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# Rankin Cycle week 2-4

advertisement ```POWERPLANT/WEEK 2-4
Course Title: POWERPLANT
Class: 4th
Lecture: 2
Week: 2-4
TEXTBOOK: POWERPLANT TECHNOLOGY
2012
(M. M. El-Wakil)
Rankin Cycle
2-1
Rankin cycle
Large electric power plants typically utilize a vapor power cycle. Regardless of the heat
source, be it nuclear or combustion of coal, oil, natural gas, wood chips, etc., the remaining
details of these plants are similar. Typically a pure working fluid, usually water, is circulated
through a cycle, and that fluid trades heat and work with its surroundings. We sketch a
typical power plant cycle for electricity generation in Fig. 2-1. The ideal Rankine cycle was
first described in 1859 by William John Macquorn Rankine, long after the steam engine was
in wide usage.
The cycle has the following steps:
• 1 → 2: isentropic compression in a pump,
• 2 → 3: isobaric heating in a boiler,
• 3 → 4: isentropic expansion in a turbine, and
• 4 → 1: isobaric cooling in a condenser.
Two variants of the T −s diagram are given in Fig. 2-2. The first is more efficient as it has
the appearance of a Carnot cycle. However, it is impractical, as it induces liquid water in
the turbine, which can damage its blades. So the second is more common.
The thermal efficiency is
𝜂=
𝑊̇𝑛𝑒𝑡
𝑄̇𝐻
=
𝑊̇𝑡𝑢𝑟𝑏𝑖𝑛𝑒 −𝑊̇𝑝𝑢𝑚𝑝
𝑄̇𝑏𝑜𝑖𝑙𝑒𝑟
(2-1)
This reduces to
𝜂=
𝑚((ℎ3 −ℎ4 )+(ℎ1 −ℎ2 ))
𝑚(ℎ3 −ℎ2 )
ℎ −ℎ
= 1 − ℎ4 −ℎ1
=1−
3
𝑞𝑜𝑢𝑡,𝑐𝑜𝑛𝑑𝑒𝑛𝑠𝑒𝑟
𝑞𝑖𝑛,𝑏𝑜𝑖𝑙𝑒𝑟
(2-2)
2
(2-3)
Note that because the Rankine cycle is not a Carnot cycle, we have
𝑞𝑜𝑢𝑡,𝑐𝑜𝑛𝑑𝑒𝑛𝑠𝑒𝑟
𝑞𝑖𝑛,𝑏𝑜𝑖𝑙𝑒𝑟
𝑇
≠ 𝑇1
(2-4)
3
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POWERPLANT/WEEK 2-4
2012
Grid
Generator
Turbine
4
Chimney
3
Condenser
Cooling tower
2
Boiler
1
Pump
Fig. 2-1: Rankine cycle schematic.
T
T
P
P
3
3
2
1
2
1
4
4
S
S
Fig. 2-2: 𝑇 − 𝑠 for two Rankine cycles.
Thus, for the pump
ℎ2 − ℎ1 = 𝑣(𝑃2 − 𝑃1 )
Since 𝑣 is nearly constant, so the integration is simple.
It might be tempting to make the Rankine cycle into a Carnot cycle as sketched in
Fig.2-3. However, it is practically difficult to build a pump to handle two-phase mixtures. The
gas phase can seriously damage the pump. Some features which could be desirable for a
Rankine cycle include:
High power output: One can enhance this by raising the fluid to a high temperature during
the combustion process or by pumping the fluid to a high pressure. Both strategies soon run
into material limits; turbine blades melt and pipes burst.
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POWERPLANT/WEEK 2-4
2012
Another strategy is to lower the condenser pressure, which means that one must maintain a
vacuum, which can be difficult.
• High thermal efficiency: The key design strategy here is lies in 1) increasing component
efficiencies, and 2) rendering the overall cycle as much like a Carnot cycle as is feasible.
Modern power plants have had revolutionary increases in overall thermal efficiency because
of enhancements, which make the process more Carnot-like.
T
P
3
2
1
4
S
Fig. 2-3: Rankine-Carnot cycle.
There are some important loss mechanisms in the Rankine cycle which inhibit efficiency.
They include:
• Turbine losses: These are the major losses. To avoid these losses requires detailed
consideration of fluid mechanics, material science, and heat transfer and is beyond
the scope of classical thermodynamics. Thermodynamics develops broad measures of
turbine efficiency such as
𝜂𝑡𝑢𝑟𝑏𝑖𝑛𝑒 =
(ℎ3 − ℎ2 )
(ℎ3 − ℎ4𝑠 )
• Pump losses: Again, fluid mechanics, machine design, and material science are required to
analyze how to actually avoid these losses. Thermodynamics characterizes them by
pump efficiency,
𝜂𝑝𝑢𝑚𝑝 =
(ℎ2𝑠 − ℎ1 )
(ℎ2 − ℎ1 )
• Heat transfer losses from components.
• Pressure drop in pipes.
• Incomplete fuel combustion.
• Pollution removal devices.
• Loss of heat to surroundings in the condenser.
One simple design strategy to make the system more Carnot-like is to use
• Reheat: a design strategy in which steam is extracted from the turbine before it is fully
expanded, then sent to the boiler again, and re-expanded through the remained
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POWERPLANT/WEEK 2-4
2012
of the turbine.
Example 2-1
Consider steam in an ideal Rankine cycle. Saturated vapor enters the turbine at 8.0𝑀𝑃𝑎.
Saturated liquid exits the condenser at 𝑃 = 0.008𝑀𝑃𝑎. Thenet power output of the cycle is
100𝑀𝑊. Find
• Thermal efficiency
• Back work ratio
• Mass flow rate of steam
• Rate of heat transfer 𝑄̇𝑖𝑛 to the fluid in the boiler
• Rate of heat transfer 𝑄̇𝑜𝑢𝑡 in the condenser
• Mass flow rate of condenser cooling water if the cooling water enters at 15℃ and exits
at 35℃.
Answer
Use the steam tables to fix the state. At the turbine inlet, one has 𝑃3 = 8.0 𝑀𝑃𝑎, and 𝑥3 =
1 (saturatedsteam). This gives two properties to fix the state, so that
𝑘𝐽
𝑘𝐽
ℎ3 = 2758 , 𝑠3 = 5.7432
𝑘𝑔
𝑘𝑔. 𝐾
𝑘𝐽
State 4 has 𝑃4 = 0.008𝑀𝑃𝑎 and 𝑠4 = 𝑠3 = 5.7432 𝑘𝑔.𝐾 so the state is fixed. From the
saturationtables, it is found then that
𝑘𝐽
𝑘𝐽
𝑠4 − 𝑠𝑓 (5.7432 𝑘𝑔.𝐾) − (0.5926 𝑘𝑔.𝐾)
𝑥4 =
=
= 0.6745
𝑘𝐽
𝑠𝑔 − 𝑠𝑓
(7.6361 𝑘𝑔.𝐾)
Note the quality is 0 ≤ 𝑥4 ≤ 1, as it must be. The enthalpy is then
𝑘𝐽
𝑘𝐽
𝑘𝐽
ℎ4 = ℎ𝑓 + 𝑥4 ℎ𝑓𝑔 = (173.88 ) + (0.6745) (2403.1 ) = 1794.8
𝑘𝑔
𝑘𝑔
𝑘𝑔
State 1 is saturated liquid at 0.008𝑀𝑃𝑎, so 𝑥1 = 0, 𝑃1 = 0.008𝑀𝑃𝑎. One then gets ℎ1 =
𝑘𝐽
𝑚3
ℎ𝑓 = 173.88 𝑘𝑔, 𝑣1 = 𝑣𝑓 = 0.0010084 𝑘𝑔 .
Now state 2 is fixed by the boiler pressure and 𝑠2 = 𝑠1 . But this requires use of the sparse
compressed liquid tables. Alternatively, the pump work is easily approximated by assuming
an incompressible fluid so that
𝑊̇
ℎ2 = ℎ1 + = ℎ1 + 𝑣1 (𝑃2 − 𝑃1 )
𝑚̇
𝑘𝐽
𝑚3
𝑘𝐽
ℎ2 = (173.88 ) + (0.0010084 ) (8000𝑘𝑃𝑎 − 8𝑘𝑃𝑎) = 181.94
𝑘𝑔
𝑘𝑔
𝑘𝑔
The net power is
𝑊̇𝑐𝑦𝑐𝑙𝑒 = 𝑊̇𝑡 + 𝑊̇𝑃
Now the first law for the turbine and pump give
𝑊𝑝̇
𝑊̇ 𝑡
= ℎ3 − ℎ4 ,
= ℎ1 − ℎ2
𝑚̇
𝑚̇
The energy input that is paid for is
𝑄̇𝑖𝑛
= ℎ3 − ℎ2
𝑚̇
The thermal efficiency is then found by
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POWERPLANT/WEEK 2-4
𝜂=
𝑊̇𝑡 + 𝑊̇𝑝 (ℎ3 − ℎ4 ) + (ℎ1 − ℎ2 )
=
ℎ3 − ℎ2
𝑄̇𝑖𝑛
𝑘𝐽
=
2012
𝑘𝐽
𝑘𝐽
𝑘𝐽
𝑘𝐽
𝑘𝐽
((2758 𝑘𝑔) − (1794.8 𝑘𝑔) + (173.88 𝑘𝑔) − (181.94 𝑘𝑔))
((2758 𝑘𝑔) − (181.94 𝑘𝑔))
= 0.371
By definition the back work ratio 𝑏𝑤𝑟 is the ratio of pump work to turbine work:
𝑊̇𝑝
𝑏𝑤𝑟 = | |
𝑊̇𝑡
ℎ1 − ℎ2
=|
|
ℎ3 − ℎ4
𝑘𝐽
= ||
𝑘𝐽
((173.88 𝑘𝑔) − (181.94 𝑘𝑔))
𝑘𝐽
𝑘𝐽
((2758 𝑘𝑔) − (1794.8 𝑘𝑔))
|
|
= 0.00837
The desired mass flow can be determined since we know the desired net power. Thus
𝑊̇𝑐𝑦𝑐𝑙𝑒
𝑚̇ =
(ℎ3 − ℎ4 ) + (ℎ1 − ℎ2 )
100 &times; 103 𝑘𝑊
𝑚̇ =
𝑘𝐽
𝑘𝐽
𝑘𝐽
𝑘𝐽
((2758 𝑘𝑔) − (1794.8 𝑘𝑔) + (173.88 𝑘𝑔) − (181.94 𝑘𝑔))
= 104.697
𝑘𝑔
𝑠
𝑘𝑔
𝑠
𝑘𝑔
) (3600 ) = 3.769 &times; 105
𝑠
ℎ𝑟
ℎ𝑟
The necessary heat transfer rate in the boiler is then
𝑄̇𝑖𝑛 = 𝑚̇(ℎ3 − ℎ2 )
𝑘𝑔
𝑘𝐽
𝑘𝐽
= (104.697 ) ((2758 ) − (181.94 )) = 269706𝑘𝑊 = 269.7𝑀𝑊
𝑠
𝑘𝑔
𝑘𝑔
In the condenser, one finds
𝑄̇𝑜𝑢𝑡 = 𝑚̇(ℎ1 − ℎ4 )
𝑘𝑔
𝑘𝐽
𝑘𝐽
= (104.697 ) ((173.88 ) − (1794.8 )) = −169705𝑘𝑊 = −169.7𝑀𝑊
𝑠
𝑘𝑔
𝑘𝑔
Note also for the cycle that one should find
𝑊̇𝑐𝑦𝑐𝑙𝑒 = 𝑄̇𝑖𝑛 + 𝑄̇𝑜𝑢𝑡 = (269.7𝑀𝑊) − (169.7𝑀𝑊) = 100𝑀𝑊
For the condenser mass flow rate now perform a mass balance:
𝑑𝐸𝐶𝑉
= 𝑄̇𝐶𝑉 − 𝑊̇𝐶𝑉 + 𝑚̇𝑐 (ℎ𝑖𝑛 − ℎ𝑜𝑢𝑡 ) + 𝑚̇(ℎ4 − ℎ1 )
𝑑𝑡
0 = 𝑚̇𝑐 (ℎ𝑖𝑛 − ℎ𝑜𝑢𝑡 ) + 𝑚̇(ℎ4 − ℎ1 )
(ℎ4 − ℎ1 )
𝑚̇𝑐 =
(ℎ𝑖𝑛 − ℎ𝑜𝑢𝑡 )
= (104.697
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POWERPLANT/WEEK 2-4
(104.697
=−
𝑘𝑔
2012
𝑘𝐽
𝑘𝐽
) ((1794.8 𝑘𝑔) − (173.88 𝑘𝑔))
𝑠
𝑘𝐽
𝑘𝐽
((62.99 𝑘𝑔) − (146.68 𝑘𝑔))
𝑘𝑔
𝑠
𝑘𝑔 3600 𝑠
𝑘𝑔
= (2027.79 ) (
) = 7.3 &times; 106
𝑠
ℎ𝑟
ℎ𝑟
= 2027
The enthalpy for the cooling water was found by assuming values at the saturated state at
the respective temperatures of 15 ℃ and 35 ℃.
2-2 Reheat
In a Rankine cycle with reheat, the steam is extracted from an intermediate stage of the
turbine and reheated in the boiler. It is then expanded through the turbine again to the
condenser pressure. One also avoids liquid in the turbine with this strategy. This generally
results in a gain in cycle efficiency. Geometrically, the behaviour on a 𝑇 − 𝑠diagram looks
more like a Carnot cycle. A schematic and 𝑇 – 𝑠diagram for the Rankine cycle with reheat is
given in Fig. 2-4.
𝑇
3
Boiler
5
3
Turbine
4
2
5
2
Pump
1
6
4
6
Condenser
𝑆
1
Figure 2-4: Rankine cycle with reheat schematic and 𝑇 − 𝑠diagram.
Example 2-2
Consider water in a Rankine power cycle with reheat. The first turbine has water enter at
𝑃3 = 8000𝑘𝑃𝑎, 𝑇3 = 480℃. The water expands to 700𝑘𝑃𝑎, undergoes reheat, and then
expands again to 8 𝑘𝑃𝑎. The mass flow rate is 𝑚̇ = 2.63 &times; 105 𝑘𝑔/ℎ𝑟. We have 𝜂𝑡 = 0.88
for each turbine, and 𝜂𝑝 = 0.88for the pump. Find the net power generated, 𝜂𝑡ℎ , and the
heat transfer to the condenser.
Answer:
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POWERPLANT/WEEK 2-4
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Let us consider the big picture first. The net specific power will be the positive effect of the
two turbines and the negative effect of the pump:
𝑤𝑛𝑒𝑡 = (ℎ3 − ℎ4 ) + (ℎ5 − ℎ6 ) + (ℎ1 − ℎ2 )
ℎ3 − ℎ4 = 𝑡𝑢𝑟𝑏𝑖𝑛𝑒 − 1
ℎ5 − ℎ6 = 𝑡𝑢𝑟𝑏𝑖𝑛𝑒 − 2
ℎ1 − ℎ2 = 𝑝𝑢𝑚𝑝
Now, the heat input for the reheat is in two stages:
𝑞𝑖𝑛 = (ℎ3 − ℎ2 ) + (ℎ5 − ℎ4 )
Lastly, the heat rejection in the condenser is
𝑞𝑜𝑢𝑡 = ℎ6 − ℎ1
Let us start at the entrance of the first turbine, at 3. We are given 𝑃3 and 𝑇3 , so we consult
thetables and find
𝑘𝐽
𝑘𝐽
ℎ3 = 3348.4 ,
𝑠3 = 6.6586
𝑘𝑔
𝑘𝑔. 𝐾
We are given 𝑃4 = 7𝑏𝑎𝑟 = 700𝑘𝑃𝑎. Now, let us get the ideal behaviour of the turbine:
𝑘𝐽
𝑠4𝑠 = 𝑠3 = 6.6586 𝑘𝑔.𝐾. At this condition, we find state 4 is a two-phase mixture. At
𝑘𝐽
𝑘𝐽
700𝑘𝑃𝑎, we find 𝑠𝑓 = 1.9922 𝑘𝑔.𝐾, 𝑠𝑔 = 6.7080 𝑘𝑔.𝐾.
So
𝑘𝐽
𝑥4𝑠
𝑘𝐽
(6.658 𝑘𝑔.𝐾) − (1.9922 𝑘𝑔.𝐾)
𝑠4𝑠 − 𝑠4
=
=
= 0.9895
𝑘𝐽
𝑘𝐽
𝑠𝑔 − 𝑠𝑓
(6.7080 𝑘𝑔.𝐾) − (1.9922 𝑘𝑔.𝐾)
We can thus get ℎ4𝑠 by consulting the tables to find
𝑘𝐽
𝑘𝐽
𝑘𝐽
ℎ4𝑠 = ℎ𝑓 + 𝑥4𝑠 ℎ𝑓𝑔 = (697.22 ) + (0.9895) (2066.3 ) = 2741.8
𝑘𝑔
𝑘𝑔
𝑘𝑔
Now
(ℎ3 − ℎ4 )
𝜂𝑡 =
(ℎ3 − ℎ4𝑠 )
So
ℎ4 = ℎ3 − 𝜂𝑡 (ℎ3 − ℎ4𝑠 )
𝑘𝐽
𝑘𝐽
𝑘𝐽
= (3348.4 ) − (0.88) ((3348.4 ) − (2741.8 ))
𝑘𝑔
𝑘𝑔
𝑘𝑔
𝑘𝐽
= 2814.6
𝑘𝑔
Now, state 5 is after the reheat, which was isobaric at 𝑃4 = 𝑃5 = 700𝑘𝑃𝑎, and the
reheating returns thetemperature to𝑇5 = 480℃. From interpolation of the superheat
𝑘𝐽
𝑘𝐽
tables, we find ℎ5 = 3361.15 𝑘𝑔,𝑠5 = 6.73695 𝑘𝑔.𝐾. After expansion in the second turbine,
𝑘𝐽
we have 𝑠𝑠6 = 𝑠5 = 6.73695 𝑘𝑔.𝐾.
And we were given 𝑃6 = 8𝑘𝑃𝑎. We consult the saturation tables to find at this pressure
𝑘𝐽
𝑘𝐽
𝑠𝑓 = 0.5926 𝑘𝑔𝐾,𝑠𝑔 = 8.2287 𝑘𝑔𝐾. Thus,
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POWERPLANT/WEEK 2-4
2012
𝑘𝐽
𝑥6𝑠
𝑘𝐽
𝑠6𝑠 − 𝑠5 (6.73695 𝑘𝑔.𝐾) − (0.5926 𝑘𝑔.𝐾)
=
=
= 0.8046
𝑘𝐽
𝑘𝐽
𝑠𝑔 − 𝑠𝑓
(8.2287 𝑘𝑔.𝐾) − (0.5926 𝑘𝑔.𝐾)
The tables then give the necessary information to compute ℎ6𝑠 :
𝑘𝐽
𝑘𝐽
𝑘𝐽
ℎ6𝑠 = ℎ𝑓 + 𝑥6𝑠 ℎ𝑓𝑔 = (173.88 ) + (0.8046) (2403.1 ) = 2107.52
𝑘𝑔
𝑘𝑔
𝑘𝑔
Now, the actual ℎ6 is found via:
ℎ6 = ℎ5 − 𝜂𝑡 (ℎ5 − ℎ6𝑠 )
𝑘𝐽
𝑘𝐽
𝑘𝐽
= (3361.15 ) − (0.88) ((3361.15 ) − (2107.52 ))
𝑘𝑔
𝑘𝑔
𝑘𝑔
𝑘𝐽
= 2257.96
𝑘𝑔
Now, the tables give us
𝑘𝐽
𝑚3
ℎ1 = ℎ𝑓 = 173.88
, 𝑣1 = 𝑣𝑓 = 0.001084
𝑘𝑔
𝑘𝑔
For the pump, we have 𝑃1 = 𝑃6 = 8𝑘𝑃𝑎 and 𝑃2 = 𝑃3 = 8000𝑘𝑃𝑎.
So
𝑤𝑠
𝜂𝑝 =
𝑤𝑝
𝑤𝑠
𝑤𝑝 =
𝜂𝑝
𝑣1 (𝑃2 − 𝑃1 )
=
𝜂𝑝
𝑚3
=
(0.001084 𝑘𝑔
0.8
= 10.83
𝑘𝐽
𝑘𝑔
So
ℎ2 = ℎ1 + 𝑤𝑝
𝑘𝐽
𝑘𝐽
= (173.88 ) + (10.83 )
𝑘𝑔
𝑘𝑔
𝑘𝐽
= 184.709
𝑘𝑔
Now, substitute all these values into 𝑤𝑛𝑒𝑡 and get
𝑤𝑛𝑒𝑡 = ((3348.4
𝑘𝐽
𝑘𝐽
𝑘𝐽
𝑘𝐽
) − (2814.6 )) + ((3361.15 ) − (2257.96 ))
𝑘𝑔
𝑘𝑔
𝑘𝑔
𝑘𝑔
+ ((173.88
𝑘𝐽
𝑘𝐽
) − (184 ))
𝑘𝑔
𝑘𝑔
= 1626.11
On a mass basis, we have
8
𝑘𝐽
𝑘𝑔
POWERPLANT/WEEK 2-4
𝑊̇ = 𝑚̇𝑤𝑛𝑒𝑡 = (2.63 &times; 105
2012
ℎ𝑟
𝑘𝐽
)(
) (1626.11 ) = 1.19 &times; 105 𝑘𝑊
𝑠
3600𝑠
𝑘𝑔
𝑘𝑔
From 𝑞𝑖𝑛 , the heat added is
𝑞𝑖𝑛 = ((3348.4
𝑘𝐽
𝑘𝐽
𝑘𝐽
𝑘𝐽
) − (184.709 )) + ((3361.15 ) − (2814.6 ))
𝑘𝑔
𝑘𝑔
𝑘𝑔
𝑘𝑔
= 3710.18
𝑘𝐽
𝑘𝑔
So the cycle’s thermal efficiency is
𝑘𝐽
𝜂𝑡ℎ
𝑤𝑛𝑒𝑡 1626.11 𝑘𝑔
=
=
𝑘𝐽 = 0.4382
𝑞𝑖𝑛
3710.18 𝑘𝑔
The heat per unit mass rejected in the condenser is from 𝑞𝑜𝑢𝑡 :
𝑞𝑜𝑢𝑡 = (2257.96
𝑘𝐽
𝑘𝐽
𝑘𝐽
) − (173.88 ) = 2084.08
𝑘𝑔
𝑘𝑔
𝑘𝑔
So the power rejected as heat is
𝑄𝐿 = 𝑚̇𝑞𝑜𝑢𝑡 = (2.63 &times; 105
𝑘𝑔
ℎ𝑟
𝑘𝐽
)(
) &times; (2084.08 ) = 1.52 &times; 105 𝑘𝑊
𝑠
3600𝑠
𝑘𝑔
Example 2-3 (H.W)
Repeat the previous analysis without reheat.
Example 2-4
A Rankine power cycle with water as the working fluid has 𝜂𝑡 = 𝜂𝑃 = 0.88. The turbine inlet
pressure and temperature are at 𝑃3 = 1200𝑝𝑠𝑖𝑎, 𝑇3 = 1000℉. The condenser pressure is
at 𝑃3 = 1 𝑝𝑠𝑖𝑎. The steam generator provides 𝑄̇𝐻 = 2 &times; 109 𝐵𝑡𝑢/ℎ𝑟. In the condenser the
cooling water enters at 60 ℉, and we wish to keep the exit cooling water temperature at
80 ℉.
Find the net power, the thermal efficiency, and the mass flow rate of cooling water, 𝑚̇𝑐𝑤 .
Answer:
We interpolate the steam tables to find (from super-heated vapour tables)
9
POWERPLANT/WEEK 2-4
ℎ3 = 1499.6
𝐵𝑡𝑢
,
𝑙𝑏𝑚
𝑠3 = 1.6297
For an isentropic turbine, we have 𝑠4𝑠 = 𝑠3 = 1.6297
4𝑠 is a two-phase mixture:
𝐵𝑡𝑢
𝑙𝑏𝑚. &deg;𝑅
𝐵𝑡𝑢
𝑙𝑏𝑚.&deg;𝑅
𝐵𝑡𝑢
𝑥4𝑠
2012
. At 𝑃4 = 1𝑝𝑠𝑖𝑎, we find state
𝐵𝑡𝑢
𝑠4 − 𝑠𝑓 (1.6297 𝑙𝑏𝑚.&deg;𝑅) − (0.1327 𝑙𝑏𝑚.&deg;𝑅)
=
=
= 0.811
𝑠𝑔 − 𝑠𝑓 (1.9779 𝐵𝑡𝑢 ) − (0.1327 𝐵𝑡𝑢 )
𝑙𝑏𝑚.&deg;𝑅
𝑙𝑏𝑚.&deg;𝑅
Thus,
ℎ4𝑠 = ℎ𝑓 + 𝑥4𝑠 ℎ𝑓𝑔 = (69.74
𝐵𝑡𝑢
𝐵𝑡𝑢
𝐵𝑡𝑢
) + (0.811) (1036
) = 909.9
𝑙𝑏𝑚
𝑙𝑏𝑚
𝑙𝑏𝑚
Now, for the actual turbine, we get
ℎ4 = ℎ3 + 𝜂𝑡 (ℎ3 − ℎ4𝑠 )
𝐵𝑡𝑢
𝐵𝑡𝑢
𝐵𝑡𝑢
= (1499.7
) − (0.88) ((499.7
) − (909.9
))
𝑙𝑏𝑚
𝑙𝑏𝑚
𝑙𝑏𝑚
𝐵𝑡𝑢
= 979.9
𝑙𝑏𝑚
Now, after the condenser, we take𝑥1 = 0, so ℎ1 = ℎ𝑓 , 𝑠1 = 𝑠𝑓 , and 𝑣1 = 𝑣𝑓 , all at 𝑃1 =
1 𝑝𝑠𝑖𝑎. These are
𝐵𝑡𝑢
𝐵𝑡𝑢
𝑓𝑡 3
ℎ1 = 69.74
, 𝑠1 = 0.1327
, 𝑣1 = 0.01614
𝑙𝑏𝑚
𝑙𝑏𝑚. &deg;𝑅
𝑙𝑏𝑚
Now
𝑤𝑝 =
𝑓𝑡 3
𝑤𝑠
𝜂𝑝
𝑙𝑏𝑓
𝑙𝑏𝑓
𝑣(𝑃4 − 𝑃3 ) (0.01614 𝑙𝑏𝑚) ((1200 𝑖𝑛2 ) − (1 𝑖𝑛2 )) 144𝑖𝑛2
𝐵𝑡𝑢
=
=
𝜂𝑝
0.88
𝑓𝑡 2 778 𝑓𝑡 𝑙𝑏𝑓
𝐵𝑡𝑢
= 4.07
𝑙𝑏𝑚
Now
ℎ2 = ℎ1 + 𝑤𝑝 = (69.74
𝐵𝑡𝑢
𝐵𝑡𝑢
𝐵𝑡𝑢
) + (4.07
) = 73.81
𝑙𝑏𝑚
𝑙𝑏𝑚
𝑙𝑏𝑚
In the boiler, we have
𝑄̇𝐻 = 𝑚̇(ℎ3 − ℎ2 )
10
POWERPLANT/WEEK 2-4
2012
𝐵𝑡𝑢
2 &times; 109
𝑄̇𝐻
ℎ𝑟
ℎ𝑟
𝑚̇ =
=
(ℎ3 − ℎ2 ) (1499.7 𝐵𝑡𝑢 ) − (73.81 𝐵𝑡𝑢 ) 3600𝑠
𝑙𝑏𝑚
= 390
𝑙𝑏𝑚
𝑙𝑏𝑚
𝑠
We also note
𝑄̇𝐻 = (2 &times; 109
𝐵𝑡𝑢
ℎ𝑟
𝐵𝑡𝑢
)
= 5.556 &times; 105
ℎ𝑟 3600 𝑠
𝑠
Now, the net power is the sum of the turbine and pump work:
𝑊̇𝑛𝑒𝑡 = 𝑚̇((ℎ3 − ℎ4 ) + (ℎ1 − ℎ2 ))
= (390
𝑙𝑏𝑚
𝐵𝑡𝑢
𝐵𝑡𝑢
𝐵𝑡𝑢
𝐵𝑡𝑢
) ((1499.6
) − (979.9
)) + ((69.74
) − (73.81
))
𝑠
𝑙𝑏𝑚
𝑙𝑏𝑚
𝑙𝑏𝑚
𝑙𝑏𝑚
= 2.01 &times; 105
𝐵𝑡𝑢
𝑠
The thermal efficiency is thus
𝐵𝑡𝑢
2.01 &times; 105 𝑠
𝑊̇𝑛𝑒𝑡
𝜂=
=
𝐵𝑡𝑢 = 0.3618
𝑄̇𝐻
5.556 &times; 105 𝑠
The cooling water and the water in the Rankine cycle exchange heat in the condenser. This
is sketched in Fig. 2-5. The first law for the heat exchanger is
Rankin cycle water
𝑚̇ℎ4
𝑚̇ℎ1
𝑇ℎ𝑜𝑡 = 80℉
𝑚̇𝑐𝑤
𝑇𝑐𝑜𝑙𝑑 = 60℉
Cooling water
𝑚̇𝑐𝑤
Fig. 2-5: Rankine cycle condenser/heat exchanger.
𝑑𝐸𝐶𝑉
= 0 = 𝑄̇𝐶𝑉 − 𝑊̇𝐶𝑉 + 𝑚̇(ℎ4 − ℎ1 ) + 𝑚̇𝑐𝑤 𝑐𝑃 (𝑇𝑐𝑜𝑙𝑑 − 𝑇ℎ𝑜𝑡 )
𝑑𝑡
𝑄̇𝐶𝑉 = 0, 𝑊̇𝐶𝑉 = 0
0 = 𝑚̇(ℎ4 − ℎ1 ) + 𝑚̇𝑐𝑤 𝑐𝑃 (𝑇𝑐𝑜𝑙𝑑 − 𝑇ℎ𝑜𝑡 )
𝑚̇(ℎ4 − ℎ1 )
= 𝑚̇𝑐𝑤
𝑐𝑃 (𝑇ℎ𝑜𝑡 − 𝑇𝑐𝑜𝑙𝑑 )
11
POWERPLANT/WEEK 2-4
=
(390
𝑙𝑏𝑚
𝑠
𝐵𝑡𝑢
𝐵𝑡𝑢
) ((979 𝑙𝑏𝑚.&deg;𝑅) − (69.74 𝑙𝑏𝑚.&deg;𝑅))
𝐵𝑡𝑢
(1.0 𝑙𝑏𝑚.&deg;𝑅) ((80℉) − (60℉))
= 17730
2-3
2012
𝑙𝑏𝑚
𝑠
Losses in Rankin cycle
• Turbine: These are typically the largest losses in the system. The turbine efficiency is
defined by
𝑤𝑡
ℎ3 − ℎ4
𝜂=
=
𝑤𝑡𝑠 ℎ3 − ℎ4𝑠
Here ℎ4𝑠 and 𝑤𝑡𝑠 are the enthalpy and work the working fluid would have achieved hadthe
process been isentropic. Note this is for a control volume.
• Pump: the losses are usually much smaller in magnitude than those for turbines.
The pump efficiency for a control volume is defined by
𝑤𝑝𝑠 ℎ2𝑠 − ℎ1
𝜂=
=
𝑤𝑝
ℎ2 − ℎ1
• Piping: pressure drops via viscous and turbulent flow effects induce entropy gains in fluid
flowing through pipes. There can also be heat transfer from pipes to the surroundings and
vice versa.
• Condenser: Losses are relatively small here.
Losses will always degrade the overall thermal efficiency of the cycle.
2-4 Regeneration
In a Rankine cycle with regeneration, some steam is extracted from the turbine and used to
pre-heat the liquid which is exiting the pump. This can lead to an increased thermal
efficiency, all else being equal. The analysis is complicated by the need to take care of more
complex mass and energy balances in some components.
2-4-1 Feedwater Heating
Feedwater heating is accomplished by heating the compressed liquid in a numberof finite
steps in heat exchangers (feed heaters) by steam that is bled from the turbineat selected
stages. (See Fig. 2-6.) Modern steam power plants use between five and eight feedwater
heating stages. None is built without feedwater heating. In a regenerative cycle, the liquid
enters the steam generator at a point below point B (Fig. 2-6). An economizer section (this is
the part of the steam generator that heats the incoming fluid between points 4 and B) is still
needed. However, it is much smaller than the one that is needed for non-regenerative
cycles. The efficiency of a well-designed Rankine cycle is the closest to the efficiency of a
Carnot cycle. The three types of feedwater heaters include:
1. Open or direct-contact type
2. Closed type with drains cascaded backward
12
POWERPLANT/WEEK 2-4
2012
3. Closed type with drains pumped forward
OPEN OR DIRECT-CONTACT FEEDWATER HEATERS
The extraction steam is mixed directly with the incoming subcooled feedwater in the open
or direct-contact feedwater heater. The mixture becomes saturated water at the extraction
steam pressure. Fig 2-6 (a, b) shows the flow diagram and corresponding T-s diagram for a
Rankine cycle using two feedwater heaters—one a low-pressure feedwater heater and the
other a high-pressure feedwater heater. (The low-pressure feedwater heater is upstream of
the high pressure feedwater heater.) Normally, modern power plants use one open-type
feedwater heater and between four and seven other heaters.
1
Turbine
2 3
4
Boiler
𝑚̇2
C
𝑚̇3
10
P
P
P
8
9
5
6
7
(a)
T
1
B
2
10
9
8
7
6
𝑚̇2
3
𝑚̇3
5
4
s
(b)
Fig 2-6 :(a) Schematic flow and (b) T-s diagrams of a non-ideal superheat Rankine cycle
with two open-type feedwater heaters.
A typical feedwater heater is shown in Fig. 2.6. The condensate “saturated water” leaves the
condenser at point 5. It is pumped to point 6 to the same pressure as extraction steam at
point 3. The subcooled water at point 6 and wet steam at point 3 mix in the low-pressure
feedwater heater to produce saturated water at point 7.
13
POWERPLANT/WEEK 2-4
2012
The amount 𝑚̇3 is sufficient to saturate the subcooled water at point 6. If the
extractionsteam at point 3 were 𝑚̇′3 (where 𝑚̇′3 &gt; 𝑚̇3 ), the flow at point 7 would be a twophase mixture that would be difficult to pump. The pressure at line 6-7 (constant) cannot be
higher than the extraction steam at point 3. Otherwise, reverse flow of condensate water
would enter the turbine at point 3.
A second pump is needed to pressurize the saturated water from point 7 to a subcooled
condition at point 8, which is the pressure of extraction steam at point 2. The steam at point
10 enter the steam generator (boiler) at its pressure. A deaerator is usually added to the
open-type feedwater heaters. The mixing process increases the surface area and liberates
non-condensable gases (e.g., N2, O2, and CO2). These gases can be vented to atmosphere.
Hence, the arrangement is called deaerating heaters or DA.
The mass balance is as follows:
Mass flow between points 1 and 2 = 1.
Mass flow between points 2 and 9 =𝑚̇2 .
Mass flow between points 2 and 3 = 1 -𝑚̇2 .
Mass flow between points 3 and 7 =𝑚̇3 .
change to 3
Mass flow between points 3 and 7 = 1 - 𝑚̇2 - 𝑚̇3 .
, 4,5,6,7
Mass flow between points 7 and 9 = 1 - 𝑚̇2 .
Mass flow between points 9 and 1 = 1.
The energy balances for the high- and low-pressure feedwater heaters, respectively, are as
follows:
𝑚̇2 (ℎ2 − ℎ9 ) = (1 − 𝑚̇2 )(ℎ9 − ℎ8 )
𝑚̇3 (ℎ3 − ℎ7 ) = (1 − 𝑚̇2 − 𝑚̇3 )(ℎ7 − ℎ6 )
Heat added
𝑞𝐴 = (ℎ1 − ℎ10 )
Turbine work
𝑤𝑡 = (ℎ1 − ℎ2 ) + (1 − 𝑚̇2 )(ℎ2 − ℎ3 ) + (1 − 𝑚̇2 − 𝑚̇3 )(ℎ3 − ℎ4 )
Pump work
|∑ 𝑤𝑝 | = (1 − 𝑚̇2 − 𝑚̇3 )(ℎ6 − ℎ5 ) + (1 − 𝑚̇2 )(ℎ8 − ℎ7 ) + (ℎ10 − ℎ9 )
≅ (1 − 𝑚̇2 − 𝑚̇3 )
Heat rejected
𝑣5 (𝑃6 − 𝑃5 )
𝑣7 (𝑃8 − 𝑃7 ) 𝑣9 (𝑃10 − 𝑃9 )
+ (1 − 𝑚̇2 )
+
𝜂𝑃
𝜂𝑃
𝜂𝑃
|𝑞𝑅 | = (1 − 𝑚̇2 − 𝑚̇3 )(ℎ4 − ℎ5 )
Net cycle work
∆𝑤𝑛𝑒𝑡 = 𝑤𝑡 − |𝑤𝑝 |
Cycle thermal efficiency
𝜂𝑡ℎ =
∆𝑤𝑛𝑒𝑡
𝑞𝐴
Work ratio
𝑊𝑅 =
𝑤𝑛𝑒𝑡
𝑤𝑇
Note that:1- In the open feedwaterheater the extracted steam from the turbine is mixed directly
with the incoming subcooled water to produce saturated steam at the extraction
steam pressure.
14
POWERPLANT/WEEK 2-4
2012
2- The amount of bled steam should equal to that would saturate the subcooled water
it is going to mix with. If it is much less it may negate the advantage of the feedwater
heater. On the other hand if it is more it will affect the turbine work by causing
losses, also it would result in two-phase mixture in the pump.
3- An open type feedwater heater is treated as mixing chambers.
4- The mass flow rate in the turbine is a variable quantity in the case of feedwater
heating.
5- Besides the condensate pump there is one additional pump per open feedwater
heater.
6- Open feedwater heaters are also called deaerating heaters or DA, as the breakup of
water in the mixing process results in non-condensable gases such as air, O2, CO2,
H2.
Example 2-5
Steam leaves boiler and enters turbine at 4 𝑀𝑃𝑎, 400℃. After expansion to 400 𝑘𝑃𝑎, some
stream is extracted for heating feedwater in an open feedwater heater.Pressure in
feedwater heater is 400 𝑘𝑃𝑎, and water leaves it at a saturated state at 400 𝑘𝑃𝑎. The
restof the steam expands through the turbine to 10 𝑘𝑃𝑎. Find the cycle efficiency.
5
Turbine
Boiler
6
7
4
Feedwater
Condenser
2
1
3
Pump/P2
Pump/P1
Fig. 2-7: Schematic for Rankine cycle with regeneration and open feedwater heating.
Answer:
• 1 → 2: compression through pump 𝑃1 ,
• 2 &amp; 6 → 3: mixing in open feedwater heater to saturated liquid state,
• 3 → 4: compression through pump 𝑃2 ,
• 4 → 5: heating in boiler,
• 5 → 6: partial expansion in turbine,
• 6 → 7: completion of turbine expansion, and
• 7 → 1: cooling in condenser.
15
POWERPLANT/WEEK 2-4
2012
From the tables, one can find
𝑘𝐽
ℎ5 = 3213.6
𝑠𝑢𝑝𝑒𝑟 𝑣𝑎𝑝𝑜𝑟 𝑎𝑡 400℃ 𝑎𝑛𝑑 4𝑀𝑃𝑎,
𝑘𝑔
𝑘𝐽
𝑘𝐽
ℎ6 = 2685.6 , ℎ1 = 191.8
𝑙𝑖𝑞𝑢𝑖𝑑 𝑎𝑡 0.01𝑀𝑃𝑎, ℎ7
𝑘𝑔
𝑘𝑔
𝑘𝐽
𝑘𝐽
= 2144.1
𝑎𝑡 0.01𝑀𝑃𝑎 𝑤𝑖𝑡ℎ 𝑥7, ℎ3 = 604.73
𝑙𝑖𝑞𝑢𝑖𝑑 𝑎𝑡 400℃
𝑘𝑔
𝑘𝑔
𝑚3
𝑚3
𝑣1 = 0.00101
, 𝑣3 = 0.001084
𝑘𝑔
𝑘𝑔
First consider the low pressure pump.
𝑘𝐽
𝑚3
ℎ2 = ℎ1 + 𝑣1 (𝑃2 − 𝑃1 ) = (191.8 ) + (0.00101 ) ((400𝑘𝑃𝑎) − (10𝑘𝑃𝑎))
𝑘𝑔
𝑘𝑔
𝑘𝐽
= 192.194
𝑘𝑔
The pump work is
𝑤𝑃1 = 𝑣1 (𝑃1 − 𝑃2 ) = (0.00101)((10𝑘𝑃𝑎) − (400𝑘𝑃𝑎))
= −0.3939
Note, a sign convention consistent with work done by the fluid is used here.
Now consider the turbine
𝑑𝑚𝑐𝑣
= 𝑚̇5 − 𝑚̇6 − 𝑚̇7
𝑑𝑡
𝑑𝑚𝑐𝑣
=0
𝑑𝑡
𝑚̇5 = 𝑚̇6 + 𝑚̇7
𝑚̇6 𝑚̇7
1=
−
𝑚̇5 𝑚̇5
change - to
+
𝑑𝐸𝑐𝑣
= 𝑄̇𝑐𝑣 − 𝑊̇𝑐𝑣 + 𝑚̇5 ℎ5 − 𝑚̇6 ℎ6 − 𝑚̇7 ℎ7
𝑑𝑡
̇
𝑑𝐸𝑐𝑣
= 0,
𝑄𝑐𝑣 = 0
𝑑𝑡
𝑊̇𝑐𝑣 = 𝑚̇5 ℎ5 − 𝑚̇6 ℎ6 − 𝑚̇7 ℎ7
On a per mass basis, we get,
𝑊̇𝑐𝑣
𝑚̇6
𝑚̇7
= ℎ5 −
ℎ6 −
ℎ
𝑚̇5
𝑚̇5
𝑚̇5 7
𝑚̇6
𝑚̇6
= ℎ5 −
ℎ6 − (1 −
)ℎ
𝑚̇5
𝑚̇5 7
𝑚̇6
𝑚̇6
= ℎ5 − ℎ6 + (1 −
) ℎ6 − (1 −
)ℎ
𝑚̇5
𝑚̇5 7
𝑚̇6
= ℎ5 − ℎ6 + (1 −
) − (ℎ6 − ℎ7 )
𝑚̇5
𝑤𝑡 =
16
add and extract
h6
POWERPLANT/WEEK 2-4
2012
Now consider the feedwater heater. The first law for this device gives
𝑑𝐸𝑐𝑣
= 𝑄̇𝑐𝑣 − 𝑊̇𝑐𝑣 + 𝑚̇2 ℎ2 + 𝑚̇6 ℎ6 + 𝑚̇3 ℎ3
𝑑𝑡
𝑑𝐸𝑐𝑣
= 0, 𝑄̇𝑐𝑣 = 0, 𝑊̇𝑐𝑣 = 0
𝑑𝑡
𝑚̇2
𝑚̇6
ℎ2 +
ℎ = ℎ3
𝑚̇3
𝑚̇3 6
𝑚̇7
𝑚̇6
ℎ2 +
ℎ =
𝑚̇5
𝑚̇5 6
𝑚̇6
𝑚̇6
(1 −
) ℎ2 +
ℎ =
𝑚̇5
𝑚̇5 6
𝑘𝐽
𝑚̇6
𝑘𝐽
𝑚̇6
𝑘𝐽
(604.73 ) = (1 −
) (192.194 ) +
(2685.6 )
𝑘𝑔
𝑚̇5
𝑘𝑔
𝑚̇5
𝑘𝑔
𝑚̇6
= 0.16545
𝑚̇5
Now get the turbine work
𝑤𝑡 = ℎ5 − ℎ6 + (1 −
= (3213.6
𝑚̇6
) (ℎ6 − ℎ7 )
𝑚̇5
𝑘𝐽
𝑘𝐽
𝑘𝐽
𝑘𝐽
) − (2685.6 ) + (1 − 0.16545) ((2685.6 ) − (2144.1 ))
𝑘𝑔
𝑘𝑔
𝑘𝑔
𝑘𝑔
= 979.908
𝑘𝐽
𝑘𝑔
Now get the work for the high-pressure pump
𝑤𝑝2
𝑚3
) ((400𝑘𝑃𝑎) − (4000𝑘𝑃𝑎))
𝑘𝑔
= −3.9024
= 𝑣3 (𝑃3 − 𝑃4 ) = (0.001084
Now
ℎ4 = ℎ3 + 𝑣3 (𝑃4 − 𝑃3 ) = (604.73
𝑘𝐽
𝑚3
) + (0.001084 ) ((4000𝑘𝑃𝑎) − (400𝑘𝑃𝑎))
𝑘𝑔
𝑘𝑔
𝑘𝐽
= 608.6
𝑘𝑔
Now get the net work
𝑊̇𝑛𝑒𝑡 = 𝑚̇5 𝑤𝑡 + 𝑚̇1 𝑤𝑝1 + 𝑚̇5 𝑤𝑝2
𝑚̇1
𝑤𝑛𝑒𝑡 = 𝑤𝑡 +
𝑤 + 𝑤𝑝2
𝑚̇5 𝑝1
𝑚̇7
= 𝑤𝑡 + (1 −
) 𝑤 + 𝑤𝑝2
𝑚̇5 𝑝1
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POWERPLANT/WEEK 2-4
2012
𝑚̇6
) 𝑤 + 𝑤𝑝2
𝑚̇5 𝑝1
𝑘𝐽
𝑘𝐽
𝑘𝐽
= (979.908 ) + (1 − 0.16545) (−0.3939 ) + (3.9024 )
𝑘𝑔
𝑘𝑔
𝑘𝑔
𝑘𝐽
= 975.677
𝑘𝑔
= 𝑤𝑡 + (1 −
Now for the heat transfer in the boiler, one has
𝑞ℎ = ℎ5 − ℎ4
𝑘𝐽
𝑘𝐽
= (3213.6 ) − (608.6 )
𝑘𝑔
𝑘𝑔
𝑘𝐽
= 2605.0
𝑘𝑔
Thus, the thermal efficiency is
𝑘𝐽
𝑤𝑛𝑒𝑡 975.677 𝑘𝑔
𝜂=
=
𝑘𝐽 = 0.375
𝑞ℎ
2605.0 𝑘𝑔
This does represent an increase of efficiency over a comparable Rankine cycle without
regeneration, which happens to be 0.369.
Closed-Type Feedwater Heater with Drains Cascaded Backward
1
T
2
B
4
𝑚̇3
𝑚̇2
8
3
10
7
12
5
6
P
9
11
(a)
18
C
POWERPLANT/WEEK 2-4
T
2012
1
B
2
8
7
6
5
11
9
3
12
10
4
s
(b)
Fig 2-8: (a) Schematic flow and (b) T-s diagrams of a non-ideal superheat Rankine cycle
with two closed-type feedwater heaters with drains cascaded backward.
This is the most commonly used type of feedwater heaters in power plants. It is a shell-and
tube heat exchanger. The feedwater passes through the tubes. On the shell side, the bled
steam transfers energy to the feedwater as it condenses. Feedwater heaters are very similar
to condensers, but they operate at higher pressures.
A boiler feedpump is usually placed after the deaerater. Figure 2.8 illustrates the flow
diagram (a) and the corresponding T-s diagram (b) of a non-ideal superheat Rankine cycle.
The cycle has two feedwater heaters of this type. Only one pump is needed. The bled steam
condenses in each feedwater heater. Then, it is fed back to the next lower-pressure
feedwater heater (it cascades from higher-pressure to lower-pressure heaters). Wet steam
at point 3 is admitted and transfers its energy to high-pressure subcooled water at point 6.
Example 2-6
In a regenerative steam cycle, employing two closed feed water heaters as shown in Fig (29) the steam is supplied to the turbine at 40 bar(4 MPa) and 500 oC and is exhausted to the
condenser at 0.035 bar(3.5kPa). The intermediate pressures are obtained such that the
saturation temperature intervals are approximately equal, giving pressure of 10(1MPa110kPa) and 1.1 bar. Calculate the amount of steam bleed at each stage, the work output of
the plant in kJ/kg of boiler steam and the thermal efficiency of the plant. Assume ideal
process where required.
Answer
From tables:
𝑘𝐽
𝑘𝐽
ℎ1 = 3445
, 𝑠1 = 7.089
= 𝑠2
𝑘𝑔
𝑘𝑔𝐾
At state 2
0.391 + 𝑥2 &times; 8.13 = 7.089
6.698
𝑥2 =
= 0.824
8.13
i.e.
ℎ2 = 112 + 0.824 &times; 2438 = 2117 𝑘𝐽/𝑘𝑔
Also,
ℎ3 = ℎ𝑓 𝑎𝑡 0.035 𝑏𝑎𝑟 = 112 𝑘𝐽/𝑘𝑔
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POWERPLANT/WEEK 2-4
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T
1
1kg
6
5
4
3
11
9
12
71kg
y1k
g
y2k
g
(1-y1)kg
8 (1-y1-y2)kg
2
10
s
(a)
(b)
Fig 2-9: T-s and flow diagrams of a non-ideal superheat Rankine cycle with two
closed-type feedwater heaters with drains cascaded backward.
𝑘𝐽
For the first stage of expansion, 1-7, 𝑠7 = 𝑠1 = 7.089 𝑘𝑔𝐾, and from tables at 10 bar 𝑠𝑔 &gt;
7.089, hence the steam is superheated at state 7.
By interpolation between 250oC and 300oC at 10 bar we have:
7.089 − 6.926
0.163
ℎ7 = 2944 + (
) (3052 − 2944) = 2944 +
&times; 108
7.124 − 6.926
0.198
i.e
ℎ7 = 3032.9 𝑘𝐽/𝑘𝑔
From the throttling process, 11-12, we have:
ℎ6 = ℎ11 = ℎ12 = 763𝑘𝐽/𝑘𝑔
For the second stage of expansion, 7-8, 𝑠7 = 𝑠8 = 𝑠1 = 7.089
1.1bar,𝑠𝑔 &gt; 7.089
𝑘𝐽
𝑘𝑔𝐾
, hence the steam is wet at state 8.
Therefore,
20
𝑘𝐽
𝑘𝑔𝐾
, and from the tables at
POWERPLANT/WEEK 2-4
2012
1.333 + (𝑥8 &times; 5.994) = 7.089, 𝑥8 = 0.961
i.e.
ℎ8 = 429 + (0.961 &times; 2251) = 2591𝑘𝐽/𝑘𝑔
For the throttling process, 9-10:
ℎ5 = ℎ9 = ℎ10 = 429 𝑘𝐽/𝑘𝑔
Applying an energy balance to the first feed heater, remembering that there is no work or
heat transfer:
𝑦1 ℎ7 + ℎ5 = 𝑦1 ℎ11 + ℎ6
ℎ6 − ℎ5
763 − 429
=
= 0.147
ℎ7 − ℎ11 3032.4 − 763
Similarly for the second heater, takingℎ4 = ℎ3 :
𝑦1 =
𝑦2 ℎ8 + 𝑦1 ℎ12 + ℎ4 = ℎ5 + (𝑦1 + 𝑦2 )ℎ9
𝑦2 (ℎ8 − ℎ9 ) + 𝑦1 ℎ12 + ℎ4 = ℎ5 + 𝑦1 ℎ9
𝑦2 (2591 − 429) + (0.147 &times; 763) + 112 = 429 + (0.147 &times; 429)
267.8
𝑦2 =
= 0.124
2162
The heat supplied to the boiler,𝑄1 per kg of boiler steam is given by:
𝑄1 = ℎ1 − ℎ6 = 3445 − 763 = 2682 𝑘𝐽/𝑘𝑔
The work output, neglecting pump work, is given by:
𝑊 = (ℎ1 − ℎ7 ) + (1 − 𝑦1 )(ℎ7 − ℎ8 ) + (1 − 𝑦1 − 𝑦2 )(ℎ8 − ℎ2 )
= (3445 − 3032.9) + (1 − 0.147)(3032.9 − 2591)
+ (1 − 0.147 − 0.124)(2591 − 2117)
= 412.1 + 376.9 + 345.5 = 1134.5 𝑘𝐽/𝑘𝑔
Then,
𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
𝑊 1134.5
=
= 0.423
𝑄1
2682
Closed vs. Open Feedwater Heaters
The open and closed feedwater heaters can be compared as follows:
1- Open feedwater heaters are simple and inexpensive and have good heat transfer
characteristics.
2- They also bring the feedwater to the saturation state.
3- For each heater, however, a pump is required to handle the feedwater.
4- The closed feedwater heaters are more complex because of the internal tubing
network, and thus they are more expensive.
5- Heat transfer in closed feedwater heaters is also less effective since the two streams
are not allowed to be in direct contact.
6- 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.
7- Most steam powerplant use a combination of open and closed feedwater heaters.
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POWERPLANT/WEEK 2-4
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2-5 The External Irreversible Rankine Cycle
External irreversibility is caused by the temperature differences between the primary heat
source (combustion gases or primary coolant in condenser) and the working fluid.
Temperature differences between condensing working fluid and the heat sink fluid
(condenser cooling water or cooling air) also cause external irreversibility.
T
a
b
B
1
Economizer
Section
4
2
3
c
d
s
Fig 2-8:External irreversibility with Rankine cycle.
Figure 2.8 illustrates the working fluid (line 4-B-1-2-3-4) in a Rankine cycle. Line a-b
represents the primary coolant in a counter flow steam generator, and line c-d represents
the heat sink fluid in a counter flow heat exchanger. If line a-b is too close to line 4-B-1, the
temperature differences between the primary coolant (represent here hot gases) and the
working fluid (represented here water vapor) would be small. Therefore, the irreversibilities
(caused by heat loss from the primary coolant) are small, but the steam generator is large
and costly (high pressure and temperature).
If line a-b is much higher than line 4-B-1 (significant temperature differences between the
primary coolant and the working fluid), the steam generator would be small and
inexpensive, but the overall temperature differences and irreversibilities would be large.
Hence,the plant efficiency would be reduced (losses to the ambient).
An examination of Fig. 2-8 reveals that a great deal of irreversibilities occur prior to the
pointof boiling (i.e., in the economizer section of the steam generator where the
temperature differences between line b-a and line 4-Bare the greatest of all during the
entire heat addition process).
The thermal efficiencies of all types of power plants suffer from this irreversibility, which
can be eliminated if the liquid is added to the boiler at point B instead of point 4.
{The process of regeneration achieves this objective by exchanging heat between the
expanding fluid in the turbine and the compressed fluid before heat addition}.
2-6 The Internally Irreversible Rankine Cycle
Internal irreversibilityis primarily the result of fluid friction, throttling, and mixing. The most
important irreversibilities in a cycle occur in turbines and pumps, and pressure losses occur
in heat exchangers, pipes, bends, valves, and so on. In turbines and pumps, the assumption
22
POWERPLANT/WEEK 2-4
2012
of adiabatic flow is still valid (the heat loss per unit mass is negligible). However, the flow is
not reversible. The entropy in both processes increases. This is illustrated in Fig. 2-9.
T
5
5’
1
P4
P1
4s
4
3
2s 2
s
Fig 2-9:AT-s diagram of an internally irreversible superheat Rankine cycle.
The turbine polytropic efficiency 𝜂𝑇 (sometimes called adiabatic or isentropic efficiency) is
given by
ℎ1 − ℎ2
𝜂𝑇 =
ℎ1 − ℎ2𝑠
Note: 𝜂𝑇 is different from the cycle thermal efficiency.
Well-designed turbines have high polytropic efficiencies. They are usually in the order of 90
percent. The presence of moisture in the steam reduces 𝜂𝑇 . Process 2-3 (Fig. 2-9) in the
condenser occurs at constant pressure and constant temperature (a two-phase
condensation process). The pump process is also adiabatic and irreversible. The entropy in
this process increases. It is a liquid (single-phase) process (3-4). The temperature and
enthalpy process (3-4) increases more than does the adiabatic and reversible process (3-4s).
Therefore, the pump absorbs more work in an irreversible process.
The pump polytropic efficiency 𝜂𝑃 (sometimes called adiabatic or isentropic efficiency) is
given by
(𝑖𝑑𝑒𝑎𝑙 𝑤𝑜𝑟𝑘)
ℎ4𝑠 − ℎ3
𝜂𝑝 =
=
(𝑎𝑐𝑡𝑢𝑎𝑙 𝑤𝑜𝑟𝑘)
ℎ4 − ℎ3
𝜂𝑃 is the reverse of 𝜂𝑇 .
The actual pump work is given by
ℎ4𝑠 − ℎ3 𝑣3 (𝑃4 − 𝑃3 )
=
𝜂𝑃
𝜂𝑃
The liquid leaving the pump is at a higher pressure P4 than the turbine inlet P5 (due to
friction throughout the system). The steam leaving the steam generator at point 5 enters
the turbine at point 1. (See Fig. 2-9.) The pressure drop between points 5 and 1 is the result
of the combined effects of friction and heat losses. Point 5′ represents the frictional effects
in the pipe connecting the steam generator (boiler) and turbine, including the turbine
throttle valve. Heat losses from that pipe reduce the entropy to 1.
𝑊𝑃 =
2-7 Efficiency and Heat Rate
23
POWERPLANT/WEEK 2-4
2012
The actual thermal efficiency of power plants is less than those computed earlier because
the analysis did not include the various auxiliaries used in a power plant and the various
irreversibilities associated with them.
The gross efficiency is calculated using the gross power of the turbine generator. This is the
power [in megawatts (MW)] that is produced before supplying the internal equipment of
the power plant (e.g., pumps, compressors, fuel-handling equipment, computers, etc.).The
net efficiency is calculated based on the net power of the plant (the gross power minus the
power needed for the internal equipment of the plant).
2-8 Cogenerations
Cogeneration is the simultaneous generation of electricity and steam (or heat) in a
powerplant. Cogeneration is recommended for industries and municipalities because it can
produce electricity more cheaply and/or more conveniently than a utility. Also, it provides
the total energy needs (heat and electricity) for the industry or municipality.
Notes:
1- I all the cycles discussed so far, the sole purpose was to convert a portion of the heat
transferred to the working fluid to work, which is the most valuable form of energy.
2- The remaining portion of the heat is rejected to the rivers, lakes, oceans, or the
atmosphere as waste heat, because its quality (or grade) is too low to be of any
practical use. Wasting a large amount of heat is a price we have to pay to produce
work, because electrical or mechanical work is the only form of energy on which
many engineering devices) such as a fan) can operate.
3- Many systems or devices, however, require energy input in the form of heat, called
PROCESS HEAT as shown in Fig 2-10.
Fig 2-10: Process heat plant.
4- Some industries that rely heavily on process heat are chemical, pulp and paper, oil
production and refining, steel making, food processing, and textile industries.
Process heat in these industries is usually supplied by steam at 5 to 7 atm (506.625
to 709.275kPa) and 150 to 200 oC.
5- Energy is usually transferred to the steam by burning coal, oil, natural gas, or
another fuel in a furnace. All the heat transferred to the steam in the boiler is used in
the process-heating units. Industries that use large amounts of process heat also
consume a large amount of electric power. Therefore, it makes economical as well as
engineering sense to use the already-existing work potential to produce power
instead of letting it go to waste.
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POWERPLANT/WEEK 2-4
2012
6- The result is a plant that produces electricity while meeting the process-heat
requirements of certain industrial processes. Such plant is called a cogeneration
plant
7- In general, cogeneration is the production of more than one useful from of energy
(such as process heat and electric power) from the same energy sources. The ideal
steam-turbine cogeneration plant described above is not practical because it cannot
adjust to the variations in power and process-heat loads.
8- The schematic of a more practical but more complex cogeneration plant is shown in
Fig 2-11. Under normal operation, some steam is extracted from the turbine at some
predetermined intermediate pressure 𝑃6 .The rest of the steam expands to the
condenser 𝑃7 and is then cooled at constant pressure.
9- At times of high demand for process heat, all the steam is routed to the processheating units and none to the condenser 𝑚7 = 0. If this not sufficient, some steam
leaving the boiler is throttled by an expansion or pressure-reducing valve to the
extraction pressure 𝑃6 and is directed to the process-heating unit. Maximum
process heating is realized when all the steam leaving the boiler passes through the
expansion valve (𝑚5 = 𝑚4 ). No power is produced in this mode.
10- When there is no demand for process heat, all the steam passes through the turbine
and the condenser (𝑚6 = 𝑚5 = 0), and the cogeneration plant operates as an
ordinary steam power plant.
11- Under optimum conditions, a cogeneration plant simulates the ideal cogeneration
plant discussed earlier. That is all the steam expands in the turbine to the extraction
pressure and continues to the process-heating unit. No steam passes through the
expansion valve or the condenser; thus, no waste heat is rejected (𝑚6 =
𝑚4 𝑎𝑛𝑑 𝑚5 = 𝑚7 = 0). Thus condition may be difficult to achieve in practice
because of the constant variations in the process-heat and power loads. But the
plant should be designed so that the optimum operation conditions are
approximated most of the time.
Fig 2-11: Cogeneration cycle.
25
POWERPLANT/WEEK 2-4
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Fig 2-12: The topping cycle: Combined gas-vapor power cycle
13. A more popular modification involves a gas power cycle topping a vapour power
cycle, which is called the combined gas-vapor cycle, or just the combined cycle Fig 212. The combined cycle of greatest interest is the gas-turbine (Brayton) cycle topping
a steam turbine (Rankine) cycle, which has a higher thermal efficiency than either of
the cycles executed individually. Gas turbine cycles typically operate at considerably
higher temperature than steam cycles. The maximum fluid temperature at the
turbine inlet is about 620oC for modern steam power plants, but over 1425oC for gasturbine plants.
2-8-1 Types of Cogeneration
The two main categories of cogeneration are
1-The topping cycle and,
2-The bottoming cycle.
The Topping Cycle, in this cycle, the primary heat source is used to generate high enthalpy
steam and electricity. Depending on process requirements, process steam at lowenthalpy is
taken from any of the following:
1- Extracted from the turbine at an intermediate stage (like feedwater heating).
2- Taken from the turbine exhaust. The turbine in this case is called a back-pressure turbine.
Process steam requirements vary widely, between 0.5 and 40 bar.
The Bottoming Cycle:In this cycle, the primary heat (high enthalpy) is used directly for
process requirements [e.g., for a high-temperature cement kiln (furnace)]. The low enthalpy
waste heat is then used to generate electricity at low efficiency. This cycle has lower
26
POWERPLANT/WEEK 2-4
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combined efficiency than the topping cycle. Thus, it is not very common. Only the topping
cycle can provide true savings in primary energy.
Summary: Arrangements of Cogeneration Plants
The various arrangements for cogeneration in a topping cycle are as follows:
1- Steam-electric power plant with a back-pressure turbine.
2- Steam-electric power plant with steam extraction from a condensing turbine.
3- Gas turbine power plant with a heat recovery boiler (using the gas turbine exhaust to
generate steam).
4- Combined steam-gas-turbine cycle power plant. The steam turbine is either of the back
pressure type or of the extraction-condensing type.
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POWERPLANT/WEEK 2-4
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Example 2-7
Consider the cogeneration plant shown in Fig. 2–13. Steam enters the turbine at 7 MPa and
500&deg;C. Some steam is extracted from the turbine at 500 kPa for process heating. The
remaining steam continues to expand to 5 kPa. Steam is then condensed at constant
pressure and pumped to the boiler pressure of 7 MPa. At times of high demand for process
heat, some steam leaving the boiler is throttled to 500 kPa and is routed to the process
heater. The extraction fractions are adjusted so that steam leaves the process heater as a
saturated liquid at 500 kPa. It is subsequently pumped to 7 MPa. The mass flow rate of
steam through the boiler is 15 kg/s. Disregarding any pressure drops and heat losses in the
piping and assuming the turbine and the pump to be isentropic, determine
a. the maximum rate at which process heat can be supplied,
b. the power produced and the utilization factor when no process heat is supplied, and
c. the rate of process heat supply when 10 percent of the steam is extracted before it
enters the turbine and 70 percent of the steam is extracted from the turbine at 500
kPa for process heating.
Solution
A cogeneration plant is considered. The maximum rate of process heat supply, the power
produced and the utilization factor when no process heat is supplied, and the rate of
process heat supply when steam is extracted from the steam line and turbine at specified
ratios are to be determined.
Analysis:
The schematic of the cogeneration plant and the T-s diagram of the cycle are shown in Fig.
2–13. The power plant operates on an ideal cycle and thus the pumps and the turbines are
isentropic; there are no pressure drops in the boiler, process heater, and condenser; and
steam leaves the condenser and the process heater as saturated liquid.
The work inputs to the pumps and the enthalpies at various states are as follows:
1
2
B
Expansion
valve
4
3
1, 2, 3
T
5
6
Process
heater
11
10
Mixing
Chamber
P2
7
10
11 7
9
8
C
9
P1
8
Fig 2-13: Schematic and T-s diagrams for example 2-7.
28
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5
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POWERPLANT/WEEK 2-4
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0.001005𝑚3
&times; (7000𝑘𝑃𝑎 − 5𝑘𝑃𝑎) = 7.03𝑘𝐽/𝑘𝑔
𝑘𝑔
0.001093𝑚3
= 𝑣7 (𝑃10 − 𝑃7 ) =
&times; (7000𝑘𝑃𝑎 − 500𝑘𝑃𝑎) = 7.10𝑘𝐽/𝑘𝑔
𝑘𝑔
ℎ1 = ℎ2 = ℎ3 = ℎ4 = 3411.4 𝑘𝐽/𝑘𝑔
ℎ5 = 2739.3 𝑘𝐽/𝑘𝑔
ℎ6 = 2073 𝑘𝐽/𝑘𝑔
ℎ7 = ℎ𝑓 𝑎𝑡 500𝑘𝑃𝑎 = 640.09 𝑘𝐽/𝑘𝑔
𝑤𝑝𝑢𝑚𝑝1,𝑖𝑛 = 𝑣8 (𝑃9 − 𝑃8 ) =
𝑤𝑝𝑢𝑚𝑝2,𝑖𝑛
ℎ8 = ℎ𝑓 𝑎𝑡 5𝑘𝑃𝑎 = 137.75 𝑘𝐽/𝑘𝑔
ℎ9 = ℎ8 + 𝑤𝑝𝑢𝑚𝑝1,𝑖𝑛 = (137.75 + 7.03) = 144.78 𝑘𝐽/𝑘𝑔
ℎ10 = ℎ7 + 𝑤𝑝𝑢𝑚𝑝2,𝑖𝑛 = (640.09 + 7.10) = 647.19 𝑘𝐽/𝑘𝑔
(a) The maximum rate of process heat is achieved when all the steam leaving the boiler
is throttled and sent to the process heater and none is sent to the turbine (that is
𝑘𝑔
𝑚̇4 = 𝑚̇7 = 𝑚̇1 = 15 𝑠 and 𝑚̇3 = 𝑚̇5 = 𝑚̇6 = 0).
Thus:
15𝑘𝑔
𝑄̇𝑃,𝑚𝑎𝑥 = 𝑚̇1 (ℎ4 − ℎ7 ) = (
) &times; (3411.4 − 640.09) = 41.5 𝑘𝑊
𝑠
The utilization factor is 100 percent in this case since no heat is rejected in the
condenser, heat losses from the piping and other components are assumed to be
negligible, and combustion losses are not considered.
(b) When no process heat is supplied, all the steam leaving the boiler passes through the
turbine and expands to the condenser pressure of 5 kPa (that is,𝑚̇3 = 𝑚̇6 = 𝑚̇1 =
15 𝑘𝑔/𝑠and 𝑚̇2 = 𝑚̇5 = 0. Maximum power is produced in this mode, which is
determined to be:
𝑘𝑔
&times; (3411.4 − 2073) = 20076 𝑘𝐽/𝑘𝑔
𝑠
𝑘𝑔
(7.03) = 105 𝑘𝑊
𝑊̇𝑝𝑢𝑚𝑝,𝑖𝑛 = 15
𝑠
𝑊̇𝑛𝑒𝑡,𝑜𝑢𝑡 = 𝑊̇𝑛𝑒𝑡,𝑜𝑢𝑡 − 𝑊̇𝑝𝑢𝑚𝑝,𝑖𝑛 = 20076 − 105 = 19971𝑘𝑊
15𝑘𝑔
𝑄̇𝑖𝑛 = 𝑚̇1 (ℎ1 − ℎ11 ) =
&times; (3411.4 − 144.78) = 48999𝑘𝑊
𝑠
𝑊̇𝑡𝑢𝑟𝑏,𝑜𝑢𝑡 = 𝑚̇(ℎ3 − ℎ6 ) = 15
Thus, the utilization factor for cogeneration plant:
𝜖𝑢 =
𝑁𝑒𝑡 𝑤𝑜𝑟𝑘 𝑜𝑢𝑡𝑝𝑢𝑡 + 𝑃𝑟𝑜𝑐𝑒𝑠𝑠 ℎ𝑒𝑎𝑡 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑
𝑊̇𝑛𝑒𝑡 + 𝑄̇𝑃 (19971 + 0)
=
=
= 0.408
𝑇𝑜𝑡𝑎𝑙 ℎ𝑒𝑎𝑡 𝑖𝑛𝑝𝑢𝑡
48999𝑘𝑊
𝑄̇𝑖𝑛
That is, 40.8 percent of the energy is utilized for a useful purpose. Notice that the utilization
factor is equivalent to the thermal efficiency in this case.
29
POWERPLANT/WEEK 2-4
2012
(c) Neglecting any kinetic and potential energy changes, an energy balance on the
process heater yields
𝐸̇𝑖𝑛 = 𝐸̇𝑜𝑢𝑡
𝑚̇4 ℎ4 + 𝑚̇5 ℎ5 = 𝑄̇𝑝,𝑜𝑢𝑡 + 𝑚̇7 ℎ7
Or
𝑄̇𝑝,𝑜𝑢𝑡 = 𝑚̇4 ℎ4 + 𝑚̇5 ℎ5 − 𝑚̇7 ℎ7
Where
𝑘𝑔
= 1.5𝑘𝑔/𝑠
𝑠
𝑘𝑔
𝑚̇5 = 0.7 &times; 15
= 10.5𝑘𝑔/𝑠
𝑠
𝑚̇7 = 𝑚̇4 + 𝑚̇5 = 12𝑘𝑔/𝑠
𝑚̇4 = 0.1 &times; 15
Thus
𝑄̇𝑝,𝑜𝑢𝑡 = 1.5
𝑘𝑔
3411.4𝑘𝐽
&times;(
) + (10.5 𝑘𝑔/𝑠) &times; (2739.3 𝑘𝐽/𝑘𝑔) − (12 𝑘𝑔/𝑠)
𝑠
𝑘𝑔
&times; (640.09𝑘𝐽/𝑘𝑔) = 26.2 𝑘𝑊
H.W
Note that 26.2 MW of the heat transferred will be utilized in the process heater. We could
also show that 11.0 MW of power is produced in this case, and the rate of heat input in the
boiler is 43.0 MW. Thus the utilization factor is 86.5 percent.
𝜖𝑢 =
𝑊̇𝑛𝑒𝑡 + 𝑄̇𝑃 (11.0 + 26.2)
=
𝑀𝑊 = 0.865
43.0
𝑄̇𝑖𝑛
Example 2-8: An Actual Steam Power Cycle
A steam power plant operates on the cycle shown in Fig. 2–14. If the isentropic efficiency of
the turbine is 87 percent and the isentropic efficiency of the pump is 85 percent, determine
(a) the thermal efficiency of the cycle and
(b) the net power output of the plant for a mass flow rate of 15 kg/s.
Solution
A steam power cycle with specified turbine and pump efficiencies is considered. The
thermal efficiency and the net power output are to be determined. The schematic of the
power plant and the T-s diagram of the cycle are shown in Fig. 2-14. The temperatures and
pressures of steam at various points are also indicated on the figure. We note that the
power plant involves steady-flow components and operates on the Rankine cycle, but the
imperfections at various components are accounted for.
a) The thermal efficiency of a cycle is the ratio of the net work output to the heat input,
and it is determined as follows:
Pump work input:
𝑤𝑝,𝑖𝑛 =
𝑤𝑠,𝑝,𝑖𝑛 𝑣1 (𝑃2 − 𝑃1 ) (0.001009𝑚3 /𝑘𝑔) &times; [(16000 − 9)𝑘𝑃𝑎]
=
=
= 19 𝑘𝐽/𝑘𝑔
𝜂𝑝
𝜂𝑝
0.85
30
POWERPLANT/WEEK 2-4
2012
Turbine work output:
𝑤𝑡,𝑜𝑢𝑡 = 𝜂𝑡 &times; 𝑤𝑠,𝑡,𝑜𝑢𝑡 = 𝜂𝑡 (ℎ3 − ℎ4𝑠 ) = 0.87 &times; (3583.1 − 2115.3)𝑘𝐽/𝑘𝑔 = 1277 𝑘𝐽/𝑘𝑔
Boiler heat input:
𝑞𝑖𝑛 = ℎ2 − ℎ1 = 3647.6 − 160.1 = 3487.5 𝑘𝐽/𝑘𝑔
Thus,
𝑤𝑛𝑒𝑡 = 𝑤𝑡,𝑜𝑢𝑡 − 𝑤𝑝,𝑖𝑛 = 1277 − 19 = 1258 𝑘𝐽/𝑘𝑔
𝜂𝑡ℎ =
𝑤𝑛𝑒𝑡
1258
=
= 0.361
𝑞𝑖𝑛
3487.5
b) The power produced by this power plant is
𝑊̇𝑛𝑒𝑡 = 𝑚̇𝑤𝑛𝑒𝑡 = 15 &times; 1258 = 18.9𝑀𝑊
Without the irreversibilities, the thermal efficiency of this cycle would be 43.0 percent
B
2
16MPa
3
15 MPa,600oC
3
T
4
10Kpa,38oC
P
1
9KPa,38oC
T
2s
C
2
1
4s
4
s
Fig 2–14: Schematic and T-s diagram for Example 2-8.
31
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