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Rankin Cycle week 2-4

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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
πœ‚=
π‘ŠΜ‡π‘›π‘’π‘‘
𝑄̇𝐻
=
π‘ŠΜ‡π‘‘π‘’π‘Ÿπ‘π‘–π‘›π‘’ −π‘ŠΜ‡π‘π‘’π‘šπ‘
π‘„Μ‡π‘π‘œπ‘–π‘™π‘’π‘Ÿ
(2-1)
This reduces to
πœ‚=
π‘š((β„Ž3 −β„Ž4 )+(β„Ž1 −β„Ž2 ))
π‘š(β„Ž3 −β„Ž2 )
β„Ž −β„Ž
= 1 − β„Ž4 −β„Ž1
=1−
3
π‘žπ‘œπ‘’π‘‘,π‘π‘œπ‘›π‘‘π‘’π‘›π‘ π‘’π‘Ÿ
π‘žπ‘–π‘›,π‘π‘œπ‘–π‘™π‘’π‘Ÿ
(2-2)
2
(2-3)
Note that because the Rankine cycle is not a Carnot cycle, we have
π‘žπ‘œπ‘’π‘‘,π‘π‘œπ‘›π‘‘π‘’π‘›π‘ π‘’π‘Ÿ
π‘žπ‘–π‘›,π‘π‘œπ‘–π‘™π‘’π‘Ÿ
𝑇
≠ 𝑇1
(2-4)
3
1
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
β„Ž2 − β„Ž1 = 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.
2
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
πœ‚π‘‘π‘’π‘Ÿπ‘π‘–π‘›π‘’ =
(β„Ž3 − β„Ž2 )
(β„Ž3 − β„Ž4𝑠 )
• 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,
πœ‚π‘π‘’π‘šπ‘ =
(β„Ž2𝑠 − β„Ž1 )
(β„Ž2 − β„Ž1 )
• 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
3
POWERPLANT/WEEK 2-4
2012
of the turbine.
Example 2-1
Consider steam in an ideal Rankine cycle. Saturated vapor enters the turbine at 8.0π‘€π‘ƒπ‘Ž.
Saturated liquid exits the condenser at 𝑃 = 0.008π‘€π‘ƒπ‘Ž. Thenet power output of the cycle is
100π‘€π‘Š. 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 π‘„Μ‡π‘œπ‘’π‘‘ 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 π‘€π‘ƒπ‘Ž, and π‘₯3 =
1 (saturatedsteam). This gives two properties to fix the state, so that
π‘˜π½
π‘˜π½
β„Ž3 = 2758 , 𝑠3 = 5.7432
π‘˜π‘”
π‘˜π‘”. 𝐾
π‘˜π½
State 4 has 𝑃4 = 0.008π‘€π‘ƒπ‘Ž and 𝑠4 = 𝑠3 = 5.7432 π‘˜π‘”.𝐾 so the state is fixed. From the
saturationtables, it is found then that
π‘˜π½
π‘˜π½
𝑠4 − 𝑠𝑓 (5.7432 π‘˜π‘”.𝐾) − (0.5926 π‘˜π‘”.𝐾)
π‘₯4 =
=
= 0.6745
π‘˜π½
𝑠𝑔 − 𝑠𝑓
(7.6361 π‘˜π‘”.𝐾)
Note the quality is 0 ≤ π‘₯4 ≤ 1, as it must be. The enthalpy is then
π‘˜π½
π‘˜π½
π‘˜π½
β„Ž4 = β„Žπ‘“ + π‘₯4 β„Žπ‘“π‘” = (173.88 ) + (0.6745) (2403.1 ) = 1794.8
π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
State 1 is saturated liquid at 0.008π‘€π‘ƒπ‘Ž, so π‘₯1 = 0, 𝑃1 = 0.008π‘€π‘ƒπ‘Ž. One then gets β„Ž1 =
π‘˜π½
π‘š3
β„Žπ‘“ = 173.88 π‘˜π‘”, 𝑣1 = 𝑣𝑓 = 0.0010084 π‘˜π‘” .
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
π‘ŠΜ‡
β„Ž2 = β„Ž1 + = β„Ž1 + 𝑣1 (𝑃2 − 𝑃1 )
π‘šΜ‡
π‘˜π½
π‘š3
π‘˜π½
β„Ž2 = (173.88 ) + (0.0010084 ) (8000π‘˜π‘ƒπ‘Ž − 8π‘˜π‘ƒπ‘Ž) = 181.94
π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
The net power is
π‘ŠΜ‡π‘π‘¦π‘π‘™π‘’ = π‘ŠΜ‡π‘‘ + π‘ŠΜ‡π‘ƒ
Now the first law for the turbine and pump give
π‘Šπ‘Μ‡
π‘ŠΜ‡ 𝑑
= β„Ž3 − β„Ž4 ,
= β„Ž1 − β„Ž2
π‘šΜ‡
π‘šΜ‡
The energy input that is paid for is
𝑄̇𝑖𝑛
= β„Ž3 − β„Ž2
π‘šΜ‡
The thermal efficiency is then found by
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POWERPLANT/WEEK 2-4
πœ‚=
π‘ŠΜ‡π‘‘ + π‘ŠΜ‡π‘ (β„Ž3 − β„Ž4 ) + (β„Ž1 − β„Ž2 )
=
β„Ž3 − β„Ž2
𝑄̇𝑖𝑛
π‘˜π½
=
2012
π‘˜π½
π‘˜π½
π‘˜π½
π‘˜π½
π‘˜π½
((2758 π‘˜π‘”) − (1794.8 π‘˜π‘”) + (173.88 π‘˜π‘”) − (181.94 π‘˜π‘”))
((2758 π‘˜π‘”) − (181.94 π‘˜π‘”))
= 0.371
By definition the back work ratio π‘π‘€π‘Ÿ is the ratio of pump work to turbine work:
π‘ŠΜ‡π‘
π‘π‘€π‘Ÿ = | |
π‘ŠΜ‡π‘‘
β„Ž1 − β„Ž2
=|
|
β„Ž3 − β„Ž4
π‘˜π½
= ||
π‘˜π½
((173.88 π‘˜π‘”) − (181.94 π‘˜π‘”))
π‘˜π½
π‘˜π½
((2758 π‘˜π‘”) − (1794.8 π‘˜π‘”))
|
|
= 0.00837
The desired mass flow can be determined since we know the desired net power. Thus
π‘ŠΜ‡π‘π‘¦π‘π‘™π‘’
π‘šΜ‡ =
(β„Ž3 − β„Ž4 ) + (β„Ž1 − β„Ž2 )
100 × 103 π‘˜π‘Š
π‘šΜ‡ =
π‘˜π½
π‘˜π½
π‘˜π½
π‘˜π½
((2758 π‘˜π‘”) − (1794.8 π‘˜π‘”) + (173.88 π‘˜π‘”) − (181.94 π‘˜π‘”))
= 104.697
π‘˜π‘”
𝑠
π‘˜π‘”
𝑠
π‘˜π‘”
) (3600 ) = 3.769 × 105
𝑠
β„Žπ‘Ÿ
β„Žπ‘Ÿ
The necessary heat transfer rate in the boiler is then
𝑄̇𝑖𝑛 = π‘šΜ‡(β„Ž3 − β„Ž2 )
π‘˜π‘”
π‘˜π½
π‘˜π½
= (104.697 ) ((2758 ) − (181.94 )) = 269706π‘˜π‘Š = 269.7π‘€π‘Š
𝑠
π‘˜π‘”
π‘˜π‘”
In the condenser, one finds
π‘„Μ‡π‘œπ‘’π‘‘ = π‘šΜ‡(β„Ž1 − β„Ž4 )
π‘˜π‘”
π‘˜π½
π‘˜π½
= (104.697 ) ((173.88 ) − (1794.8 )) = −169705π‘˜π‘Š = −169.7π‘€π‘Š
𝑠
π‘˜π‘”
π‘˜π‘”
Note also for the cycle that one should find
π‘ŠΜ‡π‘π‘¦π‘π‘™π‘’ = 𝑄̇𝑖𝑛 + π‘„Μ‡π‘œπ‘’π‘‘ = (269.7π‘€π‘Š) − (169.7π‘€π‘Š) = 100π‘€π‘Š
For the condenser mass flow rate now perform a mass balance:
𝑑𝐸𝐢𝑉
= 𝑄̇𝐢𝑉 − π‘ŠΜ‡πΆπ‘‰ + π‘šΜ‡π‘ (β„Žπ‘–π‘› − β„Žπ‘œπ‘’π‘‘ ) + π‘šΜ‡(β„Ž4 − β„Ž1 )
𝑑𝑑
0 = π‘šΜ‡π‘ (β„Žπ‘–π‘› − β„Žπ‘œπ‘’π‘‘ ) + π‘šΜ‡(β„Ž4 − β„Ž1 )
(β„Ž4 − β„Ž1 )
π‘šΜ‡π‘ =
(β„Žπ‘–π‘› − β„Žπ‘œπ‘’π‘‘ )
= (104.697
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POWERPLANT/WEEK 2-4
(104.697
=−
π‘˜π‘”
2012
π‘˜π½
π‘˜π½
) ((1794.8 π‘˜π‘”) − (173.88 π‘˜π‘”))
𝑠
π‘˜π½
π‘˜π½
((62.99 π‘˜π‘”) − (146.68 π‘˜π‘”))
π‘˜π‘”
𝑠
π‘˜π‘” 3600 𝑠
π‘˜π‘”
= (2027.79 ) (
) = 7.3 × 106
𝑠
β„Žπ‘Ÿ
β„Žπ‘Ÿ
= 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 = 8000π‘˜π‘ƒπ‘Ž, 𝑇3 = 480℃. The water expands to 700π‘˜π‘ƒπ‘Ž, undergoes reheat, and then
expands again to 8 π‘˜π‘ƒπ‘Ž. The mass flow rate is π‘šΜ‡ = 2.63 × 105 π‘˜π‘”/β„Žπ‘Ÿ. We have πœ‚π‘‘ = 0.88
for each turbine, and πœ‚π‘ = 0.88for the pump. Find the net power generated, πœ‚π‘‘β„Ž , and the
heat transfer to the condenser.
Answer:
6
POWERPLANT/WEEK 2-4
2012
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:
𝑀𝑛𝑒𝑑 = (β„Ž3 − β„Ž4 ) + (β„Ž5 − β„Ž6 ) + (β„Ž1 − β„Ž2 )
β„Ž3 − β„Ž4 = π‘‘π‘’π‘Ÿπ‘π‘–π‘›π‘’ − 1
β„Ž5 − β„Ž6 = π‘‘π‘’π‘Ÿπ‘π‘–π‘›π‘’ − 2
β„Ž1 − β„Ž2 = π‘π‘’π‘šπ‘
Now, the heat input for the reheat is in two stages:
π‘žπ‘–π‘› = (β„Ž3 − β„Ž2 ) + (β„Ž5 − β„Ž4 )
Lastly, the heat rejection in the condenser is
π‘žπ‘œπ‘’π‘‘ = β„Ž6 − β„Ž1
Let us start at the entrance of the first turbine, at 3. We are given 𝑃3 and 𝑇3 , so we consult
thetables and find
π‘˜π½
π‘˜π½
β„Ž3 = 3348.4 ,
𝑠3 = 6.6586
π‘˜π‘”
π‘˜π‘”. 𝐾
We are given 𝑃4 = 7π‘π‘Žπ‘Ÿ = 700π‘˜π‘ƒπ‘Ž. Now, let us get the ideal behaviour of the turbine:
π‘˜π½
𝑠4𝑠 = 𝑠3 = 6.6586 π‘˜π‘”.𝐾. At this condition, we find state 4 is a two-phase mixture. At
π‘˜π½
π‘˜π½
700π‘˜π‘ƒπ‘Ž, we find 𝑠𝑓 = 1.9922 π‘˜π‘”.𝐾, 𝑠𝑔 = 6.7080 π‘˜π‘”.𝐾.
So
π‘˜π½
π‘₯4𝑠
π‘˜π½
(6.658 π‘˜π‘”.𝐾) − (1.9922 π‘˜π‘”.𝐾)
𝑠4𝑠 − 𝑠4
=
=
= 0.9895
π‘˜π½
π‘˜π½
𝑠𝑔 − 𝑠𝑓
(6.7080 π‘˜π‘”.𝐾) − (1.9922 π‘˜π‘”.𝐾)
We can thus get β„Ž4𝑠 by consulting the tables to find
π‘˜π½
π‘˜π½
π‘˜π½
β„Ž4𝑠 = β„Žπ‘“ + π‘₯4𝑠 β„Žπ‘“π‘” = (697.22 ) + (0.9895) (2066.3 ) = 2741.8
π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
Now
(β„Ž3 − β„Ž4 )
πœ‚π‘‘ =
(β„Ž3 − β„Ž4𝑠 )
So
β„Ž4 = β„Ž3 − πœ‚π‘‘ (β„Ž3 − β„Ž4𝑠 )
π‘˜π½
π‘˜π½
π‘˜π½
= (3348.4 ) − (0.88) ((3348.4 ) − (2741.8 ))
π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
π‘˜π½
= 2814.6
π‘˜π‘”
Now, state 5 is after the reheat, which was isobaric at 𝑃4 = 𝑃5 = 700π‘˜π‘ƒπ‘Ž, and the
reheating returns thetemperature to𝑇5 = 480℃. From interpolation of the superheat
π‘˜π½
π‘˜π½
tables, we find β„Ž5 = 3361.15 π‘˜π‘”,𝑠5 = 6.73695 π‘˜π‘”.𝐾. After expansion in the second turbine,
π‘˜π½
we have 𝑠𝑠6 = 𝑠5 = 6.73695 π‘˜π‘”.𝐾.
And we were given 𝑃6 = 8π‘˜π‘ƒπ‘Ž. We consult the saturation tables to find at this pressure
π‘˜π½
π‘˜π½
𝑠𝑓 = 0.5926 π‘˜π‘”πΎ,𝑠𝑔 = 8.2287 π‘˜π‘”πΎ. Thus,
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POWERPLANT/WEEK 2-4
2012
π‘˜π½
π‘₯6𝑠
π‘˜π½
𝑠6𝑠 − 𝑠5 (6.73695 π‘˜π‘”.𝐾) − (0.5926 π‘˜π‘”.𝐾)
=
=
= 0.8046
π‘˜π½
π‘˜π½
𝑠𝑔 − 𝑠𝑓
(8.2287 π‘˜π‘”.𝐾) − (0.5926 π‘˜π‘”.𝐾)
The tables then give the necessary information to compute β„Ž6𝑠 :
π‘˜π½
π‘˜π½
π‘˜π½
β„Ž6𝑠 = β„Žπ‘“ + π‘₯6𝑠 β„Žπ‘“π‘” = (173.88 ) + (0.8046) (2403.1 ) = 2107.52
π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
Now, the actual β„Ž6 is found via:
β„Ž6 = β„Ž5 − πœ‚π‘‘ (β„Ž5 − β„Ž6𝑠 )
π‘˜π½
π‘˜π½
π‘˜π½
= (3361.15 ) − (0.88) ((3361.15 ) − (2107.52 ))
π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
π‘˜π½
= 2257.96
π‘˜π‘”
Now, the tables give us
π‘˜π½
π‘š3
β„Ž1 = β„Žπ‘“ = 173.88
, 𝑣1 = 𝑣𝑓 = 0.001084
π‘˜π‘”
π‘˜π‘”
For the pump, we have 𝑃1 = 𝑃6 = 8π‘˜π‘ƒπ‘Ž and 𝑃2 = 𝑃3 = 8000π‘˜π‘ƒπ‘Ž.
So
𝑀𝑠
πœ‚π‘ =
𝑀𝑝
𝑀𝑠
𝑀𝑝 =
πœ‚π‘
𝑣1 (𝑃2 − 𝑃1 )
=
πœ‚π‘
π‘š3
=
(0.001084 π‘˜π‘”
0.8
= 10.83
π‘˜π½
π‘˜π‘”
So
β„Ž2 = β„Ž1 + 𝑀𝑝
π‘˜π½
π‘˜π½
= (173.88 ) + (10.83 )
π‘˜π‘”
π‘˜π‘”
π‘˜π½
= 184.709
π‘˜π‘”
Now, substitute all these values into 𝑀𝑛𝑒𝑑 and get
𝑀𝑛𝑒𝑑 = ((3348.4
π‘˜π½
π‘˜π½
π‘˜π½
π‘˜π½
) − (2814.6 )) + ((3361.15 ) − (2257.96 ))
π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
+ ((173.88
π‘˜π½
π‘˜π½
) − (184 ))
π‘˜π‘”
π‘˜π‘”
= 1626.11
On a mass basis, we have
8
π‘˜π½
π‘˜π‘”
POWERPLANT/WEEK 2-4
π‘ŠΜ‡ = π‘šΜ‡π‘€π‘›π‘’π‘‘ = (2.63 × 105
2012
β„Žπ‘Ÿ
π‘˜π½
)(
) (1626.11 ) = 1.19 × 105 π‘˜π‘Š
𝑠
3600𝑠
π‘˜π‘”
π‘˜π‘”
From π‘žπ‘–π‘› , the heat added is
π‘žπ‘–π‘› = ((3348.4
π‘˜π½
π‘˜π½
π‘˜π½
π‘˜π½
) − (184.709 )) + ((3361.15 ) − (2814.6 ))
π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
= 3710.18
π‘˜π½
π‘˜π‘”
So the cycle’s thermal efficiency is
π‘˜π½
πœ‚π‘‘β„Ž
𝑀𝑛𝑒𝑑 1626.11 π‘˜π‘”
=
=
π‘˜π½ = 0.4382
π‘žπ‘–π‘›
3710.18 π‘˜π‘”
The heat per unit mass rejected in the condenser is from π‘žπ‘œπ‘’π‘‘ :
π‘žπ‘œπ‘’π‘‘ = (2257.96
π‘˜π½
π‘˜π½
π‘˜π½
) − (173.88 ) = 2084.08
π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
So the power rejected as heat is
𝑄𝐿 = π‘šΜ‡π‘žπ‘œπ‘’π‘‘ = (2.63 × 105
π‘˜π‘”
β„Žπ‘Ÿ
π‘˜π½
)(
) × (2084.08 ) = 1.52 × 105 π‘˜π‘Š
𝑠
3600𝑠
π‘˜π‘”
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 πœ‚π‘‘ = πœ‚π‘ƒ = 0.88. The turbine inlet
pressure and temperature are at 𝑃3 = 1200π‘π‘ π‘–π‘Ž, 𝑇3 = 1000℉. The condenser pressure is
at 𝑃3 = 1 π‘π‘ π‘–π‘Ž. The steam generator provides 𝑄̇𝐻 = 2 × 109 𝐡𝑑𝑒/β„Žπ‘Ÿ. 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, π‘šΜ‡π‘π‘€ .
Answer:
We interpolate the steam tables to find (from super-heated vapour tables)
9
POWERPLANT/WEEK 2-4
β„Ž3 = 1499.6
𝐡𝑑𝑒
,
π‘™π‘π‘š
𝑠3 = 1.6297
For an isentropic turbine, we have 𝑠4𝑠 = 𝑠3 = 1.6297
4𝑠 is a two-phase mixture:
𝐡𝑑𝑒
π‘™π‘π‘š. °π‘…
𝐡𝑑𝑒
π‘™π‘π‘š.°π‘…
𝐡𝑑𝑒
π‘₯4𝑠
2012
. At 𝑃4 = 1π‘π‘ π‘–π‘Ž, we find state
𝐡𝑑𝑒
𝑠4 − 𝑠𝑓 (1.6297 π‘™π‘π‘š.°π‘…) − (0.1327 π‘™π‘π‘š.°π‘…)
=
=
= 0.811
𝑠𝑔 − 𝑠𝑓 (1.9779 𝐡𝑑𝑒 ) − (0.1327 𝐡𝑑𝑒 )
π‘™π‘π‘š.°π‘…
π‘™π‘π‘š.°π‘…
Thus,
β„Ž4𝑠 = β„Žπ‘“ + π‘₯4𝑠 β„Žπ‘“π‘” = (69.74
𝐡𝑑𝑒
𝐡𝑑𝑒
𝐡𝑑𝑒
) + (0.811) (1036
) = 909.9
π‘™π‘π‘š
π‘™π‘π‘š
π‘™π‘π‘š
Now, for the actual turbine, we get
β„Ž4 = β„Ž3 + πœ‚π‘‘ (β„Ž3 − β„Ž4𝑠 )
𝐡𝑑𝑒
𝐡𝑑𝑒
𝐡𝑑𝑒
= (1499.7
) − (0.88) ((499.7
) − (909.9
))
π‘™π‘π‘š
π‘™π‘π‘š
π‘™π‘π‘š
𝐡𝑑𝑒
= 979.9
π‘™π‘π‘š
Now, after the condenser, we takeπ‘₯1 = 0, so β„Ž1 = β„Žπ‘“ , 𝑠1 = 𝑠𝑓 , and 𝑣1 = 𝑣𝑓 , all at 𝑃1 =
1 π‘π‘ π‘–π‘Ž. These are
𝐡𝑑𝑒
𝐡𝑑𝑒
𝑓𝑑 3
β„Ž1 = 69.74
, 𝑠1 = 0.1327
, 𝑣1 = 0.01614
π‘™π‘π‘š
π‘™π‘π‘š. °π‘…
π‘™π‘π‘š
Now
𝑀𝑝 =
𝑓𝑑 3
𝑀𝑠
πœ‚π‘
𝑙𝑏𝑓
𝑙𝑏𝑓
𝑣(𝑃4 − 𝑃3 ) (0.01614 π‘™π‘π‘š) ((1200 𝑖𝑛2 ) − (1 𝑖𝑛2 )) 144𝑖𝑛2
𝐡𝑑𝑒
=
=
πœ‚π‘
0.88
𝑓𝑑 2 778 𝑓𝑑 𝑙𝑏𝑓
𝐡𝑑𝑒
= 4.07
π‘™π‘π‘š
Now
β„Ž2 = β„Ž1 + 𝑀𝑝 = (69.74
𝐡𝑑𝑒
𝐡𝑑𝑒
𝐡𝑑𝑒
) + (4.07
) = 73.81
π‘™π‘π‘š
π‘™π‘π‘š
π‘™π‘π‘š
In the boiler, we have
𝑄̇𝐻 = π‘šΜ‡(β„Ž3 − β„Ž2 )
10
POWERPLANT/WEEK 2-4
2012
𝐡𝑑𝑒
2 × 109
𝑄̇𝐻
β„Žπ‘Ÿ
β„Žπ‘Ÿ
π‘šΜ‡ =
=
(β„Ž3 − β„Ž2 ) (1499.7 𝐡𝑑𝑒 ) − (73.81 𝐡𝑑𝑒 ) 3600𝑠
π‘™π‘π‘š
= 390
π‘™π‘π‘š
π‘™π‘π‘š
𝑠
We also note
𝑄̇𝐻 = (2 × 109
𝐡𝑑𝑒
β„Žπ‘Ÿ
𝐡𝑑𝑒
)
= 5.556 × 105
β„Žπ‘Ÿ 3600 𝑠
𝑠
Now, the net power is the sum of the turbine and pump work:
π‘ŠΜ‡π‘›π‘’π‘‘ = π‘šΜ‡((β„Ž3 − β„Ž4 ) + (β„Ž1 − β„Ž2 ))
= (390
π‘™π‘π‘š
𝐡𝑑𝑒
𝐡𝑑𝑒
𝐡𝑑𝑒
𝐡𝑑𝑒
) ((1499.6
) − (979.9
)) + ((69.74
) − (73.81
))
𝑠
π‘™π‘π‘š
π‘™π‘π‘š
π‘™π‘π‘š
π‘™π‘π‘š
= 2.01 × 105
𝐡𝑑𝑒
𝑠
The thermal efficiency is thus
𝐡𝑑𝑒
2.01 × 105 𝑠
π‘ŠΜ‡π‘›π‘’π‘‘
πœ‚=
=
𝐡𝑑𝑒 = 0.3618
𝑄̇𝐻
5.556 × 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
π‘šΜ‡β„Ž4
π‘šΜ‡β„Ž1
π‘‡β„Žπ‘œπ‘‘ = 80℉
π‘šΜ‡π‘π‘€
π‘‡π‘π‘œπ‘™π‘‘ = 60℉
Cooling water
π‘šΜ‡π‘π‘€
Fig. 2-5: Rankine cycle condenser/heat exchanger.
𝑑𝐸𝐢𝑉
= 0 = 𝑄̇𝐢𝑉 − π‘ŠΜ‡πΆπ‘‰ + π‘šΜ‡(β„Ž4 − β„Ž1 ) + π‘šΜ‡π‘π‘€ 𝑐𝑃 (π‘‡π‘π‘œπ‘™π‘‘ − π‘‡β„Žπ‘œπ‘‘ )
𝑑𝑑
𝑄̇𝐢𝑉 = 0, π‘ŠΜ‡πΆπ‘‰ = 0
0 = π‘šΜ‡(β„Ž4 − β„Ž1 ) + π‘šΜ‡π‘π‘€ 𝑐𝑃 (π‘‡π‘π‘œπ‘™π‘‘ − π‘‡β„Žπ‘œπ‘‘ )
π‘šΜ‡(β„Ž4 − β„Ž1 )
= π‘šΜ‡π‘π‘€
𝑐𝑃 (π‘‡β„Žπ‘œπ‘‘ − π‘‡π‘π‘œπ‘™π‘‘ )
11
POWERPLANT/WEEK 2-4
=
(390
π‘™π‘π‘š
𝑠
𝐡𝑑𝑒
𝐡𝑑𝑒
) ((979 π‘™π‘π‘š.°π‘…) − (69.74 π‘™π‘π‘š.°π‘…))
𝐡𝑑𝑒
(1.0 π‘™π‘π‘š.°π‘…) ((80℉) − (60℉))
= 17730
2-3
2012
π‘™π‘π‘š
𝑠
Losses in Rankin cycle
• Turbine: These are typically the largest losses in the system. The turbine efficiency is
defined by
𝑀𝑑
β„Ž3 − β„Ž4
πœ‚=
=
𝑀𝑑𝑠 β„Ž3 − β„Ž4𝑠
Here β„Ž4𝑠 and 𝑀𝑑𝑠 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
𝑀𝑝𝑠 β„Ž2𝑠 − β„Ž1
πœ‚=
=
𝑀𝑝
β„Ž2 − β„Ž1
• 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
π‘šΜ‡2
C
π‘šΜ‡3
10
P
P
P
8
9
5
6
7
(a)
T
1
B
2
10
9
8
7
6
π‘šΜ‡2
3
π‘šΜ‡3
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 π‘šΜ‡3 is sufficient to saturate the subcooled water at point 6. If the
extractionsteam at point 3 were π‘šΜ‡′3 (where π‘šΜ‡′3 > π‘šΜ‡3 ), 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 =π‘šΜ‡2 .
Mass flow between points 2 and 3 = 1 -π‘šΜ‡2 .
Mass flow between points 3 and 7 =π‘šΜ‡3 .
change to 3
Mass flow between points 3 and 7 = 1 - π‘šΜ‡2 - π‘šΜ‡3 .
, 4,5,6,7
Mass flow between points 7 and 9 = 1 - π‘šΜ‡2 .
Mass flow between points 9 and 1 = 1.
The energy balances for the high- and low-pressure feedwater heaters, respectively, are as
follows:
π‘šΜ‡2 (β„Ž2 − β„Ž9 ) = (1 − π‘šΜ‡2 )(β„Ž9 − β„Ž8 )
π‘šΜ‡3 (β„Ž3 − β„Ž7 ) = (1 − π‘šΜ‡2 − π‘šΜ‡3 )(β„Ž7 − β„Ž6 )
Heat added
π‘žπ΄ = (β„Ž1 − β„Ž10 )
Turbine work
𝑀𝑑 = (β„Ž1 − β„Ž2 ) + (1 − π‘šΜ‡2 )(β„Ž2 − β„Ž3 ) + (1 − π‘šΜ‡2 − π‘šΜ‡3 )(β„Ž3 − β„Ž4 )
Pump work
|∑ 𝑀𝑝 | = (1 − π‘šΜ‡2 − π‘šΜ‡3 )(β„Ž6 − β„Ž5 ) + (1 − π‘šΜ‡2 )(β„Ž8 − β„Ž7 ) + (β„Ž10 − β„Ž9 )
≅ (1 − π‘šΜ‡2 − π‘šΜ‡3 )
Heat rejected
𝑣5 (𝑃6 − 𝑃5 )
𝑣7 (𝑃8 − 𝑃7 ) 𝑣9 (𝑃10 − 𝑃9 )
+ (1 − π‘šΜ‡2 )
+
πœ‚π‘ƒ
πœ‚π‘ƒ
πœ‚π‘ƒ
|π‘žπ‘… | = (1 − π‘šΜ‡2 − π‘šΜ‡3 )(β„Ž4 − β„Ž5 )
Net cycle work
βˆ†π‘€π‘›π‘’π‘‘ = 𝑀𝑑 − |𝑀𝑝 |
Cycle thermal efficiency
πœ‚π‘‘β„Ž =
βˆ†π‘€π‘›π‘’π‘‘
π‘žπ΄
Work ratio
π‘Šπ‘… =
𝑀𝑛𝑒𝑑
𝑀𝑇
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 π‘€π‘ƒπ‘Ž, 400℃. After expansion to 400 π‘˜π‘ƒπ‘Ž, some
stream is extracted for heating feedwater in an open feedwater heater.Pressure in
feedwater heater is 400 π‘˜π‘ƒπ‘Ž, and water leaves it at a saturated state at 400 π‘˜π‘ƒπ‘Ž. The
restof the steam expands through the turbine to 10 π‘˜π‘ƒπ‘Ž. 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 & 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
π‘˜π½
β„Ž5 = 3213.6
π‘ π‘’π‘π‘’π‘Ÿ π‘£π‘Žπ‘π‘œπ‘Ÿ π‘Žπ‘‘ 400℃ π‘Žπ‘›π‘‘ 4π‘€π‘ƒπ‘Ž,
π‘˜π‘”
π‘˜π½
π‘˜π½
β„Ž6 = 2685.6 , β„Ž1 = 191.8
π‘™π‘–π‘žπ‘’π‘–π‘‘ π‘Žπ‘‘ 0.01π‘€π‘ƒπ‘Ž, β„Ž7
π‘˜π‘”
π‘˜π‘”
π‘˜π½
π‘˜π½
= 2144.1
π‘Žπ‘‘ 0.01π‘€π‘ƒπ‘Ž π‘€π‘–π‘‘β„Ž π‘₯7, β„Ž3 = 604.73
π‘™π‘–π‘žπ‘’π‘–π‘‘ π‘Žπ‘‘ 400℃
π‘˜π‘”
π‘˜π‘”
π‘š3
π‘š3
𝑣1 = 0.00101
, 𝑣3 = 0.001084
π‘˜π‘”
π‘˜π‘”
First consider the low pressure pump.
π‘˜π½
π‘š3
β„Ž2 = β„Ž1 + 𝑣1 (𝑃2 − 𝑃1 ) = (191.8 ) + (0.00101 ) ((400π‘˜π‘ƒπ‘Ž) − (10π‘˜π‘ƒπ‘Ž))
π‘˜π‘”
π‘˜π‘”
π‘˜π½
= 192.194
π‘˜π‘”
The pump work is
𝑀𝑃1 = 𝑣1 (𝑃1 − 𝑃2 ) = (0.00101)((10π‘˜π‘ƒπ‘Ž) − (400π‘˜π‘ƒπ‘Ž))
= −0.3939
Note, a sign convention consistent with work done by the fluid is used here.
Now consider the turbine
π‘‘π‘šπ‘π‘£
= π‘šΜ‡5 − π‘šΜ‡6 − π‘šΜ‡7
𝑑𝑑
π‘‘π‘šπ‘π‘£
=0
𝑑𝑑
π‘šΜ‡5 = π‘šΜ‡6 + π‘šΜ‡7
π‘šΜ‡6 π‘šΜ‡7
1=
−
π‘šΜ‡5 π‘šΜ‡5
change - to
+
𝑑𝐸𝑐𝑣
= 𝑄̇𝑐𝑣 − π‘ŠΜ‡π‘π‘£ + π‘šΜ‡5 β„Ž5 − π‘šΜ‡6 β„Ž6 − π‘šΜ‡7 β„Ž7
𝑑𝑑
Μ‡
𝑑𝐸𝑐𝑣
= 0,
𝑄𝑐𝑣 = 0
𝑑𝑑
π‘ŠΜ‡π‘π‘£ = π‘šΜ‡5 β„Ž5 − π‘šΜ‡6 β„Ž6 − π‘šΜ‡7 β„Ž7
On a per mass basis, we get,
π‘ŠΜ‡π‘π‘£
π‘šΜ‡6
π‘šΜ‡7
= β„Ž5 −
β„Ž6 −
β„Ž
π‘šΜ‡5
π‘šΜ‡5
π‘šΜ‡5 7
π‘šΜ‡6
π‘šΜ‡6
= β„Ž5 −
β„Ž6 − (1 −
)β„Ž
π‘šΜ‡5
π‘šΜ‡5 7
π‘šΜ‡6
π‘šΜ‡6
= β„Ž5 − β„Ž6 + (1 −
) β„Ž6 − (1 −
)β„Ž
π‘šΜ‡5
π‘šΜ‡5 7
π‘šΜ‡6
= β„Ž5 − β„Ž6 + (1 −
) − (β„Ž6 − β„Ž7 )
π‘šΜ‡5
𝑀𝑑 =
16
add and extract
h6
POWERPLANT/WEEK 2-4
2012
Now consider the feedwater heater. The first law for this device gives
𝑑𝐸𝑐𝑣
= 𝑄̇𝑐𝑣 − π‘ŠΜ‡π‘π‘£ + π‘šΜ‡2 β„Ž2 + π‘šΜ‡6 β„Ž6 + π‘šΜ‡3 β„Ž3
𝑑𝑑
𝑑𝐸𝑐𝑣
= 0, 𝑄̇𝑐𝑣 = 0, π‘ŠΜ‡π‘π‘£ = 0
𝑑𝑑
π‘šΜ‡2
π‘šΜ‡6
β„Ž2 +
β„Ž = β„Ž3
π‘šΜ‡3
π‘šΜ‡3 6
π‘šΜ‡7
π‘šΜ‡6
β„Ž2 +
β„Ž =
π‘šΜ‡5
π‘šΜ‡5 6
π‘šΜ‡6
π‘šΜ‡6
(1 −
) β„Ž2 +
β„Ž =
π‘šΜ‡5
π‘šΜ‡5 6
π‘˜π½
π‘šΜ‡6
π‘˜π½
π‘šΜ‡6
π‘˜π½
(604.73 ) = (1 −
) (192.194 ) +
(2685.6 )
π‘˜π‘”
π‘šΜ‡5
π‘˜π‘”
π‘šΜ‡5
π‘˜π‘”
π‘šΜ‡6
= 0.16545
π‘šΜ‡5
Now get the turbine work
𝑀𝑑 = β„Ž5 − β„Ž6 + (1 −
= (3213.6
π‘šΜ‡6
) (β„Ž6 − β„Ž7 )
π‘šΜ‡5
π‘˜π½
π‘˜π½
π‘˜π½
π‘˜π½
) − (2685.6 ) + (1 − 0.16545) ((2685.6 ) − (2144.1 ))
π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
= 979.908
π‘˜π½
π‘˜π‘”
Now get the work for the high-pressure pump
𝑀𝑝2
π‘š3
) ((400π‘˜π‘ƒπ‘Ž) − (4000π‘˜π‘ƒπ‘Ž))
π‘˜π‘”
= −3.9024
= 𝑣3 (𝑃3 − 𝑃4 ) = (0.001084
Now
β„Ž4 = β„Ž3 + 𝑣3 (𝑃4 − 𝑃3 ) = (604.73
π‘˜π½
π‘š3
) + (0.001084 ) ((4000π‘˜π‘ƒπ‘Ž) − (400π‘˜π‘ƒπ‘Ž))
π‘˜π‘”
π‘˜π‘”
π‘˜π½
= 608.6
π‘˜π‘”
Now get the net work
π‘ŠΜ‡π‘›π‘’π‘‘ = π‘šΜ‡5 𝑀𝑑 + π‘šΜ‡1 𝑀𝑝1 + π‘šΜ‡5 𝑀𝑝2
π‘šΜ‡1
𝑀𝑛𝑒𝑑 = 𝑀𝑑 +
𝑀 + 𝑀𝑝2
π‘šΜ‡5 𝑝1
π‘šΜ‡7
= 𝑀𝑑 + (1 −
) 𝑀 + 𝑀𝑝2
π‘šΜ‡5 𝑝1
17
POWERPLANT/WEEK 2-4
2012
π‘šΜ‡6
) 𝑀 + 𝑀𝑝2
π‘šΜ‡5 𝑝1
π‘˜π½
π‘˜π½
π‘˜π½
= (979.908 ) + (1 − 0.16545) (−0.3939 ) + (3.9024 )
π‘˜π‘”
π‘˜π‘”
π‘˜π‘”
π‘˜π½
= 975.677
π‘˜π‘”
= 𝑀𝑑 + (1 −
Now for the heat transfer in the boiler, one has
π‘žβ„Ž = β„Ž5 − β„Ž4
π‘˜π½
π‘˜π½
= (3213.6 ) − (608.6 )
π‘˜π‘”
π‘˜π‘”
π‘˜π½
= 2605.0
π‘˜π‘”
Thus, the thermal efficiency is
π‘˜π½
𝑀𝑛𝑒𝑑 975.677 π‘˜π‘”
πœ‚=
=
π‘˜π½ = 0.375
π‘žβ„Ž
2605.0 π‘˜π‘”
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
π‘šΜ‡3
π‘šΜ‡2
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:
π‘˜π½
π‘˜π½
β„Ž1 = 3445
, 𝑠1 = 7.089
= 𝑠2
π‘˜π‘”
π‘˜π‘”πΎ
At state 2
0.391 + π‘₯2 × 8.13 = 7.089
6.698
π‘₯2 =
= 0.824
8.13
i.e.
β„Ž2 = 112 + 0.824 × 2438 = 2117 π‘˜π½/π‘˜π‘”
Also,
β„Ž3 = β„Žπ‘“ π‘Žπ‘‘ 0.035 π‘π‘Žπ‘Ÿ = 112 π‘˜π½/π‘˜π‘”
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POWERPLANT/WEEK 2-4
2012
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.
π‘˜π½
For the first stage of expansion, 1-7, 𝑠7 = 𝑠1 = 7.089 π‘˜π‘”πΎ, and from tables at 10 bar 𝑠𝑔 >
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
β„Ž7 = 2944 + (
) (3052 − 2944) = 2944 +
× 108
7.124 − 6.926
0.198
i.e
β„Ž7 = 3032.9 π‘˜π½/π‘˜π‘”
From the throttling process, 11-12, we have:
β„Ž6 = β„Ž11 = β„Ž12 = 763π‘˜π½/π‘˜π‘”
For the second stage of expansion, 7-8, 𝑠7 = 𝑠8 = 𝑠1 = 7.089
1.1bar,𝑠𝑔 > 7.089
π‘˜π½
π‘˜π‘”πΎ
, hence the steam is wet at state 8.
Therefore,
20
π‘˜π½
π‘˜π‘”πΎ
, and from the tables at
POWERPLANT/WEEK 2-4
2012
1.333 + (π‘₯8 × 5.994) = 7.089, π‘₯8 = 0.961
i.e.
β„Ž8 = 429 + (0.961 × 2251) = 2591π‘˜π½/π‘˜π‘”
For the throttling process, 9-10:
β„Ž5 = β„Ž9 = β„Ž10 = 429 π‘˜π½/π‘˜π‘”
Applying an energy balance to the first feed heater, remembering that there is no work or
heat transfer:
𝑦1 β„Ž7 + β„Ž5 = 𝑦1 β„Ž11 + β„Ž6
β„Ž6 − β„Ž5
763 − 429
=
= 0.147
β„Ž7 − β„Ž11 3032.4 − 763
Similarly for the second heater, takingβ„Ž4 = β„Ž3 :
𝑦1 =
𝑦2 β„Ž8 + 𝑦1 β„Ž12 + β„Ž4 = β„Ž5 + (𝑦1 + 𝑦2 )β„Ž9
𝑦2 (β„Ž8 − β„Ž9 ) + 𝑦1 β„Ž12 + β„Ž4 = β„Ž5 + 𝑦1 β„Ž9
𝑦2 (2591 − 429) + (0.147 × 763) + 112 = 429 + (0.147 × 429)
267.8
𝑦2 =
= 0.124
2162
The heat supplied to the boiler,𝑄1 per kg of boiler steam is given by:
𝑄1 = β„Ž1 − β„Ž6 = 3445 − 763 = 2682 π‘˜π½/π‘˜π‘”
The work output, neglecting pump work, is given by:
π‘Š = (β„Ž1 − β„Ž7 ) + (1 − 𝑦1 )(β„Ž7 − β„Ž8 ) + (1 − 𝑦1 − 𝑦2 )(β„Ž8 − β„Ž2 )
= (3445 − 3032.9) + (1 − 0.147)(3032.9 − 2591)
+ (1 − 0.147 − 0.124)(2591 − 2117)
= 412.1 + 376.9 + 345.5 = 1134.5 π‘˜π½/π‘˜π‘”
Then,
π‘‡β„Žπ‘’π‘Ÿπ‘šπ‘Žπ‘™ 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
π‘Š 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
2012
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 πœ‚π‘‡ (sometimes called adiabatic or isentropic efficiency) is
given by
β„Ž1 − β„Ž2
πœ‚π‘‡ =
β„Ž1 − β„Ž2𝑠
Note: πœ‚π‘‡ 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 πœ‚π‘‡ . 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 πœ‚π‘ƒ (sometimes called adiabatic or isentropic efficiency) is
given by
(π‘–π‘‘π‘’π‘Žπ‘™ π‘€π‘œπ‘Ÿπ‘˜)
β„Ž4𝑠 − β„Ž3
πœ‚π‘ =
=
(π‘Žπ‘π‘‘π‘’π‘Žπ‘™ π‘€π‘œπ‘Ÿπ‘˜)
β„Ž4 − β„Ž3
πœ‚π‘ƒ is the reverse of πœ‚π‘‡ .
The actual pump work is given by
β„Ž4𝑠 − β„Ž3 𝑣3 (𝑃4 − 𝑃3 )
=
πœ‚π‘ƒ
πœ‚π‘ƒ
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.
π‘Šπ‘ƒ =
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 π‘š7 = 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 (π‘š5 = π‘š4 ). 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 (π‘š6 = π‘š5 = 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 (π‘š6 =
π‘š4 π‘Žπ‘›π‘‘ π‘š5 = π‘š7 = 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
2012
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
2012
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
2012
Example 2-7
Consider the cogeneration plant shown in Fig. 2–13. Steam enters the turbine at 7 MPa and
500°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
4
5
6
POWERPLANT/WEEK 2-4
2012
0.001005π‘š3
× (7000π‘˜π‘ƒπ‘Ž − 5π‘˜π‘ƒπ‘Ž) = 7.03π‘˜π½/π‘˜π‘”
π‘˜π‘”
0.001093π‘š3
= 𝑣7 (𝑃10 − 𝑃7 ) =
× (7000π‘˜π‘ƒπ‘Ž − 500π‘˜π‘ƒπ‘Ž) = 7.10π‘˜π½/π‘˜π‘”
π‘˜π‘”
β„Ž1 = β„Ž2 = β„Ž3 = β„Ž4 = 3411.4 π‘˜π½/π‘˜π‘”
β„Ž5 = 2739.3 π‘˜π½/π‘˜π‘”
β„Ž6 = 2073 π‘˜π½/π‘˜π‘”
β„Ž7 = β„Žπ‘“ π‘Žπ‘‘ 500π‘˜π‘ƒπ‘Ž = 640.09 π‘˜π½/π‘˜π‘”
π‘€π‘π‘’π‘šπ‘1,𝑖𝑛 = 𝑣8 (𝑃9 − 𝑃8 ) =
π‘€π‘π‘’π‘šπ‘2,𝑖𝑛
β„Ž8 = β„Žπ‘“ π‘Žπ‘‘ 5π‘˜π‘ƒπ‘Ž = 137.75 π‘˜π½/π‘˜π‘”
β„Ž9 = β„Ž8 + π‘€π‘π‘’π‘šπ‘1,𝑖𝑛 = (137.75 + 7.03) = 144.78 π‘˜π½/π‘˜π‘”
β„Ž10 = β„Ž7 + π‘€π‘π‘’π‘šπ‘2,𝑖𝑛 = (640.09 + 7.10) = 647.19 π‘˜π½/π‘˜π‘”
(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
π‘˜π‘”
π‘šΜ‡4 = π‘šΜ‡7 = π‘šΜ‡1 = 15 𝑠 and π‘šΜ‡3 = π‘šΜ‡5 = π‘šΜ‡6 = 0).
Thus:
15π‘˜π‘”
𝑄̇𝑃,π‘šπ‘Žπ‘₯ = π‘šΜ‡1 (β„Ž4 − β„Ž7 ) = (
) × (3411.4 − 640.09) = 41.5 π‘˜π‘Š
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,π‘šΜ‡3 = π‘šΜ‡6 = π‘šΜ‡1 =
15 π‘˜π‘”/𝑠and π‘šΜ‡2 = π‘šΜ‡5 = 0. Maximum power is produced in this mode, which is
determined to be:
π‘˜π‘”
× (3411.4 − 2073) = 20076 π‘˜π½/π‘˜π‘”
𝑠
π‘˜π‘”
(7.03) = 105 π‘˜π‘Š
π‘ŠΜ‡π‘π‘’π‘šπ‘,𝑖𝑛 = 15
𝑠
π‘ŠΜ‡π‘›π‘’π‘‘,π‘œπ‘’π‘‘ = π‘ŠΜ‡π‘›π‘’π‘‘,π‘œπ‘’π‘‘ − π‘ŠΜ‡π‘π‘’π‘šπ‘,𝑖𝑛 = 20076 − 105 = 19971π‘˜π‘Š
15π‘˜π‘”
𝑄̇𝑖𝑛 = π‘šΜ‡1 (β„Ž1 − β„Ž11 ) =
× (3411.4 − 144.78) = 48999π‘˜π‘Š
𝑠
π‘ŠΜ‡π‘‘π‘’π‘Ÿπ‘,π‘œπ‘’π‘‘ = π‘šΜ‡(β„Ž3 − β„Ž6 ) = 15
Thus, the utilization factor for cogeneration plant:
πœ–π‘’ =
𝑁𝑒𝑑 π‘€π‘œπ‘Ÿπ‘˜ π‘œπ‘’π‘‘π‘π‘’π‘‘ + π‘ƒπ‘Ÿπ‘œπ‘π‘’π‘ π‘  β„Žπ‘’π‘Žπ‘‘ π‘‘π‘’π‘™π‘–π‘£π‘’π‘Ÿπ‘’π‘‘
π‘ŠΜ‡π‘›π‘’π‘‘ + 𝑄̇𝑃 (19971 + 0)
=
=
= 0.408
π‘‡π‘œπ‘‘π‘Žπ‘™ β„Žπ‘’π‘Žπ‘‘ 𝑖𝑛𝑝𝑒𝑑
48999π‘˜π‘Š
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 = πΈΜ‡π‘œπ‘’π‘‘
π‘šΜ‡4 β„Ž4 + π‘šΜ‡5 β„Ž5 = 𝑄̇𝑝,π‘œπ‘’π‘‘ + π‘šΜ‡7 β„Ž7
Or
𝑄̇𝑝,π‘œπ‘’π‘‘ = π‘šΜ‡4 β„Ž4 + π‘šΜ‡5 β„Ž5 − π‘šΜ‡7 β„Ž7
Where
π‘˜π‘”
= 1.5π‘˜π‘”/𝑠
𝑠
π‘˜π‘”
π‘šΜ‡5 = 0.7 × 15
= 10.5π‘˜π‘”/𝑠
𝑠
π‘šΜ‡7 = π‘šΜ‡4 + π‘šΜ‡5 = 12π‘˜π‘”/𝑠
π‘šΜ‡4 = 0.1 × 15
Thus
𝑄̇𝑝,π‘œπ‘’π‘‘ = 1.5
π‘˜π‘”
3411.4π‘˜π½
×(
) + (10.5 π‘˜π‘”/𝑠) × (2739.3 π‘˜π½/π‘˜π‘”) − (12 π‘˜π‘”/𝑠)
𝑠
π‘˜π‘”
× (640.09π‘˜π½/π‘˜π‘”) = 26.2 π‘˜π‘Š
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.
πœ–π‘’ =
π‘ŠΜ‡π‘›π‘’π‘‘ + 𝑄̇𝑃 (11.0 + 26.2)
=
π‘€π‘Š = 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:
𝑀𝑝,𝑖𝑛 =
𝑀𝑠,𝑝,𝑖𝑛 𝑣1 (𝑃2 − 𝑃1 ) (0.001009π‘š3 /π‘˜π‘”) × [(16000 − 9)π‘˜π‘ƒπ‘Ž]
=
=
= 19 π‘˜π½/π‘˜π‘”
πœ‚π‘
πœ‚π‘
0.85
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POWERPLANT/WEEK 2-4
2012
Turbine work output:
𝑀𝑑,π‘œπ‘’π‘‘ = πœ‚π‘‘ × π‘€π‘ ,𝑑,π‘œπ‘’π‘‘ = πœ‚π‘‘ (β„Ž3 − β„Ž4𝑠 ) = 0.87 × (3583.1 − 2115.3)π‘˜π½/π‘˜π‘” = 1277 π‘˜π½/π‘˜π‘”
Boiler heat input:
π‘žπ‘–π‘› = β„Ž2 − β„Ž1 = 3647.6 − 160.1 = 3487.5 π‘˜π½/π‘˜π‘”
Thus,
𝑀𝑛𝑒𝑑 = 𝑀𝑑,π‘œπ‘’π‘‘ − 𝑀𝑝,𝑖𝑛 = 1277 − 19 = 1258 π‘˜π½/π‘˜π‘”
πœ‚π‘‘β„Ž =
𝑀𝑛𝑒𝑑
1258
=
= 0.361
π‘žπ‘–π‘›
3487.5
b) The power produced by this power plant is
π‘ŠΜ‡π‘›π‘’π‘‘ = π‘šΜ‡π‘€π‘›π‘’π‘‘ = 15 × 1258 = 18.9π‘€π‘Š
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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