
π‘š²
1 πΎπ‘ƒπ‘Ž
×
1𝐽
1 π‘βˆ™π‘š
×
1 𝐾𝐽
1000 𝐽
)
𝐾𝐽
WP3 = 8.9798 π‘˜π‘”
h12 = h11 + Wp3
𝐾𝐽
𝐾𝐽
h12 = 962.11 π‘˜π‘” + 8.9798π‘˜π‘”
𝐾𝐽
h12 = 971.0898 π‘˜π‘”
οƒ˜ State 13
P12 = P13 = 10MPa
𝐾𝐽
h10 = h12 = h10 =971.0898π‘˜π‘”
For m1,
(1-m)
h10
m1 =
m1 =
Ein = Eout
(m1)(h2) + (1-m1)(h9) = m1h6 + (1-m1)(h10)
(m1)(h2) – h9 – m1h9 = m1h11 + h10 – m1h10
m1h2 – m1h9 - m1h11 + m1h10 = h10 – h9
m1[(h2 – h11) + (h10 – h9)] = h10 – h9
(m1)(h2)
h2
(1-m1)
h9
h11
m1
h₁₀ – h₉
(hβ‚‚ – h₁₁) + (h₁₀ – h₉)
𝐾𝐽
π‘˜π‘”
(971.0898−707.5253)
(3202.0316−962.11)
𝐾𝐽
𝐾𝐽
+(971.0898−707.5253)
π‘˜π‘”
π‘˜π‘”
m1 = 0.1038
For m2,
h4
m2
(1-m1-m2)
h7
h8
(1-m1)
m2 =
(1−m1)(h8−h7)
Ein = Eout
(m2)(h4) + (1-m1-m2)(h7) = (1-m1)(h8)
(m2)(h4) – h7 – m1h7 – m2h7 = h8 – m1h8
m2(h4 – h7) + (1-m1)(h7) = (1-m1)(h8)
m2(h4 – h7) = (1-m1)(h8-h7)
h4 – h7
83
m2 =
𝐾𝐽
π‘˜π‘”
(1−0.1038)(697.22−174.5778)
𝐾𝐽
π‘˜π‘”
(3277.0990−174.5778)
m2 = 0.1510
QAR = (1-m1) (h3 – h2)
QAR = (1-0.1038)(3731.5-3202.0316)
QAR = 474.5096
𝐾𝐽
π‘˜π‘”
𝐾𝐽
π‘˜π‘”
QAB = h1 – h3
QAB = (3674.6-971.0898)
QAB = 2703.5102
𝐾𝐽
π‘˜π‘”
𝐾𝐽
π‘˜π‘”
QA = QAB + QAR
QA = (2703.5102+474.5096)
QA = 3178.0198
𝐾𝐽
π‘˜π‘”
𝐾𝐽
π‘˜π‘”
QR = (1-m1-m2)(h5-h6)
QR = (1-0.1038-0.1510)(2394.1041-173.88)
QR = 1654.5110
𝐾𝐽
π‘˜π‘”
𝐾𝐽
π‘˜π‘”
For πœ‚TH,
πœ‚TH = 1πœ‚TH = 1-
QR
𝑄𝐴
𝐾𝐽
π‘˜π‘”
𝐾𝐽
3178.0198
π‘˜π‘”
1654.5110
πœ‚TH = 0.4794
πœ‚TH = 47.94%
For WPtotal’,
WPtotal’= Wp1’ + Wp2’ + Wp3’
WPtotal’=
WPtotal’=
π‘Šπ‘1
πœ‚β‚š
+
π‘Šπ‘2
𝐾𝐽
0.6978
π‘˜π‘”
0.88
πœ‚β‚š
+
WPtotal’= 22.7079
+
π‘Šπ‘3
πœ‚β‚š
𝐾𝐽
π‘˜π‘”
10.3053
0.88
+
8.9798
𝐾𝐽
π‘˜π‘”
0.88
𝐾𝐽
π‘˜π‘”
84
TS DIAGRAM
86
3. Consider a steam enters a turbine at 12 MPa and 520 °C and is condensed in the condenser at a
pressure of 12 kPa. A portion of the steam is extracted from the turbine at 3 MPa for used in a
closed feed water heater while the remaining steam is reheated at the same pressure to 620 °C.
The extracted steam is completely condensed in the heater and is pumped to 12 MPa before
mixing with the feed water at the same pressure. Additionally, steam is extracted from the low
pressure turbine at a pressure of 0.4 MPa for an open feed water heater. Determine the ideal
thermal efficiency as well as the actual work done of the pump, assuming that the turbine and
pump have efficiencies of 85%.
Given:
P1 = 10 MPa
T1 = 620 °C
P2 = 2.5 MPa
(1-m1)
P3 = 2.5 MPa
πœ‚T = 88%
P4 = 0.7 MPa
(1-m1-m2)
m2
(1-m1)
P6 = 0.08 MPa
m1
πœ‚p = 88%
Required:
a) QA
b) πœ‚TH’
Solution:
οƒ˜ State 1
P1 =12 MPa
T1 = 520 °C
tsat = 324.75 °C; tsat < T1; superheated vapor
οƒ˜ State 2State 2
𝐾𝐽
S1 = S2 = 6.9587π‘˜π‘”βˆ™πΎ
P2 = 2.5 MPa
𝐾𝐽
h1 = 3401.8 π‘˜π‘”
𝐾𝐽
s1 = 6.5555π‘˜π‘”βˆ™πΎ
S (π‘˜π‘”βˆ™πΎ)
𝐾𝐽
h (π‘˜π‘”)
𝐾𝐽
6.5390
S2 = 6.5555
6.5825
2993.5
h2
3018.7
𝐾𝐽
h2 = 3003.0586 π‘˜π‘”
87
οƒ˜ State 2’
πœ‚T =
β„Žβ‚−β„Žβ‚‚′
β„Žβ‚−β„Žβ‚‚
𝐾𝐽
−β„Žβ‚‚′
π‘˜π‘”
3401.8
0.85 =
𝐾𝐽
π‘˜π‘”
(3401.8−3003.0586)
𝐾𝐽
h2’= 3062.8689 π‘˜π‘”
οƒ˜ State 3
P2 = P3 = 3MPa
T3 = 520 °C
tsat = 233.90 °C; tsat < T1; superheated vapor
𝐾𝐽
h3 = 3501.5 π‘˜π‘”
𝐾𝐽
s3 = 7.2913 π‘˜π‘”βˆ™πΎ
οƒ˜ State 4
S3 = S4 = 7.2913
𝐾𝐽
π‘˜π‘”βˆ™πΎ
P4 =0.4 MPa
𝐾𝐽
sg4 = 6.8959 π‘˜π‘”βˆ™πΎ
sg4 < s4; superheated vapor
S (π‘˜π‘”βˆ™πΎ)
𝐾𝐽
h (π‘˜π‘”)
𝐾𝐽
7.2570
2902.3
S4 = 7.2913
h4
7.2986
2923.0
𝐾𝐽
h4 = 2919.3675 π‘˜π‘”
οƒ˜ State 4’
πœ‚T =
β„Žβ‚ƒ−β„Žβ‚„′
β„Žβ‚ƒ−β„Žβ‚„
3501.5
0.85 =
𝐾𝐽
−β„Žβ‚„′
π‘˜π‘”
𝐾𝐽
π‘˜π‘”
(3501.5−2919.3675)
𝐾𝐽
h4’= 3006.6874 π‘˜π‘”
οƒ˜ State 5
𝐾𝐽
S3 = S4 = S5 = 7.2913π‘˜π‘”βˆ™πΎ
P5 =0.012 MPa
𝐾𝐽
sg5 = 8.0863 π‘˜π‘”βˆ™πΎ
h5 = hf5 + x5 (hfg5)
𝐾𝐽
𝐾𝐽
h5 = 206.92 π‘˜π‘”+ (0.8924) (2384.1π‘˜π‘”)
𝐾𝐽
h5 = 2334.4908π‘˜π‘”
sg4 > s4; mixture
88
X5 =
X5 =
S₃− Sf₃
𝑆𝑓𝑔₃
𝐾𝐽
(7.2913−0.6963)
π‘˜π‘”βˆ™πΎ
𝐾𝐽
7.390 π‘˜π‘”βˆ™πΎ
X5 = 0.8224
X5 = 82.24 %
οƒ˜ State 5’
πœ‚T =
β„Žβ‚„−β„Žβ‚…′
β„Žβ‚„−β„Žβ‚…
𝐾𝐽
0.85 =
2919.3675π‘˜π‘”−β„Žβ‚…′
𝐾𝐽
π‘˜π‘”
(2919.6375−2334.4908)
h5’= 2422.2223
𝐾𝐽
π‘˜π‘”
οƒ˜ State 6
P5 = P6 = 0.012 MPa
π‘š³
𝜐f5 = 𝜐5 = 1.0119 × 10-3 π‘˜π‘”
𝐾𝐽
hf5 = h5 = 206.92 π‘˜π‘”
οƒ˜ State 7
P4 = P7 = 0.4 MPa
WP1 = 𝜐6 (P7 – P6)
π‘š³
WP1 = 1.0119 × 10-3 π‘˜π‘” (0.4MPa – 0.012 MPa)
WP1
π‘š³
1000 πΎπ‘ƒπ‘Ž
= 1.0119 × 10-3
(0.388 MPa ×
×
π‘˜π‘”
1 π‘€π‘ƒπ‘Ž
1000
𝑁
π‘š²
1 πΎπ‘ƒπ‘Ž
1𝐽
1 𝐾𝐽
× 1 π‘βˆ™π‘š × 1000 𝐽 )
𝐾𝐽
WP1 = 0.3926 π‘˜π‘”
h7 = h6 + Wp1
𝐾𝐽
𝐾𝐽
h7 = 206.92 π‘˜π‘” + 0.3926π‘˜π‘”
𝐾𝐽
h7 =207.3126 π‘˜π‘”
Wp1’ =
Wp1’ =
π‘Šπ‘1
πœ‚β‚š
0.3926
𝐾𝐽
π‘˜π‘”
0.85
Wp1’ = 0.4619
h7’ = h6 + Wp1’
𝐾𝐽
𝐾𝐽
h7’ = 206.92 π‘˜π‘” + 0.4619π‘˜π‘”
𝐾𝐽
h7’ =207.3819 π‘˜π‘”
89
οƒ˜ State 8
P4 = P7 = P8 = 0.4 MPa
π‘š³
𝜐f4 = 𝜐8 = 1.0836 × 10-3 π‘˜π‘”
𝐾𝐽
hf4 = h8 = 604.74 π‘˜π‘”
οƒ˜ State 9
P1 = P9 = 12 MPa
WP2 = 𝜐8 (P9 – P8)
π‘š³
WP2 = 1.0836× 10-3 π‘˜π‘” (12 MPa – 0.4 MPa)
WP2
1000 πΎπ‘ƒπ‘Ž
π‘š³
= 1.0836× 10-3
(11.6 MPa ×
π‘˜π‘”
1 π‘€π‘ƒπ‘Ž
×
1000
𝑁
π‘š²
1 πΎπ‘ƒπ‘Ž
×
1𝐽
1 π‘βˆ™π‘š
×
1 𝐾𝐽
1000 𝐽
)
𝐾𝐽
WP2 = 12.5698 π‘˜π‘”
h9 = h8 + Wp2
𝐾𝐽
𝐾𝐽
h9 = 604.74 π‘˜π‘” + 12.5698 π‘˜π‘”
𝐾𝐽
h9 = 617.3098 π‘˜π‘”
Wp2’ =
Wp2’ =
π‘Šπ‘2
πœ‚β‚š
12.5698
𝐾𝐽
π‘˜π‘”
0.85
Wp2’ = 14.788
h9’ = h8 + Wp2’
𝐾𝐽
𝐾𝐽
h9’ = 604.74π‘˜π‘” + 14.788π‘˜π‘”
𝐾𝐽
h9’ =319.528 π‘˜π‘”
οƒ˜ State 10
P9 = P10 = 12 MPa
h10 = h12
οƒ˜ State 11
P2 = P11 = 3 Pa
π‘š³
𝜐f2 = 𝜐11 = 1.2165 × 10-3 π‘˜π‘”
𝐾𝐽
hf2 = h11 = 1008.42 π‘˜π‘”
οƒ˜ State 12
P10 = P12 = 12 MPa
WP3 = 𝜐11 (P12 – P11)
90
π‘š³
WP3 = 1.2165× 10-3 π‘˜π‘” (12 MPa – 3 MPa)
WP3
π‘š³
1000 πΎπ‘ƒπ‘Ž
= 1.2165 × 10-3
(9 MPa ×
×
π‘˜π‘”
1 π‘€π‘ƒπ‘Ž
1000
𝑁
π‘š²
1 πΎπ‘ƒπ‘Ž
×
1𝐽
1 π‘βˆ™π‘š
×
1 𝐾𝐽
1000 𝐽
)
𝐾𝐽
WP3 = 10.9485 π‘˜π‘”
h12 = h11 + Wp3
𝐾𝐽
𝐾𝐽
h12 = 1008.42 π‘˜π‘” + 10.9485π‘˜π‘”
𝐾𝐽
h12 = 1019.3685 π‘˜π‘”
Wp3’ =
Wp3’ =
π‘Šπ‘3
πœ‚β‚š
10.9485
𝐾𝐽
π‘˜π‘”
0.85
Wp3’ = 12.8806
h12’ = h8 + Wp3’
𝐾𝐽
𝐾𝐽
h12’ = 1008.42π‘˜π‘” + 12.8806π‘˜π‘”
𝐾𝐽
h12’ =1021.3006 π‘˜π‘”
οƒ˜ State 13
P12 = P13 = 10MPa
𝐾𝐽
h10 = h12 = h10 =971.0898π‘˜π‘”
For m1,
(1-m)
h10
m1 =
m1 =
Ein = Eout
(m1)(h2) + (1-m1)(h9) = m1h6 + (1-m1)(h10)
(m1)(h2) – h9 – m1h9 = m1h11 + h10 – m1h10
m1h2 – m1h9 - m1h11 + m1h10 = h10 – h9
m1[(h2 – h11) + (h10 – h9)] = h10 – h9
(m1)(h2)
h2
(1-m1)
h9
h11
m1
h₁₀ – h₉
(hβ‚‚ – h₁₁) + (h₁₀ – h₉)
(1019.3685−617.3098)
(3003.0586−1008.42)
𝐾𝐽
π‘˜π‘”
𝐾𝐽
𝐾𝐽
+(1019.3685−617.3098)
π‘˜π‘”
π‘˜π‘”
m1 = 0.1678
91
For m2,
Ein = Eout
(m2)(h4) + (1-m1-m2)(h7) = (1-m1)(h8)
(m2)(h4) – h7 – m1h7 – m2h7 = h8 – m1h8
m2(h4 – h7) + (1-m1)(h7) = (1-m1)(h8)
m2(h4 – h7) = (1-m1)(h8-h7)
m2
h4
(1-m1-m2)
h7
h8
(1-m1)
m2 =
(1−m1)(h8−h7)
m2 =
h4 – h7
𝐾𝐽
π‘˜π‘”
(1−0.1678)(604.74−207.3126)
𝐾𝐽
π‘˜π‘”
(3277.0990−174.5778)
m2 = 0.1220
QAR = (1-m1) (h3 – h2)
QAR = (1-0.1678)(3501.5-3003.0586)
QAR = 464.7349
𝐾𝐽
π‘˜π‘”
𝐾𝐽
π‘˜π‘”
QAB = h1 – h3
QAB = (3401.8-1019.3685)
QAB = 2382.4215
𝐾𝐽
π‘˜π‘”
𝐾𝐽
π‘˜π‘”
QA = QAB + QAR
QA = (2382.4315+464.7949)
QA = 2847.1664
𝐾𝐽
π‘˜π‘”
𝐾𝐽
π‘˜π‘”
For πœ‚TH,
πœ‚TH’ = 1m1’ =
m1’ =
𝑄𝑅′
𝑄𝐴′
h₁₀′ – h₉′
(hβ‚‚′ – h₁₁) + (h₁₀′ – h₉′)
(1021..3006−619.528)
(3062.8698−1008.42)
𝐾𝐽
π‘˜π‘”
𝐾𝐽
𝐾𝐽
+(1021.3006−319.5288)
π‘˜π‘”
π‘˜π‘”
m1’ = 0.1197
92
m2 =
m2 =
(1−m1′)(h8−h7′)
h4′ – h7′
(1−0.1197)(304.74−207.3819)
𝐾𝐽
π‘˜π‘”
𝐾𝐽
π‘˜π‘”
(3006.6874−207.3819)
m2 = 0.1250
QR’ = (1-m1’-m2’)(h5’-h6)
QR’ = (1-0.1197-0.1250)(2422.2223-206.92)
QR’ = 1673.2178
𝐾𝐽
π‘˜π‘”
𝐾𝐽
π‘˜π‘”
QA’ = (h1 – h13)+(1-m1’)( h3 – h2’)
QA’ = (3401.8-1021.3006)
QA’ = 2766.6256
𝐾𝐽
π‘˜π‘”
+(1-0.1197)(3501.5-3062.8698)
𝐾𝐽
π‘˜π‘”
𝐾𝐽
π‘˜π‘”
πœ‚TH’ = 1πœ‚TH’ = 1-
𝑄𝑅′
𝑄𝐴′
𝐾𝐽
π‘˜π‘”
𝐾𝐽
2766.6256
π‘˜π‘”
1673.2178
πœ‚TH’ = 0.3952
πœ‚TH’ = 39.52%
TS DIAGRAM
93
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