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Commentator

Join Date: Apr 2005
Location: Phoenix, AZ
Posts: 61

### Heat Dissipation

06/24/2012 11:13 PM

In this age of software making things too easy, sometimes you have to resort to old fashioned thinking it thru. I forgot a simple formula to find heat dissipation on the surface area of a hydraulic reservoir. I sure could use a little help!!

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Guru

Join Date: Mar 2007
Location: at the beach in Florida
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#1

### Re: Heat dissipation

06/24/2012 11:37 PM

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

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Commentator

Join Date: Apr 2005
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#2

### Re: Heat dissipation

06/24/2012 11:47 PM

SolarEagle,

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Join Date: Dec 2006
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#3

### Re: Heat dissipation

06/25/2012 3:56 AM

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

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Guru

Join Date: May 2007
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#4

### Re: Heat dissipation

06/25/2012 4:19 AM

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

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Join Date: Dec 2006
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#5

### Re: Heat dissipation

06/25/2012 4:13 PM

Hey Del, long time no see you purrrrrrrrrrrrrring.

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Guru

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#6

### Re: Heat Dissipation

06/25/2012 10:36 PM

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.

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#7

### Re: Heat Dissipation

06/29/2012 12:30 PM

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.