1-introduction - higher technological institute

Prof. Dr. Hesham Mostafa, HTI, Mech. Eng. Dept., Thermal Engineering, ME231, Jan 2018.

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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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


Simple Rankine cycle


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, %.


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]


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]


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


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]


Rankine cycle

with

Reheat

+

OFWH

h-s diagram for


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= 𝑚𝑠𝑡 qadd = 𝑚𝑠𝑡 (h1-h9) = [KW]

WP= 𝑚𝑠𝑡 (1-y) (h5 – h4) + 𝑚𝑠𝑡 y (h8 – h6) = [KW]

Qrej = (1-y) (h3-h4) = [ kJ/kg]


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


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 .


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


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= 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


y (h2-h7) = h6 - h5

TTD : difference between both outlet of CFWH

T7=T6 – TTD T6=T7 – TTD

h6 = Cp T6 = h7- Cp TTD



RANKINE CYCLE with (super heater + reheater + OFWH + CFWH)

Reheat

+

OFWH

+

CFWH

h-s diagram for


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


NOTE : h14= y1 h13 +(1-y1) h12


(1-y1) (h12-h10) = y1 (h4-h11)

Or

y1 h4 + (1-y)h10 = (1-y) h12 + y h11


(1-y1-y2) h8 + y2 h5 = (1-y1) h9


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.




• Bleeding steam to OFWH @ 1

bar




• Bleeding steam to CFWH @ 2

bar




• 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


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 %


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 %


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 %


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


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


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 %


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



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


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


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


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


• Expands up to saturation

line.


1182.3 4.99 1177.31 3235.31 36.38



line.


• Bleeding steam to OFWH

@ 1 bar

1150.81

4.98 1145.83 3010.4

38.06



line.


• Bleeding steam to CFWH

@ 2 bar

1128.738

4.96 1123.778 2937.84

38.25



line.


• Bleeding steam to OFWH

@ 1 bar

• Bleeding steam to CFWH

@ 2 bar

1136

4.975 1131 2939.36

38.478


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






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


NOTE : h14= y1 h13 +(1-y1) h12


(1-y1) (h12-h10) = y1 (h4-h11)

(1-y1) (522.48-426.9) = y1 (2680-535)

y1=0.0426


(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


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


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.


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


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


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


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



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


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


(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


Qcond = ��𝑐.𝑤CPw (Tc.w.o -Tc.w.i )

73256.4 = ��𝑐.𝑤*4.2 *(22 - 10)

��𝑐.𝑤= 1453.5 kg/s


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


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 )


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




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



(1-y1) ( h12 – h10 )= y1 (h4 – h11 )

(1 – y1 ) ( 547.4 – 425.9 ) = y1 ( 2760 – 561)

y1= 0.0523


(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


Deaerator

Feed water flow diagram :


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


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

-----------------------------------------


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


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

----------------------------------------------------------


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


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

----------------------------------------------------------


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


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%


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.


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 .


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


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


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


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


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


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 %


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%

-----------------------------------


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


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


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


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


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


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.

------------------------------------------------------------


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.


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.


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%


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


= 176000 -55728.7 = 120271.3 KJ/min



= 320 * 4.2 *40 = 53760 KJ/min

2) Heat loss to oil

Q2 = 𝑚𝑜𝑖𝑙 CPoil ΔToil

= 140 * 2.1 * 28 = 8232 KJ/min


3) Heat loss to water in calorimeter

Q3 = 𝑚𝑐𝑎𝑙 CPw ΔTcal

= 300 * 4.2 * 42 = 52920 KJ/kg


Q4 = Qloss - (Q1 + Q2 + Q3)

= 120271.3 – (53760 + 8232 + 52920 ) = 5359.3 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


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


= 57333 – 16971 = 40362 KJ/min



= 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


Q4 = Qloss - (Q1 + Q2 + Q3)

= 40362– (14364 + 10340 + 2277) = 13381 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 %


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


= 7525 - 1890 = 5635 KJ/min



= 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


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


Q4 = Qloss - (Q1 + Q2 + Q3)

=5635 – ( 2116.8+ 1144.08+ 227.5 ) = 2146.62 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 %


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 .


2- Water tube boiler

• high capacity

• high pressure

• super heated steam

• water inside tubes


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


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 .


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.


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


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.


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


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 .


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


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


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


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


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%


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


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


��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


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


��𝑓 (𝐶. 𝑉)

Ƞ𝑏 =2000(2724.42−147)

250 (29800)∗ 100% = 69.19 %


Example ( 5.5 ):

The following data are collected during a boiler trial:

Steam generated 100 Ton/hr, @ 60 bar & 360 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 ):


Steam generated 650 Ton/hr, @ 10 bar & D.S.S.


Fuel used = 55 Ton/hr, Fuel C.V. =40000 kJ/kg.





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%.


Find boiler efficiency and Draw heat balance sheet per 1 kgf.

Example ( 5.7 ):


Steam generated 100 Ton/hr, @ 60 bar & 360 oC.


Fuel used = 8.5 Ton/hr , Fuel C.V. =42000 kJ/kg.





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%.


Find boiler efficiency and Draw heat balance sheet per 1 kgf .


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)

-------------------------------------------------------------------------------------


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



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


-----------------------------------------------------------

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


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


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


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


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


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


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


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


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.


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


-----------------------------------------------------------------------------

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


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


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



3478000

500 ∗ 8760 = 0.794



3478000

600∗ 8760 = 0.661



3478000

600∗ 8760 = 0.661



500

600 = 0.833


----------------------------------------------------------------------------------

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


HR= I

L = 106 (

8

L + 8 + 0.008L )


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


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




344000

50 ∗ 8760 = 0.785



344000

60∗ 8760 = 0.654



344000

60∗ 8760 = 0.654



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


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


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 ]


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


𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦∗8760 =


𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦∗𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑡𝑖𝑚𝑒

operation time = 8760 hr

𝐸𝑛𝑒𝑟𝑔𝑦 = 625 ∗ 8760 ∗ 0.8 = 4380000 MW. hr

HRavg =𝐻𝑒𝑎𝑡


Heat = 9 * 4380000*103 = 3.942*1010 MJ


Fixed cost (Cf )

Cf = R.P

P = installation cost [LE/KW] * capacity [KW]

P = 8000 * 625*103 = 5*109 L.E


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


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


= 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

----------------------------------------------------------


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


HR= I

L = 106 (

8

L + 8 + 0.008L )


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


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


C) Load factor = 𝐸𝑛𝑒𝑟𝑔𝑦


376600

55 ∗ 8760 = 0.781



376600

60∗ 8760 = 0.716



376600

60∗ 8760 = 0.716



55

60 = 0.916

d) cost of KW .hr = 0.95 L.E/KW.hr


𝐸𝑛𝑒𝑟𝑔𝑦 [𝐾𝑊.ℎ𝑟]

0.95 = 𝐶𝑡

376600∗103

Ct = 357.77*106 L.E

Fixed cost (Cf )

Cf = R.P


P = 8000 * 60*103 = 480*106 L.E


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





= 3000* 8760 = 26.28*106 L.E


= 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



]=

3.21653 ∗ 1012

40000= 80.413 ∗ 106𝑘𝑔 = 80413 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


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





= 3000* 8760 = 26.28*106 L.E


= 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



]=

4.136 ∗ 1012

40000= 103.4 ∗ 106𝑘𝑔 = 103400 𝑡𝑜𝑛


𝑝𝑟𝑖𝑐𝑒 𝑜𝑓 𝑓𝑢𝑒𝑙 =365.692 ∗ 106

103400= 3536.67 𝐿𝐸/𝑡𝑜𝑛

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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.


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



= 4.822∗1010

3584500 = 13452 MJ/MW.hr = 13.452 MJ/KW.hr


Fixed cost (Cf )

Cf = R.P


P = 8000 * 600*103 = 4.8*109 L.E


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







]=

4.822 ∗ 1010 ∗ 103

42000= 1.147 ∗ 109𝑘𝑔 = 1.147 ∗ 106 𝑡𝑜𝑛

Fuel cost = 1.147 ∗ 106 * 5000 = 5735*106 L.E


= 5000 *8760 = 43.8*106 L.E


= 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


Ct = Cf + Co

Ct = 522.24*106 + 5822.6*106 = 6344.84*106 L.E



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



= 5.913∗1010

438000 = 13500 MJ/MW.hr = 13.5 MJ/KW.hr


Fixed cost (Cf )

Cf = 522.24*106 L.E






]=

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


= 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


Ct = Cf + Co

Ct = 522.24*106 + 7130.6*106 = 7652.84*106 L.E



Cost of KW/ hr = 7652.84∗106

4380000∗103 = 1.74 L.E /KW.hr

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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


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



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



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]


(Cost of KW/ hr)A = 98.448∗106

203800∗103 = 0.483 L.E /KW.hr

(Cost of KW/ hr)B = 𝐶𝑡𝐵 [𝐿.𝐸]


(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.

1-introduction - higher technological institute

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