Power plant steam condenser

1. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
9/2/2020 1 | P a g e
[power plant Laboratory II]
University of Baghdad
Name: - Saif Al-din Ali -B-

2. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
9/2/2020 2 | P a g e
TABLE OF CONTENTS
ABSTRACT.........................................................................I
INTRODUCTION..............................................................II
APPARATUS...................................................................III
Calculations and results...................................................V
DISCUSSION..................................................................VI

3. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
9/2/2020 3 | P a g e
Experiment Name: - steam condenser
1. Abstract
This experiment is conducted to evaluate the thermal operation of steam
condenser under co- current and counter current mode
2. Introduction
The condenser is a heat exchanger of the recuperative type (ie)
transport media exchanger heat either side of a dividing all the
tubes), where the construction is simple it may be possible to
arrange a constant temperature difference between the fluid
3. APPARATUS

4. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
9/2/2020 4 | P a g e
4. Calculations and results
PARALLE FLWO
CONSTANT P=2.5 bar .t=120 sec
Steam Water
NO Q 𝑚3
/ℎ𝑟 Tin 𝑐°
Tout 𝑐°
Tin 𝑐°
Tout 𝑐°
𝑚 𝑠
`
𝑑𝑚3
1 1.2 134 100 23 72 18*0.35
2 1 135 101 23 76 17*0.35
3 0.7 136 101.3 23 84 15*0.35
4 0.5 136 103 23 92 13*0.35
𝑸𝒘 = 𝒎̇ ∗ 𝑪𝒑 ∗ (𝑻𝟒 − 𝑻𝟑)̇
1. 𝑄𝑤 =
1.2
3600
∗ 4186.8 ∗ (72 − 23) = 68.844 𝑘𝑔/𝑠𝑒𝑐
2. 𝑄𝑤 =
1
3600
∗ 4186.8 ∗ (76 − 23) = 61.639 𝑘𝑔/𝑠𝑒𝑐
3. 𝑄𝑤 =
0.7
3600
∗ 4186.8 ∗ (84 − 23) = 49.6601 𝑘𝑔/𝑠𝑒𝑐
4. 𝑄𝑤 =
0.5
3600
∗ 4186.8 ∗ (92 − 23) = 40.1235 𝑘𝑔/𝑠𝑒𝑐
𝑯𝒇𝒈 = 𝑯𝒈 − 𝑯𝒇
𝐻𝑓𝑔 = 2725.53 - 563.49 = 2162.04 KJ/kg
𝐻𝑓𝑔 = 2726.87 - 567.77= 2159.1 KJ/kg
𝐻𝑓𝑔 = 2728.2 - 572.05= 2156.15 KJ/kg
𝐻𝑓𝑔 = 2728.2 - 572.07= 2156.13 KJ/kg
𝑿 =
𝑸𝒔
𝒎𝒔
𝟏𝟐𝟎
∗ 𝒉𝒇𝒈
1. 𝑋 = 68.3844 / (6.6/120) * 2162.04 = 0.602468
2. 𝑋 = 61.639/ (5.95/120) * 2159.1 = 0.575767
3. 𝑋 = 68.3844 / (5.25/120) * 2156.15 = 0.526442
4. 𝑋 = 68.3844 / (4.55/120) * 2156.13 = 0.490788
𝑳𝒎𝒕𝒅 =
(𝑻𝟏 − 𝑻𝟑) − (𝑻𝟐 − 𝑻𝟒)
𝐥𝐧 (
𝑻𝟏 − 𝑻𝟑
𝑻𝟐 − 𝑻𝟒
)
1. 𝐿𝑚𝑡𝑑 = (134-23) *(100-72)/ln ((134-23)/ (100-72)) =60.26171 𝑐°
2. 𝐿𝑚𝑡𝑑 = (135-23) *(101-76)/ln ((135-23)/ (101-76)) =61.639 𝑐°
3. 𝐿𝑚𝑡𝑑 = (136-23) *(101.3-84)/ln ((136-23)/ (101.3-84)) =50.99427 𝑐°
4. 𝐿𝑚𝑡𝑑 = (136-23) *(102-92)/ln ((136-23)/ (103-92)) =43.78636 𝑐°

5. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
9/2/2020 5 | P a g e
𝐔 = 𝐐𝐰/𝟎. 𝟏𝟐 ∗ 𝐋𝐦𝐭𝐝
1. 𝑈 = 68.3844/0.12* 60.26171=9.456585 w/sec
2. 𝑈 = 61.639/0.12* 58.01458= 8.853953 w/sec
3. 𝑈 = 49.6601/0.12* 50.99427= 8.115307 w/sec
4. 𝑈 = 40.1235/0.12* 43.78636= 7.636225 w/sec
PARALLE FLWO
CONSTANT 𝑚`
=1 𝑚3
/ℎ𝑟.t=120 sec
Steam Water
NO Ps Tin 𝑐°
Tout 𝑐°
Tin 𝑐°
Tout 𝑐°
𝑚 𝑠
`
𝑑𝑚3
1 1.5 136 99 23 72 17*0.35
2 2 144 100 23 79 19*0.35
3 2.5 148 101 23 85 20*0.35
4 3 152 102 23 92 21*0.35
𝑸𝒘 = 𝒎̇ ∗ 𝑪𝒑 ∗ (𝑻𝟒 − 𝑻𝟑)̇
1. 𝑄𝑤 =
1
3600
∗ 4186.8 ∗ (72 − 23) = 68.844 𝑘𝑔/𝑠𝑒𝑐
2. 𝑄𝑤 =
1
3600
∗ 4186.8 ∗ (79 − 23) = 65.28 𝑘𝑔/𝑠𝑒𝑐
3. 𝑄𝑤 =
1
3600
∗ 4186.8 ∗ (85 − 23) = 72.106 𝑘𝑔/𝑠𝑒𝑐
4. 𝑄𝑤 =
1
3600
∗ 4186.8 ∗ (95 − 23) = 83.736 𝑘𝑔/𝑠𝑒𝑐
𝑯𝒇𝒈 = 𝑯𝒈 − 𝑯𝒇
1. 𝐻𝑓𝑔 = 2693.11- 467.08= 2226.03KJ/kg
2. 𝐻𝑓𝑔 = 2706.7- 504.7= 2202KJ/kg
3. 𝐻𝑓𝑔 = 2716.9- 535.37= 2181.53 KJ/kg
4. 𝐻𝑓𝑔 = 2725.3- 561.47= 2163.83 KJ/kg
𝑿 =
𝑸𝒔
𝒎𝒔
𝟏𝟐𝟎
∗ 𝒉𝒇𝒈
1. 𝑋 = 56.987 / (5.95/120) * 2226.03= 0.516308
2. 𝑋 = 65.128/ (6.65/120) * 2202= 0.533716
3. 𝑋 = 72.106/ (7/120) * 2181.53 = 0.566622
4. 𝑋 = 83.736/ (7.35/120) * 2163.83 = 0.631805

6. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
9/2/2020 6 | P a g e
𝑳𝒎𝒕𝒅 =
(𝑻𝟏 − 𝑻𝟑) − (𝑻𝟐 − 𝑻𝟒)
𝐥𝐧 (
𝑻𝟏 − 𝑻𝟑
𝑻𝟐 − 𝑻𝟒
)
1. 𝐿𝑚𝑡𝑑 = (136-23) *(99-72)/ln ((136-23)/ (99-72)) =60.07 𝑐°
2. 𝐿𝑚𝑡𝑑 = (144-23) *(100-79)/ln ((144*23)/ (100-79)) =57.10148 𝑐°
3. 𝐿𝑚𝑡𝑑 = (148-23) *(101-85)/ln ((148-23)/ (101-85)) =53.022 𝑐°
4. 𝐿𝑚𝑡𝑑 = (152-23) *(103-95)/ln ((152-23)/ (103-95)) =41.86825 𝑐°
𝐔 = 𝐐𝐰/𝟎. 𝟏𝟐 ∗ 𝐋𝐦𝐭𝐝
1. 𝑈 = 56.987/0.12* 60.07 =9.456585 w/sec
2. 𝑈 = 65.128/0.12* 57.10148 = 8.853953 w/sec
3. 𝑈 = 72.106/0.12* 53.022= 8.115307 w/sec
4. 𝑈 = 83.736/0.12* 41.86825= 7.636225 w/sec
counter FLWO
CONSTANT P=2.5 bar .t=120 sec
Steam Water
NO Q 𝑚3
/ℎ𝑟 Tin 𝑐°
Tout 𝑐°
Tin 𝑐°
Tout 𝑐°
𝑚 𝑠
`
𝑑𝑚3
1 1.2 137 100 23 102 19*0.35
2 1 137 101 23 107 18*0.35
3 0.7 136 102 23 115 17*0.35
4 0.5 136 103 23 128 16*0.35
𝑸𝒘 = 𝒎̇ ∗ 𝑪𝒑 ∗ (𝑻𝟒 − 𝑻𝟑)̇
1. 𝑄𝑤 =
1.2
3600
∗ 4186.8 ∗ (106 − 23) = 110.2524 𝑘𝑔/𝑠𝑒𝑐
2. 𝑄𝑤 =
1
3600
∗ 4186.8 ∗ (115 − 23) = 97.692 𝑘𝑔/𝑠𝑒𝑐
3. 𝑄𝑤 =
0.7
3600
∗ 4186.8 ∗ (122 − 23) = 74.8972 𝑘𝑔/𝑠𝑒𝑐
4. 𝑄𝑤 =
0.5
3600
∗ 4186.8 ∗ (128 − 23) = 61.0575 𝑘𝑔/𝑠𝑒𝑐
𝑯𝒇𝒈 = 𝑯𝒈 − 𝑯𝒇
1. 𝐻𝑓𝑔 = 2729.9 - 576.26= 2153.64 KJ/kg
2. 𝐻𝑓𝑔 = 2729.9 - 576.26= 2153.64 KJ/kg
3. 𝐻𝑓𝑔 = 2728.6 - 571.97= 2156.63 KJ/kg
4. 𝐻𝑓𝑔 = 2728.6 - 571.97= 2156.36 KJ/kg

7. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
9/2/2020 7 | P a g e
𝑿 =
𝑸𝒔
𝒎𝒔
𝟏𝟐𝟎
∗ 𝒉𝒇𝒈
1. 𝑋 = 110.2524/ (6.65/120) * 2153.64 = 0.923793
2. 𝑋 = 97.692 / (6.3/120) * 2153.64 = 0.864026
3. 𝑋 = 74.8972 / (5.95/120) * 2156.63 = 0.700413
4. 𝑋 = 61.0575 / (5.6/120) * 2156.36 = 0.606676
𝑳𝒎𝒕𝒅 =
(𝑻𝟏 − 𝑻𝟒) − (𝑻𝟐 − 𝑻𝟑)
𝐥𝐧 (
𝑻𝟏 − 𝑻𝟒
𝑻𝟐 − 𝑻𝟑
)
1. 𝐿𝑚𝑡𝑑 = (137-102) *(100-23)/ln ((137-102)/ (100-23)) =53.268 𝑐°
2. 𝐿𝑚𝑡𝑑 = (137-107) *(101-23)/ln ((137-107)/ (101-23)) =50.234 𝑐°
3. 𝐿𝑚𝑡𝑑 = (136-115) *(102-23)/ln ((136-115)/ (102-23)) =43.776 𝑐°
4. 𝐿𝑚𝑡𝑑 = (136-128) *(103-23)/ln ((136-128)/ (136-128)) =31.269 𝑐°
𝐔 = 𝐐𝐰/𝟎. 𝟏𝟐 ∗ 𝐋𝐦𝐭𝐝
𝑈 = 110.2524/0.12* 60.07 =9.456585 w/sec
𝑈 = 97.692 /0.12* 50.234 = 8.853953 w/sec
𝑈 = 74.8972 /0.12* 43.776= 8.115307 w/sec
𝑈 = 61.0575 /0.12* 31.269= 7.636225 w/sec
counter FLWO
CONSTANT 𝑚`
=1 𝑚3
/ℎ𝑟.t=120 sec
Steam Water
NO Ps Tin 𝑐°
Tout 𝑐°
Tin 𝑐°
Tout 𝑐°
𝑚 𝑠
`
𝑑𝑚3
1 1.5 136 101 23 106 18*0.35
2 2 143 102 23 115 20*0.35
3 2.5 146 103 23 122 21*0.35
4 3 150 105 23 128 22*0.35

8. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
9/2/2020 8 | P a g e
𝑸𝒘 = 𝒎̇ ∗ 𝑪𝒑 ∗ (𝑻𝟒 − 𝑻𝟑)̇
1. 𝑄𝑤 =
1
3600
∗ 4186.8 ∗ (106 − 23) = 95.529 𝑘𝑔/𝑠𝑒𝑐
2. 𝑄𝑤 =
1
3600
∗ 4186.8 ∗ (115 − 23) = 106.996 𝑘𝑔/𝑠𝑒𝑐
3. 𝑄𝑤 =
1
3600
∗ 4186.8 ∗ (122 − 23) = 115.137 𝑘𝑔/𝑠𝑒𝑐
4. 𝑄𝑤 =
1
3600
∗ 4186.8 ∗ (128 − 23) = 122.115 𝑘𝑔/𝑠𝑒𝑐
𝑯𝒇𝒈 = 𝑯𝒈 − 𝑯𝒇
1. 𝐻𝑓𝑔 = 2693.11-467.08= 2226.03 KJ/kg
2. 𝐻𝑓𝑔 = 2706.7- 504.7= 2202 KJ/kg
3. 𝐻𝑓𝑔 = 2716.9- 535.37= 2181.53 KJ/kg
4. 𝐻𝑓𝑔 = 2725.3- 561.47= 2163.83 KJ/kg
𝑿 =
𝑸𝒔
𝒎𝒔
𝟏𝟐𝟎
∗ 𝒉𝒇𝒈
1. 𝑋 = 95.529/ (6.3/120) * 2226.03= 0.8259
2. 𝑋 = 106.996/ (7/120) * 2202= 0.8329
3. 𝑋 = 115.137 / (7.35/120) * 2181.53 = 0.8616
4. 𝑋 = 122.115 / (7.7/120) * 2163.83 = 0.87955
𝑳𝒎𝒕𝒅 =
(𝑻𝟏 − 𝑻𝟒) − (𝑻𝟐 − 𝑻𝟑)
𝐥𝐧 (
𝑻𝟏 − 𝑻𝟒
𝑻𝟐 − 𝑻𝟑
)
𝐿𝑚𝑡𝑑 = (136-106) *(101-23)/ln ((136-106)/ (101-23)) =50.234 𝑐°
𝐿𝑚𝑡𝑑 = (143-115) *(102-23)/ln ((143-115)/ (102-23)) =106.99 𝑐°
𝐿𝑚𝑡𝑑 = (146-122) *(103-23)/ln ((146-122) /(103-23)) =115.137 𝑐°
𝐿𝑚𝑡𝑑 = (150-128) *(105-23)/ln ((150-128) /(105-23)) =122.115 𝑐°
𝐔 = 𝐐𝐰/𝟎. 𝟏𝟐 ∗ 𝐋𝐦𝐭𝐝
𝑈 = 95.529/0.12* 60.07 =16.01 w/sec
𝑈 = 106.996 /0.12* 106.99= 18.134 w/sec
𝑈 = 115.137 /0.12* 115.137= 20.628 w/sec
𝑈 = 122.115 /0.12* 122.115= 22.3144 w/sec

9. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
9/2/2020 9 | P a g e
Qw
kg/sec
Lmtd
c`
U
w/sec
hf
KJ/kg
hg
KJ/kg
hfg
KJ/kg
x
PARALLE FLWO
CONSTANT P=2.5 bar .t=120 sec
68.3844 60.2617 9.45659 563.49 2725.53 2162.04 0.6025
61.639 58.0146 8.85395 567.77 2726.87 2159.1 0.5758
49.6601 50.9943 8.11531 572.05 2728.2 2156.15 0.5264
40.1235 43.7864 7.63623 572.07 2728.2 2156.13 0.4908
PARALLE FLWO
CONSTANT 𝒎`
=1 𝒎 𝟑
/𝒉𝒓.t=120 sec
56.987 60.0747 7.90502 467.08 2693.11 2226.03 0.5163
65.128 57.1015 9.50472 504.7 2706.7 2202 0.5337
72.106 53.0227 11.3326 535.37 2716.9 2181.53 0.5666
83.736 41.8683 16.6666 561.47 2725.3 2163.83 0.6318
counter FLWO
CONSTANT P=2.5 bar .t=120 sec
110.2524 53.2686 17.2479 576.26 2729.9 2153.64 0.9238
97.692 50.2349 16.2059 576.26 2729.9 2153.64 0.864
74.8972 43.776 14.2576 571.97 2728.6 2156.63 0.7004
61.0575 31.2692 16.272 571.97 2728.6 2156.63 0.6067
counter FLWO
CONSTANT 𝒎`
=1 𝒎 𝟑
/𝒉𝒓.t=120 sec
96.529 50.2349 16.0129 467.08 2693.11 2226.03 0.826
106.996 49.1688 18.1341 504.7 2706.7 2202 0.833
115.137 46.5127 20.6282 535.37 2716.9 2181.53 0.8617
122.115 45.6039 22.3144 561.47 2725.3 2163.83 0.8795

10. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
9/2/2020 10 | P a g e
0
5
10
15
20
25
0 20 40 60 80 100 120 140
U
w/sec
Qw kg/sec
PARALLE FLWO P=c PARALLE FLWO m=cP= counter FLWO P=c counter FLWO m=c
0
5
10
15
20
25
0 10 20 30 40 50 60 70
U
w/sec
Lmtd
c`
PARALLE FLWO P=c PARALLE FLWO m=c counter FLWO P=c counter FLWO m=c

11. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
9/2/2020 11 | P a g e
5. DISCUSSION
1 .Comment on the calculated values of heat removed by cooling water and that rejected
by steam.
Although ordinary heat exchangers may be extremely different in design and
construction and may be of the single- or two-phase type, their modes of operation
and effectiveness are largely determined by the direction of the fluid flow within
the exchanger.
The most common arrangements for flow paths within a heat exchanger are counter-
flow and parallel flow. A counter-flow heat exchanger is one in which the direction
of the flow of one of the working fluids is opposite to the direction to the flow
of the other fluid. In a parallel flow exchanger, both fluids in the heat exchanger
flow in the same direction.
Figure 2 represents the directions of fluid flow in the parallel and counter-flow
exchangers. Under comparable conditions, more heat is transferred in a counter-flow
arrangement than in a parallel flow heat exchanger.
0
5
10
15
20
25
0 0.5 1 1.5 2 2.5 3 3.5
U
w/sec
p bar
PARALLE FLWO P=2.5 c PARALLE FLWO m=c counter FLWO P=2.5 c counter FLWO m=c

12. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
9/2/2020 12 | P a g e
The temperature profiles of the two heat exchangers indicate two major disadvantages in
the parallel-flow design. First, the large temperature difference at the ends (Figure 3) causes
large thermal stresses. The opposing expansion and contraction of the construction
materials due to diverse fluid temperatures can lead to eventual material failure. Second,
the temperature of the cold fluid exiting the heat exchanger never exceeds the lowest
temperature of the hot fluid. This relationship is a distinct disadvantage if the design
purpose is to raise the temperature of the cold fluid.
The design of a parallel flow heat exchanger is advantageous when two fluids are
required to be brought to nearly the same temperature.
The counter-flow heat exchanger has three significant advantages over the parallel flow
design. First, the more uniform temperature difference between the two fluids minimizes
the thermal stresses throughout the exchanger. Second, the outlet temperature of the cold
fluid can approach the highest temperature of the hot fluid (the inlet temperature). Third,

13. Saif al-din ali Madi
Department of Mechanical Engineering/ College of Engineering/ University of Baghdad
9/2/2020 13 | P a g e
the more uniform temperature difference produces a more uniform rate of heat transfer
throughout the heat exchanger.
Whether parallel or counter-flow, heat transfer within the heat exchanger involves
both conduction and convection. One fluid (hot) convectively transfers heat to the
tube wall where conduction takes place across the tube to the opposite wall. The
heat is then convectively transferred to the second fluid. Because this process
takes place over the entire length of the exchanger, the temperature of the fluids
as they flow through the exchanger is not generally constant, but varies over the
entire length, as indicated in Figure 3. The rate of heat transfer varies along the
length of the exchanger tubes because its value depends upon the
temperature difference between the hot and the cold fluid at the point being
viewed.
2. For different values of heat removed by cooling water, draw U - 0 curve for co - current
and counter current heat exchanger
The solution was done
3. Comment on the values of U for different steam inlet pressure.
We note that when the pressure increases, the U increases, that is, the relationship is
direct and according to the scheme
4. Comment on the values of U for different cooling water inlet temperature,
We note the relationship directly, according to the scheme and the measured heat values