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

AI Thread Summary
The discussion revolves around calculating the heat required by fuel in a boiler operating at 82% efficiency, with specific temperatures for feed water and superheated steam. The initial calculations yield two different values for the heat required by fuel, leading to confusion. The correct approach involves understanding that the heat loss due to inefficiency cannot simply be added back to the energy produced at 82% efficiency. Instead, the total heat required should be derived from the energy output divided by the efficiency factor. Clarification is provided that multiplying by 1.18 incorrectly adds a portion of the efficiency value, leading to the discrepancy in results.
mt05
Messages
7
Reaction score
0
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/kgefficiency = Qsteam/Qfuel
.82 = 2628.7128/Qfuel

Qfuel = 3205.07 kj/kg3) I get two different answers for quantity of heat required by fuel. Can someone please explain why? Thank you.
 
Last edited:
Physics news on Phys.org
mt05 said:
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.


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.
 
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.
 
mt05 said:
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.

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.
 
ah yes that makes sense. Thanks for the help!
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top