Thermal Conductivity (k) question, time dimension

[Plasmech]

Sorry if this seems like a "stupid" question, but let's say I want to figure out how much energy is lost through a .05m thick aluminum plate, 1 square meter surface area, with a 100 degree K delta T. To find the energy transmitted:

Q=dQ/dt = k*A*dT/x = 237 w/(m*K) * (100K/.05m) = 474,000 watts.

OK...where does the dt (time) come into play? Is that 474,000 watts per *hour*? Here's where I got the k from:

http://en.wikipedia.org/wiki/Thermal_conductivity

Thanks for any help!

 

[IRstuff]

You're misnaming units. You say energy when you calculate energy rate

Watts = Joules/second

TTFN

FAQ731-376

 

[Twoballcane]

watts is really Jouls/sec which is BTU/HR and Cal/sec

Tobalcane
"If you avoid failure, you also avoid success."

 

[Plasmech]

Ah...well that would explain that now wouldn't it! Thanks guys.

 

[Twoballcane]

darn SI units! Messes up everythin...:+)

Tobalcane
"If you avoid failure, you also avoid success."

 

[Plasmech]

How would I convert Joules/second to horsepower? 1 watt = 756 HP, but in my calcs it seems orders of magnitude off...

 

[TGS4]

You've got the conversion backwards - 1hp=745.7W. That woudl explain your OOM issue...

 

[25362]

Regretfully, there are several definitions of HP. From the CRC Handbook (77 Ed):

1 HP W = J/s
metric 735.499
550 ft.lb/s 745.6999
electric 746.0
water 746.043
UK 745.7

and this is keeping out 1 boiler HP.

 

[ivymike]

sounds like 2 definitions to me.

 

[Plasmech]

So I figured that based on my equation above, I'm conducting [474,000 watts / 756] = 626 *HORSEPOWER* worth of energy out of my system? Whoa, something is obviously wrong here...

Plastics Industry

 

[Twoballcane]

well ... yes, 474Kw is a lot of power. What do you have that is generating or using that much power? Usally when you do converstin like this your talking about pumps or electric motors.

Tobalcane
"If you avoid failure, you also avoid success."

 

[Twoballcane]

Now that Im thinking about it, why are you solving heat transfer problems when it sounds like your application is motors?

Tobalcane
"If you avoid failure, you also avoid success."

 

[Plasmech]

I'm only converting to horspower because that is something easy to relate to. Not that watts aren't, as in 100 watt light bulbs.

Plastics Industry

 

[Twoballcane]

Hmmm...ok then why do you think 626 hp is obviously wrong?

Tobalcane
"If you avoid failure, you also avoid success."

 

[IRstuff]

some comments:
> You need to be more careful with units. Your calculated answer is 474 kW/m^2, not 474 kW. That means that the answer must be multiplied by the appropriate area. If you have a larger or smaller area than 1 m^2, the temperature rise will be different.

> Your answer must be caveated by the fact that the contact thermal resistance to the aluminum might be quite high, relative to the aluminum itself. This would detract from the calculated heat flux capability.

> It's generally difficult to establish a fixed cold-side temperature. Unless you're running chilled water at a pretty drastic flow on the other side, the temperature of the cold side will not stay put. If there's air on the cold side, unless you've got a gigantic flow, you'll be hard pressed to maintain the heat flow. Even forced air convection would limit you to well less than 50 W/m^2/K, so you'd need a delta-T of 10,000 K to match your calculated heat flow

> Your comments suggest a slightly misaligned view of the equation. The equation basically says that IF you FLOW 474 kW/m^2 through that chunk of aluminum, you'l GET a 100-K temperature rise from the cold side. Note that this is a STEADY-STATE solution, meaning that the temperature rises exists, ONLY IF you maintain that heat flow. If the heat flow drops to 47 kW, you'll only get a 10-K rise. If your application cannot sustain that kind of heat output, then that temperature rise cannot exist. Generally speaking, getting no more than about 5 K or 20 K across an aluminum structure is more the norm.

TTFN

FAQ731-376

 

[Plasmech]

Thanks IR. Actually, in my equation above, I did in fact take ito account the 1.6 sq. meters, I just forgot to type it on the actual forum post.

I guess I need to estimate the temperature of the air in contact with the vessel wall on the outside. Maybe 150F.

Plastics Industry

 

[IRstuff]

Basically, heat transfer is often viewed as a series resistance problem, with process-wall, wall-heatsink, heatsink-air, etc., thermal resistances. Aluminum, being highly thermally conductive, has a very LOW thermal resistance, so the overall thermal resistance is rarely affected by the bulk thermal resistance of the aluminum. Most good designs wind up being limited by the last step of the thermal resistance, that of the convection to air, because that's the hardest one to overcome.

TTFN

FAQ731-376

 

[Plasmech]

IR: could you look at my vacuum resistivity quesion? Thanks!

Plastics Industry