Skip to main content

Chain conveyor troubleshoot guide

  Problem Identification & Corrective Action Sl No. Problem Potential Cause Solution 1 Chain rises off from sprocket 1.Excess chain slack. 2.Excess wear at the bases of sprocket teeth. 3.Excess chain extension. 4.Foreign material stuck to the bases of sprocket teeth. 5.Reverse rotation of conveyor 6.Uneven chain tightening of chain 1.Adjust the amount of slack equally at both sides. 2.Carryout hard facing of the sprockets 3.Replace the elongated chain parts. 4.Remove the foreign material from the bases of the teeth. 5.Avoid reverse rotation of the conveyor when chain tightened is uneven 2 Chain separates poorly from the sprocket. 1.Sprocket misalignment. 2.Excess chain slack. 3.Excess wear at the bases of sprocket teeth. 4.Uneven chain tightening of chain 1-Adjust alignment. 2-Adjust the amount of slack.

Boiler calculations for Boiler operation Engineer Exam (BOE)

 

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%

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

                 READ Power plant O&M books

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

Therefore total SO2 generated = 654.54 X 2 =1309.08 Tones

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

ηAPHg = (Flue gas inlet temp.-Flue gas outlet temp.) X 100 / (Flue gas inlet temperature -Air inlet temperature)

ηAPHg =(240-150) X 100 / (240-35) =43.9%

 

APH air side efficiency

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

ηAPHa =(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.

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

ηEco = (245-165 ) X 100 / (385-215)

ηEco = 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

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

η D/A =(15-0.007) X 100 /15

η 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 =∏XD2 / 4 = 3.142 X (0.150)2/4 =0.018 M2

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==∏XD2 / 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 M2

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 oC

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

 Power plant equipments efficiency calculation

Basics of fuels & combustion


Basics of Thermodynamics


Thermal expansion in Boilers


Why & How these in Boilers


Boiler draught & Chimney


Reasons for boilers explosion


Reasons for more fuel consumption in Boilers


10-Tips to reduce LOI in Boilers


Basics questions & Answers on AFBC Boilers


Feed water heaters


Boilers troubleshooting guide


Safety valves in Boilers


IBR forms used in power plants

Comments

Frequently visited posts

Boiler feed pumps (BFP) questions & answers for interview

Boiler feed pumps (BFP) questions & answers for interview: 1-What is the function of Boiler feed pumps (BFP) in power plant? Functions: To supply the feed water to boilers To conduct the Boiler hydraulic tests To supply the desuperheating & attemperator water required for process steam lines & boilers respectively TOP-6 BEST POWER PLANT O&M BOOKS Boiler calculations for Boiler operation engineer (BOE) exam Viva Questions & answers for preparation of BOE exam & interview  2- What are the type of prime movers (drives) used for BFPs? Prime movers: LT drive (415 V) HT drive (11 KV) Turbo drive (Steam driven) 3-What are the auxiliaries associated with BFP? BFP auxiliaries Cooling water pump & lines Lube oil system ARC valve Mechanical seal flushing system Balance leak off line & its PRV 4-What are the various pipe lines connected to BFP? Questions & Answers on AFBC Boilers Suction pipe line D

Basic things you must know about nut bolts & spanners

A Bolt is a mechanical device which has head on one end of a body and a thread on the other end designed for fastening of two parts. Bolt is inserted into holes in assembly parts, it is mated with a tapped nut. Tension is normally induced in the bolt to compress the assembly by rotating the nut. This may also be done by rotation of the bolt head. A Screw is a headed and threaded bolt used without a nut. It is inserted into an internally tapped hole and tension is induced by rotation of the screw head. A Stud is a fastener with no head but it has threads at both ends of the shank. It, like a screw, has one end those screws into a tapped hole. A nut is used on the other end to create tension. Major diameter – largest diameter of thread (D) Minor diameter – smallest diameter of thread Pitch – distance between adjacent threads.       Boiler calculations for Boiler operation engineer (BOE) exam TOP-6 BEST POWER PLANT O&M BOOKS Types of nut bolts & Washer

Calculation of heat rate & efficiency of the power plant

Heat rate  is the amount of energy used by an electrical generator/power plant to generate one kilo Watt-hour (kWh) of electricity Heat rate (HR) = Heat input / Power generation =Kcal / Kwh Boiler calculations for Boiler operation engineer (BOE) exam Viva Questions & answers for preparation of BOE exam & interview Total heat input: The chemical energy available in the fuel (coal, biomass, oil, gas etc) is converted into heat energy in Boilers, this process is called as oxidation. The heat available in the fuel is measured in terms of Kcal/kg, KJ/kg or BTU units. The part of this fuel is used as useful heat and rest is lost as dry flue gas loss, moisture loss, un  burnt loss, radiation/convection losses etc.Based on Boiler efficiency this heat energy from the fuel is utilised, generally fuel heat utilisation is in the range of 60 to 90%. This heat generated in the boilers due to oxidation of fuel is used to generate high pressure & temperature steam. Thus genera

STEAM CONDENSER,VACUUM AND CALCULATIONS

A steam condenser is device or an appliance in which steam condenses and heat released by steam is absorbed by water. Heat is basically shell & tube type heat exchanger, where cooling water passes through tubes & steam condenses in shell. The functions of the condensers are: It condenses the steam exhausted from Turbine last stage Increase the thermal efficiency of the plant reducing the exhaust pressure and thereby reducing the exhaust temperature It maintains a very low back pressure on the exhaust side of the Turbine Supplies feed water to Boiler through deaerator TOP-6 BEST POWER PLANT O&M BOOKS Condenser related components: Hot well Cooling water inlet & outlet system Cooling tower Support springs or expansion neck Air Ejector system Condensate extraction system Cooling water tubes & tube sheet Vacuum breaker valve Safety valve or rapture disc Water box Air & water vent lines Types of steam condensers: Surface Co

22-Most likely Question Answer Guide for Boilers troubleshooting

TOP-6 BEST POWER PLANT O&M BOOKS 1.What are the emergencies that are expected to happen in boiler operation? Emergencies that can happen in boilers operation are: Boilers tubes leakage Failure of feed water control station Unbalanced draft in furnace Furnace explosion Blow down valve failure Feed water pumps failure Secondary combustion in super heaters and ESP Also read 16-Perfect reasons for increasing the fuel consumption of Boilers 2.List out the potential reasons for boiler tubes failure. Reasons for boiler tubes failure: Failure due to overheating Failure due to internal scales Failure due to aging Failure due to uneven expansion Failure due to ash and flue gas erosion Failure due to material defects Failure due to internal and external tube corrosion Improper or effected circulation due to opening/passing of low point drains 3.List down the sequential action taken after boiler tube leakage to restore the boiler. Following activities ar

Questions answers & calculation on Boiler draught & Chimney

1-What do you mean by the term draught? It is the pressure difference, which causes flow of gases to take places. 2-What is the function of draught in Boiler? To supply the combustion air required for proper combustion To evacuate the combustion products from the furnace or combustion chamber To evacuate the combustion products to the atmosphere So draught is the pressure difference between the inside the boiler and outside air. Inside the boiler pressure is due to combustion products (Flue gas) & outside pressure is due to fresh atmospheric air. 3-What are the different types of draughts used in power plant boilers? There are mainly two types of draughts. Natural draught & artificial draught Natural draught is generally obtained by Chimney Artificial draughts have main two types Steam jet: It is again classified into Induced & forced draught Mechanical draught: This is classified into Induced draught, forced draught & Natural draught
close