**1-Oxygen percentage in Boiler
outlet flue gas is 4.9%, then what will be the percentage of excess air?**

We have excess
air EA = O2 X 100 / (21-O2)

EA = 4.7 X
100 / (20-4.7)

EA = 30.71%

Read **Power plant eqpts.standard operating procedures**

**2-Calculate the Oxygen level
(O2) in flue gas, if excess air is 25%**

We have Excess
air EA = O2 X 100 / (21-O2)

25 = O2 X
100 / (21-O2)

O2 = 4.2%

**3-A Boiler’s combustion
system requires 5.5 kg of air for burning 1 kg of fuel, then calculate the
total air required for complete combustion if its flue gas has 4.1% of O2**

We have,

Total air = (1 +
EA/100) X Theoretical air

EA = O2 X 100 /
(21-O2)

EA = 4.1 X 100 /
(20-4.1) = 25.78%

Therefore Total
air = (1 + 25.78/100) X 5.5 = 6.92 kg of air per kg of fuel burnt

**4-A Coal fired boiler having
total heating surface area 5200 M2 produces 18 kg of steam per square meter per
hour of heating surface, then calculate the Boiler capacity in TPH**

Boiler Capacity =
(Heating g surface area X Steam generation per square meter)

Boiler Capacity =
5200 X 18 = 93600 kg/hr =93600 / 1000 = 93.6 TPH

**5-Calculate the
Theoretical air & Excess air required to burn 10 MT of coal having carbon
(C) 48%, Hydrogen (H2) 3.8%, Oxygen (O2) 8.2% & Sulphur (S) 0.6% in it. For
complete combustion operator is maintaining 4% of O2 in flue gas.**

We have Theoretical air, Th = (11.6 X %C +
34.8 X (H2-O2/8) + 4.35 X S) / 100

Th
= (11.6 X 48 + 34.8 X (3.8-8.2/100) + 4.35 X 4) / 100

Th
= 5.84 kg/kg of fuel

We have excess
air EA = O2 X 100 / (21-O2)

= 4 X
100 /(21-4) = 23.52%

Total air = (1 +
EA/100) X Theoretical air

Total air = (1 +
23.52/100) X 5.84 = 7.21 kg of air per kg of fuel burnt

**6-Calculate the Oxygen required
for complete combustion of 1 kg of Carbon**

On complete combustion, Carbon becomes carbon
dioxide

C + O2 = CO2 + Heat (8084 kcal/kg)

Write down the molecular weights of carbon,
oxygen & carbon dioxide

12 + 32 = 44

Divide by 12

1 + 2.67 = 3.67

So, 2.67 kg of oxygen is required for complete
combustion of 1 kg of Carbon

**7-Calculate the amount
of Oxygen required for complete combustion of 1 kg of Methane**

On complete combustion, Methane becomes carbon
dioxide & water

CH4 + 2O2 = CO2 + 2H2O Heat

Write down the molecular weights of Methane,
oxygen & carbon dioxide

16 + 64 = 44 + 36

Divide by 16

1 + 4 = 2.75 + 2.25

So, 4 kg of Oxygen is required for complete
combustion of 1 kg of Methane

**8-A Coal sample having
Carbon, Hydrogen, Oxygen & Sulphur percentages 50%,3.5%,8.6% & 1%
respectively, then calculate the GCV of coal.**

GCV of coal = (8080 X C + 34500 X (H2-O2/8) +
2440 X S)) / 100

GCV of coal = (8080 X 50 + 34500 X (3.5-8.6/8)
+ 2440 X 1)) / 100 = 4901 Kcal/kg

**9-A Boiler uses imported
coal to generate 100 TPH of steam, the O2 & CO2 in flue gases are 5% &
14.5% respectively. Calculate the mass of flue gas generated if following is
the ultimate analysis of fuel.**

**Carbon C = 52%**

**Hydrogen H2 = 3.25%**

**Oxygen O2 = 8.3%**

**Sulphur S = 0.3%**

**Nitrogen N2 = 1.1%**

We have Theoretical air, Th = (11.6 X %C +
34.8 X (H2-O2/8) + 4.35 X S) / 100

Th
= (11.6 X 52 + 34.8 X (3.25-8.3/100) + 4.35 X 0.3) / 100

Th
= 6.82 kg/kg of fuel

We have excess
air EA = O2 X 100 / (21-O2)

= 5 X
100 /(21-5) = 31.25%

Total air = (1 +
EA/100) X Theoretical air

Total air = (1 +
31.25/100) X 6.82 = 8.95 kg of air per kg of fuel burnt

Mass of flue gas generated Mfg = Mass of CO2
in flue gas + Mass of N2 in fuel + Mass of N2 in air + Mass of O2 in the flue
gas + Mass of SO2 in the flue gas

Mass of flue gas generated Mfg = (Carbon
percentage in fuel X Mol.weight of CO2) / Mol.weight of Carbon + 0.011 + (8.95
X 77 / 100) + ((8.95-6.82) X 23 / 100) + (0.003 X Mol.weight of SO2) /
Molecular weight of sulphur

Mass of flue gas generated Mfg = (0.52 X 44 /
12) + 0.011 + 6.89 + 0.49 + (0.003 X 64) / 32 =9.3 kg of flue gas per kg of
fuel burnt.

**10-Calculate the Sulphur
dioxide generated per day in a 150 TPH boiler, where coal burned is having 0.5%
of sulphur. Consider steam to fuel ratio 5.5 & Boiler operates on full
load for 24 hours.**

We have S + O2 = SO2

32 + 32 = 64

1 + 1 = 2

That is 1 kg of sulphur generates 2 kg of
Sulphur dioxide on complete combustion.

Total coal consumed in a day = Steam generated
in 24 hours / Steam to coal ratio

Total coal consumed in a day = 150 X 24 / 5.5

Total coal consumed in a day = 654.54
Tones/day

Total sulphur in the coal = 654.54 X 0.5% =3.27 Tones

Therefore total SO2 generated = 3.27 X 2
= 6.54 Tones in a day

**11-A 100 TPH coal fired
boiler generating 8.5 kg of flue gas (Mfg) per kg of fuel burnt at 150 Deg
(Tfg) Calculate the heat loss due to dry flue gas loss. Consider coal GCV 5000
kcal/kg & ambient air temperature 25 Deg C (Ta)**

Heat loss due to dry flue gas = Mass of flue
gas (Mfg) X Specific heat of flue gas (Cp) X (Tfg-Ta)

Heat loss due to dry flue gas = 8.5 X 0.24 X
(150-25) =255 kcal/kg (Specific heat of flue gas = 0.24 kcal/kg)

% of heat loss = 255 X 100 / Coal GCV = 25500
/ 5000 =5.1%

**12-Calculate the heat
loss due to formation of water from 3.22% hydrogen present in coal of GCV 4500
kcal/kg. Consider Boiler outlet flue gas temperature is 145 deg C & ambient
temperature 30 deg C**

Heat loss due to formation of water from hydrogen
in fuel = 9 X H2 X (584 + Specific heat of moisture (Cp) X (Tfg-Ta)

= 9 X
0.0322 X (584 + 0.45 X (145-30)) = 184.24 kcl/kg

% of
heat loss = 184.24 X 100 / Coal GCV = 18424 / 4500 =4.09%

**13-Calculate the heat
loss due to 20% (M) moisture present in coal of GCV 3900 kcal/kg. Consider
Boiler outlet flue gas temperature is 145 deg C & ambient temperature 30
deg C**

Heat loss due to moisture in fuel = M% X (584
+ Specific heat of moisture (Cp) X (Tfg-Ta)

= 0.2 X (584 + 0.45 X (145-30)) = 127.15 kcl/kg

% of
heat loss = 127.15 X 100 / Coal GCV = 18424 / 3900 =3.26%

**14-A coal fired Boiler’s
fly ash collected at APH & analyzed for unburnt. Report shows 22% of
unburnt & 750 kcal/kg GCV. Calculate the heat loss due to this unburnt.
Consider coal GCV 4700 kcal/kg & ash Percentage in coal 6%**

Total ash present in 1 kg of coal = 1 X 6% =
0.06 kg

Total unburnt present in ash = 0.06 X 25% =
0.0015 kg

Heat loss due to unburnt = 0.0015 X 750 =
1.125 kcal/kg

Percentage of heat loss = 1.125 X 100 / 4700 =
0.024%

**15-A Boiler losses have
been analyzed & found as below**

**Heat loss due to dry
flue gas (L1) = 5.2%**

**Heat loss due to
formation of water from hydrogen (L2) = 3.3%**

**Heat loss due to
moisture in fuel (L3) = 6%**

**Heat loss due to unburnt
fuel in ash (L4) = 0.02%**

**Heat loss due to
incomplete combustion (L5) = 0.8%**

**Heat loss due to
moisture present in air (L6) = 0.6%**

**Then calculate the
Boiler efficiency by indirect method**

Boiler efficiency = 100- Total losses =
100-(Sum of L1 to L6)

Boiler efficiency = 100 – (5.2 + 3.3 + 6 +
0.02 + 0.8 + 0.6) = 84.08%

**16-A coal fuel with GCV
5500 kcal/kg & having moisture 12%
& Hydrogen 3.1% in it is burnt in a Boiler with air fuel ratio
8:1.Neglecting ash, calculate the maximum possible temperature (Tfg) attained
in the furnace.Assume whole heat of combustion is given to the products of combustion.
Take specific heat of gases generated 0.24 kcal/kg & ambieant air
temperature 28 deg C (Ta)**

For calculation,
need LCV

We have LCV = HCV-(9
X H2% X 586) = 5500-(9 X 0.031 X 586) =5336.5 kcal/kg

Mass of flue gas
generated per kg of coal burnt, Mfg = mass of air per kg of coal burnt + 1

=8+1 =9 kg

Heat released by
combustion = heat absorbed by gases

5336.5 = Mfg X Cp
X (Tfg-Ta)

5336.5 = 9 X 0.24
X (Tfg-28)= 2498.6 deg C

So maximum
temperature attained is 2498.6 deg C

**17-A 200 TPH boiler
generates 190 TPH (Q1) steam at pressure 121 kg/cm2 and temperature 550 deg C
& consumes 31 TPH of coal having GCV 5200 kcal/kg. Calculate the Boiler
efficiency if it requires 25 TPH (Q3) attemperator water at temperature 125 deg
C.**

Consider feed water temperature at economizer
inlet is 210 deg C & ignore blow down loss.

Boiler efficiency = (Steam generation X
Enthalpy –(Feed water at flow X Enthalpy + Attemperator water X Enthalpy)) X
100 / (Coal consumed X Coal GCV)

Ignoring blow down water loss,

Feed water flow Q2= Total steam generated –
Attemperator water flow = 190-25 =165 TPH

Refer steam table for enthalpy values

Enthalpy of steam Hg = 830 kcal/kg

Enthalpy of feed water Hf1 = 214.34 kcal/kg

Enthalpy of attemperator water Hf2 = 125.4
kcal/kg

Boiler efficiency = (Q1 X Hg – (Q2 X Hf1 + Q3
X Hf2)) X 100 / (31 X 5200)

Boiler efficiency = (190 X 830 – (165 X 214.34
+ 25 X 125.4)) X 100 / (31 X 5200) = 73.94%

**18-Calculate the oil
consumption of a 75 TPH (Q1) oil fired Boiler having efficiency 88% &
generates steam at 65 kg/cm2 pressure & 485 deg C temperature. Consider
feed water temperature at economizer inlet 160 deg C & oil GCV 10000
kcal/kg**

Boiler efficiency = (Q1 X Hg – Q2 X Hf) / (Mf
X GCV)

Here feed water quantity is not given, so
assume feed water flow = Steam generation =Q1=Q2

Ignore blow down losses

Now, refer steam tables for enthalpy

Enthalpy of steam Hg =807 kcal/kg

Enthalpy of feed water Hf =161.3 kcal/kg

0.88 = (75 X 807 –75 X 161.3) / (Mf X 10000)

Oil consumption (Mf) = 5.5 TPH

**19-A biomass fired boiler of efficiency 60%
operates 285 days in a year , it generates 25 TPH (Q1) steam for a process at
pressure 21 kg/cm2 & 360 deg C temperature. Calculate the cost of fuel to
operate the boiler. Assume fuel GCV 2500 kcal/kg, feed water temperature 105
deg C & cost of fuel per ton is Rs 2800.**

First calculate the fuel consumption

Boiler efficiency = (Q1 X Hg – Q2 X Hf) / (Mf
X GCV)

Here feed water quantity is not given, so
assume feed water flow = Steam generation =Q1=Q2

Ignore blow down losses

Now, refer steam tables for enthalpy

Enthalpy of steam at pressure 21 kg/cm2 &
temperature 360 deg C,Hg =754 kcal/kg

Enthalpy of feed water at temperature 105 deg
C, Hf =105.3 kcal/kg

0.60 = (25 X 754 –75 X 105.3) / (Mf X 2800)

Fuel consumption (Mf) = 9.65 TPH

Cost of fuel for operating the Boiler for 285
days = 9.65 X 24 X 285 X 2800 =18.16 crore

**20-Calculate the
specific fuel consumption of a power plant. If steam Turbine consumes 4.2 Tons
of steam to generate 1 MW of power. Assume steam to fuel ratio of Boiler is
4.8.**

We have Specific fuel consumption (SFC) =
Specific steam consumption (SSC) / Steam to fuel ratio of Boiler (SFR)

SFC = 4.2 / 4.8 =0.875 kg of fuel for
generating 1 kw of power

**21-Calculate the GCV of
a bagasse sample having moisture 51% & pole 1.5%.**

We have GCV of bagasse = 4600-46 X Moisture
-12 X pole

=
4600-46 X 51-12 X 1.5 =2236 kcal/kg

**22-A coal sample contains Carbon
40%, Oxygen 8.3%, Hydrogen 3.5% and Sulphur 0.5%, Nitrogen 1.0%, then calculate
its GCV/HCV, LCV and NCV if its total moisture content is 12%.**

We
have the Theoretical formula for GCV,

GCV/HCV
= (8084 X C% + 28922 X (H2% 2 O2%/8) + 2224 X S %)/100

= (8084 X 40 + 28922 X (3.5 – 8.3/8) + 2224 X
0.5)/100

= 3968 kcal/kg

LCV
= HCV - (9 X H2 X 586)

= 3968 - (9 X 0.035 X 586)

= 3783.41 kcal/kg.

NCV
= (GCV - 10.02 X Total moisture)

= (3968 - 10.02 X 4.4) =3923.9 kcal/kg

**23-A 200 TPH coal fired boiler
is loaded up to 90% of its MCR thought the day. The steam fuel ratio (SFR) of
this Boiler is 4.8 & ash percentage in coal is 6.5%.Calculate the revenue
generated in a month by selling quantity of ash generated. Consider 30 days in
a month & cost of ash per MT Rs 100.**

Total steam generated in a day = (200
X 90 /100) X 24 =4320 tones

Total consumed = Steam generated / SFR
= 4320 / 4.8 =900 Tones

Total ash generated in a day = Coal
consumption in day X Ash % in coal

= (900 X 6.5 / 100) =58.5 MT

Total revenue generated in a month =
58.5 X 30 X 100 = Rs 175500

**24-A 150 TPH (Ms) boiler
generates steam at pressure 88 kg/cm2 & temperature 520 deg C. A feed water
at temperature 105 deg C is being used to reduce steam temperature from 450 deg
C to 380 deg C to maintain constant main steam temperature at SH
oiutlet.Calculate the water required for desuperheating.**

Attemperator
inlet steam enthalpy at pressure 88 kg/cm2 & temperature 450 deg C, Hg1
=770 kcal/kg

Attemperator
inlet steam enthalpy at pressure 88 kg/cm2 & temperature 380 deg C, Hg2
=750 kcal/kg

Attemperator
inlet steam enthalpy at temperature 105 deg C, Hf =105.11 kcal/kg

Heat
lost by steam = Heat gained by desuper heating water

Ms
X (Hg1-Hg2) = MW X (Hg2-Hf)

Mw
=150 X (770-750) / (750-105.11) = 4.88 TPH

**25-A boiler steam drum safety
valve lifts at 115 kg/cm2 and reseats at 110 kg/cm2, then calculate its**

**Blow down percentage?**

** **

Blow down of safety valve = (Set pressure – Reseat pressure) X
100/Set pressure

= (115 – 110) X 100/115 = 4.5%

** **

**Viva Questions & answers for preparation of BOE exam & interview **

**26-A boiler’s SH steam line
safety valve is set at 72 kg/cm2 & blow down rate kept 2.5%, calculate the
pressure at which safety valve reseats**

** **

Blow down of safety valve = (Set pressure – Reseat pressure) X
100/Set pressure

2.5 = (75-Reseat pressure) X 100 / 75

Reseat pressure =73.13 kg/cm2

** **

**27-A air Pre heater (APH) Flue
gas inlet/out let & Air inlet/outlet
temperatures are 240 deg C/150 deg C & 35 deg C/185 deg C, calculate the
effectiveness of APH from gas side & air side**

**APH gas side efficiency**

Î·APHg = (Flue gas inlet temp.-Flue
gas outlet temp.) X 100 / (Flue gas inlet temperature -Air inlet temperature)

Î·APHg =(240-150) X 100 / (240-35)
=43.9%

**APH air side efficiency**

Î·APHa = (Air outlet temp.Tao-Air
inlet temp.)) X 100 / (Flue gas inlet temperature -Air inlet temperature)

Î·APHa =(185-35) X 100 / (240-35) =73.17%

**28-Calculate the economiser
effectiveness, whose feed water inlet & outlet temperatures are 165 Deg C
& 245 Deg C respectively & flue gas inlet & outlet temperatures 385
deg C & 215 deg c respectively.**

Î·Eco. = (Economizer outlet feed water temperature -Economizer
inlet feed water temperature) X 100 / (Economizer inlet flue gas
temperature - Economizer inlet feed water temperature)

Î·Eco = (245-165 ) X 100 / (385-215)

Î·Eco = 47.05%

**29-A HP heater is been used to
raise the feed water temperature from 105 deg C to 150 deg C by using Turbine
bleed steam at inlet temperature 280 deg C, calculate the HP heater
effectiveness. Consider the HP heater condensate out let temperature is 140 deg
C**

HP heater effectiveness = It is calculated as temperature range of
steam X 100/ Temperature range of feed water

HP heater effectiveness = (280-140/ (150-105) = 3.1

**30-A HP heater is used to heat
100 TPH feed water from 110 deg C to 145 deg C by using bleed steam at pressure
15 kg/cm2 and temperature 320 deg C, calculate the quantity of steam required
if condensate outlet temperature is 155 deg Consider specific heat of water =1
kcal/kg**

Enthalpy of steam used at pressure 15kg/cm2 & temperature 320
deg C by referring steam table =735.29 kcal/kg

Enthalpy of condensate leaving HP heater =156.12 kcal/kg

Heat lost by steam = Heat gained by feed water

Mass of steam X (Enthalpy of steam-Enthalpy of condensate water) =
Mass of water X Cp X Rise in feed water temperature

Mass of steam = 100 X 1 X (145-110) / (735.29-156.12)

Mas of steam Ms =6.04 TPH

**31-A spray cum tray Deaerator
inlet & outlet water oxygen concentration is 15 & 0.007 ppm
respectively, calculate the Deaerator efficiency**

Î· D/A =
(Concentration of Oxygen in inlet water-Concentration of oxygen in outlet
water) X 100 /(Concentration of Oxygen in inlet water)

Î· D/A =(15-0.007) X 100 /15

Î· D/A = 99.53%

**32-In a coal based Thermal power
plant, a 10” steam line is left uninsulated around 2 meters of its length.
Because of this there is a loss of heat around 17500 kcal/hr .Calculate the
extra fuel consumption in a day to compensate this loss. Consider coal GCV
& boiler efficiency 4500 kcal/kg & 90% respectively**

We have,

Heat loss in terms of fuel = Heat loss / (Fuel GCV X Boiler
efficiency) = 17500 / (4500 X 0.9) =4.32 kg/hr

So extra coal consumption to compensate the heat loss = 4.32 X 24
=103.7 kg/day

**33-A boiler of generates 85 TPH
of steam at pressure 65 kg/cm2 & temperature 490 deg C, calculate the
velocity of this steam if it passes through 150 NB steam line.**

Velocity of steam inside the pipe line = Flow in steam line
(m3/sec) / Area of steam line (m2)

Convert steam flow 85 TPH into m3/sec

That is 85000 kg/hr / Density of steam

Refer steam table & find out density of steam at above
pressure & temperature

Density of steam = 20.13 kg/m3

Area of pipe line =∏XD^{2} / 4 = 3.142 X (0.150)^{2}/4
=0.018 M^{2}

Steam flow = 85000 / (20.13 X 3600) = 1.17 m3/sec

Then,

Velocity of steam inside the pipe line = 1.17 / 0.018 =65.16 m/sec

**34-A air preheater of heating surface
2800 M2 need to design for a Boiler of 120 TPH, the total area required for
flue gas flow is around 8.5 M2.Calculate the number of tubes & their
length. Select tube size OD 63.5mm X 2.34mm thickness.**

For finding out number of tubes, calculate the area of one tube

A==∏XD^{2} / 4

Where D is inside diameter of tube (considered flue gas passes
through tubes & air outside the tubes)

Inside diameter of tube = OD -2 X Thickness = 63.5-2 X 2.34 =58.82
mm =0.05882 meters

Area, A = 3.142 X (0.05882)^{2} /4 =0.0027 M2

Total required area for flue gas flow is 8.5 meter

Then, Number of tubes = 8.5 / 0.0027 =3148.1 nos

Take round figure =3148 Nos

Given that, total heating surface area of the tubes = 2800 M^{2}

**Length of the tube,**

(2∏RL) X Number of tubes= 2800

R = Outside diameter of tube =63.5/2 =31.75 mm = 0.03175 meter

Length of tubes, L=2800 / (2 X ∏ X 0.03175 X 3148) =4.46 meter or
4460 mm

**35-A boiler of operating
pressure 66 kg/cm2 has LHS water wall panel total length of 25 meters from
bottom header to top header, calculate the maximum thermal expansion of panel.
Consider the MOC of tube material SA210 Gr.A (Carbon steel) & ambient air
temperature 30 deg C**

We have,

Total thermal expansion
=Length of the panel X Coefficient of expansion carbon steel pipe X Operating
temperature of metal

As per boiler operating
temperature & carbon steel material maximum metal temperature will be 390
deg C

Coefficient of
expansion carbon steel pipe=11.6 X 10^{-6} m/m ^{o}C

Maximum possible thermal
expansion of LHS water wall panel=25 X 11.6 X 10^{-6} X 390 =0.113
meter =113.1 mm

**36-The
maximum permissible limit of a 75 TPH boiler’s TDS is 120 mg/l. If the TDS of
feed water is 5 mg/l and percentage of makeup water is 7%. Then calculate the
percentage of blow down and blow down water quantity.**

% of blow down = (Feed water TDS X % of makeup water)
X 100/(Boiler permissible TDS - Feed water TDS)

= (5 X 0.07) X 100 / (120 - 5)

= 0.3%

Quantity of blow down water = (0.3 X 75 / 100) = 0.225
TPH

**37-What quantity of flash steam
is produced when steam drum water operating at 70 kg/cm2 is released in CBD
tank at atmospheric pressure at the rate of 1.1 TPH**

Sensible heat at high pressure in drum water S1 = 305 kcal/kg……
Refer steam table

Sensible heat of steam at atmospheric pressure in CBD tank S2 =
100 kcal/kg…… Refer steam table

Latent heat of flash steam L = 539 kcal/kg

CBD water flow rate Q = 1.1 TPH

Percentage of flash steam produced % = (S1 - S2) X 100 / L

= (305 –
100) X 100 / 539

= 38%

Total quantity of flash steam produced per hour = (1.1 X 38 / 100)
= 0.42 TPH

**38-What is the COC of a Boiler, if boiler water has chloride 95 ppm & feed water 21 ppm.**

COC = Boiler water chloride / Feed water chloride

COC = 95 / 21 = 4.5

**39-Convert 100 TPH boiler capacity into BHP capacity**

We have, 1 BHP = 15.65 kg/hr

There fore total BHP = 100 X 1000 / 15.65 = 6389.8

**40-A Boiler of capacity 75 TPH operating at 65 kg/cm2 & temperature 490 deg C,calaculate boiler BHP.Assume feed water temperature 150 deg C**

Total heat content in out let steam = 75 X 1000 X (Enthalpy of outlet steam- Enthalpy of feed water)

=75000 X (811-151) =49500000 kcal/hr

We have 1 BHP =8436 kcal/hr

So Boiler BHP = 49500000 / 8436 =5867.7

**41-Calculate the condensate
formed, if 25 TPH steam at pressure 7 kg/cm2 & temperature 180 deg C is
supplied to a process plant situated at a distance of 500 meter from the
generation plant. The pressure & temperature at the end user point are 6.2
kg/cm2 & 170 deg C respectively.**

Enthalpy of steam generating end
= 666.74 kcal/kg

Enthalpy of steam at end user = 662.41
kcal/kg

Enthalpy of evaporation at
average pressure (7 + 6.2) / 2 =6.6 kg/cm2 = 489.95 kcal/kg

Condensate formed = 25 X (666.74-662.41)
/ 489.95 =0.22 TPH

**Read related articles**

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**Attemperator & related calculations in Boilers**

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