Calculate the efficiency of a furnace’s heat transfer process

[Gregs6799]

Homework Statement

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Hi guys.
Been looking at this question for a while and can't figure it out.

An aluminium furnace is fuelled with propane gas. The furnace is loaded up with 4 aluminium ingots each of mass 22.5 kg.

The question is:
If 2.1 kg of propane is used to complete the melting process calculate the efficiency of the furnace’s heat transfer process.

Homework Equations

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Calorific value for propane 49.93 MJ/kg.

The Attempt at a Solution

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From an earlier question I worked out the following:

660 – 20 = 640
22.5 x 0.91 x 640 = 13,104 kJ
Melting point = 13,104 kJ
321 x 22.5 = 7,222.5
Boiling point = 7,222.5 kJ
Total energy = 13,104 + 7,222.5 = 20,326.5 kJ

So far I've played around with the values:

49.93 x 22.5 = 1,123.425 MJ or kg

49.93 x 2.1 = 104.853 MJ or kg

Do I have to replace the original melting point answer with one of the above answers?

I'm very new to physics so any help very much appreciated.

Thanks.

 

[Bystander]

Step at a time: how much heat do you need; how much heat is available?

 

[haruspex]

660 – 20 = 640

True, but what do these numbers represent?

22.5 x 0.91 x 640

What is the 0.91?

Melting point = 13,104 kJ

A melting point would be a temperature, not a quantity of energy.

321

What's this?

 

[Chestermiller]

It looks like you calculated the amount of heat required to melt the aluminum bars correctly, and your second calculation (involving the 2.1 kg) gives the heat available from the propane correctly. So, what is the efficiency?

 

[Gregs6799]

Haruspex,

660 = melting temperature of aluminium, minus 20 as the ingots are stored at that temperature
0.91 = Specific heat capacity for aluminium (cp):
321 = Latent heat of melting for aluminium (L): kJ/kg

 

[Gregs6799]

Step at a time: how much heat do you need; how much heat is available?

I need 13104 kJ to melt the ingots, or 20,326.5 kJ with the boiling point, and I have 104.853 MJ (104853 kj) from the propane, so I'm assuming that its very effective to achieve the objective?

Also apologies for confusing units of energy with temperature units.

 

[haruspex]

Haruspex,

660 = melting temperature of aluminium, minus 20 as the ingots are stored at that temperature
0.91 = Specific heat capacity for aluminium (cp):
321 = Latent heat of melting for aluminium (L): kJ/kg

Ok, so in

Melting point = 13,104 kJ

presumably you meant energy to reach melting point, and in

Boiling point = 7,222.5 kJ

you meant energy to melt the aluminium. (Boiling?!)
Don't forget there are four bars.

What is the logic behind this calculation:

49.93 x 22.5 = 1,123.425 MJ or kg

and how can it result in either MJ or kg?

 

[Gregs6799]

13,104 kj is the melting point
7,222.5 kj to reach boiling point, I assumed you needed both units for the correct answer not just the melting point.
So I'm using the 13,104 x 4 = 52,416 kj?
Calorific value for propane = 49.93 MJ/kg that's why I did 49.93 x 22.5 (weight of the bars) to get 1,123.425 kg, I put MJ as was unsure if the answer was going to be weight or energy.

As I said, very new to physics.
Appreciated.

 

[Chestermiller]

That's 49.93 MJ per kg of propane, not aluminum.

 

[haruspex]

7,222.5 kj to reach boiling point,

The aluminium melts. It does not get anywhere near boiling.