Exercise-VI
A furnace melts 305 tons of copper and raises its temperature to 1600K in 6 hrs. The furnace consumes 4.2 tons/hr oil. Fuel oil analyses 85% C, 12% H and 3% O and its net calorific value is 9446 kcal/kg. Combustion air is 20% excess than theoretically required. Heat loss from the furnace to the surrounding is 50% of the heat that is required to melt copper and raise its temperature to 1600K. Calculate the inputs and outputs of energy and show them on the energy diagram
First we have to do the material balance to find out the amounts of POC and then energy carried by POC by using the sensible heat values.
Sensible heat in POC is calculated to be 2.54 x 107 kcal/hr
Sensible heat in copper is calculated to be 0.95 x 107 kcal/hr
Heat losses are 50% of the sensible heat in copper = 0.48 x 107 kcal/hr
Heat balance :
Heat input |
(kcal/hr) |
% of total |
Calorific value of fuel |
3.97 x 107 |
100 |
Heat output |
(kcal/hr) |
% of total |
Heat to POC |
2.54 x 107 |
64 |
|
Heat to copper |
0.95 x 107 |
24 |
Heat losses |
0.48 x 107 |
12 |
|
The above values are shown in the figure:
| Figure 39.2: |
Energy flow diagram |
Now consider the use of heat recovery device such as waste heat boiler. The POC enters the waste heat boiler and exit the boiler at 600K, Fifteen percent (15%) of the heat of POC is lost to the surrounding from the boiler. Show the flow of energy 
| Figure 39.3: |
Energy flow diagram showing the furnace and heat recovery system |
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