Post on 30-Apr-2022
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 1
سم الله الرحمن الرحيم
HTI, Mech. Eng. Dept., Thermal Engineering, ME 231
Prof. Dr. Hesham Mostafa
1-Introduction
Course syllabus from internal regulation:
Application of thermodynamics and heat transfer to power stations,
combustion engines, industrial plants. Emphasis is given to energy planning
and economic utilization. Cogeneration of energy in industrial systems is
needed.
References: 1) P.K.Nag , “Power Plant Engineering”, McGraw Hill, 2008.
2) M.M.El-Wakil , “Power Plant Technology”, McGraw Hill, 2015.
3) R.S.Khurmi and J.K. Jupta, A Text book of “Thermal Engineering”,1998.
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Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 2
Carnot cycle
Carnot efficiency:
ηc= 1- (T1 /T2 )
where : T1=Minimum temperature, K
T2=Maximum temperature, K
State parameters:
1. Pressure; Pa( N/m2 ).
2. Temperature; (K).
3. Specific volume; (m3/kg).
4. Internal energy; (J/kg).
5. Enthalpy; (J/kg).
6. Entropy; (J/kg.K).
--------------------------------------------------------------------
Entropy (s); It means transformation.
It measures the disorder of the molecules.
The entropy of a substance is zero at absolute zero temperature.
Change of entropy (ds);
ds =Heat supplied or rejected (dQ)
Absolute temperature (T)
ds =dQ
T
T
s
2T
T1
1
2
4
3
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 3
Simple Rankine cycle
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 4
Simple Rankine cycle Processes:
1→ 2 : isentropic expansion for steam in turbine.
2→ 3 : heat reject in condenser.
3→ 4 : isentropic compression for water in pump.
4→ 1 : heat added in boiler.
WT = m.st (h1 –h2 ) , wT = (h1 –h2 )
Wp = m.st (h4- h3 ) = , wp =(h4 –h3 ) = υ ΔP =
1
𝜌 𝛥𝑃 ≅ 0.1 (ΔP in bar)
Qadd = m.st (h1- h4) , qadd= (h1 –h4 )
Qrej= m.st (h2- h3) , q rej = (h2 –h3 )
ηth = Wnet
Q𝑎𝑑𝑑 %
Where;
m.st = mass flow rate of steam, kg/s
h1 = Specific enthalpy of steam inlet to turbine, kJ/kg.
h2 = Specific enthalpy of steam inlet to condenser, kJ/kg.
h3 = Specific enthalpy of steam inlet to pump, kJ/kg.
h4 = Specific enthalpy of steam inlet to boiler, kJ/kg.
Qadd = Heat added, kW
Qrej = Heat reject, kW
WT = Turbine power, kW.
WP = Pump power, kW.
wT = Specific work for turbine , kJ/kg.
wP = Specific work for pump , kJ/kg.
ηth = Thermal efficiency, %.
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 5
Methods of feed water treatments
Water used in boiler must be free from various impurities. The different
treatments to remove the various impurities are as follow;
1. Mechanical treatment:
Rankine cycle with modifications
1) Superheat: Superheat has an additional benefit; it results in drier steam at turbine exhaust.
This means that; turbine operation with less moisture leads to more efficient
and less prone to blade damage.
WT=𝑚𝑠𝑡 (h1-h2) = [KW] (work done by turbine)
Qadd=𝑚𝑠𝑡 (h1-h4) = [ KW] (heat added to feed water in boiler )
Wp= 𝑚𝑠𝑡 (h4-h3) = [ KW] (pump work)
Wnet=WT - WP = [KW]
Qrej= 𝑚𝑠𝑡 (h2- h3) = [KW]
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 6
Reheat :
Superheated steam expanded part of the way in a high pressure turbine,
after which it is returned back to the boiler to reheat at a constant
pressure to the same maximum temperature or near it. The reheated
steam expands in low pressure turbine to the condenser pressure.
Reheat results in drier steam at turbine exhaust, which is beneficial for
real cycle. It is found that, the efficiency is improved (by about 3%) if
the reheat pressure was about 20%- 25% of the maximum pressure.
wt= (h1 - h2)+ (h3 - h4) = [KJ/kg]
WT=𝑚𝑠𝑡 wt = 𝑚𝑠𝑡 (h1 - h2)+ 𝑚𝑠𝑡 (h3 - h4) = [KW]
qadd= (h1-h6) + (h3 – h2 ) = [KJ/kg]
Qadd=𝑚𝑠𝑡 qadd = 𝑚𝑠𝑡 (h1-h6) + 𝑚𝑠𝑡 (h3 – h2 ) = [KW]
Wp= 𝑚𝑠𝑡 (h6-h5) = [KW]
Wnet=WT-WP
Qrej= 𝑚𝑠𝑡 (h4 – h5 ) = [KW]
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 7
2) Regeneration:
The irreversibility can be eliminated if the feed water to the boiler at saturation
temperature which corresponding to boiler pressure. Turbine work is decreases
and small increase in efficiency was obtained.
There are three types of feed water heaters;
1- Open or direct contact type.
2- Closed type with drains pumped forward.
3- Closed type with drains cascaded backward.
a-Rankine cycle with open or direct contact type (OFWH)
wt= (h1-h4) + (1-y) ( h4 - h2) = [KJ/kg]
WT= 𝑚𝑠𝑡 wt = 𝑚𝑠𝑡 (h1 – h4)+ 𝑚𝑠𝑡 (1-y) (h4 – h2) = [KW]
qadd= (h1-h7) = [KJ/kg]
Qadd=𝑚𝑠𝑡 qadd = 𝑚𝑠𝑡 (h1-h7) = [KW]
Wp= 𝑚𝑠𝑡 wp = 𝑚𝑠𝑡 [(1-y) (h6 - h3)+ (h7 - h5)] = [KW]
Wnet=WT-WP = [KW]
Qrej= 𝑚𝑠𝑡 (1-y) (h2-h3) = [KW]
WHERE : y= Fraction of the steam extracted to OFWH
y
1-y
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 8
Heat balance for OFWH
h5 = h6 (1-y) + y h4
NOTE :
The open feed water heater can be added with the reheating cycle
.
From h-s diagram
wt= (h1 – h2) + (h3 -h7) +(1-y) (h7 – h4)
WT=𝑚𝑠𝑡 wt = 𝑚𝑠𝑡 (h1 – h2)+ 𝑚𝑠𝑡 (h3 -h7) + 𝑚𝑠𝑡 (1-y) (h7 – h4)
qadd= (h1-h9) + (h3 – h2) = [KJ/kg]
QAdd= 𝑚𝑠𝑡 (h1-h9) + 𝑚𝑠𝑡 (h3 – h2) = [KW]
wp=(1-y) (h6-h5)+ (h9-h8) = [KJ/kg]
WP = 𝑚𝑠𝑡 wp = 𝑚𝑠𝑡 (1-y) (h6-h5)+ 𝑚𝑠𝑡 (h9-h8) = [KW]
Qrej= 𝑚𝑠𝑡 (1-y) (h4-h5) = [KW]
Rankine cycle
with
Reheat
+
OFWH
h-s diagram for
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 9
b-Closed type with drains pumped forward.
wt= (h1 – h2) +(1-y) (h2 – h3) = [KJ/kg]
WT= 𝑚𝑠𝑡 𝑤𝑡 = 𝑚𝑠𝑡 (h1 – h2)+ 𝑚𝑠𝑡 (1-y) (h2 – h3) = [KW]
qadd= (h1-h9) = [KJ/kg]
QAdd= 𝑚𝑠𝑡 qadd = 𝑚𝑠𝑡 (h1-h9) = [KW]
WP= 𝑚𝑠𝑡 (1-y) (h5 – h4) + 𝑚𝑠𝑡 y (h8 – h6) = [KW]
Qrej = (1-y) (h3-h4) = [ kJ/kg]
Qrej= 𝑚𝑠𝑡 (1-y) (h3-h4) = [KW]
WHERE :
y= Fraction of the steam taken to CFWH
NOTE : h9= y h8 +(1-y) h7
y
1-y
y
1-y
7y) h-(1 9h
8y h
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 10
TTD = Terminal Temperature Difference.
TTD = Sat. Temp. of bleeding steam – Exit feed water temperature
TTD: is the difference in Temp. between the outlet of hot and cold fluid in heat
exchanger , and since CFWH is considered heat exchanger between the
extracted steam and feed water .
ΔT2=TTD = T6 - T7 T7=T6 - TTD
h7 = h6 – Cp *TTD = Cp (T6 – TTD)
TTD is about 2°C OR 3°C
Heat balance for CFWH
(1-y) (h7 – h5 ) = y (h2 - h6 )
NOTE:
The closed feed water heater can be also added with the reheating cycle .
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 11
From h-s diagram
wt= (h1 – h2) + (h3 -h7) +(1-y) (h7 – h4)
WT=𝑚𝑠𝑡 wt = 𝑚𝑠𝑡 (h1 – h2)+ 𝑚𝑠𝑡 (h3 -h7) + 𝑚𝑠𝑡 (1-y) (h7 – h4)
qadd= (h1-h11) + (h3 – h2) = [KJ/kg]
Qadd= 𝑚𝑠𝑡 (h1-h11) + 𝑚𝑠𝑡 (h3 – h2) = [KW]
wp=(1-y) (h6-h5)+ (h10-h8) = [KJ/kg]
Wp = 𝑚𝑠𝑡 wp = 𝑚𝑠𝑡 (1-y) (h6-h5)+ 𝑚𝑠𝑡 (h10-h8) = [KW]
Qrej= 𝑚𝑠𝑡 qrej = 𝑚𝑠𝑡 (1-y) (h4-h5) = [KW]
NOTE : h11= y h10 +(1-y) h9
ΔT2=TTD = T8 - T9 T9=T8 - TTD
h9 = h8 – Cp *TTD = Cp (T9 – TTD)
9y) h-(1
Rankine cycle
with
Reheat
+
CFWH
h-s diagram for
11h
10h y
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 12
c-Closed type with drains cascaded backward
wt= (h1 – h2) +(1-y) (h2 – h3) = [KJ/kg]
WT= 𝑚𝑠𝑡 𝑤𝑡 = 𝑚𝑠𝑡 (h1 – h2)+ 𝑚𝑠𝑡 (1-y) (h2 – h3) = [KW]
qadd= (h1-h6) = [KJ/kg]
QAdd= 𝑚𝑠𝑡 qadd = 𝑚𝑠𝑡 (h1-h6) = [KW]
wp= (h5 - h4) = [KJ/kg]
WP= 𝑚𝑠𝑡 (h5 – h4) = [KW]
qrej= (1-y) (h3-h4) + y (h8-h4) = [KJ/kg]
= (1-y) (h3-h8) + (h8 - h4) = [KJ/kg]
Qrej = 𝑚𝑠𝑡 qrej = [KW]
= 𝑚𝑠𝑡 (1-y) (h3-h4) +𝑚𝑠𝑡 y (h8-h4) = [KW]
=𝑚𝑠𝑡 (1-y) (h3-h8) + 𝑚𝑠𝑡 (h8 - h4) = [KW]
WHERE : y= Fraction of the steam taken to CFWH
Heat balance for CFWH
y (h2-h7) = h6 - h5
TTD : difference between both outlet of CFWH
T7=T6 – TTD T6=T7 – TTD
h6 = Cp T6 = h7- Cp TTD
TTD is about 2°C OR 3°C
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 13
RANKINE CYCLE with (super heater + reheater + OFWH + CFWH)
Reheat
+
OFWH
+
CFWH
h-s diagram for
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 14
From h-s diagram :
wt= (h1-h2) + (h3-h4)+(1-y1)(h4-h5) + (1-y1-y2) (h5-h6) = [kJ/kg]
WT = 𝑚𝑠𝑡 wt
= 𝑚𝑠𝑡 (h1-h2) +𝑚𝑠𝑡 (h3-h4)+ 𝑚𝑠𝑡 (1-y1)(h4-h5) + 𝑚𝑠𝑡 (1-y1-y2) (h5-h6)= [KW]
qadd= (h1-h14) +(h3-h2) = [kJ/kg]
QAdd = 𝑚𝑠𝑡 qadd =𝑚𝑠𝑡 (h1-h14) +𝑚𝑠𝑡 (h3-h2) = [KW]
wp= (1-y1-y2) (h8-h7) + (1-y1)(h10-h9) + y1 (h13-h11) = [KJ/kg]
or
wp= (1-y1-y2) (0.1ΔP8,7) + (1-y1) (0.1ΔP10,9) + y1 (0.1ΔP13,11) = [KJ/kg]
WP= 𝑚𝑠𝑡 wp = [KW]
qrej= (1-y1-y2) (h6-h7) = [kJ/kg]
Qrej= 𝑚𝑠𝑡 (1-y1-y2) (h6-h7) = [KW]
TTD : difference between both outlet of CFWH
T11=T12 – TTD T12=T11 – TTD
h12 = Cp T12 = h11 - Cp TTD
TTD is about 2°C OR 3°C
NOTE : h14= y1 h13 +(1-y1) h12
Heat balance for CFWH
(1-y1) (h12-h10) = y1 (h4-h11)
Or
y1 h4 + (1-y)h10 = (1-y) h12 + y h11
Heat balance for OFWH
(1-y1-y2) h8 + y2 h5 = (1-y1) h9
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 15
Example (1.1):
For the Rankine cycle steam at 50 bar expands in steam turbine. Condenser
pressure, 0.1 bar.
Calculate Wnet &ηth for case (6). Also, find the amount of mass flow rate of
cooling water in condenser if temperature rise in cooling water was 10 oC and
steam flow rate was 100 Ton/hr. You can Fill the following table
Case
no.
WT,
kJ/kg
WP,
kJ/kg
Wnet,
kJ/kg
qadd,
kJ/kg
ηth,
%
1 Simple cycle.
2 Superheat to 350 oC.
3 • Superheat to 350 oC.
• Expands up to saturation line.
• Reheat to to 350 oC.
4 • Superheat to 350 oC.
• Expands up to saturation line.
• Reheat to to 350 oC.
• Bleeding steam to OFWH @ 1
bar
5 • Superheat to 350 oC.
• Expands up to saturation line.
• Reheat to to 350 oC.
• Bleeding steam to CFWH @ 2
bar
6 • Superheat to 350 oC.
• Expands up to saturation line.
• Reheat to to 350 oC.
• Bleeding steam to OFWH @ 1
bar
• Bleeding steam to CFWH @ 2
bar
Solution
General givens:
Pst = 50 bar Pcon= 0.1 bar ΔTcon=10oC 𝑚𝑠𝑡 =100 ton/hr = 100
3.6 𝐾𝑔/𝑠
Find:
Wnet & ηth for each case and fill the table above
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 16
1-simple case
from tables @ P=50 bar
h1=hg=2794 kJ/kg , s1=5.973 kJ/kg.k
mark point (1) at chart using pressure
line =50 bar intersect with saturation
vapor line .
from point (1) draw vertical line (s=c)
until intersect with P=0.1 bar line .
h2=1887 kJ/kg
h3=hf @ P=0.1 bar
h3=192 kJ/kg
h4= h3 + 0.1 ΔP
h4= 192+ 0.1 (50-0.1)=196.99 kJ/kg
WNET=WT - WP
WT= h1 - h2
WT= 2794-1887=907 kJ/kg
Wp=0.1*ΔP = 0.1(50-0.1)=4.99 kJ/kg
WNET=907 - 4.99= 902.01 kJ/kg
Qadd= h1 – h4
Qadd= 2794-196.99 =2597.01 kJ/kg
ηth =WNET
Qadd
ηth = 902.01
2597.01∗ 100 = 34.73 %
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 17
2-super heat to 350°C
From super heating sables using
p=50 bar and T=350°C to find (1)
h1= 3069.3 kJ/kg
s1= 6.451 kJ/kg.k
In chart draw vertical line (s=c) untill
intersect with P=0.1 bar line to find (2)
h2= 2042 kJ/kg
(also can be found from thermo dynamics
laws ) using sf , sg , hf and hg at P=0.1 bar
x= (s1 - sf)/(sg - sf)
h2= hf + x(hg - hf)
h3= hf @ P=0.1 bar
h3= hf= 192 kJ/kg
h4= h3 + 0.1 ΔP
h4= 192+ 0.1 (50-0.1)=196.99 kJ/kg
WT= h1 - h2
WT= 3069.3-2042=1027.3 kJ/kg
Wp=0.1*ΔP = 0.1(50-0.1)=4.99 kJ/kg
WNET=WT - WP
WNET=1027.3- 4.99= 1022.31 kJ/kg
Qadd= h1 – h4
Qadd= 3069.3-196.99 =2872.31 kJ/kg
ηth =WNET
Qadd
ηth = 1022.31
2872.31∗ 100 = 35.59 %
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 18
3-super heat to 350°C and expand to saturation line then reheat to 350°C
h1= 3069.3 kJ/kg (same as previous case )
point (2) can be obtained by drawing (s=c)
line up to sat. vapor line
h2=2790 kJ/kg
h3 draw line from point (2) parallel with
pressure lines untill intersect with temp. line
T= 350°C , so h3= 3153 kJ/kg
point (4) is obtained by drawing S=C line
untill intersect with P=0.1 bar
h4 = 2250 kJ/kg
h5 = hf @ P=0.1 bar =192 kJ/kg
h6= h5 + 0.1 ΔP
h6= 192+ 0.1 (50-0.1)=196.99 kJ/kg
WT= (h1 - h2) + (h3-h4)
WT= (3069.3-2790)+(3153-2250) =1182.3 kJ/kg
Wp=0.1*ΔP = 0.1(50-0.1)=4.99 kJ/kg
WNET=WT - WP
WNET=1182.3- 4.99= 1177.31 kJ/kg
Qadd=( h1 – h6)+(h3 - h2)
Qadd= (3069.3-196.99)+(3153-2790) =3235.31 kJ/kg
ηth =WNET
Qadd
ηth = 1177.31
3235.31∗ 100 = 36.38 %
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 19
4- case(3) + bleeding steam to OFWH @ 1bar
h1=3069.3 kJ/kg , h2=2790 kJ/kg
h3= 3153 kJ/kg , h4 = 2250 kJ/kg
h5 = hf @ P=0.1 bar =192 kJ/kg
h6= h5 + 0.1 ΔP1,0.1
h6= 192+ 0.1 (1-0.1)=192.09 kJ/kg
point (7) is obtained by drawing vertical
line (s=C) from (3) until intersect with
P=1 bar line to get (h7= 2585 kJ/kg )
h8= hf @ p=1 bar = 417 kJ/kg
h9= h8 + 0.1 ΔP50,1
h9= 417+ 0.1 (50-1)=421.9 kJ/kg
HEAT BALANCE FOR O.F.W.H
h6 (1-y) +y h7 =h8
192.09*(1-y)+y*2585 =417
y=0.094
WT= (h1 - h2) + (h3-h7)+(1-y)(h7-h4)
WT= (3069.3-2790)+(3153-2585)+(1-0.094)(2585-2250) =1150.81 kJ/kg
Wp=0.1*(1-y)ΔP1,0.1+0.1*ΔP50,1 or Wp= (h6-h5)(1-y)+(h9-h8)
= 0.1(1-.094)(1-0.1)+0.1(50-1) =4.98 kJ/kg
WNET=WT - WP
WNET=1150.81- 4.98= 1145.83 kJ/kg
Qadd=( h1 – h9)+(h3 - h2)
Qadd= (3069.3-421.9)+(3153-2790) =3010.4 kJ/kg
ηth =WNET
Qadd =
1145.83
3010.4∗ 100 = 38.06 %
6y) h-(1
8h
7Y h
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 20
5-case (3) + bleeding steam to CFWH @ P=2 bar
h1=3069.3 kJ/kg
h2=2790 kJ/kg
h3= 3153 kJ/kg
h4 = 2250 kJ/kg
h5 = hf @ P=0.1 bar =192 kJ/kg
h6= h5 + 0.1 ΔP50,0.1
h6= 192+ 0.1 (50-0.1)=196.99 kJ/kg
point (7) is obtained by drawing vertical
line (s=C) from (3) until intersect with
P=2 bar line to get (h7= 2702 kJ/kg )
h8= hf @ p=2 bar = 505 kJ/kg
h10= h8 + 0.1 ΔP50,2
h10= 505+ 0.1 (50-2)=509.8 kJ/kg
h9 = h8 – T.T.D * Cp T.T.D (Terminal temperature difference) = 3 or 2°C
= 505-3*4.2 =492.4 kJ/kg
HEAT BALANCE FOR C.F.W.H
h6 (1-y) +y h7 = y h8 + (1-y) h9
196.99*(1-y)+y*2702 =505*y +(1-y)*492.4
y=0.1185
h11=(1-y)*h9+y*h10
=(1-.1185)*492.4+ 0.1185*509.8 = 494.46 kJ/kg
6y) h-(1
9y) h-(1
y8h
10Y h
9y) h-(1 11h
7Y h
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 21
WT= (h1 - h2) + (h3-h7)+(1-y)(h7-h4)
WT= (3069.3-2790)+(3153-2702)+(1-0.1185)(2702-2250)
=1128.738 kJ/kg
Wp=0.1*(1-y)ΔP50,0.1+y*0.1*ΔP50,1 or Wp= (h6-h5)(1-y)+(h10-h8)*y
= 0.1(1-0.1185)(50-0.1)+0.1185*0.1(50-2) =4.96 kJ/kg
WNET=WT - WP
WNET=1128.738- 4.96= 1123.778 kJ/kg
Qadd=( h1 – h11)+(h3 - h2)
Qadd= (3069.3-494.46)+(3153-2790) =2937.84 kJ/kg
ηth =𝑊𝑛𝑒𝑡
𝑄𝑎𝑑𝑑
= 1123.738
2937.84∗ 100 = 38.25 %
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 22
5-super heat to 350°C then expant to sat. line and super heat to 350°C
+ bleeding steam to OFWH and to CFWH @ 1 bar and 2 bar respectively
h1=3069.3 kJ/kg (chart)
h2=2790 kJ/kg (chart)
h3= 3153 kJ/kg (chart)
h4 = 2250 kJ/kg (chart)
h5 = hf @ P=0.1 bar =192 kJ/kg
h6= h5 + 0.1 ΔP1,0.1
h6= 192+ 0.1 (1-0.1)=192.09 kJ/kg
h7= 2702 kJ/kg (chart)
h8= 2585 kJ/kg (chart)
h9= hf @ p=1 bar = 417 kJ/kg
h10= hf @ p=2 bar = 505 kJ/kg
h11= h9 + 0.1 ΔP50,1
h11= 417+ 0.1 (50-1)=421.9 kJ/kg
h12 = h10 – T.T.D * Cp T.T.D (Terminal temperature difference) = 3 or 2°C
= 505 - 3*4.2 =492.4 kJ/kg
h13= h10 + 0.1 ΔP50,2
h13= 505+ 0.1 (50-2)=509.8 kJ/kg
HEAT BALANCE FOR C.F.W.H
h11 (1-y1) +y1 h7 = y1 h8 + (1-y1) h9
421.9*(1-y1)+y1*2702 =505*y1 +(1-y1)*492.4
y1=0.0311
11) h1y-(1
12h )1y-(1
1y10 h
7h 1Y
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 23
HEAT BALANCE FOR O.F.W.H
h6 (1-y1-y2) +y2 h8 =(1-y1) h9
192.09*(1-0.0311-y2)+y2*2585 =417 (1-0.0311)
y2=0.091
h14= y1 h13 +(1-y1) h12
=(0.0311)*509.8+ (1-0.0311)*492.4
= 492.94 kJ/kg
WT= (h1 - h2) + (h3-h7) + (1-y1)(h7-h8)+ (1-y1-y2)(h7-h4)
WT= (3069.3-2790)+(3153-2702)+(1-0.0311)(2702-2585)
+(1-0.0311-0.091)(2585 -2250)
=1136 kJ/kg
Wp=(1-y1-y2)*0.1 *ΔP1,0.1+(1-y1)*0.1*ΔP50,1+y1* 0.1 ΔP50,2
= (1-0.0311-0.096)*0.1*(1-0.1)+(1-0.0311)*0.1(50-1)+ 0.0311*0.1*(50-2)
= 4.975 kJ/kg
or Wp= (h6-h5) (1-y1-y2)+(h11-h9)(1-y1)+(h13-h10)(y1)
= (192.09-192)(1-0.0311-0.096)+(421.9-417)(1-0.0311)+0.0311(509.8-505)
= 4.975 kJ/kg
WNET=WT - WP
WNET=1136- 4.975= 1131 kJ/kg
Qadd=( h1 – h14)+(h3 - h2)
Qadd= (3069.3-492.94)+(3153-2790) =2939.36 kJ/kg
𝑊𝑛𝑒𝑡
𝑄𝑎𝑑𝑑th =η
1131
2939.36∗ 100 = 38.478 % =
6) h2y-1y-1(
9) h1y-(1
8h 2Y
13h 1Y
12) h1y-(1 14H
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 24
condenser
��𝑠𝑡 = 100 ton/hr
= 100∗1000
3600= 27.778 𝑘𝑔/𝑠
c.wTΔ cp 𝑚𝑐.𝑤) =5h-4) (h2y-1y-(1𝑚𝑠𝑡
*4.2*10 𝑚𝑐.𝑤192) =-.096)(2250-.0311-27.778 (1
4277.21 ton /hr=1188.11 kg/s = 𝑚𝑐.𝑤
with previous with values at each case Fill the table
Case
no.
WT, kJ/kg WP,
kJ/kg
Wnet,
kJ/kg
qadd,
kJ/kg
ηth, %
1 Simple cycle. 907 4.99 902.01 2597.01 34.73
2 Superheat to 350 oC. 1027.3 4.99 1022.31 2872.31 35.59
3 • Superheat to 350 oC.
• Expands up to saturation
line.
• Reheat to to 350 oC.
1182.3 4.99 1177.31 3235.31 36.38
4 • Superheat to 350 oC.
• Expands up to saturation
line.
• Reheat to to 350 oC.
• Bleeding steam to OFWH
@ 1 bar
1150.81
4.98 1145.83 3010.4
38.06
5 • Superheat to 350 oC.
• Expands up to saturation
line.
• Reheat to to 350 oC.
• Bleeding steam to CFWH
@ 2 bar
1128.738
4.96 1123.778 2937.84
38.25
6 • Superheat to 350 oC.
• Expands up to saturation
line.
• Reheat to to 350 oC.
• Bleeding steam to OFWH
@ 1 bar
• Bleeding steam to CFWH
@ 2 bar
1136
4.975 1131 2939.36
38.478
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 25
Example (1.2): For the modified Rankine cycle 180 Ton/hr of steam at 100 bar, 400 oC expands in high
pressure turbine up to the saturation line. Then it is reheated to the same
maximum temperature and expanded in low pressure turbine up to condenser
pressure, 0.1 bar. Bleeding steam at 2.5 bar & 1 bar to closed feed water heater
and open feed water heater respectively. Isentropic efficiency for turbine is
90%. Draw schematic diagram and h-s diagram.
Also Find thermal efficiency ,WT, Qadd , WNET
Solution
Givens :
𝑚𝑠𝑡 = 180 ton/hr = 180∗1000
3600= 50 𝑘𝑔/𝑠
Pst=100 bar , Tst=400 °C
Pcond=0.1 bar
PCFWH=2.5 bar ; POFWH= 1 bar
ηisen= 90%
From steam super heat tables
@ Pst=100 bar , Tst=400 °C
h1=3097 kJ/kg
h2=2803 kJ/kg (chart)
h3=3235 kJ/kg (chart)
h4= 2680 kJ/kg (chart)
h5=2532 kJ/kg (chart)
h6=2210 kJ/kg (chart)
h7= hf @ 0.1 bar = 192 kJ/kg (tables)
h8= h7+ΔP1,0.1 = 192 + 0.1 (1-0.1) = 192.09 kJ/kg
h9= hf @ 1 bar = 417 kJ/kg (tables )
h10= h9 + ΔP100,1=417 + 0.1(100-1) = 426.9 kJ/kg
h11 = hf @ 2.5 bar = 535 kJ/kg ; T12 = Tsat =127.4°C (tables)
T12 = T11 – TTD = 127.4 – 3 =124.4 °C
h12 = CP*T12 = 4.2 *124.4 =522.48 kJ/kg
or
h12= h11 – CP *TTD = 535 – 3*4.2 = 522.4 kJ/kg
h13= h11+0.1 ΔP1,0.1 = 535+ 0.1(100-2.5) =544.75 kJ/kg
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 26
NOTE : h14= y1 h13 +(1-y1) h12
Heat balance for CFWH
(1-y1) (h12-h10) = y1 (h4-h11)
(1-y1) (522.48-426.9) = y1 (2680-535)
y1=0.0426
Heat balance for OFWH
(1-y1-y2) h8 + y2 h5 = (1-y1) h9
(1-0.0426 -y2) 192.09 + 2532*y2= (1-0.0426) 417
y2= 0.092
h14= y1 h13 +(1-y1) h12 h14= 0.426 *544.75 +(1-0.0426)*522.48
=523.4 KJ /kg
wt,isen= (h1-h2) + (h3-h4)+(1-y1)(h4-h5) + (1-y1-y2) (h5-h6)
(3097-2803)+(3235-2680)+(1-0.0426)(2680-2532)
+ (1-0.0426-0.092)(2532-2210)
wt,isen = 1269.354 KJ /kg
wt = ηisen *wt,isen = 0.9*1269.354 =1142.4186 KJ /kg
qadd= (h1-h14) +(h3-h2)
(3097-523.4) + (3235-2803) = 3005.6 kJ/kg
wp= (1-y1-y2) (h8-h7) + (1-y1)(h10-h9) + y1 (h13-h11)
=(1-0.0426-0.092) (192.09-192) + (1-0.0426)(426.9-417)
+ 0.0426 (544.75- 535) = 9.971 KJ /kg
or
wp= (1-y1-y2) 0.1(ΔP8,7) + (1-y1) 0.1(ΔP10,9) + y1 0.1(ΔP13,11)
=(1-0.0426-0.092)* 0.1*(1-0.1) + (1-0.0426)*0.1*(100-1)
+ 0.0426 *0.1*(100-2.5) = 9.971 KJ /k
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 27
wnet =wt - wp
= 1142.4186 -9.971 = 1132.4476 KJ /kg
𝜂𝑡ℎ =𝑤𝑛𝑒𝑡
𝑞𝑎𝑑𝑑
𝜂𝑡ℎ =1132.4476
3005.6 *100% = 37.677%
WT = 𝑚𝑠𝑡 wt = 50 * 1142.4 186 = 57120.93 KW
WP= 𝑚𝑠𝑡 wp = [KW]
= 50 *9.971 = 498.55 KW
WNET =WT – WP
=57120.93-498.55 = 56622.38 KW
QAdd = 𝑚𝑠𝑡 qadd = 50*3005.6 = 150280 KW
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 28
Drawing the major components of steam power plant.
• Steam generator (boiler).
• Steam turbine.
• Steam condenser plant .
• Circulating pump.
• Accessories;
Feed water heaters [closed and open (like deaerator) ]
Cooling Tower.
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 29
2-steam condenser plant
Pc = Psteam + Pair
Psteam : steam partial pressure
Pair : air partial pressure
Pc : condenser pressure
Vaccum efficiency
Ƞvac = 𝑃𝑎𝑡𝑚−𝑃𝑐
𝑃𝑎𝑡𝑚−𝑃𝑠𝑡𝑒𝑎𝑚
Condenser efficiency
Ƞcond = 𝑎𝑐𝑡𝑢𝑎𝑙 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑟𝑖𝑠𝑒 𝑜𝑓 𝑐𝑜𝑜𝑙𝑖𝑛𝑔 𝑤𝑎𝑡𝑒𝑟
𝑚𝑎𝑥. 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑓𝑜𝑟 𝑐𝑜𝑜𝑙𝑖𝑛𝑔 𝑤𝑎𝑡𝑒𝑟 =
𝑇𝑐𝑤𝑜−𝑇𝑐𝑤𝑖
𝑇𝑠 − 𝑇𝑐𝑤𝑖
Where: Ts @ Pc
Pg : gauge pressure
Pabs : absolute pressure
Pvac : vaccum pressure
Patm = atmospheric pressure
Pabs = Patm + Pg
Pabs = Patm - Pvac
Atm
Absolute datum
Pg
Patm
Pabs
Pvac
Pabs
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 30
2-Steam condensers
a-Open or jet condenser
Heat balance
��𝑐.𝑤
wh
��𝑠𝑡
sth c.wh��𝑐.𝑤+ sth��𝑠𝑡= o) h��𝑐.𝑤+��𝑠𝑡(
cT p= C oh
= condensate temperaturec T
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 31
b-closed (surface) condenser
Equations :
Qcond = ��𝑠𝑡 (hst – hcond.)
= ��𝑐.𝑤CPw (Tc.w.o -Tc.w.i )
= U Am ΔTm
hcond = CPw Tcondensate
CPw = 4.2 kJ/kg.C
��𝑐.𝑤= ρ u a N
Where ρ : water density [kg/m3]
u: water velocity inside tube [m/s]
a: tube cross sectional area [m2]
a= 𝜋
4 di
2
N: number of tubes
Am = π dm L N M
dm =𝑑𝑜+𝑑𝑖
2
M: number of paths
ΔTm = ΔT1−ΔT2
ln(ΔT1ΔT2
)
ΔT1 = Tsat – Tc.w.i
ΔT2 = Tsat – Tc.w.o
id
od
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 32
Example (2.1):
a) Draw and mention the major components of steam condensing plant.
b) In a condenser test the following observations were taken:
Vacuum reading = 690 mm Hg. Barometric reading = 750 mm Hg.
Mean condenser temperature = 35 oC. Hot well temperature = 29 oC.
Inlet cooling water temperature = 12 oC.
Outlet cooling water temperature = 22 oC.
Steam flow rate (m.st ) = 1250 kg/hr.
Find 1.) Vacuum efficiency. 2.) Condenser efficiency. 3) Mass of air per 1 m3
of condenser volume. 4.) Mass flow rate of cooling water if dryness fraction
(x) for inlet steam was 0.85
-----------------------------------------------
Example (2.2) :
Design a surface condenser from the following:
Steam pressure is 0.1 bar ; x=0.95
Steam flow rate 100 ton/hr
Inlet and outlet Cooling water temperatures are 8°C and 26°C respectively , the
overall heat transfer coefficient is 3.5 KW/m2 °C , cooling water velocity in
condenser tubes is 1.5 m/s . Condenser tube diameters are 30 mm and 34 mm.
Solution
Givens:
Pcond.=0.1 bar ; x=0.95
��𝑠𝑡=100 ton/hr = 100∗1000
3600 kg/s
Tc.w.i=8°C , Tc.w.o=26°C
U=3.5 KW/m2 °C
u =1.5 m/s
di=30 mm ; do=34 mm
Qcond = ��𝑠𝑡 (hst – hcond.)
from tables @Pcond=0.1 bar
, KJ/kg 2392= fgh 192 KJ/kg ; = fh ;C °45.8 =satT
fg+ x h f= h sth
= 192 + 0.95 (2392) = 2464.4 KJ/kg
hcond.= 192 KJ/kg
Qcond = 100∗1000
3600 (2464.4 – 192) = 63122.22 KW
Qcond = ��𝑐.𝑤CPw (Tc.w.o -Tc.w.i )
63122.22 = ��𝑐.𝑤*4.2 *(26 - 8)
��𝑐.𝑤= 834.95 kg/s
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 33
��𝑐.𝑤= ρ u a N
834.95= 1000 *1.5*𝜋
4 (30*10-3)* N
N = 788 tube
ΔTm = ΔT1−ΔT2
ln(ΔT1ΔT2
)
ΔT1 = 45.8– 8 = 37.8 °C
ΔT2 = 45.8 – 26 = 19.8 °C
ΔTm = 37.8−19.8
ln(37.8
19.8)
= 27.83°C
Qcond = U Am ΔTm
63122.22 = 3.5 * Am* 27.83
Am = 648 m2
Am = π dm L N M
648 = π * (0.03+0.034
2) * L*788*1
L = 8.179 m
-----------------------------------------------------------
=45.8 °C
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 34
Example (2.3 ):
For the modified Rankine cycle 150 Ton/hr of steam at 60 bar, 350 oC expands
in high pressure turbine up to the saturation line. Then it is reheated to the
same maximum temperature and expanded in low high pressure turbine up to
condenser pressure, 0.08 bar. Bleeding steam at 2 bar to closed feed water
heater. Design a surface condenser (find condenser length and number of
tubes) if the overall heat transfer coefficient was 0.5 kW/m2 oC. Condenser
tube diameters are 28 mm and 32 mm. Cooling water velocity inside condenser
tube was 0.2 m/s. Inlet and outlet cooling water temperatures are 10 oC and 22 oC respectively
Solution
Givens:
Pst=60 bat , Tst =350 °C
Pcond.=0.08 bar
��𝑠𝑡=150 ton/hr = 150∗1000
3600 kg/s
Tc.w.i=10°C , Tc.w.o=22°C
U=0.5 KW/m2 °C
u =0.2 m/s
di=28 mm ; do=32 mm
bleed steam @ 2 bar
since only Required is making
design for condenser
h4 = 2190 KJ/kg (chart)
h5 = hf @ 0.08 bar =174 KJ/kg
h6 = h5 + 0.1 ΔP60, 0.08
h6 = 174 + 0.1 ( 60 – 0.08 ) =180 KJ/kg
h7 = 2635 KJ/kg ( chart)
h8 =hf @ 2 bar = 505 KJ/kg
h9 = h8 - CP * TTD
h9 = 505 – 4.2*3 =492.4 KJ/kg
Heat balance for CFWH
(1-y) ( h9 – h6 )= y (h7 – h8 )
(1 – y ) ( 492.4 – 180 ) = y ( 2635 – 505)
y= 0.1279
Qcond = ��𝑠𝑡(1-y) (h4 – h5)
= 150∗1000
3600*(1-0.1279)* (2190 -174) = 73256.4 KW
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 35
Qcond = ��𝑐.𝑤CPw (Tc.w.o -Tc.w.i )
73256.4 = ��𝑐.𝑤*4.2 *(22 - 10)
��𝑐.𝑤= 1453.5 kg/s
��𝑐.𝑤= ρ u a N
1453.5= 1000 *0.2*𝜋
4 (28*10-3)* N
N = 11803 tube
ΔTm = ΔT1−ΔT2
ln(ΔT1ΔT2
)
ΔT1 = 41.5 -10 = 31.5 °C
ΔT2 = 41.5 – 22 = 19.5 °C
ΔTm = 31.5−19.5
ln(31.5
19.5)
= 25°C
Qcond = U Am ΔTm
73256.4 = 0.5 * Am* 25
Am = 5860.512 m2
Am = π dm L N M
5860.512 = π * (0.028+0.032
2) * L*11803*1
L = 5.26 m
-----------------------------------------------------------
=41.5°C
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 36
Closed Feed Water Heaters
Equations :
Qcond = ��𝑏 (hst – hw.)
= ��𝑓.𝑤CPw (Tf.w.o -Tf.w.i ) = ��𝑓.𝑤(hf.w.o -hf.w.i)
= U Am ΔTm
Where ��𝑏 : Bleed steam flow rate [kg/s]
Tf.w.o= Tw – TTD
hf.w.o= hw – Cp TTD = CP Tf.w.o
TTD ≈ 2 or 3 °C
��𝑓𝑤= ρ u a N
Where ρ : water density [kg/m3]
u: water velocity inside tube [m/s]
a: tube cross sectional area [m2]
a= 𝜋
4 di
2
N: number of tubes
Am = π dm L N M
dm =𝑑𝑜+𝑑𝑖
2
M: number of paths
ΔTm = ΔT1−ΔT2
ln(ΔT1ΔT2
)
ΔT1 = Tsat – Tf.w.i
ΔT2 = Tsat – Tf.w.o = TTD
id
od
(��𝑓𝑤 , hf.w.i , Tf.w.i )
(��𝑓𝑤 , hf.w.o , Tf.w.o)
(��𝑏 , hw ,TW)
(��𝑏 , hst )
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 37
Example (2.4 ):
For the modified Rankine cycle 200 Ton/hr of steam at 90 bar, 400 oC expands
in high pressure turbine up to the saturation line. Then it is reheated to the
same maximum temperature and expanded in low pressure turbine up to
condenser pressure, 0.1 bar. Bleeding steam at 3 bar & 1 bar to closed feed
water heater and open feed water heater respectively.
Draw surface condenser in details, and Find mass flow rate of cooling water
for steam condenser if temperature rise in cooling water was 10 oC. Design a
closed feed water heater (find its length and number of tubes) if the overall
heat transfer coefficient was 500 W/m2 oC. Tube diameters are 20 mm and 24
mm. Water velocity inside (CFWH) tube was 0.25 m/s. Terminal Temperature
Difference was 3 oC. Number of passes =4.
Solution
Drawings can be found in lectures
Givens:
Pst=90 bar Tst =400 °C Pcond.=0.1 bar
��𝑠𝑡=200 ton/hr = 200∗1000
3600 kg/s
ΔTc.w=10°C U=500 W/m2 °C = 0.5 KW/m2 °C
u =0.25 m/s di=20 mm do=24 mm
bleed steam to CFWH @ 3 bar bleed steam to OFWH @ 1 bar
TTD=3°C M=4
h4 = 2760 KJ/kg (chart)
h5 = 2575 KJ/kg (chart)
h6 = 2242 KJ/kg (chart)
h7 = hf @ 0.1 bar =192 KJ/kg
h8 = h7 + 0.1 ΔP1 ,0.08
h8 = 192 + 0.1 ( 1 – 0.1 ) =192.09 KJ/kg
h9 =hf @ 1 bar = 417 KJ/kg
h10 = h9 + 0.1 ΔP90 ,1
h10 = 417 + 0.1 ( 90 – 1 ) =425.9 KJ/kg
h11 = hf @ 3 bar = 561 KJ/kg
h12 = h11 - CP * TTD
h12 = 561 – 4.2*3 = 547.4 KJ/kg
h13 = h11 + 0.1 ΔP90 ,3
h13 = 561 + 0.1 ( 90 – 3 ) =469.7 KJ/kg
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 38
Heat balance for CFWH
(1-y1) ( h12 – h10 )= y1 (h4 – h11 )
(1 – y1 ) ( 547.4 – 425.9 ) = y1 ( 2760 – 561)
y1= 0.0523
Heat balance for OFWH
(1-y1-y2)h8 + y2 h5 = (1-y1) h9
( 1- 0.0523 - y2) 192.09 + y2 2575 = (1- 0.0523) 417
y2 = 0.0894
𝑚𝑠𝑡 (1-y1-y2) (h6-h7 ) = 𝑚𝑐𝑤 CPw (ΔTcw )
200 ( 1- 0.0523 – 0.0894 )(2242 – 192) =𝑚𝑐𝑤 *4.2*10
𝑚𝑐𝑤 = 8378.73 ton/hr = 2327.4 kg/s (cooling water flow rate)
QCFWH = 𝑚𝑠𝑡 y1 (h4 – h11) = 𝑚𝑠𝑡 (1-y1) (h12 – h10 )
= 200
3.6 * 0.0523 (2760 – 561 ) =6389.31 KW
𝑚𝑠𝑡 (1-y1) = ρ u a N
200
3.6 ( 1- 0.0523 ) = 1000 * 0.25*
𝜋
4 (0.02)2 * N
N=670 tube
QCFWH =U Am ΔTm
ΔTm = ΔT1−ΔT2
ln(ΔT1ΔT2
)
ΔT1 = Tsat -T10
T10 =ℎ10
𝐶𝑃=
425.9
4.2= 101.4°C
ΔT1 = 133.5 -101.4 = 32.1 °C
ΔT2 =T11 – T12 = TTD = 3°C
ΔTm = 32.1−3
ln(32.1
3) = 12.27°C
Qcond = U Am ΔTm
6389.31 = 0.5 * Am* 12.27
Am = 1041.45 m2
Am = π dm L N M
1041.45 = π * (0.020+0.024
2) * L*670*4
L = 5.62 m
=133.5
°C
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 39
Deaerator
Feed water flow diagram :
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 40
Evaporators
Single stage evaporator
assume no losses
��𝒔𝒕 (hst – ho) = ��𝒘 (hg/p1 – hw) hst @ Pst and Xst
ho = hf @ Pst
hg/p1 = hg @ P1
hw = CPw Twi
multi stage evaporator
First stage
��𝒔𝒕 (hst – ho1) = ��𝒘1 (hg/p1 – hi,1)
Second stage
��𝑤1 (hg/P1 – ho2) = ��𝑤2 (hg/P2 – hi,2) ho1 = hf @ Pst (unless mention else )
hg/P1 = hg @ P1
ho2 = hf @P1 (unless mention else )
hg/P2 =hg@P2
h1 = CP Tw1
Note :
If mentioned Condensate subcooled temperature: the outlet temperature of the
condensed steam is lower than Tsat with given Tsub value i.e (To = Tsat – Tsub) .
𝑚𝑠𝑡
ho
1 P
stP
𝑚𝑠𝑡
sth
𝑚𝑤 ,hg
𝑚��,hw
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 41
Example (2.5 ):
A single stage evaporator receives heating steam at P =2 bar , x=0.95 and with
flow rate 1 kg/s . Inlet raw water at 1 bar , 20 °C .
Find amount of liberated (generated) vapour .
Solution
Givens:
Pst = 2 bar , Xst = 0.95
��𝑠𝑡= 1 kg/s
P1= 1 bar , T1 = 20 °C
Find ��𝒘= ??
assume no losses
��𝒔𝒕 (hst – ho) = ��𝒘 (hg/p1 – hw)
From tables@ Pst = 2 bar , x=0.95
hf = 505 KJ/kg , hfg = 2202 KJ/kg
hst = hf + x hfg
= 505 + 0.95 *2202 = 2596.9 KJ/kg
ho = hf @2 bar = 505 KJ/kg
hg/p1 = hg @ 1 bar = 2675 KJ/kg
hw = CPw Twi
= 4.2* 20 =84 KJ/kg
1*( 2596.9 – 505 ) = ��𝒘 (2675 – 84 )
��𝒘 = 0.807 kg/s
-----------------------------------------
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 42
Example (2.6 ):
Calculate the make-up water (@ 25 oC ) required for a double effects
evaporator per hour. If heating steam flow rate was 100 kg/hr and its
conditions @ 2.5 bar, x=0.9. Water pressure inside first and second effects of
evaporator are 2 bar & 1.5 bar respectively. Condensate subcooled from first
and second effects of evaporator are 2 oC and 1 oC respectively.
Solution
Givens:
Pst = 2.5 bar , Xst = 0.9
��𝑠𝑡= 100 kg/hr
P1= 2 bar ,P2 = 1.5 bar , T1 = 25 °C
Tsub1 = 2°C Tsub 2 = 1°C Twi = 25°C
Note :
Condensate subcooled temperature: means that the outlet temperature of the
condensed steam is lower than Tsat with given Tsub value i.e (To = Tsat – Tsub) .
Find ��𝒘𝟏 = ??
��𝒘𝟐= ??
First stage : assume no losses
��𝒔𝒕 (hst – ho1) = ��𝑤1 (hg/p1 – hi)
From tables@ Pst = 2.5 bar , x=0.9
hf = 535 KJ/kg , hfg = 2182 KJ/kg , Tsat = 127.4 °C
hst = hf + x hfg
= 535 + 0.9 *2182 = 2498.8 KJ/kg
ho1 = CP To1
To1 = Tsat – Tsub1 = 127.4 -2 =125.4 °C
ho1 = 4.2 * 125.4= 526.68 KJ/kg
or
ho1 = hf – CP Tsub1 = 535- 4.2*2 = 526.6 KJ/kg
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 43
hg/p1 = hg @ 2 bar = 2707 KJ/kg
hi = CPw Twi
= 4.2* 25 =105 KJ/kg
100*( 2498.8 – 526.68 ) = ��𝑤1 (2707 – 105 )
��𝒘 = 75.79 kg/hr
Second stage : assume no losses
��𝑤1 (hg/p1 – ho2) = ��𝒘𝟐 (hg/p2 – hi)
hg/p1 = 2707 KJ/kg
ho2 = CP To2
@ p1 = 2 bar Tsat =120.2 °C , hf = 505 KJ/kg
To2 = Tsat – Tsub2 = 120.2 -1 =119.2 °C
ho1 = 4.2 * 119.2= 500.64 KJ/kg
or
ho2 = hf /P1 – CP Tsub1 = 505- 4.2*1 = 500.8 KJ/kg
hg/p2 = hg @ 1.5 bar = 2693 KJ/kg
hi = CPw Twi
= 4.2* 25 =105 KJ/kg
75.79*( 2707 – 500.64 ) = ��𝑤2 (2693 – 105 )
��𝒘𝟐 = 64.61 kg/hr
----------------------------------------------------------
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 44
Example (2.7 ):
Calculate the make-up water for a double stage effects evaporator per hour .
If heating steam @ 2.5 bar ; x=0.9 and flow rate was 400 kg/hr .
1st effect 2nd effect
Pressure 1.5 bar 1 bar
Temperature inlet 20°C 20°C
Condensate subcooled 3°C zero
Solution
Givens:
Pst = 2.5 bar , Xst = 0.9
��𝑠𝑡= 400 kg/hr
P1= 1.5 bar ,P2 = 1 bar , Twi,1 =Twi,2 = 20 °C
Tsub1 = 3°C Tsub 2 = zero
Find ��𝒘𝟏 = ??
��𝒘𝟐= ??
First stage : assume no losses
��𝒔𝒕 (hst – ho1) = ��𝑤1 (hg/p1 – hi,1)
From tables@ Pst = 2.5 bar , x=0.9
hf = 535 KJ/kg , hfg = 2182 KJ/kg , Tsat = 127.4 °C
hst = hf + x hfg
= 535 + 0.9 *2182 = 2498.8 KJ/kg
ho1 = CP To1
To1 = Tsat – Tsub1 = 127.4 -3 =124.4 °C
ho1 = 4.2 * 124.4= 522.48 KJ/kg
or
ho1 = hf – CP Tsub1 = 535- 4.2*3 = 522.4 KJ/kg
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 45
hg/p1 = hg @ 1.5 bar = 2693 KJ/kg
hi,1 = CPw Twi,1
= 4.2* 20 =84 KJ/kg
400*( 2498.8 – 522.48 ) = ��𝑤1 (2693 – 84 )
��𝒘 = 303 kg/hr
Second stage : assume no losses
��𝑤1 (hg/p1 – ho2) = ��𝒘𝟐 (hg/p2 – hi,2)
hg/p1 = 2693 KJ/kg
ho2 = hf @ P1 = 467 KJ/kg (Tsub2 = zero then To2=Tsat )
hg/p2 = hg @ 1 bar = 2675 KJ/kg
hi,2 = CPw Twi = 4.2* 20 =84 KJ/kg
303*( 2693 – 467 ) = ��𝑤2 (2675 – 84 )
��𝒘𝟐 = 260.31 kg/hr
----------------------------------------------------------
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 46
3-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;
ηco = WT+QH
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 steam the heat added per unit total
energy output is;
e
ηe
+1−e
ηh
Where;
e= electrical fraction of total energy output = WT
WT+QH
ηe = electrical plant efficiency.
ηh = steam (or heat) generator efficiency.
The combined efficiency for separate generation is given as;
ηc = 1
e
ηe +
1−e
ηh
ηc : combined efficiency for two separate electrical and thermal plant
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 47
cogeneration is beneficial : if ηco > ηc
the cogeneration plant efficiency ηco = WT+QH
Qadd exceeds that of the
combined efficiency for separate generation ηc = 1
e
ηe +
1−e
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 .
From the previous figure :
WT = 30 unit
QH = 45 unit
Qadd =100 unit
ηe = 31 %
ηh= 80%
e = WT
WT+QH =
30
30+45 = 0.4 ; 1-e = 0.6
ηco = WT+QH
Qadd =
30+45
100 = 75%
ηc = 1
0.4
0.31 +
0.6
0.8 = 49%
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 48
Types of cogeneration
There are two categories of cogeneration;
1.) The topping cycle:
Process steam pressure requirements vary between 0.5 bar and 40 bar.
Therefore, primary heat at high temperature and low temperature are used.
It is possible to generate the required power and make available the required
quantity of exhaust steam at the desired low heating temperature. Exhaust
steam from turbine is utilized for process heating in which case is called back
pressure turbine. The process heating was replacing the condenser of the
ordinary Rankine cycle.
Most process applications required steam at low grade temperature.
There are several arrangements for cogeneration in topping cycle as;
a) Steam- electrical power-plant with a back pressure turbine.
b) Steam- electrical power-plant with steam extraction from turbine.
c) Gas turbine power plant with a heat recovery boiler.
d) Combined steam-gas-turbine-cycle power-plant.
Arrangement a: suitable for low electrical demand compared with heat demand.
Arrangement d: suitable for high electrical demand compared with heat demand.
Arrangement c: lies in between.
Arrangement b: suitable over a wide range of ratios.
2.) The bottoming cycle
Primary heat is used at high temperature directly for process heat requirements
(as in cement kiln). The low temperature waste heat is used to generate
electricity at low efficiency.
Only the topping cycle can provide true saving in primary energy.
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 49
From figure:
you can notice that the total fuel used in cogeneration plant =(100 unit fuel ) is
lower than the total fuel used in two separate plant to produce same amount of
heat and electricity (91+56 =147 unit fuel) , although the output is the same.
And so ηco = 75% while ηc = 51%
Here ηe and ηh for separate electrical and thermal plant are given but to find it
in the problems :
ηh : separate thermal plant efficiency is the boiler efficiency
because in thermal plant , assuming no pressure drop (ideal ) .
then the only component in the cycle is the boiler used to generate heat .
𝛈𝐡 = 𝛈𝐛𝐨𝐢𝐥𝐞𝐫
ηe : separate electrical plant efficiency can be obtained by substituting
the process with CFWH .
i.e : the same amount of bled steam extracted to the process heat
in the cycle will be used to increase temperature of the feed water
in closed feed water heater before going to the boiler .
Finding ηh , ηe and ηc will be explained further in the examples .
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 50
Example ( 3.1 ):
In textile factory required 10 ton/hr of steam for process heating @3 bar dry
saturated and 1000 Kw of power for which aback pressure turbine with 70%
internal efficiency is to be used. find steam condition @ inlet to the turbine .
Solution
Given
��𝑠𝑡 = 10 𝑡𝑜𝑛/ℎ𝑟
PH=3 bar (dry saturated)
Power generated = 1000 Kw
ηs,T =70%
��𝑠𝑡= 2.778 kg/s
power =��𝑠𝑡(h1-h2,a)
h2,a = hg @ 3 bar because process heat
inlet is dry saturated
h2,a=2725 KJ/kg
1000=2.778 (h1-2725)
h1= 3085 KJ/kg
ηs,T =ℎ1− ℎ2,𝑎
ℎ1− ℎ2,𝑠
0.7 =3085− 2725
3085− ℎ2,𝑠
h2,s =2570.7 KJ/kg
To get point(1)(steam inlet conditions):
a- draw h1=3085 KJ/kg (horizontal line )
b- draw h2,s=2570.7 KJ/kg ( line)
c- from intersection of h2,s line with pressure line P=3 bar mark point (2,s).
d- draw vertical line from point (2,s) until intersect with h1 line , this will be point (1) then read the values of pressure and temperature from the P,T lines steam condition @ inlet to the turbine P1 = 37.5 bar , T1= 344 °C
----------------------------------------------------
s
2,a
sth
s2,h
d
b
a
c
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 51
Example ( 3.2 ):
In a cogeneration plant (combined power and process heat), the boiler
generates 21 Ton/hr of steam @17 bar and 230 oC. A part of the steam goes to
a process heater which consumes 133 kW, the steam leaving the process heater
@ 17 bar, x=0.96 being throttled (at constant enthalpy) to 3.5 bar. The
remaining steam flows through high pressure turbine which exhausts at a
pressure of 3.5 bar. The exhaust steam mixes with the process steam before
entering the low pressure turbine. The low pressure turbine develops 1333 kW.
At the end of expansion, steam goes to condenser @ pressure is 0.3 bar and
x=0.92. Draw h-s diagram and schematic diagram. Also, determine;
1) The steam quality at the exhaust of high pressure turbine. 2) The power
developed by high pressure turbine.
3) The isentropic efficiency of high pressure turbine. 4) Cogeneration plant
efficiency.
Solution
GIVENS
mst =21 ton/hr = 5.8333 kg/s
Pst=17 bar Tst=230 °C Qh=133 KW X2=0.96 WLPT=1333 KW
Pcond=0.3bar x6=0.92
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 52
h1=2865 KJ/kg (super heated steam tables @ 17 bar and 230 °C )
from tables@17 bar , x=0.96
hf=872 KJ/kg hfg=1923 KJ/kg
h2=hf+X(hfg)
=872 + 0.96 (1923) = 2718 KJ/kg
Qh=m ( h1-h2)
133= m (2865 – 2718 )
m = 0.904 kg/s
h3=h3 (throttling @ constant enthalpy)
from point (1) on the chart draw (s=c) line until P=3.5 bar to get point (4,s)
h4,s=2570 KJ/kg
from tables @ P=0.3 bar and x=0.92
hf=289 KJ/kg hfg=2336KJ/kg
h6= hf+X(hfg)
= 289 + 0.92 *2336 =2438 KJ/kg
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 53
WLPT= mst (h5 -h6)
1333= 5.833 (h5-2438 )
h5=2666.5 KJ/kg
Now mark point (5) @ intersection of( P=3.5 bar line ) with h5=2666.5 KJ/kg
then draw vertical line (s=c) until intersection with P=0.3 bar to get point(6,s)
h6,s = 2280 KJ/kg
h7=hf @ 0.3 bar =289 KJ/kg
h8=h7+0.1 ΔP17,0.3
h8 = 289 + 0.1 (17-0.3) =290.67 KJ/kg
Heat balance
mst h5=�� h3 +(mst - �� )h4
5.833*2666.5 = 0.904*2718 + (5.833-0.904)*h4
h4=2657 KJ/kg
(1) - X4=ℎ4,𝑎−ℎ𝑓@3.5
ℎ𝑓𝑔@3.5 =
2657−5842148
= 0. 965
(The steam quality at the exhaust of high pressure turbine =0.965 )
(2) WHPT= (mst - �� ) (h1-h4)
=(5.833-0.904)(2865 -2657) = 1025 KW
ηisen =ℎ1−ℎ4
ℎ1−ℎ4,𝑠
(3) -ηisen =2865−2657
2865−2570 *100= 70.5%
ηco=𝑄ℎ+𝑊ℎ𝑝𝑇+𝑊𝑙𝑝𝑇
𝑄𝑎𝑑𝑑
Qadd=mst (h1-h8)
=5.833(2865 – 290.67 ) =15016 KW
(4)- ηco=133+1025+1333
15016 *100= 16.58%
��3h
4)h �� -st m(
m5h st
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 54
Example (3.3 ):
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.
Compute:
1) The steam flow rate in boiler in Ton/hr.
2) Input heat to boiler.
3) Heat reject in 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 efficiencies.
Solution
Givens:
WT = 5.6 MW QH=1.2 MW
Pst=50 bar , Tst = 500°C
Pcond. = 0.1 bar
Pproces = 2 bar
From chart
h1 @ 50 bar & 500 C =3433 KJ/kg
h5=2645 KJ/Kg
h2=2210 KJ/Kg
From tables
h6=hf @ 2 bar = 505 KJ/Kg k
h3=hf @ 0.1 bar = 192 KJ/Kg
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 55
QH= m (h5-h6)
1.2*103= m (2645-505)
m=0.56 Kg/sec
WT = mst (h1- h5) + (mst-m) (h5- h2)
5.6*103 = mst (3433-2645) + (mst -0.56) (2645-2210)
mst =4.778 Kg/sec
2) Input heat to boiler
Qadd = (mst-m) (h1 – h3 ) + m (h1 – h6)
Qadd = (4.778+0.56) (3433-192)+0.56 (3433-505)
Qadd = 15310.32 KW = 15.31032 MW
3) Heat reject in condenser
Qrej = (mst-m) (h2 – h3)
Qrej = (4.778+0.56) (2210-192)
Qrej =8511.93 KW
4) Rate of fuel burnt in boiler
ηb = 𝑄𝑎𝑑𝑑
𝑄𝑇 =
𝑄𝑎𝑑𝑑
��𝑓 𝐶.𝑉
0.88= 15310.32
��𝑓 25000
mf = 0.695 kg/sec = 2.505 ton/hr
𝜂𝑐𝑜 =𝑊𝑇+𝑄𝐻
𝑄𝑎𝑑𝑑=
5.6+1.2
15.31032= 44.41 %
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 56
For combined efficiency ηc :
We can use same flow rate and using that bled steam is used for heating feed
water (CFWH) instead of process heat .
In conventional cycle that will produce
same amount of power (5.6 MW)
we will use same enthalpy values to find
new Qadd for separate cycle
h8=h6 – CP TTD
= 505 – 4.2* 3 =492.4 KJ/kg
Qadd= ��𝑠𝑡 (h1-h8)
Qadd= 4.778 (3433 – 492.4)=14050 KW = 14.05 MW
ηe = 𝑊𝑇
𝑄𝑎𝑑𝑑 =
5.6
14.050 = 39.8%
e= 𝑊𝑇
𝑊𝑇+𝑄𝐻 =
5.6
5.6+1.2 = 0.823
1-e = 0.177
𝜂𝑐 =1
𝑒𝜂𝑒
+1 − 𝑒
𝜂ℎ
𝜂𝑐=1
0.823
.398+
0.177
0.88
= 44%
-----------------------------------
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 57
Example ( 3.4 ):
A steam power plant produces 500 MW, with inlet steam to high pressure
turbine at 100 bar, 500 oC and condensation at 0.1 bar. It has one stage of
reheat at 8 bar, which raises the steam temperature back to 500 oC. 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/m2 oC. Tube diameters are 24 mm
and 28 mm. Water velocity inside (CFWH) tube was 1 m/s. Terminal
Temperature Difference was 3 oC. Number of passes =4.
b) A cogeneration plant is considered; the heating load is supplied by extracting
the same amount of steam which flow in CFWH at the reheat pressure, then it
condensed in process heater to saturated liquid at 8 bar and pumped to the
boiler. The remaining steam at 8 bar is reheated and then expanded in the low
pressure turbine. 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 efficiencies.
Givens
Wt=500 MW=500,000 KW
Steam inlet to H.P.T @ P1= 100 bar,T1= 500 oC
Pcond.=0.1 bar
Preheat=8 bar to T=500°C
solution
a)from super heating tables
@ P1= 100 bar,T1= 500 oC
h1=3373 kJ/kg
from chart draw vertical (s=c) line to
intersect with pressure line P=8 bar
h2=2740 kJ/kg
h3=hf @ 8 bar = 721 kJ/kg
elevate with 8 bar line untill reaching
Treheat=500°C to get h4 = 3480 kJ/kg
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 58
from point (4) draw (s=c) line to P=0.1 bar
h5 =2490 KJ/kg
h6= hf @ 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 T.T.D (Terminal temperature difference) = 3°C
= 721-3*4.2 =708.4 kJ/kg
or h9= Cp*T9 T9=T3 – T.T.D T3=Tsat. @ P=8 bar = 170.4 °C
HEAT BALANCE FOR C.F.W.H
h7 (1-y) +y h2 = y h3 + (1-y) h9
202*(1-y)+y*2740 =721*y +(1-y)*708.4
y=0.2
h10=(1-y)*h9+y*h8
=(1-0.2)*708.4+ 0.2*730.2 = 712.76 kJ/kg
WT= 𝑚𝑠𝑡 (h1 - h2) + 𝑚𝑠𝑡 (1-y)(h4-h5)
500*1000= 𝑚𝑠𝑡 (3373-2740)+ 𝑚𝑠𝑡 (1-0.2)(2480-2490)
𝑚𝑠𝑡 =350.877 kg/s (mass flow rate to H.P.T)
Qadd=𝑚𝑠𝑡 ( h1 – h10)+ 𝑚𝑠𝑡 (1-y)(h4 - h2)
Qadd= 350.877*(3373-721.76)+350.877*(1-0.2) (3480-2740) =1141136 KW
ηth =𝑊𝑛𝑒𝑡
𝑄𝑎𝑑𝑑=
500∗1000
1141136∗ 100 = 43.81%
7y) h-(1
9y) h-(1
y3h
8Y h
9y) h-(1 10h
2Y h
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 59
Design a closed feed water heater
Qh = 𝑦 ∗ 𝑚𝑠𝑡 (h2-h3) = U Am ΔTm
=0.2 * 350.877*(2740-721)=141684 KW = 141.684 MW
�� = 𝑚𝑠𝑡 (1 − 𝑦) = ρ u a N
350.877*(1-0.2) =1000*1*𝜋
4*(24*10-3)2 *N
N= 620 tube
T9=T3 – T.T.D T3=Tsat. @ P=8 bar
T9=170.4 – 3 =167.4 °C
T7=h7/Cp = 202/4.2 = 48.085 °C
ΔT1= Tsat -T7 =170.4 - 48.085= 122.3 °C
ΔT2= Tsat -T9= T.T.D = 3 °C
ΔTm=𝛥𝑇1−𝛥𝑇2
ln(𝛥𝑇1𝛥𝑇2
) =
122.3−3
ln (122.3
3) = 32.17 °C
Qh = U Am ΔTm 141684 = 1.5*Am* 32.17
Am= 2936.15 m2
Am= π dm L N M
dm=𝑑𝑜+ 𝑑𝑖
2
2936.15=π*0.028+ 0.024
2*L*620*4
L=14.19 m
1TΔ
2TΔ
C°=170.4satT
9TΔ
7TΔ
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 60
b)
same as (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
but since the bled steam isn’t to be used
to heat the water as (a) , it goes to
process heat then pumped to boiler after process heat is finished at sat. liquid
h9=(1-y)*h7+y*h8
=(1-0.2)*202+ 0.2*730.2 = 307.64 kJ/kg
1) Qrej= 𝑚𝑠𝑡 (1-y)(h5-h6)
= 350.877(1-0.2)(2490 – 192) =645039 KW
ηboiler= 𝑄𝑢𝑠𝑒
𝑄𝑇=
𝑄𝑎𝑑𝑑
𝑄𝑇
0.88 = 𝑄𝑎𝑑𝑑
𝑚𝑓 25∗103
Qadd =𝑚𝑠𝑡 (h1-h9) + 𝑚𝑠𝑡 (1-y)(h4-h2)
= 350.877 (3373-307.64)+350.877 (1-0.2) (3480-2740) =1283257 KW
2) 𝑚𝑓 = 1283257
0.88 ∗ 25∗103 = 58.33 𝐾𝑔/𝑠 =209.988 ton/hr
8Y h
7y) h-(1 9h
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 61
ηco=𝑄ℎ+𝑊𝑇
𝑄𝑎𝑑𝑑
ηco=500∗1000+141684
1283257*100= 50%
ηe = 43.81 %
ηh = 88 %
e= 𝑊𝑇
𝑊𝑇+𝑄𝐻 =
500
500+141.684 = 0.78
1-e = 0.22
𝜂𝑐 =1
𝑒𝜂𝑒
+1 − 𝑒
𝜂ℎ
𝜂𝑐= 1
0.78
.4381+
0.22
0.88
= 49%
ηco > ηc
-------------------------------------------------------
Example ( 3.4 ):
a) Define cogeneration and mention in detail its types.
b) A textile factory required 10 Ton/hr of steam for process heat @ 3 bar,
DSS, and 1 MW of power for which a back pressure turbine is to be used.
Find Steam condition @inlet to the turbine, and cogeneration efficiency.
------------------------------------------------------------
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 62
4-Heat Balance Sheet for Internal Combustion Engine
One of thermodynamic test for internal combustion engines is heat balance
sheet, which taken the required measurements after the engine has reached the
steady state conditions.
Also, other important tests like;
• Indicated Mean Effective Pressure.
• Indicated power and thermal efficiency.
• Engine speed and temperature.
• Brake torque, brake power and mechanical efficiency.
• Fuel consumption, air consumption, and volumetric efficiency.
Indicated power (IP)
Indicated power (IP) is the power actually developed by the engine cylinder;
IP = ((MEP)*105*(L A n)k )/60 Watt
Where;
MEP = Mean Effective Pressure, bar.
L = Stroke length, m.
A = Piston area (cross section), m2.
n = number of working strokes per minute,
n = N/2 for 4 stroke engine
n = N for 2 stroke engine
N: RPM
k = number of cylinders, -
Heat Balance Sheet for Internal Combustion Engine:
Heat balance sheet is done during a certain time (one minute);
• Heat supplied by the fuel; Qt = 𝐦𝐟. (C.V. ), kJ/min.
Where;
mf. = Mass flow rate of supplied fuel, kg/min.
C.V. = Lower calorific value of fuel, kJ/kg.
• Heat absorbed in I. P. produced;
IP = (MEP)*105*(L A n)k, kJ/min.
• Heat losses
∑Qloss = QT – I. P. , kJ/min.
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 63
1- Heat rejected to cooling water:
Q1 = mcw. Cpcw ( Tcw,o –Tcw,i ) , kJ/min.
Where;
mcw. = mass flow rate of cooling water, kg/min.
Cpcw = specific heat at constant pressure for cooling water, kJ/kg.oC.
Tcw,i = Cooling water inlet temperature, oC.
Tcw,o = Cooling water outlet temperature, oC.
2- Heat carried away by exhaust gasses:
Q2 = mg. Cpg ( Tg,o –Tg,i ) , kJ/min.
Where;
mg. = mass flow rate of exhaust gasses, kg/min.
Cpg = specific heat at constant pressure for exhaust gasses, kJ/kg.oC.
Tg,i = exhaust gasses inlet temperature, oC.
Tg,o = exhaust gasses outlet temperature, oC.
3- Unaccounted losses:
There are some of heat due to friction leakage, radiation,…..etc., which can not
be determine experimentally. Then;
Q3 = ∑Qloss – (Q1 + Q2 ) , kJ/min.
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 64
Example (4.1 ):
In an internal combustion engine: Indicated power developed =18 kW.
Cooling water flow rate=12 kg/min. Temperature rise of cooling water=25 oC.
Exhaust gas flow rate was 4 kg/min. Temperature rise of exhaust gas 220 oC.
Fuel consumption= 6 kg/hr, Fuel C.V. =44000 kJ/kg. Take: Cpg =1.1 kJ/kg.oC.
, Cpcw =4.2 kJ/kg.oC.
Draw heat balance sheet for engine per 1 min.
Solution
Givens:
I.P = 18 KW
𝑚𝑐𝑤 = 12 kg/min , ΔTcw =25 °C
𝑚𝑔 =4 kg/min , ΔTg =220 °C
𝑚𝑓 = 6 kg/hr =0.1 kg/min , C.V. = 44000 kJ/kg
Cpg =1.1 kJ/kg.oC , Cpcw =4.2 kJ/kg.oC.
QT = 𝑚𝑓 C.V.
= 0.1 * 44000 = 4400 KJ/min
I.P = 18 KW
= 18*60 =1080 KJ/min
∑Qloss = QT – I. P
= 4400 -1080 = 3320 KJ/min
1) Heat loss to cooling water
Q1 = 𝑚𝑐𝑤 CPcw ΔTcw
= 12 * 4.2 * 25 = 1260 KJ/min
2) Heat loss to exhaust (flue gases)
Q2 = 𝑚𝑔 CPg ΔTg
= 4 * 1.1 * 220 = 968 KJ/min
3) Unaccounted heat loss
Q3 = Qloss - (Q1 +Q2)
= 3320 – (1260 + 968 ) = 1092 KJ/min
Heat balance sheet :
Equation Value(KJ/min) %
QT (total ) QT = 𝑚𝑓 C.V. 4400 100%
I.P (indicated power) 1080 24.54%
∑Qloss (total loss) ∑Qloss = QT – I. P 3320 75.45%
Q1 (cooling water) Q1 = 𝑚𝑐𝑤 CPcw ΔTcw 1260 28.63%
Q2 (exhaust loss ) Q2 = 𝑚𝑔 CPg ΔTg 968 22%
Q3 (unaccounted) Q3 = ∑Qloss – (Q1 +Q2) 1092 24.81%
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 65
Example ( 4.2 ):
The following data are collected during a trial on the 6 cylinder, four stroke
diesel engine (360 mm bore and 500 mm stroke) has the following data:
Engine speed=500 rpm. Fuel consumption= 240 kg/hr,
Fuel C.V. =44000 kJ/kg.
Jacket cooling water = 320 kg/min.
Rise in cooling water temperature =40 oC.
Piston cooling oil =140 kg/min, Cpoil = 2.1 kJ/kg oC.
Temperature rise of oil = 28 oC.
All heat of exhaust gases is absorbed in calorimeter; circulating water in gas
calorimeter is 300 kg/min. with temperature rise 42 oC. Mean effective
pressure = 7.3 bar.
Draw heat balance sheet.
solution Givens:
K= 6 ; 4 stroke ; D=0.36 m ; L = 0.5 m
N= 500 rpm MEP= 7.3 bar = 730 kpa
𝑚𝑓 = 240 kg/hr = 4 kg/min ; C.V. = 44000 kJ/kg
𝑚𝑐𝑤 = 320 kg/min ; ΔTcw =40 °C
𝑚𝑜𝑖𝑙 = 140 kg/min ; ΔToil =28 °C
𝑚𝑐𝑎𝑙. =300 kg/min ; ΔTcal =42 °C
Cpoil =2.1 kJ/kg.oC
QT = 𝑚𝑓 C.V.
= 4 * 44000 = 176000 KJ/min
I.P = (MEP)*102*(L A n)k
=(MEP)*102*(L*𝜋
4(D)2 *
𝑁
2)k
= (7.3)*102*(0.5) *𝜋
4(0.36)2*
500
2)* 6 = 55728.7 KJ/min
∑Qloss = QT – I. P
= 176000 -55728.7 = 120271.3 KJ/min
1) Heat loss to cooling water
Q1 = 𝑚𝑐𝑤 CPcw ΔTcw
= 320 * 4.2 *40 = 53760 KJ/min
2) Heat loss to oil
Q2 = 𝑚𝑜𝑖𝑙 CPoil ΔToil
= 140 * 2.1 * 28 = 8232 KJ/min
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 66
3) Heat loss to water in calorimeter
Q3 = 𝑚𝑐𝑎𝑙 CPw ΔTcal
= 300 * 4.2 * 42 = 52920 KJ/kg
4) Unaccounted heat loss
Q4 = Qloss - (Q1 + Q2 + Q3)
= 120271.3 – (53760 + 8232 + 52920 ) = 5359.3 KJ/min
Heat balance sheet :
Equation Value(KJ/min) %
QT (total ) QT = 𝑚𝑓 C.V. 176000 100%
I.P (indicated power) I.P =(MEP)*102*(L A n)k 55728.7 31.66 %
∑Qloss (total loss) ∑Qloss = QT – I. P 120271.3 68.34 %
Q1 (cooling water) Q1 = 𝑚𝑐𝑤 CPcw ΔTcw 53760 30.54 %
Q2 (exhaust loss ) Q2 = 𝑚𝑔 CPg ΔTg 8232 4.47%
Q3 (loss to oil ) Q3=𝑚𝑜𝑖𝑙 CPoil ΔToil 52920 30.06 %
Q4 (unaccounted) Q4 = ∑Qloss – (Q1 +Q2 +Q3) 5359 3.04 %
Example ( 4.3 ):
The following data is given for a 4-stroke , 4-cylinder diesel engine :
Diameter of cylinder is 35 cm ; piston stroke 40 cm ,speed of engine315 rpm,
indicated mean effective pressure 7 bar , fuel consumption is 80 kg/hr .
calorific value of the fuel 43000 KJ/kg ,
air consumption is 30 kg/min
cooling water flow rate 90 kg/min , rise in cooling water temperature 38°C
piston cooling oil used 45 kg/min , rise cooling oil 23°C
exhaust gas temperature =322°C , ambient air temp. =22°C
CPg= 1.1 KJ/kg. °C , CPoil = 2.2 KJ/kg. °C
Draw :heat balance sheet per minute .
Solution
Givens:
K= 4 ; 4 stroke
D=0.35 m ; L = 0.4 m
N= 315 rpm MEP= 7 bar = 700 kpa
𝑚𝑓 = 80 kg/hr ; C.V. = 43000 kJ/kg
𝑚𝑎𝑖𝑟 = 30 kg/min
𝑚𝑐𝑤 = 90 kg/min , ΔTcw =38 °C
𝑚𝑜𝑖𝑙 = 45 kg/min , ΔToil =23 °C
CPoil =2.2 kJ/kg.oC , CPg =1.1 kJ/kg.oC.
Texh =322 oC , Tamb= 22 oC
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 67
QT = 𝑚𝑓 C.V.
= 80
60* 43000 = 57333 KJ/min
I.P = (MEP)*102*(L A n)k
=(MEP)*102*(L*𝜋
4(D)2 *
𝑁
2)k
= (7)*102 * (0.4) * 𝜋
4(0.35)2 *(
315
2 )* 4 = 16971 KJ/min
∑Qloss = QT – I. P
= 57333 – 16971 = 40362 KJ/min
1) Heat loss to cooling water
Q1 = 𝑚𝑐𝑤 CPcw ΔTcw
= 90 *4.2 *38 = 14364 KJ/min
2) Heat carried away by exhaust gasses Q3 = 𝑚𝑔 CPg (Tg,o – Tg,i)
Tg,i = Tamb = 22 °C (because exhaust gases go to ambient air )
Since exhaust gases are produced by combustion of fuel with air
𝑚𝑔 = 𝑚𝑓 + 𝑚𝑎𝑖𝑟
= 80
60 + 30 =
94
3 = 31.333 kg/min
Q3= 31.333* 1.1 *(322- 22) = 10340 KJ/kg
3) Heat loss to oil
Q2 = 𝑚𝑜𝑖𝑙 CPoil ΔToil
= 45 * 2.2 * 23 = 2277 KJ/min
4) Unaccounted heat loss
Q4 = Qloss - (Q1 + Q2 + Q3)
= 40362– (14364 + 10340 + 2277) = 13381 KJ/min
Heat balance sheet :
Equation Value(KJ/min) %
QT (total ) QT = 𝑚𝑓 C.V. 57333 100%
I.P (indicated power) I.P =(MEP)*102*(L A n)k 16971 29.6 %
∑Qloss (total loss) ∑Qloss = QT – I.P 40362 70.39 %
Q1 (cooling water) Q1 = 𝑚𝑐𝑤 CPcw ΔTcw 14364 25.05 %
Q2 (exhaust loss ) Q2 = 𝑚𝑔 CPg (Tgo – Tamb) 10340 18.03%
Q3 (loss to oil ) Q3=𝑚𝑜𝑖𝑙 CPoil ΔToil 2277 3.97 %
Q4 (unaccounted) Q4 = ∑Qloss – (Q1 +Q2 +Q3) 13381 23.33 %
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 68
Example ( 4.4 ):
The following observation were made during test on an oil engine with indicated
power 31.5 KW , fuel used 10.5 kg/hr , calorific value of the fuel 43000 KJ/kg .
Jacket circulating water 540 kg/hr , rise in cooling water temperature 56°C .
Exhaust gases passed through exhaust gas calorimeter .
For finding heat absorbed by calorimeter. water circulated through Calorimeter
with 454 kg/hr , rise in temperature of calorimeter water 36°C .
Temperature of exhaust gases leaving calorimeter 82°C,ambient temperature17°C
Air to fuel ratio 19:1
solution Givens:
IP = 31.5 KW
𝑚𝑓 = 10.5 kg/hr = 0.175 kg/min ; C.V. = 43000 kJ/kg
𝑚𝑐𝑤 = 540 kg/hr ; ΔTcw =56 °C
𝑚𝑐𝑎𝑙. =454 kg/hr ; ΔTcal =36 °C
Texh =82 oC ; Tamb= 17 oC
Cpg =1 kJ/kg.oC
𝐴 𝐹⁄ =19 : 1
QT = 𝑚𝑓 C.V.
= 10.5
60 * 43000 = 7525 KJ/min
I.P =31.5 *60= 1890 KJ/min
∑Qloss = QT – I. P
= 7525 - 1890 = 5635 KJ/min
1) Heat loss to cooling water
Q1 = 𝑚𝑐𝑤 CPcw ΔTcw
= 540
60 * 4.2 *56 = 2116.8 KJ/min
2) Heat loss to water in calorimeter
Q2 = 𝑚𝑐𝑎𝑙 CPw ΔTcal
= 454
60* 4.2 * 36 = 1144.08 KJ/kg
3) Heat carried away by exhaust gasses Q3 = 𝑚𝑔 CPg (Tg,o – Tg,i)
Tg,i = Tamb = 22 °C (because exhaust gases go to ambient air )
Tg,o = Texh = 82 °C
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 69
𝐴
𝐹=
19
1=
��𝑎𝑖𝑟
��𝑓
19 ∗ 𝑚𝑓 = 𝑚𝑎𝑖𝑟
Since exhaust gases are produced as result of combusting of fuel with air
𝑚𝑔 = 𝑚𝑎𝑖𝑟 + 𝑚𝑓
𝑚𝑔 = 19 ∗ 𝑚𝑓 + 𝑚𝑓
𝑚𝑔 = 19 ∗ 10.5 + 10.5 = 210 𝑘𝑔/ℎ𝑟
Q3= 210
60*1 *(82- 17) = 227.5 KJ/kg
4) Unaccounted heat loss
Q4 = Qloss - (Q1 + Q2 + Q3)
=5635 – ( 2116.8+ 1144.08+ 227.5 ) = 2146.62 KJ/min
Heat balance sheet :
Equation Value(KJ/min) %
QT (total ) QT = 𝑚𝑓 C.V. 7525 100%
I.P (indicated power) 1890 25.11 %
∑Qloss (total loss) ∑Qloss = QT – I. P 5635 74.88 %
Q1 (cooling water) Q1 = 𝑚𝑐𝑤 CPcw ΔTcw 2116.8 28.13 %
Q2 (calorimeter ) Q2=𝑚𝑐𝑎𝑙 CPw ΔTcal 227.5 3.02 %
Q3 (exhaust loss ) Q3 = 𝑚𝑔 CPg ΔTg 1144.08 15.2%
Q4 (unaccounted) Q4 = ∑Qloss – (Q1 +Q2 +Q3) 2146.62 28.52 %
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 70
5-Steam Generator (Boiler)
a- Classification of boilers
1- Fire tube boiler
• Low capacity .
• Low pressure .
• Dry saturated steam (D.S.S) .
• Fire (flue gases)inside tubes .
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 71
2- Water tube boiler
• high capacity
• high pressure
• super heated steam
• water inside tubes
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 72
Heat Balance Sheet for Steam Generator
Evaporation rate = 𝑡𝑜𝑡𝑎𝑙 𝑠𝑡𝑒𝑎𝑚 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑
𝑡𝑜𝑡𝑎𝑙 𝑓𝑢𝑒𝑙 𝑏𝑢𝑟𝑛𝑡 (𝑢𝑠𝑒𝑑)
= 𝑡𝑜𝑡𝑎𝑙 𝑠𝑡𝑒𝑎𝑚 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑
𝐺𝑟𝑎𝑡𝑒 𝑎𝑟𝑒𝑎
= 𝑡𝑜𝑡𝑎𝑙 𝑠𝑡𝑒𝑎𝑚 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑
𝑓𝑢𝑟𝑛𝑎𝑛𝑐𝑒 𝑣𝑜𝑙𝑢𝑚𝑒
Evaporation capacity for boiler :-
• Feed water temperature
• Working pressure
• Fuel
• Final condition of steam
Equivalent evaporation (me):
𝑚𝑒 =𝑚𝑠(ℎ𝑠𝑡 − ℎ𝑓𝑤)
2257
2257 : is laten heat of vaporization @ 1 bar
Equivalent evaporation is defined as : the quantity of D.S.S that could be
generated by the boiler per unit time from the water @100°C
to steam @ 100°C .
Boiler efficiency :-
Ƞ𝑏 =��𝑠𝑡(ℎ𝑠𝑡 − ℎ𝑓𝑤)
��𝑓 (𝐶. 𝑉)=
𝑄𝑢𝑠𝑒𝑓𝑢𝑙
𝑄𝑇
Ƞ𝑏 =𝑚𝑠(ℎ𝑠𝑡 − ℎ𝑓𝑤)
(𝐶. 𝑉)
Where : 𝑚𝑠 =��𝑠𝑡
��𝑓 [kgst/kgf]
Total losses : ∑Qloss = QT - Quse
Quseful : useful heat , KW
QT : total input heat , KW
C.V : fuel calorific value , KJ/kg
��𝑠𝑡 : steam flow rate , kg/s
��𝑓 : fuel flow rate , kg/s
ℎ𝑠𝑡 : specific enthalpy for steam , KJ/kg
ℎ𝑓𝑤 : specific enthalpy for water , KJ/kg
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 73
Boiler trial to determine :
• Generating capacity
• Thermal efficiency
• Heat balance sheet for boiler
Heat balance sheet for boiler per 1kgfuel :
Total input heat : (per 1 kgfuel)
QT = ��𝑓 ( 𝐶.𝑉)
��𝑓 = 1* C.V = [KJ /kgf ] (same unit of C.V )
Useful heat (per 1 kgfuel) :
Quse = ��𝑠𝑡
��𝑓 (hst – hfw ) = [KJ /kgf ]
= ms ( hst – hfw ) = [KJ /kgf ]
= 𝑘𝑔𝑠𝑡
𝑘𝑔𝑓∗
𝐾𝐽
𝑘𝑔𝑠𝑡 =
𝐾𝐽
𝑘𝑔𝑓
Total losses
∑Qloss = QT - Quse = [KJ /kgf ]
Losses in boilers :-
1) Dry gas loss:
Q1 = mg CPg (Tg,o = Tg,i) = [KJ /kgf ]
mg = (𝐴
𝐹)𝑎𝑐𝑡 + mc
(𝐴
𝐹)𝑎𝑐𝑡 =
𝐶% 𝑁%
33 (𝐶𝑂2%+𝐶𝑂%)
Where :
mg : mass of flue gas per 1 kgfuel , [kgflue gas / kgf ]
mc :mass of carbon per 1 kgfuel , [kgcarbon / kgf ]
Tg,i : inlet gas temperature ,°C
Tg,o : outlet gas temperature ,°C
CPg : specific heat for flue gas , [KJ/kg. °C]
(𝐴
𝐹)𝑎𝑐𝑡 :actual air to fuel ratio
C% : percentage of carbon mass in fuel .
N% : percentage of nitrogen in flue gas .
CO% : percentage of CO in flue gas .
CO2% : percentage of CO2 in flue gas .
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 74
2) Un burned fuel :
Q2 = mun burned C.V = [KJ /kgf ]
mun burned = mass of unburned fuel
mun burned = ��𝑢𝑛
𝑚𝑓 = [𝑘𝑔𝑢𝑛 𝑏𝑢𝑟𝑛𝑒𝑑
𝑓𝑢𝑒𝑙 /𝑘𝑔𝑓𝑢𝑒𝑙]
3) Moisture loss in fuel :
Q3 = (mm + 9 H2 ) (hs – hw )
Where :
mm : mass of moisture in fuel per 1 kgfuel , [kg / kgf ]
H2 :mass of hydrogen in fuel per 1 kgfuel , [kg / kgf ]
ℎ𝑠 : specific enthalpy of liberated super heated steam in flue gas @Tg,o
and partial pressure of steam in flue gas , [KJ/kg ]
ℎ𝑤: specific enthalpy for water @ Tg,i (boiler house temperature), KJ/kg
ℎ𝑤= CPw Tg,i
CPg : specific heat for flue gas , [KJ/kg. °C]
4) Incomplete combustion :
Q4 = 𝐶𝑂%
𝐶𝑂2% +𝐶𝑂% mc * 24000 = [kg / kgf ]
Where :
mc :mass of carbon per 1 kgfuel , [kgcarbon / kgf ]
CO% : percentage of CO in flue gas .
CO2 % : percentage of CO2 in flue gas .
5) Moisture loss in combustion air :
Q5 = 1.926 (𝐴
𝐹)𝑎𝑐𝑡 H (Tg,o – Tg,i )
H :specific humidity of combustion air , [𝑘𝑔𝑤𝑎𝑡𝑒𝑟 𝑣𝑎𝑝𝑜𝑟
/ 𝑘𝑔𝑑𝑟𝑦 𝑎𝑖𝑟]
6) Thermal radiation loss and other unaccounted loss :
Q6 = ∑Qloss – (Q1 +Q2 +Q3 +Q4 + Q5 ) = [kg / kgf ]
NOTE:
1- ms , mg and mun burned are not mass flow rates .
2- Un burned fuel loss may not exist if mun burned equal zero .
3- Incomplete combustion also doesn’t exist if CO% in flue gas was zero .
4- If mm (mass of moisture in fuel per 1 kgfuel ) equal zero use all previous
equation directly to obtain the boiler losses , but if it have a value
mentioned in the problem . Every term of the (analysis of fuel by mass)
and calorific value C.V exist in one of the equations ( percentage or
mass) should be multiplied by (1-mm ) as will be explained.
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 75
Analysis of fuel and flue gas :
Mass analysis for fuel : C% , H2% , and Ash
Flue gas analysis by volume : CO2% , CO% , N2% , and O2%
In the previous boiler equations , if there was [CO2% , CO% , N2% or C% ]
with the percentage sign (%) then the number in (Mass analysis for fuel) or
(Flue gas analysis by volume) is used as given .
But if the formulas contains [ H2 , mc ] without percentage sign (%) then the
given numbers in (Mass analysis for fuel) must be divided by 100 before using
it in the equations .
H2 = 𝐻2%
100 , mc =
𝐶%
100
Mass of moisture in fuel (mm)
1- If is exists in the problem as mentioned before then . Every term of the
(analysis of fuel by mass) and calorific value C.V exist in one of the
equations ( percentage or mass) should be multiplied by (1-mm ).
(C%)act = C% (1-mm)
(mc )act = mc (1-mm) = 𝐶%
100 (1-mm)
(H2 )act = H2 (1-mm) = 𝐻2%
100 (1-mm)
(C.V)act = C.V (1-mm )
Mass analysis of fuel No mass of moisture With mass of moisture (mm)
Carbon percentage C% C% (1-mm)
Carbon mass in fuel (C% / 100 ) = mc (C% / 100 ) (1-mm) = mc (1-mm)
Hydrogen mass in fuel (H2% / 100) = H2 (H2% / 100) (1-mm) = H2(1-mm)
Calorific value C.V C.V (1-mm)
Formulas will change due to (mm ) :
QT = (1-mm)* C.V
mg = (𝐴
𝐹)𝑎𝑐𝑡 + mc (1-mm)
(𝐴
𝐹)𝑎𝑐𝑡 =
𝐶% (1−𝑚𝑚) 𝑁%
33 (𝐶𝑂2%+𝐶𝑂%)
Q2 = mun burned C.V
Q3 = (mm + 9 H2 (1-mm) ) (hs – hw )
Q4 = 𝐶𝑂%
𝐶𝑂2% +𝐶𝑂% mc (1-mm)* 24000
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 76
Explanations Boiler losses :
1- dry gas loss :
after the flow gases outlet from the preheater (the last stage that flue
gases are used to heat combustion air ) but still hot with temperature
Tg,o > 100°C and this temperature is considered loss because it wasn’t
useful to through this thermal energy in the flue gases to the ambient
(boiler house temperature ).
2- Un burned fuel :
From the name we can predict that some of the fuel didn’t burn so this is
loss because we didn’t benefit from all the fuel in the boiler .
3- Moisture loss in fuel :
First of all the fuel contains some moisture as (mm) or as (H2) which
forms H2O after burning the fuel with air .
This amount of water means that there is mass in the fuel isn’t applicable
to burn generating heat because water (moisture) doesn’t have calorific
value thus the moisture in fuel isn’t benifical .
when we burn this fuel containing moisture , the moisture evaporates
and becomes super heated steam due to combustion of the fuel, so this
moisture actually absorbed some of the heat from burning the fuel and
this carries away heat in the form of its latent heat, and instead of using
this heat to evaporate the water in the boiler drum , it were used to
evaporate the moisture in the fuel it self
4- Incomplete combustion :
in complete combustion all the carbon become CO2 and having CO in
the combustion product means that the combustion wasn’t complete
because amount of air wasn’t enough .
5- Moisture loss in combustion air :
Loss due to moisture in air (H2O), the air used for combusting the fuel is
taken from the ambient so the amount of moisture in air .
Vapour in the form of humidity in the incoming air, is superheated as it
passes through the boiler. Since this heat passes up the stack
6- Un accounted loss :
Loss due to surface radiation, convection to the surrounding and other
unaccounted sources.
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 77
Boiler Plant :
1) Heat balance for super heater
mfg CPg (Tfg,i – Tfg,o)s.heater = ms (hsuper -hDSS )
mfg : total mass of flue gas per 1 kgf
mfg = (𝐴
𝐹)𝑎𝑐𝑡+ 1 =[ kgfg/kgf ]
Tfg,i : super heater flue gas inlet temp.
Tfg,o : super heater flue gas outlet temp.
hsuper :outlet super heater specific enthalpy of steam.
hDSS :intlet super heater specific enthalpy of steam.
2) Heat balance for economizer
mfg CPg (Tfg,i – Tfg,o)eco = ms CPw (Tfw,o -Tfw,i)
Tfg,i : economizer flue gas inlet temp.
Tfg,o : economizer flue gas outlet temp.
Tfw,o : economizer water outlet temp.
Tfg,i : economizer water inlet temp.
3) Heat balance for air preheater
mfg CPg (Tfg,i – Tfg,o)air preheater = mair CPair (Ta,o – Ta,i )
mair : mass of air per 1 kgf = air to fuel ratio (𝐴
𝐹)𝑎𝑐𝑡
T1 :air preheater flue gas inlet temp.
T2 : air preheater flue gas outlet temp.
Tair,i :air preheater inlet temp. of air .
Tair,o : air preheater outlet temp. of air .
CPair : specific heat for air
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 78
Notes :
1- There is two (mass of flue gas per 1 kgf ) were mentioned in previous
equation either in boiler losses formulas or in the heat balance for (super
heater , economizer and air preheater )
a- mg : mass of flue gases per 1kgf in boiler losses equations .
mg = = (𝐴
𝐹)𝑎𝑐𝑡+ mc = kgflue /kgf
b- mfg : mass of flue gases per 1kgf in heat balance equations .
mfg = (𝐴
𝐹)𝑎𝑐𝑡+ 1 = kgflue /kgf
2- If there is (mm) :
(𝐴
𝐹)𝑎𝑐𝑡 =
𝐶% (1−𝑚𝑚) 𝑁%
33 (𝐶𝑂2%+𝐶𝑂%)
mg = (𝐴
𝐹)𝑎𝑐𝑡+ mc (1-mm) = kgflue /kgf
mfg = (𝐴
𝐹)𝑎𝑐𝑡+ 1 = kgflue /kgf
So what is the difference between both mg and mfg ?
mg : is used in calculating the dry gas loss because rest of components
like moisture present in fuel the. The losses due to these components have not
been included in the dry flue gas loss since they are separately calculated as a
wet flue gas loss.
mfg : because all the component of fuel (carbon , hydrogen and moisture )
combust with air producing flue gas which is used with it’s components
in super heater , reheater , economizer and air preheater .
At last mg doesn’t include moisture mass (mm) and (H2 ) which forms H2O in
combustion product unlike mfg which include all the flue gas components .
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 79
Example (5)
The following data collected during a boiler trial per hour .
• Steam generated = 640 kg/hr
• Fuel used =55 kg/hr
• Temperature of feed water =50 °C
• Outlet steam pressure = 10 bar (i.e D.S.S )
• Boiler room temperature = 30°C
• Fuel calorific value = 40 MJ/kgf
• Flue gas outlet temperature = 150 °C
Composition of fuel oil by mass C% =85% H2% = 13% Ash =2%
Flue gas analysis by volume CO2 % = 12.5% ; CO% = 0.5 % ; N2% = 82% ; O2% = 5%
Partial pressure of water vapor carried by flue gas =0.1 bar
Calculate : boiler efficiency and draw heat balance sheet .
Cpw = 4.2 KJ/kg oC , CPair = 1 KJ/kg oC and CPg = 1.1 KJ/kg oC
Solution
Givens:
��𝑠𝑡 = 640 kg/hr ��𝑓 = 55 kg/hr
Tfwi =50 oC Psteam = 10 bar (DSS)
Tg,i =30 o
C Tg,o =150 oC
C.V= 40 MJ/kgf Pp= 0.1 bar
QT = 1* C.V = 40000 KJ/Kgf
Quse =ms (hst – hfw ) = ��𝑠𝑡
��𝑓 (hst – hfw )
From tables @Pst = 10 bar hst = hg = 2778 KJ/Kg hfw = Cp* Tfw
=4.2 * 50=210 KJ/Kg
Quse = 640
55 (2778 – 210 ) = 29882.18 KJ/Kgf
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 80
ηboiler = 𝑄𝑢𝑠𝑒
𝑄𝑇
= 29882.18
40000 * 100 = 74.7 %
∑Qloss = QT - Quse
= 40000 - 29882.18=10117.82 KJ/Kgf
Losses in boiler
1- Dry gas loss:
Q1 = mg CPg (Tg,o = Tg,i)
mg = (𝐴
𝐹)𝑎𝑐𝑡 + mc
(𝐴
𝐹)𝑎𝑐𝑡 =
𝐶% 𝑁2%
33 (𝐶𝑂2% + 𝐶𝑂%)
(𝐴
𝐹)𝑎𝑐𝑡 =
85% 82%
33 (0.5%+12.5%) = 16.247
mg=16.247 + .85= 17.097 Kg/Kgf
Q1= 17.097*1.1*(180-30) =2256.81 KJ/Kgf
2- Moisture loss in fuel: -
Q2 = (mm + 9 H2 ) (hs – hw )
From superheated tables @ p= 0.1 bar & Tgo=150 oC
hs= 2783 KJ/Kg
hw = CPw Tg,i
= 4.2 *30 = 126 KJ/kg
Q2 = (0 + 9*0.13) (2783 – 126) =3108.69 KJ/Kgf
3- Incomplete combustion
Q3 = 𝐶𝑂%
𝐶𝑂2% +𝐶𝑂% mc * 24000
= 0.5%
12.5 % +0.5% *0.85 * 24000 = 784.61 KJ/Kgf
4- Thermal radiation loss and other unaccounted loss
Q4 = ∑Qloss – (Q1 +Q2 +Q3 )
=10117.82 - (2256.81+ 3108.69 + 784.61) = 3967.71 KJ/Kgf
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 81
Heat balance sheet
Equation Value (Kj/kgf) %
QT 1* C.V 40000 100%
Quse ms (hst – hfw) 29882.18 74.7 %
∑Qloss QT - Quse 10117.82 25.29 %
Q1 mg CPg (Tg,o = Tg,i) 2256.81 5.64 %
Q2 (mm + 9 H2 ) (hs – hw ) 3108.69 7.77 %
Q3 𝐶𝑂%
𝐶𝑂2% +𝐶𝑂% mc * 24000 784.61 1.96 %
Q4 ∑Qloss – (Q1 +Q2 +Q3 ) 3967.71 9.919 %
------------------------------------------------
Example (5)
A boiler generates 20,000kg/hr at 40 bar and 400°C . feed water enters
economizer at 100 °C . mass flow rate of fuel =2000 kg/hr ,
fuel C.V= 40000 KJ/kg ,moisture in fuel = 4%, boiler room temp.=35°C
economizer flue gas inlet and outlet temp. are 400°C, 250°C respectively.
flue gas inlet and outlet temp. from air preheater are 250°C , 150°C .
specific humidity = 0.008 kgw/kgair and partial pressure of water vapor in
flue gas is 0.1 bar .
mass analysis for dry fuel :
C%=83% , H2% = 14% ,and Ash=3%
mass analysis by volume:
CO2%=12% , CO%=1% , N2%=80% , and O2%=7%
Take Cpw = 4.2 KJ/kg oC , CPair = 1.005 KJ/kg oC and CPg = 1.15 KJ/kg oC
Find:
1- boiler efficiency .
2- heat balance sheet .
3- outlet water temp. from economizer .
4- outlet air temp. for air preheater .
Solution
Given :
��𝑠𝑡 = 20000 kg/hr , Pst = 2000 bar Tst = 400°C
Tfwi =100 oC ��𝑓 = 2000 kg/hr
C.V= 40000 kJ/kgf , mm = 4% , TBR =35 oC
Economizer : Tfg,i = 400°C , Tfg,o = 250°C Air preheater : Tfg,i = 250°C , Tfg,o =150°C
H=0.008 kgw/kgair , Pp= 0.1 bar
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 82
Because there is mm = 4%
QT= (1-mm)*C.V
= (1-0.04)*40000= 38400 KJ/Kgf
Quse = ��𝑠𝑡
��𝑓 (hst – hfw )
From superheated tables @ p=40 bar & T= 400 oC
hst = 3214 KJ/Kg
hfw =CPw Tfwi
= 4.2*100= 420 KJ/Kg
Quse = 20000
2000 (3214 – 420) = 27940 KJ/Kg
∑Qloss = QT - Quse
= 38400 - 27940 =10460 KJ/Kg
ηboiler = 𝑄𝑢𝑠𝑒
𝑄𝑇
= 27940
38400 * 100 % =72.76%
Losses in boiler
1- Dry gas loss:
Q1 = mg CPg (Tg,o - Tg,i)
mg = (𝐴
𝐹)𝑎𝑐𝑡 + mc (1-mm)
(𝐴
𝐹)𝑎𝑐𝑡 =
𝐶% (1 − 𝑚𝑚) 𝑁2%
33 (𝐶𝑂2% + 𝐶𝑂%)
(𝐴
𝐹)𝑎𝑐𝑡 =
83% ∗(1−0.04)∗ 80%
33 (12%+1%) = 14.858
mg=14.858 + 0.83* (1-0.04)=15.654 Q1=15.654*1.15*(150-35)=2070.24 KJ/Kgf
2- Moisture loss in fuel: -
Q2 = (mm + 9 H2 (1-mm)) (hs – hw )
From superheated tables @ Pp= 0.1 bar & Tg,o=150 o
C
hs= 2783 KJ/Kg
hw =CPw Tg,i ; Tg,i = TBR = 35°C
hw = 4.2 * 35 = 147 KJ/kg
Q2 = (0.04+ 9*0.14 (1- 0.04) ) (2783- 147)=3293.95 KJ/Kgf
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 83
3- Moisture loss in combustion air :
Q3 = 1.926 (𝐴
𝐹)𝑎𝑐𝑡 H (Tg,o – Tg,i )
Tg,o = Tfg,o (from air preheater) Q3 =1.926 * 14.858 * 0.008 * (150-35) =26.32 KJ/Kgf
4- Incomplete combustion
Q4 = 𝐶𝑂%
𝐶𝑂2% +𝐶𝑂% * mc*(1 - mm) * 24000
= 1%
12% +1% 0.83*(1-0.04) * 24000 = 1471 KJ/Kgf
5- Thermal radiation loss and other unaccounted loss
Q5 = ∑Qloss – (Q1 +Q2 +Q3+Q4 )
=10460 - (2070.24 + 3293.95 + 26.32+1471) = 3598.74 KJ/Kgf
HEAT BALANCE SHEET
Equation Value Percentage
QT (1-mm)*C.V 38400 100%
Quse ��𝑠𝑡
��𝑓 (hst – hfw ) 27940 72.76%
∑Qloss QT - Quse 10460 27.24%
Q1 mg CPg (Tg,o - Tg,i) 2070.24 5.4%
Q2 (mm + 9 H2 (1-mm)) (hs – hw ) 3293.95 8.6%
Q3 1.926 (𝐴
𝐹)𝑎𝑐𝑡 H (Tg,o – Tg,i ) 26.32 0.068%
Q4 𝐶𝑂%
𝐶𝑂2% +𝐶𝑂% * mc*(1 - mm) * 24000 1471 3.83%
Q5 Q5 = ∑Qloss – (Q1 +Q2 +Q3+Q4 ) 3598.74 9.37%
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 84
Heat balance for economizer
mfg = (𝐴
𝐹)𝑎𝑐𝑡 + 1
=14.858 + 1 = 15.858
mfg CPg (Tg,o - Tg,i) = ms CPw (Tec,o –Tec,i )
15.858*1.15*(400-250) = 20000
2000*4.2*(Teco -100)
Tec,o=165.13 oC
Heat balance for economizer
mfg CPg (Tg,o - Tg,i) = mair CPair (Tair,o-Tair,i)
15.858*1.15*(250-150)= 14.858*1.005*( Tair,o-35)
Tair,o=157.12 oC
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 85
Example ( 5.3 ):
For a boiler plant consists of boiler, economizer and super-heater:
Coal used = 675 kg/hr. Fuel C.V =29800 kJ/kg. Steam Pressure = 14 bar.
Water evaporated= 5000 kg/hr.
Feed water temperature entering and leaving economizer are 35 oC and 135 oC
respectively.
Dryness fraction of steam leaving boiler = 0.98.
Temperature of steam leaving super-heater = 320 oC.
Draw boiler plant and water tube boiler.
Calculate: overall efficiency of the plant, and percentage of the available heat
utilized in the economizer, boiler, and super-heater.
Solution
Given:
��𝑓= 675 kg/hr ; C.V = 29800 kJ/kg
Pst = 14 bar ; X = 0.98
��𝑠𝑡= 5000 kg/hr
Economizer : Tfw,i = 35 oC ; Tfw,o = 135 oC
Super heater : Tsuper,o = 320 oC
Ƞ𝑏 =��𝑠𝑡(ℎ𝑠𝑡 − ℎ𝑓𝑤)
��𝑓 (𝐶. 𝑉)
From tables @Pst = 14 bar and x=0.98
hf = 830 KJ/kg hfg = 1960 KJ/kg
hst = hf + x hfg
= 830 + 0.98*1960 = 2750.8 KJ/kg
hfw = CPw Tfw,o
= 4.2 *135 =567 KJ/kg
Ƞ𝑏 =5000(2750.8−567)
675 (29800)∗ 100% = 54.28%
Qeconomizer = ms CPw (Tfw,o -Tfw,i )
= 5000
675 *4.2 * (135 -35 ) = 3111.11 KJ/kgf
Qboiler = ms ( hst – hfw )
= 5000
675 * ( 2750.8 – 567 ) = 16176.29 KJ/kgf
Qsuper.heater = ms (hsuper – hst )
From super heated tables @Pst = 14 bar and T=320°C
hsuper = 3085 KJ/kg
Qsuper.heater = 5000
675 * ( 3085 - 2750.8 ) = 2475.55 KJ/kgf
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 86
Example ( 5.4 ):
The following observations are taken during a boiler trial:
Coal used = 250 kg/hr. Fuel C V =29800 kJ/kg.
Water evaporated= 2000 kg/hr.
Steam Pressure = 12 bar. Dryness fraction = 0.97.
Feed water temperature = 35 oC.
Calculate: equivalent evaporation and boiler efficiency.
Solution
Given :
��𝑓= 250 kg/hr C.V = 29800 kJ/kg
��𝑠𝑡= 2000 kg/hr
Pst = 12 bar X = 0.97
Tfw = 35 °C
𝑚𝑒 =𝑚𝑠(ℎ𝑠𝑡 − ℎ𝑓𝑤)
2257
From tables @Pst = 12 bar and x=0.97
hf = 798 KJ/kg hfg = 1986 KJ/kg
hst = hf + x hfg
= 798 + 0.97*1986 = 2724.42 KJ/kg
hfw = CPw Tfw
= 4.2 *35 =147 KJ/kg
ms = ��𝑠𝑡
��𝑓 =
2000
250 = 8
𝑚𝑒 =8 ∗ (2724.42 − 147)
2257= 9.135 𝑘𝑔𝑠𝑡/𝑘𝑔𝑓
Ƞ𝑏 =��𝑠𝑡(ℎ𝑠𝑡 − ℎ𝑓𝑤)
��𝑓 (𝐶. 𝑉)
Ƞ𝑏 =2000(2724.42−147)
250 (29800)∗ 100% = 69.19 %
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 87
Example ( 5.5 ):
The following data are collected during a boiler trial:
Steam generated 100 Ton/hr, @ 60 bar & 360 oC.
Feed water temperature = 110 oC.
Fuel used = 8.5 Ton/hr , Fuel C.V. =42000 kJ/kg.
Boiler house temperature = 37 oC.
Partial pressure for vapor carried with flue gas=0.09 bar.
Flue gas outlet temperature from boiler= 115 oC.
Specific humidity for air in flue gas , H= 0.075 kg/kgdry air .
Mass analysis for fuel was: C=84%, H2=13%, and Ash=3%.
Flue gas analysis by volume was:CO2= 13% ,CO=1% ,N2=82 % , and O2= 4%.
Take Cpg=1.15 kJ/kg oC , Cpw=4.2 kJ/kg oC.
Draw heat balance sheet per 1 kgf . Also, draw boiler plant and water tube
boiler.
Example ( 5.6 ):
The following data are collected during a boiler trial:
Steam generated 650 Ton/hr, @ 10 bar & D.S.S.
Feed water temperature = 50 oC.
Fuel used = 55 Ton/hr, Fuel C.V. =40000 kJ/kg.
Boiler house temperature = 35 oC.
Partial pressure for vapor carried with flue gas=0.1 bar.
Flue gas outlet temperature from boiler= 100 oC.
Specific humidity for air in flue gas , H= 0.077 kg/kgdry air .
Mass analysis for fuel was: C=85%, H2=13%, and Ash=2%. Flue gas analysis by volume was:CO2= 13% ,CO=1% , N2=82 % ,and O2= 4%.
Take Cpg=1.15 kJ/kg oC , Cpw=4.2 kJ/kg oC.
Find boiler efficiency and Draw heat balance sheet per 1 kgf.
Example ( 5.7 ):
The following data are collected during a boiler trial:
Steam generated 100 Ton/hr, @ 60 bar & 360 oC.
Feed water temperature = 110 oC.
Fuel used = 8.5 Ton/hr , Fuel C.V. =42000 kJ/kg.
Boiler house temperature = 37 oC.
Partial pressure for vapor carried with flue gas=0.09 bar.
Flue gas outlet temperature from boiler= 115 oC.
Specific humidity for air in flue gas , H= 0.075 kg/kgdry air .
Mass analysis for fuel was: C=84%, H2=13%, and Ash=3%. Flue gas analysis by volume was: CO2= 13%, CO=1%, N2=82 % , and O2=
4%.
Take Cpg=1.15 kJ/kg oC , Cpw=4.2 kJ/kg oC.
Find boiler efficiency and Draw heat balance sheet per 1 kgf .
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 88
6-Station performance
Plant performance is defined the input output curve which derived from tests
as;
I= fn(L) I = a + a1 L + a2 L2 + a3 L3 + …..
The slope for I/O curve @ the given load is defined as Incremental Rate.
Physically the IR is the amount of additional energy required to produce an
added unit of output at any given load.
𝐈𝐑 = 𝐝𝐈
𝐝𝐋
The efficiency curve is defined as;
𝛈 = 𝐨𝐮𝐭𝐩𝐮𝐭
𝐈𝐧𝐩𝐮𝐭
𝛈 = 𝐋 × 𝐂𝐨𝐧𝐬𝐭.
𝐈 × 𝟏𝟎𝟎
Heat rate is the reciprocal of efficiency;
𝐇𝐑 = 𝐈
𝐋
𝐇𝐑 =𝐚
𝐋 + 𝐛 + 𝐜𝐋 + 𝐝𝐋𝟐 + ⋯ ..
Maximum efficiency can be found @ Minimum HR:
𝐝𝐇𝐑
𝐝𝐋=
𝐝 (𝐈𝐋)
𝐝𝐋= 𝟎
=𝐋𝐝𝐈 − 𝐈𝐝𝐋
𝐋𝟐 = 𝟎.
L dI=I dL
𝐝𝐈
𝐝𝐋=
𝐈
𝐋
IR= HRmin
For maximum efficiency the heat rate is minimum
Also @Ƞmax heat rate = incremental rate (min)
-------------------------------------------------------------------------------------
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 89
Example (6.1 ):
Find maximum efficiency and minimum heat rate from the following I/O
curve:
I=2.4*106 [100+2L +0.0004L3] , where L in MW and I in kJ/hr.
Solution
HR = I
L
HR=2.4*106 [100
L +2 + 0.0004L2]
At min HR or max. Efficiency:
d(HR)
dL =
d(2.4∗106
[100L
+2 +0.0004L2
])
dL=0
2.4*106 [−100
L2 + 0.0008L]=0
[−100
L2 + 0.0008L]=0
100
L2 = 0.0008L
L3= 100
0.0008 = 125000
Therefore; L= √1250003
= 50 MW
I@L=50 = 2.4*106 [100+2*50 +0.0004*503] = 1.68*109 600*106 kJ/kg
η = L × Const.
I=
50 × 3600 ∗ 1000
600 ∗ 106= 30%
HR=2.4*106 [100
50 +2 + 0.0004*502] =12*106 kJ/MW.hr = 12 MJ/kW.hr
HRmin = 12 MJ/kW. hr
η max = 30 %
------------------------------
Example (6.2):
Find maximum efficiency and minimum heat rate from the following I/O
curve:
I=106 [16+5L +0.02L3] , where L in MW and I in kJ/hr.
Also, Draw I//O curve , η – L curve, HR – L curve, and IR – L curve.
Solution
I=106 [16+5L +0.02L3]
IR = dI
dL
= 106 [5 +0.06L2]
HR= I
L
=106 [16
L+5 +0.02L2]
η = L × Const.
I
NOTE :
the Const. in efficiency equ. is
to convert MW to KJ/hr
or convert KJ/hr to MW
MW =MJ
s=
kJ∗1000hr
3600
MW= KJ
hr∗ 3600 ∗ 1000
MW= kJ
hr *3.6*106
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 90
At min HR or max. Efficiency:
d(HR)
dL =
d(106 [16
L+5 +0.02L2])
dL = 0
106 [−16
L2 + 0.04L]=0
[−16
L2 + 0.04L]=0
16
L2 = 0.04L
L3= 100
0.04 = 400
Therefore; L= √4003
= 7.368 MW
I@L=7.368 = 106 [16+5*7.368 +0.02*7.3683] = 60.839*106 kJ/kg
η = L × Const.
I=
7.368 × 3600 ∗ 1000
60.839 ∗ 106= 43.59%
HR=106 [16
7.368 +5+ 0.02*7.3682] =8257301 kJ/MW.hr = 8.2573 MJ/kW.hr
HRmin = 8.2573 MJ/kW. hr
η max = 43.59 %
L , MW 0 2 4 6 8 10
I , MJ/hr 16000 26160 37280 50320 66240 86000
IR,MJ/KW.hr 5 5.24 5.96 7.16 8.84 11
HR , MJ/KW.hr Ꚙ 13.08 9.32 8.386667 8.28 8.6
η, % 0 27.52 38.62 42.92 43.47 41.86
0
10000
20000
30000
40000
50000
60000
70000
80000
90000
100000
0 2 4 6 8 10 12
I/O curve
0
5
10
15
20
25
30
35
40
45
50
0 2 4 6 8 10 12
η- L curve
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 91
-----------------------------------------------------------
Example ( 6.3 ):
A 20 MW station has the following I/O curve;
I= 106[30 + 0.5L+0.65 L2+ 0.01L3] , where L in MW and I in kJ/hr
Find the increase in input to increase the output from 7 MW to 9 MW.
Solution
• From I/O curve
I@ L=7 MW = 106[30 + 0.5L+0.65 L2+ 0.01L3]
=106[30 + 0.5(7)+0.65 (7)2+ 0.01(7)3]
= 68.78 × 106 kJ/hr.
I@ L=9 MW = 106[30 + 0.5L+0.65 L2+ 0.01L3]
=106[30 + 0.5(9)+0.65 (9)2+ 0.01(9)3]
= 94.44 × 106 kJ/hr.
The increase in input to increase output 2 MW = (94.44 – 68.78)× 106
= 25.66× 106 kJ/hr.
• From IR curve:
IR @L= 8MW = dI/dL
=106[ 0.5 + 1.3 L+ 0.03 L2] =12.82× 106 kJ/MW.hr
Therefore;
The increase in input to increase output 2 MW =2×12.82× 106
= 25.64× 106 kJ/hr.
-----------------------------------------------
0
2
4
6
8
10
12
14
16
18
20
0 2 4 6 8 10 12
HR - L curve
0
2
4
6
8
10
12
0 2 4 6 8 10 12
IR-L Curve
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 92
Example (6.4 ):
A 20 MW station has the following I/O curve;
I= 106[30 + 0.5L+0.65 L2+ 0.01L3] , where L in MW and I in kJ/hr
Find the average heat rate of this station for a day when it was operating at a
load 20 MW for 12 hr. and was kept hot at zero load for the remaining 12 hr.
Also, Compute the value of HRav at load factor =1 (ie. The same energy were
produced for the day at a constant 24-hr load).
Solution
I@ L=0 MW = 106[30 + 0.5L+0.65 L2+ 0.01L3]
= 30 × 106 kJ/hr.
I@ L=20 MW = 106[30 + 0.5L+0.65 L2+ 0.01L3]
=106[30 + 0.5(20)+0.65 (20)2+ 0.01(20)3]
= 380 × 106 kJ/hr.
Energy = L1 * Time1 + L2*Time2
Energy = 20×12 +0×12 = 240 MW.hr.
Heat = I1 * Time + I2 * Time2
Heat = [ 30 × 12+ 380 × 12] × 106 = 4920×106 kJ
HRav = Heat
Energy =
4920∗106
240 = 20.5×106
kJ/MW.hr = 20.5 MJ/KW.hr
Case (2) same energy for day at a constant 24-hr load and load factor = 1
Load factor = Energy
load∗duration
1 = 240
load∗24
Load = 10 MW (const. load for 24-hr)
L@L=10 MW = 106 [30+ 0.5*10 +0.65*(20)2 +0.01*(10)3 ] = 110*106 KJ
Heat = I * Time
Heat = 110*106 * 24 = 2.64*109 KJ/hr
HRav = Heat
Energy =
2.64∗109
240 = 11×106
kJ/MW.hr = 11 MJ/KW.hr
--------------------------------------------------------
20
load
20 MW 20 MW
Time,hr
20 MW
12
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 93
Example (6.5):
Derive the required condition to devise a load between two units for most
economical operation.
Devise a load between the following two units for most economical operation:
Unit (a): I= 106[10+5L+0.2L2] , Lmax=10 MW, where L in MW and I in kJ/hr.
Unit (b): I= 106[10+6L+0.02L3] , Lmax=10 MW, where L in MW and I in kJ/hr.
solution
For most economica operation Ic = minimum
dIC
dIa= 0 ; IC = Ia + Ib
dIa
dIa+
dIb
dIa= 0
dIa
dIa+
dIb
dIa∗
dIb
dIb= 0
dIa
dIa+
dIb
dIb∗
dIb
dIa= 0
IC = Ia + Ib
dIc
dIa=
dIa
dIa+
dIb
dIa
dIc
dIa= 0
0 =dIa
dIa+
dIb
dIa
dIb
dIa= −
dIa
dIa
dIb
dIa= −1
From 1 ,and 2
dIa
dIa+
dIb
dIb∗ (−1) = 0
dIa
dIa=
dIb
dIb
IRa = IRb
For most economical operation.
Incremental rate of (a) = incremental rate of (b)
1
2
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 94
Ia= 106[10+5L+0.2L2]
Ib= 106[10+6L+0.02L3] ,
IR= dI
dL
dIa = 106 [ 5+0.4 L]
dIb = 106 [ 6 + 0.06 L2 ]
L (MW) 0 2 4 6 8 10
IRa (KJ/MW.hr) 5000000 5800000 6600000 7400000 8200000 9000000
IRb (KJ/MW.hr) 6000000 6240000 6960000 8160000 9840000 12000000
Draw IRa and IRb curves then collect load of the two curves to get the IRc
curve .
From the curves distribute the total load on the two stations .
Lc 0 2 4 6 8 10 12 14 16 18 20
La 0 2 3 3.4 4.2 5.5 6.7 8 9.2 10 10
Lb 0 0 1 2.6 3.8 4.5 5.3 6 6.8 8 10 ----------------------------------------------------------
0
1000000
2000000
3000000
4000000
5000000
6000000
7000000
8000000
9000000
10000000
11000000
12000000
0 2 4 6 8 10 12 14 16 18 20 22
bIR aIR
CIR
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 95
Example ( 6.6 ):
State the required rules for most economical operation in power stations. And,
Tabulate a capacity scheduling for the following units;
Number 1 2 3 4 5 6
Capacity,
MW
20 30 40 40 50 50
Order of
efficiency
5 4 2 1 3 6
Solution
Load,MW 1 2 3 4 5 6 Total,MW
40 40 40 80
80 40 40 50 130
110 30 40 40 50 160
130 20 30 40 40 50 180
180 20 30 40 40 50 50 230
------------------------------------------
Example ( 6.7):
Tabulate a capacity scheduling for the following units :
Number 1 2 3 4 5 6 7
Capacity, MW 20 30 40 40 40 50 50
Order of efficiency 7 6 3 2 1 4 5
Solution
Load,MW 1 2 3 4 5 6 7 Total,MW
40 40 40 80
80 40 40 40 120
120 40 40 40 50 170
170 40 40 40 50 50 220
200 30 40 40 40 50 50 250
220 20 30 40 40 40 50 50 270
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 96
7) Load curves
To design power plant it is important to know the following conditions for
energy supply;
1) Maximum demand.
2) Total energy required.
3) Distribution of energy demand.
7.1 Maximum Demand for Consumer:
a) Every consumer has connected load and maximum demand. The relation
between them is defined as;
Demand factor = 𝐌𝐚𝐱𝐢𝐦𝐮𝐦 𝐝𝐞𝐦𝐚𝐧𝐝
𝐂𝐨𝐧𝐧𝐞𝐜𝐭𝐞𝐝 𝐥𝐨𝐚𝐝 ≤1
Experience shows that: Demand factor for hotels was about 25% and for
refrigeration plants was about 90 %.
b) Experience shows that, the maximum demand of individual consumers do not
occur simultaneously but are spread out over a period of time. The time
distribution of maximum demands for group of consumers is defined as;
Group Diversity factor = 𝐒𝐮𝐦 𝐨𝐟 𝐢𝐧𝐝𝐢𝐯𝐢𝐝𝐮𝐚𝐥 𝐦𝐚𝐱𝐢𝐦𝐮𝐦 𝐝𝐞𝐦𝐚𝐧𝐝𝐬
𝐌𝐚𝐱𝐢𝐦𝐮𝐦 𝐝𝐞𝐦𝐚𝐧𝐝 𝐨𝐟 𝐭𝐡𝐞 𝐠𝐫𝐨𝐮𝐩 ≥1
c) The peak demand of a system is made up of the individual demands of the
group of consumers. The diversity is measured by;
Peak diversity factor = 𝐌𝐚𝐱𝐢𝐦𝐮𝐦 𝐝𝐞𝐦𝐚𝐧𝐝 𝐨𝐟 𝐭𝐡𝐞 𝐜𝐨𝐧𝐬𝐮𝐦𝐞𝐫 𝐠𝐫𝐨𝐮𝐩
𝐃𝐞𝐦𝐚𝐧𝐝 𝐨𝐟 𝐭𝐡𝐞 𝐜𝐨𝐧𝐬𝐮𝐦𝐞𝐫 𝐠𝐫𝐨𝐮𝐩 𝐚𝐭 𝐭𝐢𝐦𝐞 𝐨𝐟 𝐬𝐲𝐬𝐭𝐞𝐦 𝐩𝐞𝐚𝐤 ≥1
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 97
Example (7.1):
It is required to add a demand of new housing development to lines of public
utility.
a) Domestic-load : There are 1000 apartments each one having 30 KW connected
load, with the following factors:
Demand factor=0.5, Group diversity factor=3.5, and Peak diversity factor=1.5.
b) Commercial-load: like stores and services as; Store number Connected load , kW Demand factor
Mosque 1 150 0.6
Church 1 150 0.56
Laundry 1 50 0.68
Theatre 1 300 0.5
Hospital 1 500 0.67
Bookstore 3 25 for each 0.66
Clothing store 5 40 for each 0.55
Different store services 20 50 for each 0.7
Commercial-load group diversity factor= 1.5, and peak diversity factor=1.1
Find increase in peak demand on the total system resulting from adding this
development. Assume line loss is to be 5% of the delivered energy.
Sol.
a) Domestic-load: Max. Demand per apartment = Demand factor × Connected load
= 0.5 × 30 = 15 kW
Max. Demand of 1000 apartments = 𝐒𝐮𝐦 𝐨𝐟 𝐢𝐧𝐝𝐢𝐯𝐢𝐝𝐮𝐚𝐥 𝐦𝐚𝐱𝐢𝐦𝐮𝐦 𝐝𝐞𝐦𝐚𝐧𝐝𝐬
𝐆𝐫𝐨𝐮𝐩 𝐝𝐢𝐯𝐞𝐫𝐬𝐢𝐭𝐲 𝐟𝐚𝐜𝐭𝐨𝐫
= (1000 × 15) / 3.5 = 4285.7 kW
Demand of 1000 apartments at time of system peak
= 𝐌𝐚𝐱𝐢𝐦𝐮𝐦 𝐝𝐞𝐦𝐚𝐧𝐝 𝐨𝐟 𝐭𝐡𝐞 𝐜𝐨𝐧𝐬𝐮𝐦𝐞𝐫 𝐠𝐫𝐨𝐮𝐩
𝐏𝐞𝐚𝐤 𝐝𝐢𝐯𝐞𝐫𝐬𝐢𝐭𝐲 𝐟𝐚𝐜𝐭𝐨𝐫
= (4285.7/ 1.4) = 3061.2 kW
c) Commercial-load: Store number Max. Demand=Connected load × Demand factor
Mosque 1 150×0.6
Church 1 150 ×0.56
Laundry 1 50 ×0.68
Theatre 1 300 ×0.5
Hospital 1 500 ×0.67
Bookstore 3 75 × 0.66
Clothing store 5 200 × 0.55
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 98
Different store services 20 1000 × 0.7
Total of commercial max demands = 1552.5 kW
Max. Demand of commercial group= 𝐒𝐮𝐦 𝐨𝐟 𝐢𝐧𝐝𝐢𝐯𝐢𝐝𝐮𝐚𝐥 𝐦𝐚𝐱𝐢𝐦𝐮𝐦 𝐝𝐞𝐦𝐚𝐧𝐝𝐬
𝐆𝐫𝐨𝐮𝐩 𝐝𝐢𝐯𝐞𝐫𝐬𝐢𝐭𝐲 𝐟𝐚𝐜𝐭𝐨𝐫
= (1552.5) / 1.5 = 1035 kW
Commercial demand at time of system peak
= 𝐌𝐚𝐱𝐢𝐦𝐮𝐦 𝐝𝐞𝐦𝐚𝐧𝐝 𝐨𝐟 𝐭𝐡𝐞 𝐜𝐨𝐧𝐬𝐮𝐦𝐞𝐫 𝐠𝐫𝐨𝐮𝐩
𝐏𝐞𝐚𝐤 𝐝𝐢𝐯𝐞𝐫𝐬𝐢𝐭𝐲 𝐟𝐚𝐜𝐭𝐨𝐫
= (1035 /1.1) = 940.9 kW
Total demand at time of system peak = 3061.2 + 940.9 = 4002.1 kW
Total increase in max. demand at station bus = 4002.1 1.05 = 4202.2 kW
-----------------------------------------------------------------------------
Load curves
Load curve is a graph which represents the variation of electrical demand with
time.
Chronological load curve. Load duration curve
One year = 8760 hrs
A.M.= After Mid-night.
P.M. = Previous Mid-night.
Load, kW
Time, hr
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 99
The annual factors are defined as follow;
1) Annual Load factor measure the variation of load over the operating
time;
Load factor = 𝐄𝐧𝐞𝐫𝐠𝐲 𝐩𝐫𝐨𝐝𝐮𝐜𝐞𝐝
𝐌𝐚𝐱.𝐥𝐨𝐚𝐝 ×𝐨𝐩𝐞𝐫𝐚𝐭𝐢𝐧𝐠 𝐓𝐢𝐦𝐞
2) The use of the generating plant over one year (8760 hrs) is measured by:
Capacity factor = 𝐄𝐧𝐞𝐫𝐠𝐲 𝐩𝐫𝐨𝐝𝐮𝐜𝐞𝐝
𝐂𝐚𝐩𝐚𝐜𝐢𝐭𝐲 ×𝟖𝟕𝟔𝟎
3) The use of the generating plant over the operating time is measured by:
Use factor = 𝐄𝐧𝐞𝐫𝐠𝐲 𝐩𝐫𝐨𝐝𝐮𝐜𝐞𝐝
𝐂𝐚𝐩𝐚𝐜𝐢𝐭𝐲 ×𝐨𝐩𝐞𝐫𝐚𝐭𝐢𝐧𝐠 𝐓𝐢𝐦𝐞
• Load factor and Use factor become identical when the peak load is equal
to the capacity of the plant over the operating time.
• Load factor, capacity factor, and Use factor become identical when the
peak load is equal to the capacity of the plant over 8760 hrs.
4) The utilization factor measure the use of the total installed capacity of
the plant:
Utilization factor = 𝐌𝐚𝐱.𝐥𝐨𝐚𝐝
𝐂𝐚𝐩𝐚𝐜𝐢𝐭𝐲
• Low utilization factor means the plant is used only for stand-by purpose.
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 100
Example (7.2 ):
A 300 MW thermal power station is to supply the power to a system having
maximum and minimum demands 240 MW and 180 MW respectively in a
year. Assuming annual load duration curve is to be a straight line between
maximum and minimum values.
Compute: load factor, capacity factor, use factor and utilization factor.
Solution
Lmax = 240 MW
Lmin = 180 MW
Energy = area under curve
Energy = 0.5(240+180)*8760
= 1839600 MW.hr
Load factor = 𝐸𝑛𝑒𝑟𝑔𝑦
𝐿𝑚𝑎𝑥 ∗ 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑡𝑖𝑚𝑒 =
1839600
240 ∗ 8760 = 0.875
Capacity factor = 𝐸𝑛𝑒𝑟𝑔𝑦
𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 ∗ 8760 =
1839600
300∗ 8760 = 0.7
Use factor = 𝐸𝑛𝑒𝑟𝑔𝑦
𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 ∗ 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑡𝑖𝑚𝑒 =
1839600
300∗ 8760 = 0.7
Utilization factor = 𝐿𝑚𝑎𝑥
𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 =
240
300 = 0.8
---------------------------------------------------------------------------------
Example (7.3)
The daily load is defined as
Time ,hr 0- 6 6-8 8-12 12-14 14-18 18-24
Load , KW 40 50 60 50 80 40
Find load factor and draw load curve and load duration curve .
Solution
Load factor = 𝑒𝑛𝑒𝑟𝑔𝑦
𝐿𝑚𝑎𝑥∗𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑡𝑖𝑚𝑒
Energy = 40*(6) + 50* (2)+60*(4)+50*(2)+ 80*(4) + 40*(6) = 1240 KW.hr
Lmax from table = 80 MW
Load factor = 1240
80∗24 = 0.645
180 MW
240 MW
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 101
-----------------------------------------------------------------------------
Example ( 7.5 ):
Steam power plant 600 MW capacity carries the following loads:
Duration Time, hr 1500 1500 2500 1500 1760
Load, MW 500 450 400 350 300
Plant performance curve is given by: I=103[1500+8L]+0.01L3
Where; L in MW and I in MJ/hr.
a. Draw load duration curve, and I-L curve.
b. Find maximum efficiency, min. heat rate and average heat rate.
c. Find load factor, capacity factor, use factor and utilization factor.
Solution
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 102
Duration Time, hr 1500 1500 2500 1500 1760
Load, MW 500 450 400 350 300
Input , MJ/hr 6750000 6011250 5340000 4728750 4170000
HR= I
L = 103 (
1500
L+8)+ 0.01L2
At min HR or max. Efficiency:
d(HR)
dL =
d(103(1500
L+8)+0.01𝐿2
dL = 0
103 (−1500
𝐿2)+ 0.02 L = 0
1500000
L2 = 0.02L
L3= 1500
0.02 = 75*106
Therefore; L= √75 ∗ 1063 = 421.716 MW
I@L=421.716 = 103[1500+8*421.716]+0.01(421.716)3 = 5.6237*106 MJ
Ƞ𝑚𝑎𝑥 = L × Const.
I=
421.716 × 3600
5.6237 ∗ 106= 26.99%
HR=103 (1500
421.716+8)+ 0.01(421.716)2 =13335 MJ/MW.hr = 13.335 MJ/kW.hr
Enenrgy = 1500*500+1500*450+2500*400+1500*350+1760*300
= 3478000 MW.hr
Heat= 6750000*1500 +6011250*1500 +5340000*2500+ 4728750*1500
+417000 0*300 = 46924200000 MJ
HRavg = 𝐻𝑒𝑎𝑡
𝐸𝑛𝑒𝑟𝑔𝑦
= 46924200000
3478000 = 13491. MJ/MW.hr = 13.491 MJ/KW.hr
Load factor = 𝐸𝑛𝑒𝑟𝑔𝑦
𝐿𝑚𝑎𝑥 ∗ 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑡𝑖𝑚𝑒 =
3478000
500 ∗ 8760 = 0.794
Capacity factor = 𝐸𝑛𝑒𝑟𝑔𝑦
𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 ∗ 8760 =
3478000
600∗ 8760 = 0.661
Use factor = 𝐸𝑛𝑒𝑟𝑔𝑦
𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 ∗ 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑡𝑖𝑚𝑒 =
3478000
600∗ 8760 = 0.661
Utilization factor = 𝐿𝑚𝑎𝑥
𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 =
500
600 = 0.833
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 103
----------------------------------------------------------------------------------
Example ( 7.6 ):
Steam power plant 60 MW capacity and load curve is defined as:
Time, hr 0-1500 1500-5000 5000-6000 6000-7000 7000-8760
Load, MW 35 45 50 40 25
I/O curve; I=106[8+8L+0.008L2] , where L in MW and I in kJ/hr.
a. Draw load curve, load duration curve, and I-L curve.
b. Find maximum efficiency, average heat rate.
c. Find load factor, capacity factor, use factor and utilization factor.
solution
1500 3500 1000 1000 1760
Load, MW 35 45 50 40 25
Input , KJ/hr 297800000 384200000 428000000 340800000 213000000
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 104
HR= I
L = 106 (
8
L + 8 + 0.008L )
At min HR or max. Efficiency:
d(HR)
dL =
d(106 (8
L +8+ 0.008L )
dL = 0
106 ( −8
𝐿2 + 0.008 )= 0
8
L2 = 0.008
L2= 8
0.008 = 1000
Therefore; L= √1000 = 31.62 MW
I@L=31.62 = 106[8+8*31.62+0.008(421.716)2 ] = 368958595 kJ/hr
η = L × Const.
I=
31.62 ∗ 3600 ∗ 1000
368958595= 42.32%
η max = 42.32 %
Enenrgy = 35*1500+3500*45+50*1000+ 40*1000+25*1760 = 344000 MW.hr
Heat= 2978*105*1500 +3842*105*3500 +4280*105*1000+ 3408*105*1000
+2130*105*1760 = 2.93508*1012 KJ
HRavg = 𝐻𝑒𝑎𝑡
𝐸𝑛𝑒𝑟𝑔𝑦 =
2.93508∗1012
344000 = 8.532*106 kJ/MW.hr = 8.532 MJ/KW.hr
0
50000000
100000000
150000000
200000000
250000000
300000000
350000000
400000000
450000000
500000000
0 10 20 30 40 50 60
I-L curve
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 105
Load factor = 𝐸𝑛𝑒𝑟𝑔𝑦
𝐿𝑚𝑎𝑥 ∗ 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑡𝑖𝑚𝑒 =
344000
50 ∗ 8760 = 0.785
Capacity factor = 𝐸𝑛𝑒𝑟𝑔𝑦
𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 ∗ 8760 =
344000
60∗ 8760 = 0.654
Use factor = 𝐸𝑛𝑒𝑟𝑔𝑦
𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 ∗ 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑡𝑖𝑚𝑒 =
344000
60∗ 8760 = 0.654
Utilization factor = 𝐿𝑚𝑎𝑥
𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 =
50
60 = 0.833
------------------------------------------------------------------------------------
Example (7.7 ):
The load duration curve carried out by the base unit having a capacity 18 MW
and stand by unit having a capacity 20 MW.
Base unit Stand by unit
Capacity , MW 18 20
Max. load, MW 18 12
Operating time, hr 8760 2190
Energy, MW.hr 101350 7350
Compute: load factor, capacity factor, use factor and utilization factor
Solution
Equation Base unit Stand by unit
Load factor 𝐸𝑁𝐸𝑅𝐺𝑌
𝐿𝑚𝑎𝑥 ∗ 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑡𝑖𝑚𝑒
101350
18∗8760 = 0.642
735012 ∗ 2190
= 0.279
Capacity factor , 𝐸𝑁𝐸𝑅𝐺𝑌
𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 ∗ 8760
101350
18∗8760 = 0.642
735020 ∗ 8760
= 0.0419
Use factor 𝐸𝑁𝐸𝑅𝐺𝑌
𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 ∗ 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑡𝑖𝑚𝑒
101350
18∗8760 = 0.642
735020 ∗ 2190
= 0.0419
Utilization factor 𝐿𝑚𝑎𝑥
𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦
18
18 =1
12
20= 0.6
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 106
8-Power Economic
To produce electricity one of the following proposal may be achieved :
1. Erection
2. Extension
3. Replacement
P:investment
S: accumulated sum
i: interest
n: no. of years
S=P(1+i)n
S= A + A(1+i) + A(1+i)2 + …………+ A(1+i)n-1
S (1+i) = A(1+i) + A(1+i)2 + A(1+i)3 + …………+ A(1+i)n
-
S i = A (1+ I )n – A
𝑆 =𝐴 ( 1+𝑖 )𝑛−𝐴
𝑖=
𝑨 [( 𝟏+𝒊 )𝒏−𝟏]
𝒊
𝑎𝑙𝑠𝑜 S= P ( 1+ i)n
………………
……
P
P(1+i) 2P(1+i) 3P(1+i) nP(1+i)
A A A A
………………
……
1
2
2 1
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 107
𝑨 [( 𝟏+𝒊 )𝒏−𝟏]
𝒊 = P ( 1+ i)n
𝐴 =𝑃 [𝑖 (1 + 𝑖)𝑛]
[(1 + 𝑖)𝑛 − 1]=
𝑃 [𝑖 (1 + 𝑖)𝑛 + 𝑖 − 𝑖]
[(1 + 𝑖)𝑛 − 1]
A = 𝑃 [𝑖 [(1+𝑖)𝑛−1]
(1+𝑖)𝑛−1+
𝑖
(1+𝑖)𝑛−1 ]
A = 𝑃 [𝑖 + 𝑖
(1+𝑖)𝑛−1]
i : interest 𝑖
(1+𝑖)𝑛−1 : depreciation
Annual total cost and cost of KW/hr :
1- Fixed cost (Cf )
Cf = R.P
Where :
P = installation cost * capacity
R: fixed change rate
R = i + depreciation + fixed taxes + fixed insurance
2- Operation cost ( Co )
Co = fuel cost + labor cost + operation cost + maintenance cost
+ other costs
Fuel cost = mf * price of ton
𝑚𝑓 =𝐻𝑒𝑎𝑡
𝐶. 𝑉
Labor cost = labor price * 8760
Operating cost = operating taxes * Energy
Total annual cost (Ct ): Ct = Cf + Co
Cost of KW/ hr = 𝐶𝑡
𝐸𝑛𝑒𝑟𝑔𝑦 = [ L.E /KW.hr ]
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 108
Example ( 8.1 ):
Steam power plant 625 MW capacity
Capacity factor = Use factor =0.8,
Average heat rate HRav = 9 MJ/kW.hr
Fuel price 3550 LE/Ton. Fuel C. V. =39420 kJ/kg.
Installation cost 8000 LE/kW.
Labor cost 5000 LE/hr. Operating taxes=0.01 LE/kW.hr.
All other costs= 9×106 LE.
Fixed taxes and insurance are 0.5% and 0.2% respectively.
Money interest, i=8%, n=20 years.
Calculate) cost of kW.hr.
Solution
Capacity factor = Use factor
𝐸𝑛𝑒𝑟𝑔𝑦
𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦∗8760 =
𝐸𝑛𝑒𝑟𝑔𝑦
𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦∗𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑡𝑖𝑚𝑒
operation time = 8760 hr
𝐸𝑛𝑒𝑟𝑔𝑦 = 625 ∗ 8760 ∗ 0.8 = 4380000 MW. hr
HRavg =𝐻𝑒𝑎𝑡
𝐸𝑛𝑒𝑟𝑔𝑦
Heat = 9 * 4380000*103 = 3.942*1010 MJ
Annual total cost and cost of KW/hr :
Fixed cost (Cf )
Cf = R.P
P = installation cost [LE/KW] * capacity [KW]
P = 8000 * 625*103 = 5*109 L.E
R = i + depreciation + fixed taxes + fixed insurance
Depreciation = 𝑖
(1+𝑖 )𝑛−1
= 0.08
(1+0.08 )20−1 = 0.0218
R = 0.08 + 0.0218 + 0.005+0.002= 0.1088
Cf = 5*109 * 0.1088 = 544*106 L.E
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 109
Operation cost ( Co )
Co = fuel cost + labor cost + operation cost + other costs
Fuel cost = mf [ton] * price of ton
𝑚𝑓 =𝐻𝑒𝑎𝑡 [ KJ ]
𝐶. 𝑉 [𝐾𝐽𝑘𝑔
]=
3.942 ∗ 1010 ∗ 103
39420= 1 ∗ 109𝑘𝑔 = 1 ∗ 106 𝑡𝑜𝑛
Fuel cost = 1 ∗ 106 * 3550 = 3550*106 L.E
Labor cost = labor price * 8760
= 5000 *8760 = 43.8*106 L.E
Operating cost = operating taxes [LE/kw.hr] * Energy [KW.hr]
= 0.01 * 4.38*106 *103 = 43.8*106 L.E
Co = 3550*106 + 43.8*106 + 43.8*106 + 9*106 = 3646.6*106 L.E
Total annual cost (Ct ):
Ct = Cf + Co
Ct = 544*106 + 3646.6*106 = 4190.6*106 L.E
Cost of KW/ hr = 𝐶𝑡 [𝐿.𝐸]
𝐸𝑛𝑒𝑟𝑔𝑦 [𝐾𝑊.ℎ𝑟] = [ L.E /KW.hr ]
Cost of KW/ hr = 4190.6∗106
4.38∗106∗103 = 0.956 L.E /KW.hr
----------------------------------------------------------
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 110
Example (8.2 ):
Steam power plant 60 MW capacity and load curve is defined as;
Time, hr 0-1500 1500-4000 4000-6000 6000-7000 7000-8760
Load, MW 35 45 55 40 35
I/O curve; I=106[8+8L+0.008L2] , where L in MW and I in kJ/hr.
a) Draw load curve, load duration curve, and I-L curve.
b) Find maximum efficiency, average heat rate.
c) Calculate load factor, capacity factor and use factor
Fuel C V =40 MJ/kg. Cost of kW.hr=0.95 LE.
Installation cost=8000 LE/kW. Labor cost=3000 LE/hr.
Fixed taxes and insurance are 0.5% and 0.2% respectively, i=9%, n=20 year,
operating taxes=0.01 LE/kW.hr, All other costs=5×106 LE.
d) Find the required fuel price in LE/Ton.
e) For the previous data except the load is constant and equal to 55 MW all the
year (8760 hr), Find the required fuel price in LE/Ton
Solution
a)
Time, hr 1500 2500 2000 1000 1760
Load, MW 35 45 55 40 35
INPUT , KJ/hr 297800000 384200000 472200000 340800000 297800000
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 111
HR= I
L = 106 (
8
L + 8 + 0.008L )
At min HR or max. Efficiency:
d(HR)
dL =
d(106 (8
L +8+ 0.008L )
dL = 0
106 ( −8
𝐿2 + 0.008 )= 0
8
L2 = 0.008
L2= 8
0.008 = 1000
Therefore; L= √1000 = 31.62 MW
I@L=31.62 = 106[8+8*31.62+0.008(421.716)2 ] = 368958595 kJ/hr
η = L × Const.
I=
31.62 ∗ 3600 ∗ 1000
368958595= 42.32%
η max = 42.32 %
Enenrgy = 35*1500+2500*45+55*000+ 40*1000+35*1760 = 376600 MW.hr
Heat= 2978*105*1500 +3842*105*2500 +4722*105*2000+ 3408*105*1000
+2978*105*1760 = 3.21653*1012 KJ
HRavg = 𝐻𝑒𝑎𝑡
𝐸𝑛𝑒𝑟𝑔𝑦 =
3.21653∗1012
376600 = 8.54*106 kJ/MW.hr = 8.54 MJ/KW.hr
0
50000000
100000000
150000000
200000000
250000000
300000000
350000000
400000000
450000000
500000000
0 10 20 30 40 50 60
Chart Title
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 112
C) Load factor = 𝐸𝑛𝑒𝑟𝑔𝑦
𝐿𝑚𝑎𝑥 ∗ 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑡𝑖𝑚𝑒 =
376600
55 ∗ 8760 = 0.781
Capacity factor = 𝐸𝑛𝑒𝑟𝑔𝑦
𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 ∗ 8760 =
376600
60∗ 8760 = 0.716
Use factor = 𝐸𝑛𝑒𝑟𝑔𝑦
𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 ∗ 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑡𝑖𝑚𝑒 =
376600
60∗ 8760 = 0.716
Utilization factor = 𝐿𝑚𝑎𝑥
𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 =
55
60 = 0.916
d) cost of KW .hr = 0.95 L.E/KW.hr
Cost of KW/ hr = 𝐶𝑡 [𝐿.𝐸]
𝐸𝑛𝑒𝑟𝑔𝑦 [𝐾𝑊.ℎ𝑟]
0.95 = 𝐶𝑡
376600∗103
Ct = 357.77*106 L.E
Fixed cost (Cf )
Cf = R.P
P = installation cost [LE/KW] * capacity [KW]
P = 8000 * 60*103 = 480*106 L.E
R = i + depreciation + fixed taxes + fixed insurance
Depreciation = 𝑖
(1+𝑖 )𝑛−1
= 0.09
(1+0.09)20−1 = 0.0195
R = 0.09 + 0.0195 + 0.005+0.002= 0.1165
Cf = 480*106 * 0.1165 = 55.92*106 L.E
Ct = Cf + Co
Co =357.77*106 - 55.92*106 = 301.85*106 L.E
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 113
Operation cost ( Co )
Co = fuel cost + labor cost + operation cost + other costs
Labor cost = labor price * 8760
= 3000* 8760 = 26.28*106 L.E
Operating cost = operating taxes [LE/kw.hr] * Energy [KW.hr]
= 0.01 * 376600*103 = 3.766*106 L.E
301.85*106 = fuel cost + 26.28*106 + 3.766*106 + 5*106
Fuel cost = 266.804*106 L.E
𝑚𝑓 =𝐻𝑒𝑎𝑡 [ KJ ]
𝐶. 𝑉 [𝐾𝐽𝑘𝑔
]=
3.21653 ∗ 1012
40000= 80.413 ∗ 106𝑘𝑔 = 80413 𝑡𝑜𝑛
Fuel cost = mf [ton] * price of ton
𝑝𝑟𝑖𝑐𝑒 𝑜𝑓 𝑓𝑢𝑒𝑙 =266.804 ∗ 106
80413= 3317.9 𝐿𝐸/𝑡𝑜𝑛
e) If load is const and equal 55 MW for 8760 hr
Energy = 55 *8760 = 481800 MW.hr
Input @ 55 MW = 472200000 KJ/hr Heat = 472200000 *8760 = 4.136*1012 KJ
Cost of KW/ hr = 𝐶𝑡 [𝐿.𝐸]
𝐸𝑛𝑒𝑟𝑔𝑦 [𝐾𝑊.ℎ𝑟]
0.95 = 𝐶𝑡
481800∗103
Ct = 457.71*106 L.E
Cf = 55.92*106 L.E
Ct = Cf + Co
Co =457.71*106- 55.92*106 = 401.79*106 L.E
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 114
Operation cost ( Co )
Co = fuel cost + labor cost + operation cost + other costs
Labor cost = labor price * 8760
= 3000* 8760 = 26.28*106 L.E
Operating cost = operating taxes [LE/kw.hr] * Energy [KW.hr]
= 0.01 * 481800*103 = 4.818*106 L.E
401.79*106 = fuel cost + 26.28*106 + 4.818*106 + 5*106
Fuel cost = 365.692*106 L.E
𝑚𝑓 =𝐻𝑒𝑎𝑡 [ KJ ]
𝐶. 𝑉 [𝐾𝐽𝑘𝑔
]=
4.136 ∗ 1012
40000= 103.4 ∗ 106𝑘𝑔 = 103400 𝑡𝑜𝑛
Fuel cost = mf [ton] * price of ton
𝑝𝑟𝑖𝑐𝑒 𝑜𝑓 𝑓𝑢𝑒𝑙 =365.692 ∗ 106
103400= 3536.67 𝐿𝐸/𝑡𝑜𝑛
----------------------------------------------------------------------------------
Example ( 8.3 ):
Steam power plant 600 MW capacity carries the following loads:
Duration Time, hr 1500 1500 2500 1500 1760
Load, MW 500 450 425 350 325
Plant performance curve is given by: I=103[1500+8L]+0.01L3, where L in MW
and I in MJ/hr.
Fuel price 5000 LE/Ton. Fuel C. V. =42000 kJ/kg.
Installation cost 8000 LE/kW.
Labor cost 5000 LE/hr. Operating taxes=0.01 LE/kW.hr.
All other costs= 8×106 LE.
Fixed taxes and insurance are 0.5% and 0.2% respectively.
Money interest, i=8%, n=20 years.
Calculate Average heat rate (HRav ) and cost of kW.hr in this case and in case
of the load is constant and equal to 500 MW all along the year.
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 115
Solution
Duration Time, hr 1500 1500 2500 1500 1760
Load, MW 500 450 425 350 325
Input , MJ/hr 6750000 6011250 5667656.25 4728750 4443281.25
Enenrgy = 1500*500+1500*450+2500*425+1500*350+1760*325
= 3584500 MW.hr
Heat= 6750000*1500 +6011250*1500 + 5667656.25*2500+ 4728750*1500
+4443281.25 *1760 = 4.822*1010 MJ
HRavg = 𝐻𝑒𝑎𝑡
𝐸𝑛𝑒𝑟𝑔𝑦
= 4.822∗1010
3584500 = 13452 MJ/MW.hr = 13.452 MJ/KW.hr
Annual total cost and cost of KW/hr :
Fixed cost (Cf )
Cf = R.P
P = installation cost [LE/KW] * capacity [KW]
P = 8000 * 600*103 = 4.8*109 L.E
R = i + depreciation + fixed taxes + fixed insurance
Depreciation = 𝑖
(1+𝑖 )𝑛−1
= 0.08
(1+0.08 )20−1 = 0.0218
R = 0.08 + 0.0218 + 0.005+0.002= 0.1088
Cf = 4.8*109 * 0.1088 = 522.24*106 L.E
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 116
Operation cost ( Co )
Co = fuel cost + labor cost + operation cost + other costs
Fuel cost = mf [ton] * price of ton
𝑚𝑓 =𝐻𝑒𝑎𝑡 [ KJ ]
𝐶. 𝑉 [𝐾𝐽𝑘𝑔
]=
4.822 ∗ 1010 ∗ 103
42000= 1.147 ∗ 109𝑘𝑔 = 1.147 ∗ 106 𝑡𝑜𝑛
Fuel cost = 1.147 ∗ 106 * 5000 = 5735*106 L.E
Labor cost = labor price * 8760
= 5000 *8760 = 43.8*106 L.E
Operating cost = operating taxes [LE/kw.hr] * Energy [KW.hr]
= 0.01 * 3584500 *103 = 35.845*106 L.E
Co = 5735*106 + 43.8*106 + 35.845*106 + 8*106 = 5822.6*106 L.E
Total annual cost (Ct ):
Ct = Cf + Co
Ct = 522.24*106 + 5822.6*106 = 6344.84*106 L.E
Cost of KW/ hr = 𝐶𝑡 [𝐿.𝐸]
𝐸𝑛𝑒𝑟𝑔𝑦 [𝐾𝑊.ℎ𝑟] = [ L.E /KW.hr ]
Cost of KW/ hr = 6344.84∗106
3584500∗103 = 1.77 L.E /KW.hr
If load is constant and equal 500 MW all year
Energy = 500*8760 =4380000 MW.hr
I@500 MW = 6750000 MJ/hr Heat = 6750000 * 8760 =5.913*1010 MJ
HRavg = 𝐻𝑒𝑎𝑡
𝐸𝑛𝑒𝑟𝑔𝑦
= 5.913∗1010
438000 = 13500 MJ/MW.hr = 13.5 MJ/KW.hr
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 117
Fixed cost (Cf )
Cf = 522.24*106 L.E
Operation cost ( Co )
Co = fuel cost + labor cost + operation cost + other costs
Fuel cost = mf [ton] * price of ton
𝑚𝑓 =𝐻𝑒𝑎𝑡 [ KJ ]
𝐶. 𝑉 [𝐾𝐽𝑘𝑔
]=
5.913 ∗ 1010 ∗ 103
42000= 1.407 ∗ 109𝑘𝑔 = 1.407 ∗ 106 𝑡𝑜𝑛
Fuel cost = 1.407 ∗ 106 * 5000 = 7035*106 L.E
Labor cost = 5000 *8760 = 43.8*106 L.E
Operating cost = operating taxes [LE/kw.hr] * Energy [KW.hr]
= 0.01 * 4380000*103 = 43.8*106 L.E
Co = 7035*106 + 43.8*106 + 43.8*106 + 8*106 = 7130.6*106 L.E
Total annual cost (Ct ):
Ct = Cf + Co
Ct = 522.24*106 + 7130.6*106 = 7652.84*106 L.E
Cost of KW/ hr = 𝐶𝑡 [𝐿.𝐸]
𝐸𝑛𝑒𝑟𝑔𝑦 [𝐾𝑊.ℎ𝑟] = [ L.E /KW.hr ]
Cost of KW/ hr = 7652.84∗106
4380000∗103 = 1.74 L.E /KW.hr
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Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 118
Example (8.4)
Choose the most economical one from the following two plants A,B to perform
the following .
Duration , hr 500 3000 1500 2000 1760
Load , MW 50 40 20 10 5
Plant (A) Plant (B)
Fuel C.V , MJ/kg 28 36
Fuel price L.E/ton 1000 1200
Labor cost 2000 , L.E/hr 48000 , L.E/day
HRavg , MJ/KW.hr 9 8.5
Installation cost, LE/kW 1000 1000
For the two plants n= 15 year
All other cost = 8*106 L.E
Repair cost = 8.5 LE/ton of fuel used
i=8%
operating taxes = 0.01 of annual operating cost
fixed annual insurance = 0.2%
Solution
Fixed cost (Cf )
Cf A = RA.PA
Cf B = RB.PB
P = installation cost [LE/KW] * capacity [KW]
because no capacity is given in problem , take capacity = Lmax = 50 MW
PA = 1000 * 50*103 = 50*106 L.E
PB = 1000 * 50*103 = 50*106 L.E
RA=RB = i + depreciation + fixed insurance
Depreciation = 𝑖
(1+𝑖 )𝑛−1 =
0.08
(1+0.08 )15−1 = 0.0368
R = 0.08 + 0.0368 +0.002= 0.1188
Cf A=Cf B = 50*106 * 0.1188 = 5.94*106 L.E
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 119
Operation cost ( Co )
Co = fuel cost + labor cost + operation cost+ maintenance cost + other costs
Energy = (500*50)+(3000*40)+(1500*20)+(2000*10)+(1760*5) =
203800 MW.hr
H.RavgA = 𝐻𝑒𝑎𝑡𝐴
𝑒𝑛𝑒𝑟𝑔𝑦
HeatA = 9*203800*103 = 1834200 MJ
H.RavgA = 𝐻𝑒𝑎𝑡𝐵
𝑒𝑛𝑒𝑟𝑔𝑦
HeatB = 8.5 * 203800*103 = 1732300 MJ
(Fuel cost)A = mfA [ton] * (price of ton)A
(Fuel cost)B = mfB[ton] * (price of ton)B
𝑚𝑓𝐴 =𝐻𝑒𝑎𝑡𝐴 [ MJ ]
(𝐶. 𝑉)𝐴 [𝑀𝐽𝑘𝑔
]=
1834200 ∗ 103
28= 65507 ∗ 103 𝑘𝑔 = 65507 𝑡𝑜𝑛
𝑚𝑓𝐵 =(𝐻𝑒𝑎𝑡)𝐵 [ KJ ]
(𝐶. 𝑉)𝐵 [𝐾𝐽𝑘𝑔
]=
1732300 ∗ 103
36= 48119 ∗ 103 𝑘𝑔 = 48119 𝑡𝑜𝑛
(Fuel cost)A = 65507 * 1000 = 65.507*106 L.E
(Fuel cost)B = 48119* 1200 = 57742*106 L.E
(Labor cost)A = (labor price)A * 8760
= 2000 *8760 = 17.52*106 L.E
(Labor cost)B = (labor price[L.E/day])B * 365
= 48000 * 365 = 17.52*106 L.E
(maintenance cost )A = repair cost [L.E/ ton fuel] * mFA
= 8.5 *65507 = 556809 L.E
(maintenance cost )B = repair cost [L.E/ ton fuel] * mfB
= 8.5 * 48119 = 409011 L.E
(Operating cost)A.B = 0.01 Co
CoA = 65.507*106 + 17.52*106 +0.01 CoA + 556809 + 8*106
CoA = 92.508*106 L.E
CoB = 57.742*106 + 17.52*106 +0.01 CoB + 409011+ 8*106
CoB = 84.516*106 L.E
Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018. Page 120
Total annual cost (Ct ):
CtA = CfA + CoA
CtA = 5.94*106 + 92.508*106 = 98.448*106 L.E
CtB = CfB + CoB
CTa = 5.94*106 + 84.516*106 = 90.456*106 L.E
(Cost of KW/ hr)A = 𝐶𝑡𝐴 [𝐿.𝐸]
𝐸𝑛𝑒𝑟𝑔𝑦 [𝐾𝑊.ℎ𝑟] = [ L.E /KW.hr ]
(Cost of KW/ hr)A = 98.448∗106
203800∗103 = 0.483 L.E /KW.hr
(Cost of KW/ hr)B = 𝐶𝑡𝐵 [𝐿.𝐸]
𝐸𝑛𝑒𝑟𝑔𝑦 [𝐾𝑊.ℎ𝑟] = [ L.E /KW.hr ]
(Cost of KW/ hr)B = 90.456∗106
203800∗103 = 0.4438 L.E /KW.hr
Plant (b) is more economical than plant (A)
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Acknowledgement:
Greet Thanks to: Abd-elrahman Esam Attia.
Also, Thanks to: Mohamed El-Agamee.