Part - a using the following data draw up a

Question

PROBLEM ON HEAT BALANCE OF SOAKING PIT NO -15
PART - A . Using the following data, draw up a heat balance for a steel ingot soaking pit for a twenty hours test period during which 130 tons of cold steel ingots were heated from 20 oc to 1220^oc.

Data -
Mean temperature of -
Air after recuperator = 620 ^oc
Blast furnace gas ( fuel) after recuperator = 450 ^oc
Flue gases after soaking pit =1050 ^oc
Flue gases after air recuperator = 770 ^oc
Flue gases after B.F. gas recuperators = 530 ^oc
Average B.F . gas consumption = 2805 m³/hr
Air /fuel gas ratio = 0.8
Flue gas /fuel gas ratio = 1.65
Net calorific value of B.F . gas =860 kcal /Nm³
Mean specific heats ( kcal/Nm3 . ^oc) are as follows-
Air - 0.301
B .F. gas - 0.310
Flue gas - 0.358
Steel - 0.180 kcal /kg .^oc

( all volumes are measured at 1 atm , 20 ^oc )-
Scale formation = 1.5% ( by weight)
Scale formation reaction is -
3Fe +2O₂ = Fe₃O₄ , ? H₂O ^oc = -- 266841 kcal
Atomic weight of iron = 56 , structural loss = 270 kw

Part ( b) - indicate the ways in which overall thermal efficiency of the soaking pit could be improved. Also , calculate the thermal efficiency of both the air and gas recuperates using the above data.

SOLUTION - datum ( reference temperature) = 20^oc

Part ( a) - volume of B.F . gas burnt in 20 hrs = 20 *2805 = 56100 m³
Heat input items-
Hence , heat of combustion of B.F. gas = volume * calorific value of B.F. Gas = 56100*860 =48246000 kcal
Heat picked by air in preheater = 56100*0.8*0.301* ( 620 -20 ) = 8105328 kcal
Heat picked by B.F. gas in gas preheater =56100 *0.310 *( 450 -20) = 7478130 kcal
Heat produced due to oxide of steel = 130000 *0.015 *266841/3*56 = 3097261 kcal
( since , 3Fe , i.e. , 3*56 kg iron oxidation produces 266841 kcal of heat)
Hence, standard heat of iron oxidation = 266841/3*56 kcal /kg
So, total heat input = 48246000 +8105328+ 7478130 +3097261 = 66926719 kcal

Heat output items-

Heat to steel = 130000*0.180* (1220-20 ) =28080000 kcal
Heat supplied to air recuperator through flue gas = 56100 *1.65*0.358 * ( 1050-770) =9278715 kcal
Heat supplied to B.F. gas recuperator through flue gas = 56100 *1.65*0.358*(770-530) = 7953185 kcal
Heat out from chimney through flue gas = 56100 *1.65* 0.358*( 530 -20) = 16900517 kcal
Structural heat loss in 20 hours = 270*20*860 = 4644000 kcal

( since 1 kwh= 860 kcal)

Hence , unaccounted ( by difference ) =70302 kcal

So total heat output = 66926719 kcal
Part - b the thermal; efficiency of soaking pit can be improved by-
- Limiting the oxidation of steel.
- Reducing the heating time by hotter ingot charging.
- Using continuous casting to avoid ingot manufacture.
- Further recovery of heat from waste gas , e.g. by a waste heat boiler to generate steam .
- Reducing the volume of waste gases or by obtaining better heat recovery from the recuperators since
- Thermal efficiency of air recuperator =( heat picked up by air in recuperator/heat supplied to air recuperator ) *100 = 8105328/9978715*100 =87.9%
And thermal efficiency of gas recuperator =( heat picked up by gas to preheater/ heat supplied to B.F. gas recuperator )*100 = 7478130/7953185*100 =94.1 %