# band / cylinder style heating element wattage calculations

The question concerns band or cylinder heating elements, which are designed to clamp around a hollow cylinder and heat whatever is flowing through it.

I'm looking for information on how to calculate the wattage required to heat room-temperature air (let's say 75F) to 475F if the air is flowing at 100CFM through a 4" diameter tube which is 10" in length. The voltage is 120 but 240 is a possibility as a plan B.

• I answered a similar question to this here: electronics.stackexchange.com/questions/321802/…
– MCG
Aug 9, 2017 at 13:08
• That should help get you started. Finding exact equations is difficult due to all the variables involved
– MCG
Aug 9, 2017 at 13:09

Start by calculating the power required just to heat the air. Power is energy per time. A useful unit of power is Watts, which is one Joule per second. Let's therefore figure out how many Joules it takes to heat one second's worth of air.

To do these calculations more easily, we convert to more sensible units. 75 °F to 475 °F is 400 °F rise, which is a 222 °C rise.

Next we need to find the mass of air that is being heated. You say the flow rate is 100 CFM, but didn't specify if this is before or after the heating. I'll pick before the heating. 100 cubic feet = 2832 liters. 2832 l/minute = 47.2 l/s.

We need to convert that to a mass flow rate. 75 °F is 297 °K. A mole of ideal gas has a volume of 22.4 l at 273 °K, so 24.4 l at your intake temperature. Air is basically N2 with a molecular weight of 28, and O2 with a molecular weight of 32. Since N2 dominates, let's say air averages to a molecular weight of 29. That means a mole of "air" has a mass of 29 grams. The 47.2 l we are trying to raise the temperature of therefore has a mass of 56 g.

So now we have reduced the problem to finding the energy it takes to heat 56 g of air by 222 °C. Heat capacity of "air" is a squishy value since air isn't a constant. By digging around on the net, it seems 1.03 J/(g °C) is about right for this case. (222 °C)(56 g)(1.03 J)/(g °C) = 12.8 kJ. That's the energy required each second, so the power is 12.8 kW. With the various approximations and slop factors, three digits implies too much accuracy, so 13 kW.

The above is only the power going into heating the air, and doesn't account for any losses. If you insulate the outside of the pipe and heater well, let's say you need 15 kW. At 120 V, that would require 125 A. That's rather unwieldy, so I'd go to plan B and use 240 V, which requires 63 A. That's still a lot. At 15 kW, you should at least look into using 3-phase power directly.

The wattage is easy. First convert to SI units to make life simpler, then it's simply heat capacity (J/k/kg) * temperature increase (K) * mass flow rate (kg/s) - giving J/s, namely, Watts.

The heat transfer, however, is not so simple. You have a very small contact area (the wall of a 10" cylinder) and poor contact between the gas and the wall - some texts would use the tube diameter, 4", as the characteristic length. I suspect you're going to have to massively rethink the heat exchanger to increase the contact area and reduce the characteristic length (distance the gas is away from the wall). For example, many small diameter tubes (shell and tube HX) or given it's electric, directly heated wire matrix inside the tube.

Which is a much bigger topic, and not EE.