Overall thermal efficiency on gross c.v.

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Reference no: EM13943730

COMBUSTION PROBLEMS NO 11
QUESTION 11 - Steam boiler plant in a petroleum refinery has an overall thermal efficiency on gross C.V. of the fuel of 85 % and an out put of 90720 kg of steam per hour. The steam is produced at a pressure of 17 kg /cm2 and is superheated to 230^oc from feed water at 50^oc.

Refinery gas is used to fire the boiler which has the following volumetric composition in percentage -
H₂ =21,CH₄ = 15 , C₂H₆ =15 , C₃H₆ = 36, N₂ =9, H₂S = 2, CO₂ =1, CO =1

FIND OUT -

A - Volumetric flow rate of refinery gas required to fire the boiler plant.
B- Volumetric capacity of induced draught fan assuming that refinery gas is burnt with 20 % excess air and exit flue gas temperature from the boiler plant at 205^oc.

GIVEN -
GASES H₂ CH₄ CO C₂H₆ C₃H₆ H₂S
GROSS C.V.

kcal/NM₃ 3026 8900 3026 17177 20470 6257

SOLUTION - Gross C.V. of refinery gas -
=21/100*3026 +1/100*3026+2/100*6257+15/100*8900+15/100*17177+36/100*20470=12072 kcal/Nm3

Enthalpy of steam at 17 kg/cm² and 230^oc =683 kcal /kg

Heat content of steam produced =90720*683 = 61961760 kcal
Heat in feed water =90720 *1*(50-0) = 4536000 kcal
Total heat to be supplied by fuel ( i.e. refinery gas ) for steam generation = 61961760-4536000 = 57425760 kcal
Heat produced by burning refinery gas = heat gained by steam
Heat produced by gas *85/100 = 57425760
( thermal efficiency of boiler = 85%)
Heat produced by gas = 57425760/0.85 =67559717 kcal

Since , C.V. of refinery gas =12072 kcal /Nm3
So , volume of refinery gas required = 67559717/12072 =5596 Nm3/hr
SO , volumetric fuel (i.e. refinery gas ) firing rate =5596 Nm3/hr

Part - B - I.D. fan capacity is computed by calculating the total volume of wet flue gases at the boiler furnace exit temperature.
Total wet flue gas volume per Nm3 of refinery gas on being burnt with 20% excess air is found as below-
Constituents quantity combustion reaction O2 required for

Nm³ combustion
Nm³
H₂ 0.21 H₂+1/2O₂ =H₂O 0.105
CO 0.01 CO + ½ O₂ = CO₂ 0.005
H₂S 0.02 2H₂S +2O₂ = 2H20 +SO₂ 0.020
CH₄ 0.15 CH₄ +2O₂ =CO₂+2H₂O 0.300
C2H₆ 0.15 2C₂H₆ +7O₂ =4CO₂+6H₂O 0.525
C₃H₆ 0.36 2C₃H₆ +9O₂ = 6CO₂+6H₂0 1.620

N₂ 0.09 INCOMBUSTIBLE
CO₂ 0.01 INCOMBUSTIBLE
TOTAL 1.00 2.575

PRODUCTS OF COMBUSTION ,NM₃
CO₂ O₂ N₂ H₂O SO₂
0.01 0.21
0.15 0.02 0.01
0.30 0.45
1.08 1.08
0.09
0.01

TOTAL - 1.55 0.51 11.62 2.06 0.01
O₂ From N₂ from
Excess air total air

TOTAL 1.55 0.51 11.62 2.06 0.01
Theoretical oxygen requirement for burning 1NM³ of refinery gas = 2.575 Nm³
Theoretical air requirement = 2.575*100/21 = 12.26 Nm3
Actual air used = (100 +20/100)*theoretical air = 1.2 *12.26 =14.71 Nm³
Excess air = 14.21- 12.26 =2.45 Nm³
O₂ present in excess air = 2.45*21/100 = 0.51 Nm³
N₂ present in actual air used = 14.21*79/100 =11.62 Nm³

So, total wet flue gas volume produced by combustion of 1m3 of refinery gas with 20 % excess air = 1.55 +0.51+11.62+2.06 +0.01 = 15.75 Nm³/Nm³ of refinery gas

Refinery gas burning rate = 5596 Nm³/hr
Total flue gas production rate = 5596*15.84 = 88640 Nm3/hr
Capacity of I.D.fan required = 88640 Nm³/hr ( at N.T.P I.E at 0^oc and 760 mm Hg)
Since , the flue gas leaves at 205^oc , the actual capacity of I.D. fan is calculated using charle's law-

V1/T1=V2/T2
88640/(273+0) =V2/(273+205)
V2 = 88640 *(273+205) /273 = 155201 Nm³ at 205 oc and 760 mmHg

Reference no: EM13943730

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