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 230oc from feed water at 50oc.
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 205oc.
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 230oc =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 0oc and 760 mm Hg)
Since , the flue gas leaves at 205oc , 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
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