Heat Dissipation

Something like this?

0.001 X Surface Area X Difference between oil and air temperature

If the oil temperature is 140 degrees, and the air temperature is 75 degrees, how much heat will a reservoir with 20 square feet of surface area dissipate?

Surface Area = 20 square feet
Temperature Difference = 140 degrees - 75 degrees = 65 degrees
0.001 X Surface Area X Temperature Difference = 0.001 X 20 X 65 = 1.3 horsepower

Note: 1 HP = 2,544 BTU per Hour

SolarEagle,

Thank you for your assistance.

Btu/hr divided by Latent Heat of Steam = lbs/hr.

In this age of software making things too easy, sometimes you have to resort to old fashioned thinking it thru.
Ain't that the truth... Soetime the software just gets you to the wrong answer quicker.
Now where's my pencil and squared pad?
Del

I think SolarEagle is uncharacteristically wrong in his explanation.

How much heat will be dissipated is quite complex but if we ignore the radiated component, assume the hydraulic oil is at a constant temperature and there is forced air flow over the surface, then the formula comes down to:

Q = ΔT/R, where R is the thermal resistance of the wall.

and R = x/Ak, where x = thickness, A = Area of surface and k = thermal conductivity.

I think that your correction is WRONG:

there are in the reservoir 3 heat exchange zones:

- oil to wall by convection

- through wall by conduction

- wall to air by convection again.

Thermal resistance of the first 2 are quite low in comparison with the last which also detriorates versus time due to dirt deposits on external wall surface.

Due to low temperature differences radiation which proportional to T^4 is very low in comparison with convection which is proportional to T^1.

If the wall is not in an induced air flow then only free convection occurs with coefficients around 8 to 15 W/m²°K. Now if the wall is steel with a thicknes of let say 6 mm and a condcutivity of 50 W/°Km you can compare the thermal resistances:

wall conduction t/(A*λ) = 6e-3/50=1.2e-4 °K/W and the one for convection 1/(α*A)= 0.067 to 0.125 °k/W. You understand that the difference is such that it can be considered that only outside convection is the criterion.