# Calculating heat req. by fuel in boiler, interpolation needed.

1. Nov 25, 2011

### mt05

1) Feed water enters a boiler at 172.52 degrees Celsius. The boiler produces super heated steam at 5600kpa and 472 degrees Celsius. The boiler efficiency is 82%. Find the Heat required by fuel.

2)
I interpolated the value given for feed water and super heated steam.

h@5600kpa 472 degrees Celsius = 3359.0128 kj/kg
hf@ 172.52 degrees Celsius = 730.30 kj/kg

So...

delta h = 3359.0128 kj/kg - 730.30kj/kg
delta h = 2628.7128 kj/kg (this is the energy that the boiler would produce running at 100% to make the given steam right?)

So...

Qfuel = (2628.7128 kj/kg)(1.18) (I'm multiplying by 1.18 to make up for the 18% efficiency difference)
Qfuel = 3101.88 kj/kg

efficiency = Qsteam/Qfuel
.82 = 2628.7128/Qfuel

Qfuel = 3205.07 kj/kg

3) I get two different answers for quantity of heat required by fuel. Can someone please explain why? Thank you.

Last edited: Nov 25, 2011
2. Nov 25, 2011

### rock.freak667

The temperatures that you are given are based on the 82% efficiency.

82% will give dh as 2628.7128 kJ/kg
100% will give 2628.718/0.82

which is the same as your last method.

3. Nov 25, 2011

### mt05

hmm sorry, I still am not following

dh = 2628.718 (this is the amount of heat made by the boiler to make 5600kpa and 472 degrees Celsius of steam) boiler running at a 82% efficiency.

18% of heat from the fuel is lost correct?

So why couldn't I take 2628.718 and multiply that by 1.18 to make up the missing 18% of heat lost by the fuel thus giving me the total heat of the fuel required.

I just don't understand why I can't do this.

4. Nov 26, 2011

### rock.freak667

Because what you'd be doing is this 2628.718(1)+2628.718(0.18), you'd be adding the value at the 82% efficiency to 18% of the 82% efficiency value.Which is incorrect.

5. Nov 26, 2011

### mt05

ah yes that makes sense. Thanks for the help!