Heating Element-Wattage requirement

[PNA]

I am trying to calculate what wattage heating element should be used to heat air from 65 deg F to 190 deg F in 3 to 5 minutes. The volume of air is 20 cubic feet. It is totally enclosed and insulated. There is an internal circulating fan to "move" air around the chamber.

I have quickly calculated that i would need 23400 watts running for 5 minutes to achieve operating temp.

This is based appx 50% losses through internal chamber insulation etc.

Am i way off base here?

Thanks

Paul

 

[btrueblood]

You're off by a bit.

q = m * Cp * dT/dt (assuming no losses, high heat transfer rate, etc.)

dT = 190-65 = 125 F = 69.4 C
dt = 5 min. = 300 sec.
Cp for air = 1.0 kJ/kg-C
m = 20 ft^3 * .0283 m^3/ft^3 * 1.1 kg/m^3 = .62 kg

q = (.62)(1.0)(69.4)/300 = .14 kJ/sec or 144 Watts. But before I ordered an element, I'd also consider how long/how much heat it takes to warm up the real-world walls of a 20 ft^3 container.

 

[PNA]

thanks for the reply
i thought the cp for air was 1.009 kJ/kg Kelvin not C
We have a similar heating application and it took 18 minutes to heat 0.21 lbs of air from 65 to 190 using 1000 watt element

Paul

 

[JStephen]

A degree K is equal to a degree C.

 

[Montemayor]

A degree Kelvin is not equal to a degree Celcius. This is not possible. There is a magnitude of difference of 274.15 degrees between them.

I think what is being thought of is that a Unit difference in Kelvins is equal to the unit difference in Celcius.

 

[IRstuff]

There is no such thing as a "degree Kelvin," or "degree K." The SI unit for thermodynamic temperature is the kelvin, symbol K:

http://physics.nist.gov/cuu/Units/units.html

Per NIST, the units degree Celsius (ºC) and K have the same numerical magnitude, given t/ºC = T/K - 273.15

The degree Celsius is the only exception to the non-capitalization of the spelled-out unit.

TTFN

 

[MortenA]

LOL

 

[quark]

In IP units, the standard equation (i.e at standard conditions) for sensible heat rate is 1.08xcfmxdT, but generally considered as 1.1xcfmxdT.

If you are considering to heat in 3 minutes then,
1.08x(20/3)x125 = 900 btu/hr or 264W (or 1.1x(20/3)x125 = 269W)

For 5 minutes, it is 1.08x(20/5)x125 = 158W

The small increment in the value of the wattage calculated by btrueblood may be due to the change in density value at standard conditions. It should be 1.2kg/cu.mtr or 0.075lb/cu.ft.

 

[PNA]

I appreciate everyones responses!

So a 1500 watt heater running for 5 minutes would be required under ideal conditions?

Paul

 

[btrueblood]

quark: yup, I eye-balled the "mean" density, since it varies as the air heats up; he stated a volume, not a mass, of air in the OP, but it makes more sense to assume a fixed mass at room temp. (e.g. closing a sealed oven door) like you did.

PNA - yes, again under "ideal" conditions. Again, real conditions will require that your container walls reach nearly the endpoint temperature too, not just the air (there's a lot more surface area to the walls than the heating element), so the real wattage requirement will be much higher, as your experience shows.

 

[PNA]

Thankyou.

I think i have the whole thing figured out and put it to an excel worksheet. It has been confusing because of going to different eng websites and a whole lot of textbooks, everyone uses terms slightly different or incorrectly.

I figured that it requires about 4 times the wattage requirement to heat the interior surface area and quite a bit more for the product to come up to temp.

The hard thing is to explain the numbers and why it is impractical cost wise and design wise to speed up a process too much.

Regards

Paul

 

[btrueblood]

PNA -

use the same formula to calculate the time and wattage for the walls and product.