Flat Horizontal Plate 2

[meca]

I have a flat horizontal stainless steel plate (10 ft x 10 ft) that is 1,000 Deg. F. I am interested in calculating the temperature at a point 10 ft directly below the center of this plate. The plate is in 70 Deg. F air, and there is no forced convection.

I have found some information in Heat Transfer books that allow me to estimate a heat transfer coeffient for natural convection on a horizontal plate facing down. However, I am unclear as to how I use this heat transfer coefficient to determine the temperature at a given point. Can anyone point me in the right direction?

Thanks,

Chris

 

[gunsmith]

I´m not sure that is the right way to go. You could try the following for an approximation. The quantity of heat transmitted between the plate and the air without radiation is:

Q = alpha*A*dt

Q = quantity of heat (W)
alpha = heat transfer coefficient (W/m^2 K)
calculate the alpha for unforced convection
A = area of the plate (m^2)
dt = difference of temperature between plate and air (K)

Putting in this Q in another formula which gives the heat transfer through a wall by a heat conductance coefficient and let the material of this wall be our surrounding air:

Q = lambda*A*dt/delta

lambda = heat conductance coefficient ((W/m K)

lambda for air is 0.023 at 273 K and 0.038 at 473 K and changes further on with growing temperature

A = intersection (area of our plate) (m^2)
dt = difference of temperature (K)
delta = thickness of the wall (in our case distance to plate) (m)

Converting the formula to dt we have:

dt = Q*delta/lambda/A

Knowing that the temperature we are looking for is lower than the temperature of the plate and predicting that the air in contact with plate has nearly the same temperature than the plate, we get the searched temperature from

t = tp - dt

tp = temperature of the plate (K)

So we have a solution - but I have my doubts that this is correct according to the laws of physics and engineering.

Andreas

 

[ramblin]

A few comments,

1) any object you put in the room, including the walls, will be subject to radiation heat transfer from the plate. in this case (1000 F), the radiation would probably be an overriding or, at the very least, a significant factor.

2) if you're just looking for air temp and neglecting radiation, a big factor would be the rate at which heat can be transferred from the air to the walls and out of the room (if the room were adiabatic and the surface maintained at 1000 F, eventually the air would also be 1000F). if you can come up with a heat transfer coefficient and area for heat losses than it's probably not such a bad assumption to assume that the room would achieve a constant temp. This temp would act as a sink for the plate heat transfer and a source for the heat loss through the wall. Of course the second source of heat to the wall would be radiation from the plate.

3) the horizontal plate would create some buoyancy driven cells that would tend to distribute hot air at the top of the room and cooler air at the bottom but like i said it's probably not far off to assume that the total air delta temp is small

4) a cfd solution could work well

 

[gunsmith]

Forget the second part of my previous post. It is completely nonsense. The result is -210000 °C as temperature difference. Only the heat calculated from the first formula seems to be right. The result is 16.46 kW.

Does anybody know how to kill my previous post in this forum. Would be a nice help.

Andreas

 

[IRstuff]

If you red flag your post, the moderator will delete as necessary

TTFN

 

[IRstuff]

Some thing that is missing in the problem description is the distance to the walls, their thickness and/or the charateristics of the final heat sink. Are the walls held at 70F?

Is the plate held at 1000F? If so, then the steady state heat flow can be calculated using the natural convection coupled with the blackbody emission. If not, then we're talking a quite complex transient problem.


A 1000F blackbody emits 0.475 W/cm^2, while a 70F blackbody emits 0.043 W/cm^2.

TTFN

 

[gunsmith]

To IRstuff:

Are You sure of the 0.475 W/cm^2. I got 24.94 kW/m^2 - that is nearly 5 times higher.

Radiation coefficient of the black body is 5.77 W/m^2/K^4 and our temperature is 810.8 K (=1000 F). Introducing that in the Boltzmann law gives

E = C*(T/100)^4

E = energy emitted (W/m^2)
C = radiation coefficient of black body (W/m^2/K^4)
T = temperature (K)

E = 5.77*(810.8/100)^4 = 24936 W/m^2

using the coefficient for gray body of 4.25 W7m^2/K^4 the result is 18367 W/m^2. Or inserting 1.7 W/m^2/K^4 (radiation coefficient of polished steel) gives 7346 W/m^2. Correct me, if the mistake is in my calculation. I did not deal with this matter since 10 years.

Andreas

 

[meca]

Thanks to all for the help. Based upon your posts, I realize the radiation is a more significant component than I realized.

The plate is actually in an elevated structure that is open (no walls); However, there are pipes and other reflective objects around. For simplicity, I thought I could get an order of magnitude estimate of the temperature by ignoring the cooling due to wind and the increased radiation due to reflection from other objects.

 

[IRstuff]

Sorry, forgot to add back 273 Kelvin, so it is 810.8 K, which would be significantly higher emission.

TTFN