# How much heat will be transfered ,from heating element to surrounding?

1. Mar 15, 2013

### avinashj

I am using an heating element to heat the air in a closed container. Joule heating will give me the "heat generated at heating element." I want to know how much of it will be transfered to the surrounding air. Should I be using heat transfer due to all three modes ?

2. Mar 15, 2013

### Staff: Mentor

All of it.

3. Mar 16, 2013

### avinashj

thanks russ. Actually i want a heat transfer of 1.2 kJ from the heating element to the surrounding air (which is at rest) in given time of 30 seconds. So, in what direction should i proceed?

4. Mar 16, 2013

### Staff: Mentor

Do you have the heating element already? Do you know what it is rated for? Check if it matches the output you want (do you know how to convert 1.2 kJ in 30 sec into watts?).

5. Mar 16, 2013

### USAGeorge

V/O=A AxV=watts

1 watt=3.6 kj/h

6. Mar 16, 2013

### avinashj

No i don't have any specific heating element. I want to decide upon it from the mentioned condition(1.2 KJ in 30 sec). My concern being -- "Even if equivalent heat is generated at the heating element, what will be the rate of that heat transfer from heating element to the surrounding air?" .
Thank you.

7. Mar 17, 2013

### Staff: Mentor

Well, all of the heat generated will be transferred to the surrounding air, so what you need is to select a heating element with the correct heat dissipation rate/electrical energy usage rate.

8. Mar 17, 2013

### sophiecentaur

It isn't really clear what the actual question is. Eventually, all the power supplied to the box will turn up to heat the surrounding air (if this means the room). But probably the most important question is what temperature will the air in the box reach and that will depend upon the insulation of the box. What is the intended function of the box? It can only delay the heat transfer, so why have it there in the first place if you want to heat the room as fast as possible?
This all makes a massive difference to the answer. The cheapest way to heat a room would probably be to buy a proprietary heater - possibly the convection type.

9. Mar 17, 2013

### CWatters

What will happen is this....

ALL the power (eg calculated using Power = Volts * Amps) will go into heating the air in the box. This is because electrical heating elements are 100% efficient. It's true that some elements give out light at various frequencies but that will also be turned into heat when it hits the box walls. Even if you used a fan heater the energy used by the fan eventually ends up as heat. I suppose if it was a very noisy fan some energy would escape the box as noise but otherwise you can consider the element as 100% efficient. This is one reason why you shouldn't believe all you read in adverts for fancy electric rads that claim to be "more efficient" or "reduce home heating bills".

The air temperature inside the box will start to rise until the power going into the box (as calculated above) matches the losses through the box walls. At that point the temperature will no longer rise.

10. Mar 18, 2013

### avinashj

@sophiecentaur -- The actual question is , What will be the rate of heat transfer from the coil to air ? air being inside a close cylinder of approx 30 cms in height and 15 cm in dia.
I am interested in "rate of heat transfer" because I want the temperature of air to reach 600 in 30 secs from room temp.
P.S -- neglecting heat loss from cylinder.

Thank you.

11. Mar 18, 2013

### sophiecentaur

This is essentially a practical, rather than a theoretical problem, I think.
A 2.4kW heater will supply the right amount of energy in 30s but you wouldn't know that the temperature distribution would be like inside the enclosure. Are you sure that your 1.2kW figure includes the power needed to heat up the element itself and the walls of the cylinder in the required time? A 'typical' heater element takes several seconds to reach operating temperature, depending on what type it is, of course. How critical is the 30s time? Is there any reason why you couldn't use a fan to get a good temperature distribution? Relying on natural convection would make it much harder to predict what would happen. I would be inclined to use a bigger heater and control it with a thermostat.