9- Cogeneration

Cogeneration is defined as; the simultaneous generation of electricity and steam (or heat) in
a single power plant.
Cogeneration plant is a plant that producing both electrical power and process heat
simultaneously.
Examples are chemical industries, paper mills, and places that use district heating.
From an energy resource point of view, cogeneration is beneficial only if it saves primary
energy when compared with separate generation of electricity and steam (or heat).
The cogeneration plant efficiency ( ηco ) is given by;

WT +QH
ηco =
Qadd
where,
WT = Electrical Energy generated, kW
QH = Heat energy in process steam, kW
Qadd = Heat added to the plant, kW

For separate generation of electricity and heat added per unit total energy output is;
QT = WT + QH
Q WT Q
QT = + QH
T QT T
1= e + (1-e)
The combined efficiency for separate generation is given as;
1
ηc = e 1−e
+
ηe ηh

Where;
WT
e = electrical energy fraction of total energy output = W
T +QH
1-e: heat generated fraction of total energy output
ηe = electrical plant efficiency.
ηh = steam (or heat) generator efficiency.
ηc: combined efficiency for two separate electrical and thermal plant

cogeneration is beneficial: if
ηco > ηc
WT +QH
the cogeneration plant efficiency ηco = exceeds that of the combined efficiency for
Qadd
1
separate generation ηc = e 1−e
+
ηe ηh

ηe : efficiency of electrical plant producing same electrical output power as the electrical split
(part) in cogeneration plant.
ηh : efficiency of thermal plant producing same thermal (heat) output power as the thermal
split (part) in cogeneration plant.

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From the previous figure:
WT = 30 unit
QH = 45 unit
Qadd =100 unit
ηe = 31 %
ηh = 80%
WT 30
e =W = 30+45 = 0.4; 1-e = 0.6
T +QH
WT +QH 30+45
ηco = = = 75%
Qadd 100
1
ηc = 0.4 0.6 = 49%
+
0.31 0.8
Example 9-1
In a cogeneration plant, the power load is 5.6 MW and the heating load is 1.2 MW.
Steam is generated at 50 bar, 500 oC and isentropic expansion in a turbine to a condenser at
0.1 bar. The heating load is supplied by extracting steam from turbine at 2 bar, which
condensed in process heater to saturated liquid at 2 bar and then pumped to the boiler. Neglect
pump work. Draw Schematic diagram, and h-s diagram and Compute:
1) The steam flow rate in boiler in Ton/hr.
2) Input heat to boiler.
3) Heat rejects through the condenser.
4) Rate of fuel burnt in boiler in Ton/hr if boiler efficiency is 88% and coal C.V. is 25
MJ/kg.
5) Electric plant efficiency.
6) Process heat efficiency.
7) Cogeneration and Combined Heat and Power (CHP) system efficiencies.
Solution

Givens:
WT = 5.6 MW QH =1.2 MW
Pst=50 bar, Tst = 500°C
Pcond. = 0.1 bar Pproces = 2 bar

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From chart
h1 (at 50 bar & 500°C) = 3433 KJ/kg
h5=2645 kJ/kg
h2=2210 kJ/kg
From tables
h6=hf @ 2 bar = 505 kJ/kg
h3=hf @ 0.1 bar = 192 kJ/kg
QH = ṁ (h5-h6)
1.2×10 = ṁ (2645-505)
3

ṁ=0.56 Kg/sec
WT = ṁst (h1- h5) + (ṁst-ṁ) (h5- h2)
5.6×10 = ṁst (3433-2645) + (ṁst -0.56) (2645-2210)
3

ṁst =4.778 kg/sec

Input heat to boiler
Qadd = (ṁst-ṁ) (h1 – h3) + ṁ (h1– h6) = (4.778 - 0.56) (3433-192) + 0.56 (3433-505)
Qadd = 15310.32 kW = 15.31032 MW
Heat rejected in condenser
Qrej = (ṁst-ṁ) (h2 – h3) = (4.778 - 0.56) (2210-192)
Qrej = 8511.93 kW
Rate of fuel burnt in boiler
Qadd Q
ηb= = ṁ add
QT C.V
f
15310.32
0.88= ṁ
f × 25000
ṁf = 0.695 kg/sec = 2.505 ton/hr
W T + QH 5.6 + 1.2
ηco = = 15.31032 = 44.41 %
Qadd
For combined efficiency (ηc):
The Same flow rate is utilized, and the same fraction of bled steam is used for heating feed
water (CFWH) instead of process heat.
Heat balance for C.F.W.H
(ṁst - ṁ) (h8- h3) = ṁ (h5-h6)
(4.778-0.56) (h8 – 192) = 1200 kW
h8= 476.5 KJ/kg
Qadd = ṁst (h1-h8) = 4.778(3433– 476.5) =14126 KW
= 14.126 MW
WT 5.6
ηe = Q =14.126 = 39.64 %
add
WT 5.6
e=W =5.6+1.2 = 0.823
T +QH
1-e = 0.177
ηb = ηh = 0.88
1 1
ηCHP = e 1−e ηCHP = 0.823 0.177 = 43.92 %
+ +
ηe ηh 0.3964 0.88

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Example 9-2
A steam power plant, the produces 500 MW, with inlet steam to high pressure turbine at 100
bar, 500oC and condensation at 0.1 bar. It has one stage of reheat at 8 bar, which raises the
steam temperature back to 500oC.
Draw Schematic diagram, and h-s diagram for the following cases.
a) One closed feed water heater receives bled steam at the reheat pressure, and the remaining
steam is reheated and then expanded in the low-pressure turbine.
Calculate mass flow rate of steam inlet to H.P. turbine and cycle efficiency.
Design a closed feed water heater (find its length and number of tubes) if the overall heat
transfer coefficient was 1.5 kW/m2oC. Tube diameters are 24 mm and 28 mm. Water
velocity inside (CFWH) tube was 1 m/s. Terminal Temperature Difference was 3oC.
Number of passes = 4
b) A cogeneration plant is considered; the heating load is supplied by extracting an amount
of steam at the reheat pressure where th isobaric condensation of heating steam occurs
through the process heater to saturated liquid and pumped to the boiler. The remaining
steam at 8 bar is reheated and then expanded in the low-pressure turbine. By ignoring the
pump work, Compute: 1) Heat reject in condenser. 2) Rate of fuel burnt in boiler in
Ton/hr if boiler efficiency is 88% and coal C.V. is 25 MJ/kg. 3) Electric plant efficiency.
4) Process heat efficiency. 5) Cogeneration and Combined Heat and Power (CHP)
efficiencies. 6) Mention the system that is more beneficial and why?

Givens
Wt=500 MW=500,000 KW
Steam inlet to H.P.T at P1= 100 bar & T1= 500 oC
Pcond. = P5 = P6 = 0.1 bar
Preheat= P2 = 8 bar to T=500°C
solution
a)
From chart
h1=3373 kJ/kg
(Check the value using tables)
h2=2740 kJ/kg
h4 = 3480 kJ/kg
h5 =2490 KJ/kg
From steam tables
h3=hf at 8 bar = 721 kJ/kg
h6= hf at P=0.1 bar =192 kJ/kg
h7=h6 + 0.1 ΔP100,0.1
h7=192+ 0.1 (100-0.1) = 202 kJ/kg
h8=h3 + 0.1 ΔP100,8
h8=721+ 0.1 (100-8) = 730.2 kJ/kg
h9 = h3 – T.T.D × Cp = 721-3×4.2 =708.4 kJ/kg
or h9= Cp×T9 T9=T3 – T.T.D T3=Tsat. at P=8 bar = 170.4 °C

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Heat balance for C.F.W.H y h2
h7 (1-y) + y h2 = y h3 + (1-y) h9
202×(1-y) + y × 2740 = 721×y + (1-y)×708.4
(1-y) h9
y = 0.2
(1-y) h7
h10 = (1-y)×h9 + y×h8
=(1-0.2)×708.4 + 0.2×730.2 = 712.76 kJ/kg Y h3
WT= ṁst (h1 - h2) + ṁst (1-y)(h4-h5)
500×1000= ṁst 3373-2740)+ ṁst (1-0.2)(2480-2490) y h8
ṁst = 350.877 kg/s (mass flow rate to H.P.T) h10
Qadd= ṁst (h1– h10) + ṁst (1-y) (h4 - h2)
Qadd= 350.877×(3373-721.76)+350.877×(1-0.2) (3480-2740) =1141136 kW
(1-y) h9
𝑊 500∗1000
ηth =𝑄 𝑛𝑒𝑡 = × 100 = 𝟒𝟑. 𝟖𝟏%
𝑎𝑑𝑑 1141136

Design a closed feed water heater

Qh = 𝑦ṁst (h2-h3) = U Am ΔTm
Qh = 0.2 × 350.877 × (2740-721) =141684 kW = 141.684 MW
ṁ = ṁst (1 − 𝑦) = ρ u a N
𝜋
ṁ = 350.877× (1-0.2) =1000 × 1 × 4 × (24×10-3)2 × N
N= 620 tube
T9=T3 – T.T.D T3=Tsat. at P=8 bar
T9=170.4 – 3 =167.4 °C
T7=h7/Cp = 202/4.2 = 48.085°C Tsat=170.
ΔT1= Tsat -T7 =170.4 - 48.085= 122.3°C 4°C ΔT
ΔT2= Tsat -T9= T.T.D = 3 °C
ΔT1 −ΔT2 122.3−3
ΔTm= ΔT = 122.3 = 32.17 °C ΔT
ln( 1 ) ln (
3
)
ΔT2

Qh = U Am ΔTm
141684= 1.5 × Am× 32.17
Am= 2936.15 m2
Am= π dm L N M
do + di
dm= 2
0.028+ 0.024
2936.15 = π × × L × 620 × 4
2

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L=14.19 m
b) Cogeneration and combined system
From the previous section (a)
h1=3373 KJ/kg
h2=2740 KJ/kg
h3=721 KJ/kg
h4=3480 KJ/kg
h5=2490 KJ/kg
h6=192 KJ/kg
h7= 202 KJ/kg
h8=730.2 KJ/kg
In cogeneration system the bled steam isn’t
to be used as process heat then pumped back to boiler as saturated liquid (i.e., CFWH
is replaced by a process heater)
h9 = (1-y) ×h7 + y×h8 = (1-0.2) × 202 + 0.2 × 730.2 = 307.64 kJ/kg
Qrej= ṁst (1-y) (h5-h6) = 350.877(1-0.2) (2490 – 192) = 645039 kW
Quse Qadd
ηboiler= =
QT QT
Qadd y h8
0.88 =ṁ 3
f × 25×10
h9
Qadd = ṁst (h1-h9) + ṁst (1-y) (h4-h2)
(1-y) h7
= 350.877 (3373-307.64) + 350.877 (1-0.2) (3480-2740) =1283257 kW
1283257
2) ṁ f = 0.88 × 25×103 = 58.33 kg/s = 209.988 ton/hr
By ignoring pump work the Wnet = WT
WT +Qh
ηco = Qadd
(500×1000)+141684
ηco = × 100= 50 %
1283257
ηe = 43.81 %
ηh = 88 %
𝑊𝑇 500
e=𝑊 =500+141.684 = 0.78
𝑇 +𝑄𝐻
1-e = 0.22
1
ηc = e 1−e
+ η
η e h
1
ηCHP = 0.78 0.22 = 49 %
+
.4381 0.88

ηco > ηCHP

Cogeneration system is beneficial compared to CHP system

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