Working out the rate of heat loss from a radiator

In summary, the conversation is discussing the correct equation to use for calculating the rate of radiant heat transfer, which is determined to be Stefan's law. It is also mentioned that the net rate of heat transfer should include the amount radiated back from the room. The participants also discuss the acceptable amount of significant figures to use in the answer, with one suggesting 2000 watts and another suggesting 1770 watts. There is also a disagreement about whether the radiator is already at a steady average temperature or not. It is concluded that the rate of heat transferred to the room is equal to the rate at which heat is transferred to the radiator from the circulating water. The conversation ends with a question about how to calculate the surface area of the radiator without
  • #1
Bolter
262
31
Homework Statement
Working out rate of heat loss from radiator
Relevant Equations
Power = energy/time
Here is the Q below

Screenshot 2019-12-06 at 12.32.23.png

This is what I have tried so far. I ended up getting a value of 1772.4 watts

Is this the correct way of doing this problem?
 

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  • #2
Not even close. What is the equation for the rate of radiant heat transfer? There is also going to be natural connective heat transfer from the radiator.
 
  • #3
Chestermiller said:
Not even close. What is the equation for the rate of radiant heat transfer? There is also going to be natural connective heat transfer from the radiator.
Sorry I'm a bit unsure to which equation I need to use for radiant heat transfer? Is it Stefan's law?
 
  • #4
Bolter said:
Sorry I'm a bit unsure to which equation I need to use for radiant heat transfer? Is it Stefan's law?
Yes, but don’t forget to include the amount radiated back from the room, to get the net rate of heat transfer.
 
  • #5
Chestermiller said:
Yes, but don’t forget to include the amount radiated back from the room, to get the net rate of heat transfer.
This was all I was able to think of but I don't know how to proceed further from this?

IMG_3422.JPG

Also how I would I work out the surface area (i.e. surface area of sphere) if I am not given a radius?
 
  • #6
It seems to me that @Bolter has the correct answer. For steady-state conditions, the rate of heat loss by the radiator must equal the rate that heat is transferred to the radiator by the circulating water. It is not necessary to worry about the various ways in which the radiator loses heat (such as radiation, convection, etc.).
 
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  • #7
TSny said:
It seems to me that @Bolter has the correct answer. For steady-state conditions, the rate of heat loss by the radiator must equal the rate that heat is transferred to the radiator by the circulating water. It is not necessary to worry about the various ways in which the radiator loses heat (such as radiation, convections, etc.).
So 1772.4 watts would be an acceptable ans?
 
  • #8
Bolter said:
So 1772.4 watts would be an acceptable ans?
I believe so. But @Chestermiller is an expert in thermodynamics. So, I don't feel comfortable without his blessing :oldsmile:

Also, if your instructor cares about significant figures, then you might want to round off your answer to an appropriate number of significant figures.
 
  • #9
TSny said:
I believe so. But @Chestermiller is an expert in thermodynamics. So, I don't feel comfortable without his blessing :oldsmile:

Also, if your instructor cares about significant figures, then you might want to round off your answer to an appropriate number of significant figures.
I think my instructor is possibly looking for this answer as I haven't yet covered heat loss through radiation, convection etc.

I'll just keep it into a minimum of 3 sig figs so 1770 watts
 
  • #10
Bolter said:
I'll just keep it into a minimum of 3 sig figs so 1770 watts
OK. The volume flow rate is only given with one significant figure. If you use the rough rules of thumb for significant figures that are given in many introductory courses, then the answer should have only one significant figure: 2000 W. But, I suspect that your answer of 1770 W is fine.
 
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  • #11
TSny said:
It seems to me that @Bolter has the correct answer. For steady-state conditions, the rate of heat loss by the radiator must equal the rate that heat is transferred to the radiator by the circulating water. It is not necessary to worry about the various ways in which the radiator loses heat (such as radiation, convection, etc.).
Maybe I am mistaken, but I was assuming that the radiator is already operating at temperature, and he was not looking at the amount of heat necessary to bring the radiator up to temperature (and certainly not in 1 second). He was interested in the rate of heat transfer to the room.
 
  • #12
Bolter said:
So 1772.4 watts would be an acceptable ans?
I totally disagree. See my previous post.
 
  • #13
Bolter said:
This was all I was able to think of but I don't know how to proceed further from this?

View attachment 253758
Also how I would I work out the surface area (i.e. surface area of sphere) if I am not given a radius?
Yes, this is correct, except that you should be using absolute temperature rather than centigrade temperatures. You get the area just by looking at the surface geometry of the radiator. Just estimate it roughly from what you see.
 
  • #14
Chestermiller said:
Maybe I am mistaken, but I was assuming that the radiator is already operating at temperature...
Yes, the radiator has already reached a steady average temperature. So, the internal energy of the radiator itself is not changing. Thus the rate at which heat is transferred to the radiator from the water equals the rate at which the radiator transfers heat to the room.
 
  • #15
TSny said:
Yes, the radiator has already reached a steady average temperature. So, the internal energy of the radiator itself is not changing. Thus the rate at which heat is transferred to the radiator from the water equals the rate at which the radiator transfers heat to the room.
Oops. I misinterpreted the question. Yes, you are of course correct.
 
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Related to Working out the rate of heat loss from a radiator

1. How is the rate of heat loss from a radiator calculated?

The rate of heat loss from a radiator is calculated using the formula Q = U x A x ΔT, where Q is the rate of heat loss in watts, U is the overall heat transfer coefficient, A is the surface area of the radiator, and ΔT is the temperature difference between the radiator and its surroundings.

2. What factors affect the rate of heat loss from a radiator?

The rate of heat loss from a radiator is affected by several factors including the surface area of the radiator, the material it is made of, the temperature difference between the radiator and its surroundings, and the presence of insulation or other barriers that may affect heat transfer.

3. How does the material of the radiator impact the rate of heat loss?

The material of the radiator can significantly impact the rate of heat loss due to differences in thermal conductivity. Materials with higher thermal conductivity, such as metals, will transfer heat more efficiently and result in a higher rate of heat loss compared to materials with lower thermal conductivity.

4. Does the size of the radiator affect the rate of heat loss?

Yes, the size of the radiator, specifically the surface area, directly affects the rate of heat loss. A larger surface area means more heat can be transferred to the surrounding air, resulting in a higher rate of heat loss.

5. How does insulation affect the rate of heat loss from a radiator?

Insulation can significantly reduce the rate of heat loss from a radiator by acting as a barrier to heat transfer. This is because insulation materials have low thermal conductivity, meaning they do not transfer heat easily. Therefore, a radiator with insulation will have a lower rate of heat loss compared to one without insulation.

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