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Bolter
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Sorry I'm a bit unsure to which equation I need to use for radiant heat transfer? Is it Stefan's law?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.
Yes, but don’t forget to include the amount radiated back from the room, to get the net rate of heat transfer.Bolter said:Sorry I'm a bit unsure to which equation I need to use for radiant heat transfer? Is it Stefan's law?
This was all I was able to think of but I don't know how to proceed further from this?Chestermiller said:Yes, but don’t forget to include the amount radiated back from the room, to get the net rate of heat transfer.
So 1772.4 watts would be an acceptable ans?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.).
I believe so. But @Chestermiller is an expert in thermodynamics. So, I don't feel comfortable without his blessingBolter said:So 1772.4 watts would be an acceptable ans?
I think my instructor is possibly looking for this answer as I haven't yet covered heat loss through radiation, convection etc.TSny said:I believe so. But @Chestermiller is an expert in thermodynamics. So, I don't feel comfortable without his blessing
Also, if your instructor cares about significant figures, then you might want to round off your answer to an appropriate number of significant figures.
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.Bolter said:I'll just keep it into a minimum of 3 sig figs so 1770 watts
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.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.).
I totally disagree. See my previous post.Bolter said:So 1772.4 watts would be an acceptable ans?
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.Bolter said:This was all I was able to think of but I don't know how to proceed further from this?
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Also how I would I work out the surface area (i.e. surface area of sphere) if I am not given a radius?
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.Chestermiller said:Maybe I am mistaken, but I was assuming that the radiator is already operating at temperature...
Oops. I misinterpreted the question. Yes, you are of course correct.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.
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.
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.
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.
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.
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.