# Homework Help: Convection or Radiation

1. Jan 31, 2012

### moatasim23

A person is standing near heater .How is he getting heat?

2. Jan 31, 2012

### gmax137

Both. Plus, some conduction through the air.

3. Jan 31, 2012

### moatasim23

It came in my test as Multiple Choice Question and there was no option for ALL.So i expect it has some definite answer.Please help me sort it out.Bcz it is confusing me.Thanks

4. Jan 31, 2012

### LURCH

I think it depends on what kind of heater, but nearly all heaters have a fan.

5. Jan 31, 2012

Sounds like your teacher just doesn't know how to make test questions. Quite literally all three modes of heat transfer apply here. I suspect he/she was looking for convection though.

6. Jan 31, 2012

### Born2bwire

Yeah, it would in some way be both (I would think that conduction by air is mostly just another way of saying convection in this case since the heated air will flow from the heater to the person, very little actual conduction will be taking place. I think you would have to assume a more or less uniform room temperature for the heat transfer from the air to be conduction). But I would say that the dominant mode would depend upon the type of heater. Take a radiator, that is mainly convection but something like an electric heater can put out a large amount of radiative infrared heat.

7. Jan 31, 2012

Movement of air as a result of temperature differences is a result of buoyancy, so the forces are parallel to gravity. Sure, in this case the fluids are arranged so that the strata are unstable and you would end up with horizontal movement that way, but all that hot air would end up heading upward anyway. It is an oversimplification to say that the air will move from the heater to the person in general.

Also, with a uniform room temperature distribution, the heat transfer would be zero. You can't transfer heat without a temperature difference.

8. Feb 1, 2012

### Born2bwire

The movement of heated air in a room does not generally just float up, due to the confinement of the room it will disperse about the ceiling and so forth plus there are the existing airflows in a room brought about from the ventilation system and pressure differences. But if you have a uniformily heated environment, then there isn't going to be much flow of heat due to the air currents. So anyone stepping into the room and absorbing heat (or cooling off) will probably experience more conduction heat transfer.

Either way, this is meant to be a simplification because it appears to be high school level at most.

Last edited: Feb 1, 2012
9. Feb 1, 2012

### jetwaterluffy

Probably what it should have said was how is he mostly getting heat, then. And the answer would be convection. Radiation is slow.

10. Feb 1, 2012

And the simplification would be that it doesn't move from the heater to the person. Natural convection is a buoyant phenomenon. The lateral movement is due to other phenomenon, for example the Rayleigh-Taylor instability, so I would contend that the oversimplified version is that the air won't move laterally from hot to cold directly as a result of the heater. It only does so because of more advanced concepts.

11. Feb 4, 2012

### moatasim23

How could it be?I thought convection was slow bcz in it the air has first to be heated and then rise up.So that convection cycle starts and the room gets heated up.In this way he gets heat.While through radiation he will get heat directly.Isnt it so?

12. Feb 5, 2012

### jetwaterluffy

Not really. Look at how dimly glass or metal glows when heated to ridiculously hot temperatures. That is the only way the energy is getting out. Convection may be slow, but radiation is slower still.

13. Feb 5, 2012

### LawrenceC

Take a look at the arithematic expressions for radiation and convection boundary conditions. For radiation it is

q = sigma * epsilon * (T^4 - Tamb^4) where

T is source temperature in absolute degrees
Tamb is temperature of body being bombarded radiation, absolute
sigma is Stephan Boltzmann constant
epsilon is emissivity - assume about 0.9
q heat exhanged per unit area-time

For natural convection a common expression from heat transfer texts (Holeman) is:

q = c * (T - Tamb)^1.333
where c is a constant 0.19
T is source temperature
Tamb is ambient temperature
q heat exhanged per unit area-time

Plug in some numbers and see what you get.