Simple heat transfer calculation - actively heated aluminum block in ambient air

[ScottThomasCarter]

Hello all - I'm struggling to calculate the rate of heat loss from a heated aluminum block into ambient air (non forced air, just simple convection/radiation). The heat comes from electric cartridge heaters embedded in the block of aluminum. I need to confirm the power (Watts) required to maintain the desired temp of 160 °C of the aluminum block while exposed to ambient air at 25 °C. This will be a steady-state calculation once the desired temperature (160 C) has been achieved. Several assumptions are made for simplicity, as this exercise is simply to ensure we have adequate wattage on the cartridge heaters embedded in the block.

Here are the specifics:

mass of the aluminum block (m = 399 kg)

volume of the block (V = .147 mˆ3)

exposed surface area of the block (all faces) (A = 2.78 mˆ2)

block temp (160 °C)

ambient air temp (25 °C)

so ΔT = 135°C

specific heat capacity of aluminum [c = .91 kJ/(kg * °C)]

specific heat capacity of atmospheric air [c = 1.006 kJ/(kg * °C)]

thermal transmittance of aluminum [205 Watts/(mˆ2 * °C)]


Using Q=mcΔT = 49057 kJ (Q = energy required to satisfy ΔT) I've already calculated the time (t in seconds) required to heat the block with the installed power (P = 18 kW).

t=Q/P = 2725 seconds (.76 hours)



To maintain this temperature my hunch is to use Q=U*A*ΔT where Q = heat power lost in dissipation (watts), U = thermal transmittance of aluminum = 205 watts/mˆ2*°C and ΔT = 135°C, but this yields a power loss/requirement of 76 kW, which doesn't seem right.

I can't seem to figure out what I'm missing. I'm trying to keep it simple and make simplifying assumptions, but the numbers just don't seem to add up.

One other variable I left out but I am seeing mentioned often is the thickness of the material. A good working number will be .12 meter as the shortest distance from the center of the block to the surface and the longest is .6m (it's a rectangular block). These are half thickness values, as mentioned they are from the centroid of the block. I don't know where/how this value would come into play, but I think it does.

I looked at the Stefan-Boltzmann equation j=σTˆ4, and as I understand it it is expressed as joules per second per square meter ( = watts/mˆ2). Using the universally defined σ = 5.670373 x 10ˆ-8 W/(mˆ2*Tˆ4) I'm left with determining T. What is thermodynamic temperature T? Is that ΔT or the absolute temp in Kelvin?

It's confusing to me that this equation (Stefan-Boltzmann) doesn't appear to be material specific. There is no input to the equation which would distinguish between ceramic (for example) and copper. Would their j value be the same?

Apologies if this is too deep, but a crash course would be appreciated that is specific to this problem.

I've attached my simple calculation spreadsheet if it helps.


Any guidance would be appreciated.

 

[SwinnyGG]

Disregard the Stefan-Boltzmann equation - what you've listed there is actually a law which describes radiation from a black body. It's not applicable to what you're doing.

You're on the correct path with Q=U*A*ΔT- your problem is calculating U. Thermal conductivity of aluminum (your 205 W/mK) describes how much thermal energy will transmit through the aluminum itself- not how much will be transmitted between an aluminum surface and are moving past it.

Determining the U value for a situation like this is, in short, complicated. The wikipedia article actually breaks it down pretty effectively.

https://en.wikipedia.org/wiki/Heat_transfer_coefficient#Overall_heat_transfer_coefficient

 

[IRstuff]

You can brute-force it; assume that the coefficient is around 7 W/m^2-K. Plug and chug and get about 2.6 kW

As for Stefan-Boltzman, you use both temperatures,

Q = emissivity * area * σ * (T_hot^4 - T_air^4)

​

The fancier approach uses different emissivities for your block vs. its surroundings.


The effective heat transfer coefficient is [σ * (T_hot^4 - T_air^4)]/ΔT = ~2.9 W/m^2-K

The first principles natural convection coefficient is also around 3.2 W/m^2-K, so 7W/m^2-K is a bit high, but gives you margin, since there are few places that are truly natural convection.


TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

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[goutam_freelance]

Without going into too much details about your calculation I wish to point out the following:
1. The quantity 'U' in maintaining mold temperature should be convective heat transfer coefficient, not 'Thermal transmittance of aluminum ' as indicated by you.
2. Calculation of U is little complicated. But as rough estimate you can consider air heat transfer coefficient values indicated in engineeringtoolbox

https://www.engineeringtoolbox.com/convective-heat-transfer-d_430.html.

With small air velocity you get U~25 W/m^2-K. So power requirement will come down to about 9.3 kW. (I have not checked all of your calculations)



Engineers, think what we have done to the environment !

https://www.linkedin.com/in/goutam-das-59743b30/

 

[prex]

I don't want to dispute the engineering toolbox site, but it doesn't seem to state correct values (and doesn't mention the radiative contribution).
From long ago, I know that a ballpark figure for heat exchange coef in calm ambient air is 10 w/sqm K (including roughly 50% contribution from radiation), and that the same can go up to 20 in a moderate windy situation. If I recall it correctly, these figures come from Perry's Chemical Eng.Handbook.
As your block is quite large, the heat exchange on the vertical walls could go over the 10 w/sqm K figure, but if you take an overall 15, you'll be certainly on the safe side.
So your installed power is more than sufficient for maintaning the temperature.


prex
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