Thermal Conductivity (k) question, time dimension
Thermal Conductivity (k) question, time dimension
(OP)
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!
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!





RE: Thermal Conductivity (k) question, time dimension
Watts = Joules/second
TTFN
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RE: Thermal Conductivity (k) question, time dimension
Tobalcane
"If you avoid failure, you also avoid success."
RE: Thermal Conductivity (k) question, time dimension
RE: Thermal Conductivity (k) question, time dimension
Tobalcane
"If you avoid failure, you also avoid success."
RE: Thermal Conductivity (k) question, time dimension
RE: Thermal Conductivity (k) question, time dimension
RE: Thermal Conductivity (k) question, time dimension
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.
RE: Thermal Conductivity (k) question, time dimension
RE: Thermal Conductivity (k) question, time dimension
Plastics Industry
RE: Thermal Conductivity (k) question, time dimension
Tobalcane
"If you avoid failure, you also avoid success."
RE: Thermal Conductivity (k) question, time dimension
Tobalcane
"If you avoid failure, you also avoid success."
RE: Thermal Conductivity (k) question, time dimension
Plastics Industry
RE: Thermal Conductivity (k) question, time dimension
Tobalcane
"If you avoid failure, you also avoid success."
RE: Thermal Conductivity (k) question, time dimension
> 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.
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RE: Thermal Conductivity (k) question, time dimension
I guess I need to estimate the temperature of the air in contact with the vessel wall on the outside. Maybe 150F.
Plastics Industry
RE: Thermal Conductivity (k) question, time dimension
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RE: Thermal Conductivity (k) question, time dimension
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