# Calculate the efficiency of a furnace’s heat transfer process

• Gregs6799
In summary: It will boil only if you get it up to 2519°C.I see thanks, I thought you needed both units for the correct answer, so just the melting point then?In summary, the question is asking for the efficiency of a furnace that is using propane gas as fuel to melt 4 aluminum ingots with a total mass of 22.5 kg. The calorific value for propane is given as 49.93 MJ/kg. Based on earlier calculations, the amount of heat required to melt the aluminum ingots is 13,104 kJ, and the amount of heat available from the propane is 104.853 MJ. The efficiency of the furnace's heat transfer process can be calculated by dividing the heat available by

## 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.

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

Gregs6799 said:
660 – 20 = 640
True, but what do these numbers represent?
Gregs6799 said:
22.5 x 0.91 x 640
What is the 0.91?
Gregs6799 said:
Melting point = 13,104 kJ
A melting point would be a temperature, not a quantity of energy.
Gregs6799 said:
321
What's this?

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?

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

Bystander said:
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.

Gregs6799 said:
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
Gregs6799 said:
Melting point = 13,104 kJ
presumably you meant energy to reach melting point, and in
Gregs6799 said:
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:
Gregs6799 said:
49.93 x 22.5 = 1,123.425 MJ or kg
and how can it result in either MJ or kg?

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.

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

Gregs6799 said:
7,222.5 kj to reach boiling point,
The aluminium melts. It does not get anywhere near boiling.

## 1. What is the formula for calculating furnace efficiency?

The formula for calculating furnace efficiency is the ratio of the useful heat output to the total heat input. This can be expressed as efficiency = (useful heat output / total heat input) x 100%.

## 2. How do you determine the useful heat output of a furnace?

The useful heat output of a furnace can be determined by measuring the amount of heat that is transferred to the desired area or object. This can be done using a heat flow meter or by measuring the temperature difference between the inlet and outlet of the furnace.

## 3. What factors affect the efficiency of a furnace's heat transfer process?

The efficiency of a furnace's heat transfer process can be affected by factors such as the design and size of the furnace, the type and quality of insulation, the type of fuel used, and the maintenance and operation of the furnace.

## 4. How can you improve the efficiency of a furnace's heat transfer process?

To improve the efficiency of a furnace's heat transfer process, you can make sure the furnace is properly sized for the space it is heating, use high-quality insulation, regularly clean and maintain the furnace, and use high-efficiency fuels or alternative sources of energy.

## 5. What is a good efficiency rating for a furnace?

A good efficiency rating for a furnace depends on the type of furnace and its intended use. Generally, a furnace with an efficiency rating of 90% or higher is considered to be highly efficient. However, it is important to consider other factors such as cost, maintenance, and environmental impact when determining the best efficiency rating for a furnace.