Thermodynamics 2

Precious Arlene Villaroza-Melendrez

62
 Table of Contents

Module 7: The Reheat Cycle 64

Introduction 64
Learning Objectives 64
Lesson 1 The Reheat Cycle 65
Assessment 7 86
Summary 87

Module 8: The Ideal Reheat Engine 88

Introduction 88
Learning Objectives 88
Lesson 1. The Ideal Reheat Engine 88
Assessment 8 103
Summary 104

Module 9: The Ideal Regenerative Rankine Cycle 105

Introduction 105
Learning Objectives 105
Lesson 1. The Ideal Regenerative Rankine Cycle 106
Assessment 9 114
Summary 115

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 MODULE 7
THE REHEAT CYCLE

Introduction

The work-producing device that best fits into the definition of a heat engine is the steam
power plant, which is an external-combustion engine. That is, combustion takes place outside
the engine, and the thermal energy released during this process is transferred to the steam
as heat (Boles and Cengel, 2004).

Learning Outcomes

At the end of this module, students should be able to:

1. Understand the process of reheat cycle ; and

2. Understand the schematic diagram of reheat cycle.

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Lesson 1. The Reheat Cycle

Reheater
1
2 3

Engine
𝑮𝒆𝒏𝒆𝒓𝒂𝒕𝒐𝒓

4
𝑺𝒕𝒆𝒂𝒎

𝑮𝒆𝒏𝒆𝒓𝒂𝒕𝒐𝒓
Condenser

𝑷𝒖𝒎𝒑

Figure 7.1 Diagrammatic Layout for Reheating Plant (Melegrito, Pasamonte, Siapno & Sta. Maria, 2008)

How can we take advantage of the increased efficiencies at higher boiler pressures
without facing the problem of excessive moisture at the final stages of the turbine? (Boles and
Cengel, 2004).

Two possibilities come to mind (Boles and Cengel, 2004) :

1. Superheat the steam to very high temperatures before it enters the turbine. This would be
the desirable solution since the average temperature at which heat is added would also
increase, thus increasing the cycle efficiency. This is not a viable solution, however, since it
requires raising the steam temperature to metallurgically unsafe levels.

65
2. Expand the steam in the turbine in two stages and reheat it in between. In other words, modify
the simple ideal Rankine cycle with a reheat process. Reheating is a practical solution to the
excessive moisture problem in turbines, and it is commonly used in modern steam power plants.

* One final observation about multiple stages of reheat in a cycle should be made. A second stage
of reheat would involve expanding the superheated high-pressure steam to an intermediate
pressure; reheating and expanding steam to a low-pressure; and reheating and expanding steam
to condenser pressure. This process could remove a good portion of the void between the
superheat horns, but the effect would be smaller than the effect of the initial stage. If a third stage
is added, even more of the void could be filled, but the marginal gain would be even smaller. In
the limit, almost Carnot efficiency in the superheat portion of the process could be achieved. Large
units with high capacity factors and/or expensive fuels frequently employ more than one stage of
reheat. However, this is an expensive process and can only be justified by extensive systems
analysis (Boles and Cengel, 2004).

3
1

5 4

S
Figure 7.2 Diagrammatic Layout for Reheating Plant (Melegrito et al., 2008).

Cycle analysis, 𝑸𝑨 (Melegrito, Pasamonte, Siapno & Sta. Maria, 2008).

66
 Energy balance:

ℎ1

𝑄𝐴

ℎ𝐵
Figure 7.3 Heat Addded

𝐸𝑖𝑛 = 𝐸𝑜𝑢𝑡
𝑄𝐵 + ℎ𝐵 = ℎ1
𝑄𝐵 = ℎ1 − ℎ𝐵

Steam Generator (Melegrito et al., 2008).

67
 Energy balance:
𝑄𝑅𝐻

Reheater

ℎ2 ℎ3

Figure 7.4 Heat Reheater

𝐸𝑖𝑛 = 𝐸𝑜𝑢𝑡
𝑄𝑅𝐻 + ℎ2 = ℎ3
𝑄𝑅𝐻 = ℎ3 − ℎ2

𝑄𝐴𝑇 = 𝑄𝐴 = 𝑄𝐵 + 𝑄𝑅𝐻
𝑄𝐴𝑇 = 𝑄𝐴 = ℎ1 − ℎ𝐵 + ℎ3 − ℎ2

For a given number of stages of reheating,

𝑛

𝑄𝐴 = 𝑄𝐵𝑜𝑖𝑙𝑒𝑟 + ∑ 𝑄𝑅𝑒ℎ𝑒𝑎𝑡𝑒𝑟
1

Where n = number of reheaters

B. Heat Rejection, 𝑸𝑹 (Melegrito et al., 2008).

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 Energy balance:

ℎ4

𝑄𝑅
Condenser

ℎ5

Figure 7.5 Condenser

𝐸𝑖𝑛 = 𝐸𝑜𝑢𝑡
ℎ4 = ℎ5 + 𝑄𝑅
𝑄𝑅 = ℎ4 − ℎ5

C. Engine Work, W (Melegrito et al., 2008).

Energy balance:

ℎ1 ℎ2 ℎ3
𝑊
Engine

ℎ4

Figure 7.6 Engine Work

𝐸𝑖𝑛 = 𝐸𝑜𝑢𝑡
ℎ1 + ℎ3 = ℎ2 + ℎ4 + 𝑊
𝑊 = ℎ1 − ℎ2 + ℎ3 − ℎ4

69
 Another means of determining engine work is by getting the sum of the work done by the
steam during the different stages of expansion.

𝑊 = 𝑊1−2 + 𝑊3−4
𝑊 = ℎ1 − ℎ2 + ℎ3 − ℎ4

D. Pump Work, 𝑾𝒑 (Melegrito et al., 2008).

ℎ5
ℎ𝐵

𝑷𝒖𝒎𝒑
𝑊𝑃
Figure 7.7 Pump Work

Approximate pump work

𝑊𝑝 ≈ 𝑣𝑓5 (𝑝𝐵 − 𝑝5 )

Exact pump work

Energy balance:

𝐸𝑖𝑛 = 𝐸𝑜𝑢𝑡
𝑊𝑝 + ℎ5 = ℎ𝐵

Net Cycle Work, 𝑾𝒏𝒆𝒕 (Melegrito et al., 2008).

𝑊𝑛𝑒𝑡 = 𝐸𝑛𝑔𝑖𝑛𝑒 𝑊𝑜𝑟𝑘 − 𝑃𝑢𝑚𝑝 𝑊𝑜𝑟𝑘

𝑊𝑛𝑒𝑡 = ℎ1 − ℎ2 + ℎ3 − ℎ4 − 𝑊𝑝

Another Method

𝑊𝑛𝑒𝑡 = 𝑄𝐴 − 𝑄𝑅
𝑊𝑛𝑒𝑡 = ℎ1 − ℎ2 + ℎ3 − ℎ4 − 𝑊𝑝

Thermal Efficiency, 𝒆𝒄 (Melegrito et al., 2008).

𝑊𝑛𝑒𝑡
𝑒𝑐 =
𝑄𝐴
𝑊𝑛𝑒𝑡 = ℎ1 − ℎ2 + ℎ3 − ℎ4 − 𝑊𝑝

𝑄𝐴 = ℎ1 − ℎ𝐵 + ℎ3 − ℎ2

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But ℎ𝐵 = ℎ5 + 𝑊𝑝

𝑄𝐴 = ℎ1 − (ℎ5 + 𝑊𝑝) + ℎ3 − ℎ2

ℎ1 − ℎ5 − 𝑊𝑝 + ℎ3 − ℎ2
𝑒𝑐 =
ℎ1 − ℎ2 + ℎ3 − ℎ4 − 𝑊𝑝

G. Steam Rate , (Melegrito et al., 2008).

3600 𝑘𝑔
𝑚𝑐 = ,
𝑊𝑛𝑒𝑡 𝑘𝑤ℎ

Examples

In a reheat cycle steam at 8.0 MPa and 485 ℃ enters the turbine and expands to
1.4 MPa. At this point, the steam is withdrawn and passed through a reheater. It re-enters
the turbine at 1.3 MPa and 720 ℃. Expansion now occurs to the condenser pressure of
0.006 MPa. For the cycle 1 and 1 kg of steam determine (a) 𝑄𝐴𝑇 , (b) 𝑊𝑛𝑒𝑡 , (c) 𝑒𝐶 , for the
engine determine (d) W, (e) 𝑒𝑐 and (f) the steam flow for an engine output of 40, 000 Kw
(Melegrito et al., 2008).

Given:  1.3 MPa and 720 ℃.

Reheater: Condenser:
 8.0 MPa and 485 ℃  0.006 MPa
Turbine: Mass:
 1.4 MPa  1 kg

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 Diagram:

Reheater 1
𝑄𝑅𝐻
1 2
3 𝑊𝑇

Engine
𝑮𝒆𝒏𝒆𝒓𝒂𝒕𝒐𝒓

4
𝑺𝒕𝒆𝒂𝒎

𝑮𝒆𝒏𝒆𝒓𝒂𝒕𝒐𝒓
Condenser 𝑄𝑅
𝑄𝐴

5
𝑷𝒖𝒎𝒑 𝑊𝑃

Figure 7.8 Diagram Layout for Reheating Plant (Melegrito et al., 2008).

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T
1
3

8.0 𝑀𝑃𝑎 𝑎𝑛𝑑 485 ℃

1.4 𝑀𝑃𝑎
2

B 0.006 𝑀𝑃𝑎

5 4

S
Figure 7.9 Temperature vs Entropy Diagram (Melegrito et al., 2008).

Note:
 𝑠1 = 𝑠2
 𝑠3 = 𝑠4

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 Data (Steam Table) :

 Point 1

Table 7.1 Interpolation At table 3 (Vapor/Superheated), (Melegrito et al., 2008).

𝒎𝟑 𝒌𝑱
Temperature Sp. Volume ( ) Internal Energy Enthalpy (kJ/kg) Entropy ( )
𝒌𝒈 𝒌𝒈−𝑲
(kJ/kg)

480 40.34 3025.7 3348.4 6.6586

485 𝑥1 𝑥2 𝑥3 𝑥4

490 41.05 3045.1 3373.4 6.6916

(40.34 + 41.05) 𝒎𝟑
= 40.695 =𝑣
2 𝒌𝒈
(3025.7 + 3045.1) kJ
= 3035.4 = 𝑢
2 kg
(3348.4 + 3373.4) kJ
= 3361.05 = ℎ
2 kg
(6.6586 + 6.6916) 𝑘𝐽
= 6.6751 =𝑠
2 𝑘𝑔 − 𝐾
𝑘𝐽
ℎ1 = ℎ 𝑎𝑡 8.0 𝑀𝑃𝑎 𝑎𝑛𝑑 485 ℃ = 3361
𝑘𝑔
 Point 2

𝒌𝑱
Table 7.2 Interpolation At table 3, 𝑷𝟐 = 𝟏. 𝟒 𝑴𝑷𝒂 @ 𝒔𝟏 = 𝟔. 𝟔𝟕𝟓𝟏 𝒌𝒈−𝑲
(Vapor/Superheated),
(Melegrito et al., 2008).

𝒎𝟑 𝒌𝑱
Temperature Sp. Volume ( 𝒌𝒈 ) Internal Energy Enthalpy (kJ/kg) Entropy (𝒌𝒈−𝑲)
(kJ/kg)

230 155.60 2661.8 2879.6 6.6539

𝒙𝟏 𝑥2 𝑥3 𝑥4 6.6751

235 157.60 2671.0 2891.7 6.6778

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 𝑘𝐽
𝑠2 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑠1 = 6.6751
𝑘𝑔 − 𝐾

6.6539 − 6.6751 230 − 𝑥1 155.60 − 𝑥2 2661.8 − 𝑥3 2879.6 − 𝑥4

= = = =
6.6539 − 6.6778 230 − 235 155.60 − 157.60 2661.8 − 2671.0 2879.6 − 2891.7

𝑘𝐽
ℎ2 = ℎ 𝑎𝑡 1.4 𝑀𝑃𝑎 𝑎𝑛𝑑 𝑠2 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑠1 = 2891
𝑘𝑔

 Point 3

Table 7.3 Interpolation At table 3, 𝟏. 𝟑 𝑴𝑷𝒂 𝒂𝒏𝒅 𝟕𝟐𝟎 ℃ (Vapor/Superheated), (Melegrito et

al., 2008).

𝒎𝟑 𝒌𝑱
Temperature Sp. Volume ( 𝒌𝒈 ) Internal Energy Enthalpy (kJ/kg) Entropy (𝒌𝒈−𝑲)
(kJ/kg)

700 407 3474.9 3922.5 8.2287

720 𝑥1 𝑥2 𝑥3 𝑥4

750 428.1 3566.6 4037.6 8.3440

700 − 720 407 − 𝑥1 3474.9 − 𝑥2 3922.5 − 𝑥3 8.2287 − 𝑥4

= = = =
700 − 750 407 − 428.1 3474.9 − 3566.6 3922.5 − 4037.6 8.2287 − 8.3440
𝑘𝐽
𝑠3 = 8.27482
𝑘𝑔 − 𝐾
𝑘𝐽
ℎ3 = ℎ 𝑎𝑡 1.3 𝑀𝑃𝑎 𝑎𝑛𝑑 720 ℃ = 3968
𝑘𝑔

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 Point 4

Table 7.4 Saturated: Pressure Table 2, (0.006 MPa) -Comparison (Melegrito et al., 2008).

Pressure sg

𝑘𝐽
0.006 8.3304
𝑘𝑔−𝐾

Saturated 𝑘𝐽
8.27482
𝑘𝑔 − 𝐾
Pressure

 Therefore the given data is saturated pressure. See the T-S Diagram at Figure 7.9

Since the problem has a given pressure of 0.006 MPa (Condenser) and
based from the cycle and T-S diagram the 𝑠4 and 𝑠3 is EQUAL . The only
way to get the other properties is to GO AT TABLE 3.

Figure 7.10 Table 3 Vapor

If you noticed at table 3 Figure 7.10, the italicize values are all saturated.

76
 𝒌𝑱
Table 7.5 Interpolation At table 3, 𝟎. 𝟎𝟎𝟔 𝑴𝑷𝒂 𝒂𝒏𝒅 𝟖. 𝟐𝟕𝟒𝟖𝟐 𝒌𝒈−𝑲
(Vapor/Superheated),
(Melegrito et al., 2008).

𝒎𝟑 𝒌𝑱
Temperature Sp. Volume ( ) Internal Energy Enthalpy (kJ/kg) Entropy (𝒌𝒈−𝑲)
𝒌𝒈
(kJ/kg)

25 407 3474.9 3922.5 8.2614

𝒙𝟏 𝑥2 𝑥3 𝑥4 8.2748

30 428.1 3566.6 4037.6 8.2926

𝑘𝐽
𝑠4 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑠3 = 8.2748
𝑘𝑔 − 𝐾

𝑘𝐽
ℎ4 = ℎ 𝑎𝑡 0.006 𝑀𝑃𝑎 𝑎𝑛𝑑 𝑠4 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑠3 = 2526
𝑘𝑔

 Point 5

Table 7.6 Saturated: Pressure Table 2, (0.006 MPa) -Comparison (Melegrito et al., 2008).

Pressure hf vf

0.006 𝑘𝐽 𝑚3
151.53 0.0010064 𝑘𝑔
𝑘𝑔

𝑘𝐽
ℎ5 = ℎ𝑓 𝑎𝑡 0.006 𝑀𝑃𝑎 = 151.53
𝑘𝑔
𝑚3
𝑣𝑓5 = 𝑣𝑓 𝑎𝑡 0.006 𝑀𝑃𝑎 = 0.0010064
𝑘𝑔

𝑚3 𝑘𝐽
𝑊𝑝 = 𝑣𝑓5 (𝑝𝐵5 − 𝑝5 ) = 0.0010064 (8000 − 6) = 8.05
𝑘𝑔 𝑘𝑔

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 Point B

𝑘𝐽
ℎ𝐵5 = ℎ5 + 𝑊𝑝 = 151.53 + 8.05 = 159.58
𝑘𝑔

Solutions:

(a)
ℎ1

𝑄𝐴 Reheater 1
𝑄𝑅𝐻
2

ℎ𝐵 3

Figure 7.11 Heat Added and Reheater

𝑄𝐴 𝑇 = 𝑄𝐴 + 𝑄𝑅𝐻

𝑄𝐴𝑇 = ℎ1 − ℎ𝐵5 + ℎ3 − ℎ2

= 3361 − 159.58 + 3968 − 2891

𝑘𝐽
𝑄𝐴𝑇 = 4278.4
𝑘𝑔

78
(b)

1 2
3 𝑊𝑇

Engine

Figure 7.12 Engine/Turbine

𝑊𝑇 = ℎ1 − ℎ2 + ℎ3 − ℎ4

= 3361 − 2891 + 3968 − 2526

𝑘𝐽
= 1912
𝑘𝑔

𝑘𝐽
𝑊𝑛𝑒𝑡 = 𝑊 − 𝑊𝑝 = 1912 − 8,05 = 1904
𝑘𝑔

𝑊 1904
(c) 𝑒𝑒 = = = 0.445 𝑜𝑟 44.5 %
𝐸𝑐 4278.5

𝑘𝐽
(d) 𝑊 = 𝑊𝑇 = 1912 𝑘𝑔

(e) 𝐸𝑐 = ℎ1 − ℎ2 + ℎ3 − ℎ𝑓5

= 3361 − 2891 + 3968 − 151.53

𝑘𝐽
= 4286.5
𝑘𝑔

𝑊 1912
𝑒𝑒 = = = 0.446 𝑜𝑟 44.6%
𝐸𝑐 4286.5

3600 3600 𝑘𝑔
(f) 𝑚 = = 1912 = 1.88 𝑘𝑤ℎ
𝑊

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 𝑘𝑔
𝑠𝑡𝑒𝑎𝑚 𝑓𝑙𝑜𝑤 = (40,000 𝑘𝑤) (1.88 )
𝑘𝑤ℎ

𝑘𝑔 𝑘𝑔
75200 𝑜𝑟 20.89
ℎ 𝑠

2.)

A reheater cycle with two stages or reheating is executed with steam expanding initially
from 20 MPa and 540 ℃. The two reheater pressure are 3.8 MPa and 0.97 MPa, and the steam
leaves each reheater at 540 ℃. Condensation occurs at 60 ℃. (a) For the ideal cycle and 1 kg/s
of steam find 𝑄𝐴𝑇 and 𝑒𝑐 . (b) For the engine and 1 kg/s of steam find W and 𝑒𝑐 (Melegrito et al.,
2008).

Given:

Steam Generator:
 20 MPa and 540 ℃
Reheater: 2
 3.8 MPa and 0.97 MPa
 540 ℃
Condenser:
 60 ℃

80
 Diagram:

Reheater 1 Reheater 2
𝑄𝑅𝐻1 𝑄𝑅𝐻
1
𝑄𝑅𝐻2
2 4 5
3 𝑊𝑇

Engine
𝑮𝒆𝒏𝒆𝒓𝒂𝒕𝒐𝒓

6
𝑺𝒕𝒆𝒂𝒎
𝑮𝒆𝒏𝒆𝒓𝒂𝒕𝒐𝒓
Condenser 𝑄𝑅
𝑄𝐴

𝑷𝒖𝒎𝒑 𝑊𝑃
Figure 7.13 Diagram Layout for Reheating Plant (Melegrito et al., 2008).

1 3
81
T

5
20 𝑀𝑃𝑎 𝑎𝑛𝑑 540 ℃

3.8 𝑀𝑃𝑎
2
0.97 𝑀𝑃𝑎 4

B 0.01994 𝑀𝑃𝑎

7 6

S
Figure 7.14 Temperature vs Entropy Diagram (Melegrito et al., 2008).

Note:

Data (Steam Table) :

𝑘𝐽
ℎ1 = ℎ 𝑎𝑡 20 𝑀𝑃𝑎 𝑎𝑛𝑑 540 ℃ = 3363.5
𝑘𝑔
𝑘𝐽
ℎ2 = ℎ 𝑎𝑡 3.8 𝑀𝑃𝑎 𝑎𝑛𝑑 𝑠2 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑠1 = 2916
𝑘𝑔
𝑘𝐽
ℎ3 = ℎ 𝑎𝑡 3.8 𝑀𝑃𝑎 𝑎𝑛𝑑 540 ℃ = 3538.8
𝑘𝑔
𝑘𝐽
ℎ4 = ℎ 𝑎𝑡 0.97 𝑀𝑃𝑎 𝑎𝑛𝑑 𝑠4 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑠3 = 3100
𝑘𝑔
𝑘𝐽
ℎ5 = ℎ 𝑎𝑡 0.97 𝑀𝑃𝑎 𝑎𝑛𝑑 540℃ = 3565.9
𝑘𝑔
𝑘𝐽
ℎ6 = ℎ 𝑎𝑡 0.01994 𝑀𝑃𝑎 𝑎𝑛𝑑 𝑠6 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑠5 = 2604
𝑘𝑔
𝑘𝐽
ℎ7 = ℎ𝑓 𝑎𝑡 60 ℃ = 251.13
𝑘𝑔
𝑚3
𝑣𝑓7 = 𝑣𝑓 𝑎𝑡 60 ℃ = 0.0010172
𝑘𝑔
𝑚3 𝑘𝐽
𝑊𝑝 = 𝑣𝑓7 (𝑝𝐵7 − 𝑝7 ) = 0.0010172 (20000 − 19.94) = 20.32
𝑘𝑔 𝑘𝑔
𝑘𝐽
ℎ𝐵7 = ℎ7 + 𝑊𝑝 = 251.13 + 20.32 = 271.45
𝑘𝑔

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Solutions:

(a)

ℎ1
𝑄𝑅𝐻1 𝑄𝑅𝐻2

𝑄𝐴 Reheater 1 Reheater 1

2 4
3 5
ℎ𝐵

Figure 7.15 Heat Added and Reheater (Melegrito et al., 2008).

𝑄𝐴𝑇 = 𝑄𝐴 + 𝑄𝑅𝐻1 + 𝑄𝑅𝐻2

𝑄𝐴𝑇 = ℎ1 − ℎ𝐵7 + ℎ3 − ℎ2 − ℎ5 − ℎ4
= 3363.5 − 271.45 + 3538.8 − 2916 + 3565.9 − 3100
𝑘𝐽
= 4108.0
𝑘𝑔
𝑘𝐽 𝑘𝑔
𝑄𝐴𝑇 = (4180.8 ) (1 )
𝑘𝑔 𝑠
𝑘𝐽
= 4180.8
𝑠

83
(b)

1 2 4

3 5 𝑊𝑇

Engine

Figure 7.15 Engine (Melegrito et al., 2008).

𝑊𝑇 = ℎ1 − ℎ2 + ℎ3 − ℎ4 + ℎ5 − ℎ6

𝑊𝑇 = ℎ1 − ℎ2 + ℎ3 − ℎ4

= 3365.3 − 2916 + 3538.8 − 3100 + 3565.9 − 2604

𝑘𝐽
𝑊𝑇 = 1848.2
𝑘𝑔

𝑘𝐽 𝑘𝑔
𝑊 = (1848.2 𝑘𝑔) (1 𝑠
)

𝑘𝐽
𝑊𝑛𝑒𝑡 = 𝑊 − 𝑊𝑝 = 1848.2 − 20.32 = 1827.9
𝑠

𝑊 1827.9
𝑒𝑒 = = = 0.4337 𝑜𝑟 43.337%
𝐸𝑐 4180.8

84
(g) 𝐸𝑐 = ℎ1 − ℎ7 + ℎ3 − ℎ2 + ℎ5 − ℎ4

𝐸𝑐 = ℎ1 − ℎ2 + ℎ3 − ℎ4 + ℎ5 − ℎ7

= 3363.5 − 2916 + 3538.3 − 3100 + 3565.9 − 251.13

𝑘𝐽
= 4201.1
𝑘𝑔

𝑘𝐽 𝑘𝑔
𝐸𝑐 = (4201.1 ) (1 )
𝑘𝑔 𝑠
𝑘𝐽
= 4201.1
𝑠

𝑊 1848.2
𝑒𝑒 = = = 0.4399 𝑜𝑟 43.99%
𝐸𝑐 4201.1

85
 Assessment Task 7

Show the COMPLETE SOLUTIONS and DIAGRAM (T-s) (Cycle) including the switching
of table and interpolations/extrapolations: NO Calculator Technique and Applications.

1. A smaller power plant produces steam at 3 MPa, 600o C in the boiler. It keeps
the condenser at 45o C by transfer of 10 MW out as heat transfer. The first turbine
section expands to 500 kPa and then flow is reheated followed by the expansion
in the low-pressure turbine. Find the reheat temperature so the turbine output is
saturated vapor. For this reheat find the total turbine power output and the boiler
heat transfer.
2. Consider an ideal steam reheat cycle where steam enters the high-pressure
turbine at 3.0 MPa, 400°C, and then expands to 0.8 MPa. It is then reheated to
400°C and expands to 10 kPa in the low-pressure turbine. Calculate the cycle
thermal efficiency and the moisture content of the steam leaving the low-pressure
turbine.
3. A smaller power plant produces 25 kg/s steam at 3 MPa, 600o C in the boiler. It
cools the condenser with ocean water so the condenser exit is at 45o C. There is
a reheat done at 500 kPa up to 400o C and then expansion in the low-pressure
turbine. Find the net power output and the total heat transfer in the boiler.
4. A steam power plant operates on the reheat Rankine cycle. Steam enters the
high-pressure turbine at 12.5 MPa and 550°C at a rate of 7.7 kg/s and leaves at
2 MPa. Steam is then reheated at constant pressure to 450°C before it expands
in the low-pressure turbine. The isentropic efficiencies of the turbine and the pump
are 85 percent and 90 percent, respectively. Steam leaves the condenser as a
saturated liquid. If the moisture content of the steam at the exit of the turbine is
not to exceed 5 percent, determine (a) the condenser pressure, (b) the net power
output, and (c) the thermal efficiency

86
 Summary (Melegrito et al., 2008)

Heat Added 𝑄𝐵 = ℎ1 − ℎ𝐵 * Depend to the no. of reheater

Heat Reheater 𝑄𝑅𝐻 = ℎ3 − ℎ2* Depend to the no. of reheater
Heat Rejection 𝑄𝑅 = ℎ4 − ℎ5 * Depend to the no. of reheater
Engine Work 𝑊 = ℎ1 − ℎ2 + ℎ3 − ℎ4 * Depend to the no. of reheater
Pump Work 𝑊𝑝 ≈ 𝑣𝑓5 (𝑝𝐵 − 𝑝5 ) * Depend to the no. of reheater
Net Cycle Work, 𝑊𝑛𝑒𝑡 𝑊𝑛𝑒𝑡 = 𝐸𝑛𝑔𝑖𝑛𝑒 𝑊𝑜𝑟𝑘 − 𝑃𝑢𝑚𝑝 𝑊𝑜𝑟𝑘
𝑊𝑛𝑒𝑡
Thermal Efficiency, 𝑒𝑐 𝑒𝑐 =
𝑄𝐴

References

Boles, Michael A. and Cengel, Yunus A., (2004).Thermodynamics an Engineering

Approach, McGraw Hill Co.. New York . 5 th ed.

Melegrito, Raymundo M., Pasamonte, Nelson M., , Renato M. and, Hipolito B.

(2008). Thermodynamics 2. National Book Store. Mandaluyong City.

87
 MODULE 8
THE IDEAL REHEAT ENGINE

Introduction

The higher a boiler pressure and the lower a condenser pressure, the lower the quality of
the turbine exhaust that results. Not even superheat may be sufficient to remedy the situation.
If a final quality of 98% in the turbine exhaust was desired, the superheat temperature would
have to extend to over 1,500° F (Boles and Cengel, 2004).

Learning Outcomes

At the end of this module, students should be able to:

1. Understand the property of process of Boiler system; and

2. Understand the function of Boiler system.

Lesson 1. The Ideal Reheat Engine

The ideal reheat engine ignores the pressure drop in the reheater. The engine is an ideal one
whether there is or there in no pressure drop in the reheater for as the expansion process is an
isentropic one (Melegrito, Pasamonte, Siapno & Sta. Maria, 2008).

Engine Analysis
A. Work, W (Melegrito et al., 2008).
𝑊 = ℎ1 − ℎ2 + ℎ3 − ℎ4

B. Steam Rate, 𝒎𝒆 , (Melegrito et al., 2008).

3600
𝑚𝑒 =
𝑊

88
 C. Energy Chargeable Against the Engine, 𝑬𝑪 , (Melegrito et al., 2008).
General equation which is applicable to a reheat engine only.
𝐸𝑐 = ℎ1 − ℎ5 + 𝑄𝑅𝐻
𝐸𝑐 = ℎ1 − ℎ5 + ℎ3 − ℎ2

D. Thermal Efficiency, 𝒆𝒆 , (Melegrito et al., 2008)

𝑊 ℎ1 − ℎ2 + ℎ3 − ℎ4
𝑒𝑒 = =
𝐸𝑐 ℎ1 − ℎ5 + ℎ3 − ℎ2

E. Heat Rate, HR , (Melegrito et al., 2008)

𝐻𝑅 = 𝑚𝑒 𝐸𝑐

Examples

In a reheat cycle steam at 8.0 MPa and 485 ℃ enters the turbine and expands to
1.4 MPa. At this point, the steam is withdrawn and passed through a reheater. It re-enters
the turbine at 1.3 MPa and 720 ℃. Expansion now occurs to the condenser pressure of
0.006 MPa. For the engine determine (a) W, (b) 𝑒𝑐 and (c) the steam flow for an engine
output of 40, 000 Kw (Melegrito et al., 2008).

Given:  1.3 MPa and 720 ℃.

Reheater: Condenser:
 8.0 MPa and 485 ℃  0.006 MPa
Turbine: Mass:
 1.4 MPa  1 kg

89
 Diagram:

Reheater 1
𝑄𝑅𝐻
1 2
3 𝑊𝑇

Engine
𝑮𝒆𝒏𝒆𝒓𝒂𝒕𝒐𝒓

4
𝑺𝒕𝒆𝒂𝒎

𝑮𝒆𝒏𝒆𝒓𝒂𝒕𝒐𝒓
Condenser 𝑄𝑅
𝑄𝐴

5
𝑷𝒖𝒎𝒑 𝑊𝑃

Figure 8.1 Diagram Layout for Reheating Plant (Melegrito et al., 2008).

90
T
1
3

8.0 𝑀𝑃𝑎 𝑎𝑛𝑑 485 ℃

1.4 𝑀𝑃𝑎
2

B 0.006 𝑀𝑃𝑎

5 4

S
Figure 8.2 Temperature vs Entropy Diagram (Melegrito et al., 2008).

Note:
 𝑠1 = 𝑠2
 𝑠3 = 𝑠4

91
 Data (Steam Table) :

 Point 1

Table 2.1 Interpolation At table 3 (Vapor/Superheated), (Melegrito et al., 2008).

𝒎𝟑 𝒌𝑱
Temperature Sp. Volume ( ) Internal Energy Enthalpy (kJ/kg) Entropy ( )
𝒌𝒈 𝒌𝒈−𝑲
(kJ/kg)

480 40.34 3025.7 3348.4 6.6586

485 𝑥1 𝑥2 𝑥3 𝑥4

490 41.05 3045.1 3373.4 6.6916

(40.34 + 41.05) 𝒎𝟑
= 40.695 =𝑣
2 𝒌𝒈
(3025.7 + 3045.1) kJ
= 3035.4 = 𝑢
2 kg
(3348.4 + 3373.4) kJ
= 3361.05 = ℎ
2 kg
(6.6586 + 6.6916) 𝑘𝐽
= 6.6751 =𝑠
2 𝑘𝑔 − 𝐾
𝑘𝐽
ℎ1 = ℎ 𝑎𝑡 8.0 𝑀𝑃𝑎 𝑎𝑛𝑑 485 ℃ = 3361
𝑘𝑔
 Point 2

𝒌𝑱
Table 8.2 Interpolation At table 3, 𝑷𝟐 = 𝟏. 𝟒 𝑴𝑷𝒂 @ 𝒔𝟏 = 𝟔. 𝟔𝟕𝟓𝟏 𝒌𝒈−𝑲
(Vapor/Superheated),
(Melegrito et al., 2008).

𝒎𝟑 𝒌𝑱
Temperature Sp. Volume ( 𝒌𝒈 ) Internal Energy Enthalpy (kJ/kg) Entropy (𝒌𝒈−𝑲)
(kJ/kg)

230 155.60 2661.8 2879.6 6.6539

𝒙𝟏 𝑥2 𝑥3 𝑥4 6.6751

235 157.60 2671.0 2891.7 6.6778

92
 𝑘𝐽
𝑠2 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑠1 = 6.6751
𝑘𝑔 − 𝐾

6.6539 − 6.6751 230 − 𝑥1 155.60 − 𝑥2 2661.8 − 𝑥3 2879.6 − 𝑥4

= = = =
6.6539 − 6.6778 230 − 235 155.60 − 157.60 2661.8 − 2671.0 2879.6 − 2891.7

𝑘𝐽
ℎ2 = ℎ 𝑎𝑡 1.4 𝑀𝑃𝑎 𝑎𝑛𝑑 𝑠2 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑠1 = 2891
𝑘𝑔

 Point 3

Table 8.3 Interpolation At table 3, 𝟏. 𝟑 𝑴𝑷𝒂 𝒂𝒏𝒅 𝟕𝟐𝟎 ℃ (Vapor/Superheated), (Melegrito et

al., 2008).

𝒎𝟑 𝒌𝑱
Temperature Sp. Volume ( 𝒌𝒈 ) Internal Energy Enthalpy (kJ/kg) Entropy (𝒌𝒈−𝑲)
(kJ/kg)

700 407 3474.9 3922.5 8.2287

720 𝑥1 𝑥2 𝑥3 𝑥4

750 428.1 3566.6 4037.6 8.3440

700 − 720 407 − 𝑥1 3474.9 − 𝑥2 3922.5 − 𝑥3 8.2287 − 𝑥4

= = = =
700 − 750 407 − 428.1 3474.9 − 3566.6 3922.5 − 4037.6 8.2287 − 8.3440
𝑘𝐽
𝑠3 = 8.27482
𝑘𝑔 − 𝐾
𝑘𝐽
ℎ3 = ℎ 𝑎𝑡 1.3 𝑀𝑃𝑎 𝑎𝑛𝑑 720 ℃ = 3968
𝑘𝑔

93
 Point 4

Table 8.4 Saturated: Pressure Table 2, (0.006 MPa) -Comparison (Melegrito et al., 2008).

Pressure sg

𝑘𝐽
0.006 8.3304
𝑘𝑔−𝐾

Saturated 𝑘𝐽
8.27482
𝑘𝑔 − 𝐾
Pressure

 Therefore, the given data is saturated pressure. See the T-S Diagram at Figure 7.9

Since the problem has a given pressure of 0.006 MPa (Condenser) and
based from the cycle and T-S diagram the 𝑠4 and 𝑠3 is EQUAL. The only
way to get the other properties is to GO AT TABLE 3.

Figure 8.3 Table 3 Vapor

If you noticed at table 3 Figure 7.10, the italicize values are all saturated.

94
 𝒌𝑱
Table 8.5 Interpolation At table 3, 𝟎. 𝟎𝟎𝟔 𝑴𝑷𝒂 𝒂𝒏𝒅 𝟖. 𝟐𝟕𝟒𝟖𝟐 (Vapor/Superheated),
𝒌𝒈−𝑲
(Melegrito et al., 2008).

𝒎𝟑 𝒌𝑱
Temperature Sp. Volume ( 𝒌𝒈 ) Internal Energy Enthalpy (kJ/kg) Entropy (𝒌𝒈−𝑲)
(kJ/kg)

25 407 3474.9 3922.5 8.2614

𝒙𝟏 𝑥2 𝑥3 𝑥4 8.2748

30 428.1 3566.6 4037.6 8.2926

𝑘𝐽
𝑠4 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑠3 = 8.2748
𝑘𝑔 − 𝐾

𝑘𝐽
ℎ4 = ℎ 𝑎𝑡 0.006 𝑀𝑃𝑎 𝑎𝑛𝑑 𝑠4 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑠3 = 2526
𝑘𝑔

 Point 5

Table 8.6 Saturated: Pressure Table 2, (0.006 MPa) -Comparison (Melegrito et al., 2008).

Pressure hf vf

0.006 𝑘𝐽 𝑚3
151.53 0.0010064
𝑘𝑔
𝑘𝑔

𝑘𝐽
ℎ5 = ℎ𝑓 𝑎𝑡 0.006 𝑀𝑃𝑎 = 151.53
𝑘𝑔
𝑚3
𝑣𝑓5 = 𝑣𝑓 𝑎𝑡 0.006 𝑀𝑃𝑎 = 0.0010064
𝑘𝑔

𝑚3 𝑘𝐽
𝑊𝑝 = 𝑣𝑓5 (𝑝𝐵5 − 𝑝5 ) = 0.0010064 (8000 − 6) = 8.05
𝑘𝑔 𝑘𝑔

95
 Point B

𝑘𝐽
ℎ𝐵5 = ℎ5 + 𝑊𝑝 = 151.53 + 8.05 = 159.58
𝑘𝑔

Solutions:

(h)
ℎ1

𝑄𝐴 Reheater 1
𝑄𝑅𝐻
2

ℎ𝐵 3

Figure 8.4 Heat Added and Reheater

𝑄𝐴 𝑇 = 𝑄𝐴 + 𝑄𝑅𝐻

𝑄𝐴𝑇 = ℎ1 − ℎ𝐵5 + ℎ3 − ℎ2

= 3361 − 159.58 + 3968 − 2891

𝑘𝐽
𝑄𝐴𝑇 = 4278.4
𝑘𝑔

96
(a)

1 2
3 𝑊𝑇

Engine

Figure 8.5 Engine/Turbine

𝑊𝑇 = ℎ1 − ℎ2 + ℎ3 − ℎ4

= 3361 − 2891 + 3968 − 2526

𝑘𝐽
= 1912
𝑘𝑔
𝑘𝐽
𝑊 = 𝑊𝑇 = 1912
𝑘𝑔

(b) 𝐸𝑐 = ℎ1 − ℎ2 + ℎ3 − ℎ𝑓5

= 3361 − 2891 + 3968 − 151.53

𝑘𝐽
= 4286.5
𝑘𝑔

𝑊 1912
𝑒𝑒 = = = 0.446 𝑜𝑟 44.6%
𝐸𝑐 4286.5

3600 3600 𝑘𝑔
(c) 𝑚 = = = 1.88
𝑊 1912 𝑘𝑤ℎ

𝑘𝑔
𝑠𝑡𝑒𝑎𝑚 𝑓𝑙𝑜𝑤 = (40,000 𝑘𝑤) (1.88 )
𝑘𝑤ℎ

97
 𝑘𝑔 𝑘𝑔
75200 𝑜𝑟 20.89
ℎ 𝑠

2.)

A reheater cycle with two stages or reheating is executed with steam expanding initially
from 20 MPa and 540 ℃. The two reheater pressure are 3.8 MPa and 0.97 MPa, and the steam
leaves each reheater at 540 ℃. Condensation occurs at 60 ℃. (A) For the engine and 1 kg/s of
steam find W and 𝑒𝑐 (Melegrito et al., 2008).

Given:

Steam Generator:
 20 MPa and 540 ℃
Reheater: 2
 3.8 MPa and 0.97 MPa
 540 ℃
Condenser:
 60 ℃

98
 Diagram:

𝑄𝑅𝐻1 Reheater 1 Reheater 2

𝑄𝑅𝐻
1
𝑄𝑅𝐻2
2 4 5
3 𝑊𝑇

Engine
𝑮𝒆𝒏𝒆𝒓𝒂𝒕𝒐𝒓

6
𝑺𝒕𝒆𝒂𝒎
𝑮𝒆𝒏𝒆𝒓𝒂𝒕𝒐𝒓
Condenser 𝑄𝑅
𝑄𝐴

𝑷𝒖𝒎𝒑 𝑊𝑃
Figure 8.6 Diagram Layout for Reheating Plant (Melegrito et al., 2008).

99
T
1 3
5
20 𝑀𝑃𝑎 𝑎𝑛𝑑 540 ℃

3.8 𝑀𝑃𝑎
2
0.97 𝑀𝑃𝑎 4

B 0.01994 𝑀𝑃𝑎

7 6

S
Figure 8.7 Temperature vs Entropy Diagram (Melegrito et al., 2008).

Note:

Data (Steam Table) :

𝑘𝐽
ℎ1 = ℎ 𝑎𝑡 20 𝑀𝑃𝑎 𝑎𝑛𝑑 540 ℃ = 3363.5
𝑘𝑔
𝑘𝐽
ℎ2 = ℎ 𝑎𝑡 3.8 𝑀𝑃𝑎 𝑎𝑛𝑑 𝑠2 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑠1 = 2916
𝑘𝑔
𝑘𝐽
ℎ3 = ℎ 𝑎𝑡 3.8 𝑀𝑃𝑎 𝑎𝑛𝑑 540 ℃ = 3538.8
𝑘𝑔
𝑘𝐽
ℎ4 = ℎ 𝑎𝑡 0.97 𝑀𝑃𝑎 𝑎𝑛𝑑 𝑠4 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑠3 = 3100
𝑘𝑔
𝑘𝐽
ℎ5 = ℎ 𝑎𝑡 0.97 𝑀𝑃𝑎 𝑎𝑛𝑑 540℃ = 3565.9
𝑘𝑔
𝑘𝐽
ℎ6 = ℎ 𝑎𝑡 0.01994 𝑀𝑃𝑎 𝑎𝑛𝑑 𝑠6 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑠5 = 2604
𝑘𝑔
𝑘𝐽
ℎ7 = ℎ𝑓 𝑎𝑡 60 ℃ = 251.13
𝑘𝑔
𝑚3
𝑣𝑓7 = 𝑣𝑓 𝑎𝑡 60 ℃ = 0.0010172
𝑘𝑔
𝑚3 𝑘𝐽
𝑊𝑝 = 𝑣𝑓7 (𝑝𝐵7 − 𝑝7 ) = 0.0010172 (20000 − 19.94) = 20.32
𝑘𝑔 𝑘𝑔
𝑘𝐽
ℎ𝐵7 = ℎ7 + 𝑊𝑝 = 251.13 + 20.32 = 271.45
𝑘𝑔

100
Solutions:

(c)

1 2 4

3 5 𝑊𝑇

Engine

Figure 8.8 Engine (Melegrito et al., 2008).

𝑊𝑇 = ℎ1 − ℎ2 + ℎ3 − ℎ4 + ℎ5 − ℎ6

𝑊𝑇 = ℎ1 − ℎ2 + ℎ3 − ℎ4

𝑊𝑇 = 3365.3 − 2916 + 3538.8 − 3100 + 3565.9 − 2604

𝑘𝐽
𝑊𝑇 = 1848.2
𝑘𝑔

𝑘𝐽 𝑘𝑔
𝑊 = (1848.2 ) (1 )
𝑘𝑔 𝑠

(i) 𝐸𝑐 = ℎ1 − ℎ7 + ℎ3 − ℎ2 + ℎ5 − ℎ4

𝐸𝑐 = ℎ1 − ℎ2 + ℎ3 − ℎ4 + ℎ5 − ℎ7

= 3363.5 − 2916 + 3538.3 − 3100 + 3565.9 − 251.13

𝑘𝐽
= 4201.1
𝑘𝑔

101
 𝑘𝐽 𝑘𝑔
𝐸𝑐 = (4201.1 ) (1 )
𝑘𝑔 𝑠
𝑘𝐽
𝐸𝑐 = 4201.1
𝑠

𝑊 1848.2
𝑒𝑒 = = = 0.4399 𝑜𝑟 43.99%
𝐸𝑐 4201.1

102
Assessment Task 8

Show the COMPLETE SOLUTIONS and DIAGRAM (T-s) (Cycle) including the switching
of table and interpolations/extrapolations: NO Calculator Technique and Applications.

1. A smaller power plant produces 25 kg/s steam at 3 MPa, 600o C in the boiler. It
cools the condenser with ocean water so the condenser exit is at 45o C. There
is a reheat done at 500 kPa up to 400o C and then expansion in the low
pressure turbine. For the engine find W and 𝑒𝑐 .
2. Consider an ideal steam reheat cycle where steam enters the high-pressure
turbine at 3.0 MPa, 400°C, and then expands to 0.8 MPa. It is then reheated to
400°C and expands to 10 kPa in the low-pressure turbine. For the engine find W
and 𝑒𝑐 .
3. The reheat pressure effect the operating variables and thus turbine performance.
Repeat Problem 2 twice, using 0.6 and 1.0 MPa for the reheat pressure For the
engine find W and 𝑒𝑐 .
4. The effect of a number of reheat stages on the ideal steam reheat cycle is to be
studied. Repeat Problem 2 using two reheat stages, one stage at 1.2 MPa and
the second at 0.2 MPa, instead of the single reheat stage at 0.8 MPa. For the
engine find W and 𝑒𝑐 .

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 Summary

𝑊 = ℎ1 − ℎ2 + ℎ3 − ℎ4 * Depend to the no. of

Work, W
reheater
3600
Steam Rate, 𝑚𝑒 𝑚𝑒 =
𝑊
Energy Chargeable Against the Engine, 𝐸𝐶 𝐸𝑐 = ℎ1 − ℎ5 + 𝑄𝑅𝐻 * Depend to the no. of
reheater
Thermal Efficiency, 𝑒𝑒 𝑊
𝑒𝑒 =
𝐸𝑐

Heat Rate, HR 𝐻𝑅 = 𝑚𝑒 𝐸𝑐

References

Boles, Michael A. and Cengel, Yunus A., (2004).Thermodynamics an Engineering

Approach, McGraw Hill Co.. New York . 5 th ed.

Melegrito, Raymundo M., Pasamonte, Nelson M., , Renato M. and, Hipolito B.

(2008). Thermodynamics 2. National Book Store. Mandaluyong City.

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 MODULE 9
THE IDEAL REGENERATIVE RANKINE CYCLE

Introduction

Another way of increasing the thermal efficiency of the Rankine cycle is regeneration.
During a regeneration process, liquid water (feedwater) leaving the pump is heated by steam
bled off the turbine at some intermediate pressure in devices called feedwater heaters. The
two streams are mixed in open feedwater heaters, and the mixture leaves as a saturated liquid
at the heater pressure. In closed feedwater heaters, heat is transferred from the steam to the
feedwater without mixing (Boles and Cengel, 2004).

Learning Outcomes

At the end of this module, students should be able to:

1. Understand the property of process of the ideal regenerative Rankine cycle; and

2. Understand the schematic diagram of regenerative Rankine cycle.

105
Lesson 1. The Ideal Regenerative Rankine Cycle

Figure 9.1 The First Part of The Heat-Addition Process in The Boiler Takes Place at Relatively
Low Temperatures.

A careful examination of the T-s diagram of the Rankine cycle redrawn in Figure 9.1
reveals that heat is transferred to the working fluid during process 2-2 at a relatively low
temperature. This lowers the average heat addition temperature and thus the cycle efficiency.
To remedy this shortcoming, we look for ways to raise the temperature of the liquid leaving the
pump (called the feedwater) before it enters the boiler (Boles and Cengel, 2004).

One such possibility is to transfer heat to the feedwater from the expanding steam in a
counterflow heat exchanger built into the turbine, that is, to use regeneration. This solution is
also impractical because it is difficult to design such a heat exchanger and because it would
increase the moisture content of the steam at the final stages of the turbine. A practical
regeneration process in steam power plants is accomplished by extracting, or “bleeding,” steam
from the turbine at various points. This steam, which could have produced more work by
expanding further in the turbine, is used to heat the feedwater instead. The device where the
feedwater is heated by regeneration is called a regenerator, or a feedwater heater (FWH) (Boles
and Cengel, 2004).

Regeneration not only improves cycle efficiency, but also provides a convenient means of
deaerating the feedwater (removing the air that leaks in at the condenser) to prevent corrosion
in the boiler. It also helps control the large volume flow rate of the steam at the final stages of
the turbine (due to the large specific volumes at low pressures). Therefore, regeneration has

106
been used in all modern steam power plants since its introduction in the early 1920s. A
feedwater heater is basically a heat exchanger where heat is transferred from the steam to the
feedwater either by mixing the two fluid streams (open feedwater heaters) or without mixing
them (closed feedwater heaters). Regeneration with both types of feedwater heaters is
discussed below (Boles and Cengel, 2004).

Open Feedwater Heaters An open (or direct-contact) feedwater heater is basically a

mixing chamber, where the steam extracted from the turbine mixes with the feedwater exiting
the pump. Ideally, the mixture leaves the heater as a saturated liquid at the heater pressure.
The schematic of a steam power plant with one open feedwater heater (also called single-stage
regenerative cycle) and the T-s diagram of the cycle are shown in Figure 9.2 In an ideal
regenerative Rankine cycle, steam enters the turbine at the boiler pressure (state 5) and
expands isentropically to an intermediate pressure (state 6). Some steam is extracted at this
state and routed to the feed water heater, while the remaining steam continues to expand
isentropically to the condenser pressure (state 7). This steam leaves the condenser as a
saturated liquid at the condenser pressure (state 1) (Boles and Cengel, 2004).

The condensed water, which is also called the feedwater, then enters an isentropic pump,
where it is compressed to the feedwater heater pressure (state 2) and is routed to the feedwater
heater, where it mixes with the steam extracted from the turbine. The fraction of the steam
extracted is such that the mixture leaves the heater as a saturated liquid at the heater pressure
(state 3). A second pump raises the pressure of the water to the boiler pressure (state 4). The
cycle is completed by heating the water in the boiler to the turbine inlet state (state 5). In the
analysis of steam power plants, it is more convenient to work with quantities expressed per unit
mass of the steam flowing through the boiler. For each 1 kg of steam leaving the boiler, y kg
expands partially in the turbine and is extracted at state 6. The remaining (1 - y) kg expands
completely to the condenser pressure. Therefore, the mass flow rates are different in different
components. If the mass flow rate through the boiler is m . , for example, it is (1 - y)m . through
the condenser. This aspect of the regenerative Rankine cycle should be considered in the
analysis of the cycle as well as in the interpretation of the areas on the T-s diagram. In light of
Figure 9.2, the heat and work interactions of a regenerative Rankine cycle with one feed water
heater can be expressed per unit mass of steam flowing through the boiler as follows (Boles
and Cengel, 2004):

107
 Figure 9.2 The Ideal Regenerative Rankine Cycle with An Open Feedwater Heater

the thermal efficiency of the Rankine cycle increases as a result of regeneration. This is because
regeneration raises the average temperature at which heat is transferred to the steam in the boiler
by raising the temperature of the water before it enters the boiler. The cycle efficiency increases
further as the number of feedwater heaters is increased. Many large plants in operation today use
as many as eight feedwater heaters. The optimum number of feedwater heaters is determined
from economical considerations. The use of an additional feedwater heater cannot be justified
unless it saves more from the fuel costs than its own cost (Boles and Cengel, 2004).

Closed Feedwater

108
 Heaters Another type of feedwater heater frequently used in steam power plants is the
closed feedwater heater, in which heat is transferred from the extracted steam to the feedwater
without any mixing taking place. The two streams now can be at different pressures, since they
do not mix. The schematic of a steam power plant with one closed feedwater heater and the T-s
diagram of the cycle are shown in Figure 9.3. In an ideal closed feedwater heater, the feedwater
is heated to the exit temperature of the extracted steam, which ideally leaves the heater as a
saturated liquid at the extraction pressure. In actual power plants, the feedwater leaves the heater
below the exit temperature (Boles and Cengel, 2004).

Figure 9.3 The Ideal Regenerative Rankine Cycle with A Closed Feedwater Heater

109
 Figure 9.4 A Steam Power Plant with One Open and Three Closed Feedwater Heaters

of the extracted steam because a temperature difference of at least a few degrees is required for
any effective heat transfer to take place. The condensed steam is then either pumped to the
feedwater line or routed to another heater or to the condenser through a device called a trap. A
trap allows the liquid to be throttled to a lower pressure region but traps the vapor. The enthalpy
of steam remains constant during this throttling process. The open and closed feedwater heaters
can be compared as follows. Open feedwater heaters are simple and inexpensive and have good
heat transfer characteristics. They also bring the feedwater to the saturation state. For each
heater, however, a pump is required to handle the feedwater. The closed feedwater heaters are
more complex because of the internal tubing network, and thus they are more expensive. Heat
transfer in closed feedwater heaters is also less effective since the two streams are not allowed
to be in direct contact. However, closed feedwater heaters do not require a separate pump for
each heater since the extracted steam and the feedwater can be at different pressures. Most
steam power plants use a combination of open and closed feedwater heaters, as shown in Figure
9.4 (Boles and Cengel, 2004).

110
Example

1. The Ideal Regenerative Rankine Cycle Consider a steam power plant operating on the
ideal regenerative Rankine cycle with one open feedwater heater. Steam enters the
turbine at 15 MPa and 600°C and is condensed in the condenser at a pressure of 10 kPa
(Boles and Cengel, 2004).

Figure 9.5 A Schematic and T-s Diagram

Some steam leaves the turbine at a pressure of 1.2 MPa and enters the open feedwater
heater. Determine the fraction of steam extracted from the turbine and the thermal
efficiency of the cycle (Boles and Cengel, 2004).

Solution

A steam power plant operates on the ideal regenerative Rankine cycle with one
open feedwater heater. The fraction of steam extracted from the turbine and the thermal
efficiency are to be determined (Boles and Cengel, 2004).

Assumptions 1

Steady operating conditions exist. 2 Kinetic and potential energy changes are
negligible (Boles and Cengel, 2004).

111
Analysis

The schematic of the power plant and the T-s diagram of the cycle are shown in
Fig. 10–18. We note that the power plant operates on the ideal regenerative Rankine
cycle. Therefore, the pumps and the turbines are isentropic; there are no pressure drops
in the boiler, condenser, and feedwater heater; and steam leaves the condenser and the
feedwater heater as saturated liquid. First, we determine the enthalpies at various states:
(Boles and Cengel, 2004).

112
 The energy analysis of open feedwater heaters is identical to the energy analysis
of mixing chambers. The feedwater heaters are generally well insulated (Q = 0), and they
do not involve any work interactions (W = 0). By neglecting the kinetic and potential
energies of the streams, the energy balance reduces for a feedwater heater to (Boles and
Cengel, 2004).

𝑚
where y is the fraction of steam extracted from the turbine (𝑚6). Solving for y and
5

substituting the enthalpy values, we find (Boles and Cengel, 2004).

113
 Assessment Task 8

Show the COMPLETE SOLUTIONS and DIAGRAM (T-s) (Cycle) including the switching
of table and interpolations/extrapolations: NO Calculator Technique and Applications.

1. A steam power plant operates with a boiler output of 20 kg/s steam at 2 MPa,
600°C. The condenser operates at 50°C dumping energy to a river that has
an average temperature of 20°C. There is one open feedwater heater with
extraction from the turbine at 600 kPa and its exit is saturated liquid. Find the
mass flow rate of the extraction flow. If the river water should not be heated
more than 5°C how much water should be pumped from the river to the heat
exchanger (condenser)?

114
Summary

Reference

Boles, Michael A. and Cengel, Yunus A., (2004).Thermodynamics an Engineering

Approach, McGraw Hill Co.. New York . 5 th ed.

115