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hydraulic cooling
- Thread starter chatoonuk
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Tain't just the temp diff. To maintain a temperature the Heat input energy = heat output energy. Then you need a chiller capable of shedding that much energy to the atmosphere or river while maintaining the cool water temp.
It starts with the specific heat of water, the flow rate, the water temp rise. It can be confounded by pressure drops VS the effective removal of the boundary layer film coefficient. A narrow fast moving flow of water is best.
It starts with the specific heat of water, the flow rate, the water temp rise. It can be confounded by pressure drops VS the effective removal of the boundary layer film coefficient. A narrow fast moving flow of water is best.
Montemayor
Chemical
chatoonuk:
Your question isn't clear as to what you want to calculate, but I'll make some assumptions. I assume you want to calculate the flowrate of the cooling water used to cool a stream of hot hydraulic oil. If so, the formula for the heat transfered should be resolved first:
Q = W Cp (T2 - T1)
Q = heat transfered, Btu/hr
W = oil mass flow rate, lb/hr
Cp = Specific Heat of the hydraulic oil, Btu/lb-oF
T2 = hot oil inlet temperature, oF
T1 = product cool oil temperature, oF
With the heat load calculated, you then resolve the same equation for the water flow rate:
W = Q/Cp (t2 - t1)
W = water mass flow rate, lb/hr
Cp = Specific Heat for water (1.0 Btu/lb-oF)
t2 = product hot water temperature out of cooler, oF
t1 = inlet cool water temperature to cooler, oF
You don't need any figure. The average Specific heats for hydraulic oil that I use are:
at 100 oF = 0.450 Btu/lb-oF
at 200 oF = 0.500 Btu/lb-oF
You don't give us the temperature range you are using, so I'm not giving any more Specific heat data. If your range is between 100 and 200 oF, use the average value.
This should give you the answer I think you're looking for.
Art Montemayor
Spring, TX
Your question isn't clear as to what you want to calculate, but I'll make some assumptions. I assume you want to calculate the flowrate of the cooling water used to cool a stream of hot hydraulic oil. If so, the formula for the heat transfered should be resolved first:
Q = W Cp (T2 - T1)
Q = heat transfered, Btu/hr
W = oil mass flow rate, lb/hr
Cp = Specific Heat of the hydraulic oil, Btu/lb-oF
T2 = hot oil inlet temperature, oF
T1 = product cool oil temperature, oF
With the heat load calculated, you then resolve the same equation for the water flow rate:
W = Q/Cp (t2 - t1)
W = water mass flow rate, lb/hr
Cp = Specific Heat for water (1.0 Btu/lb-oF)
t2 = product hot water temperature out of cooler, oF
t1 = inlet cool water temperature to cooler, oF
You don't need any figure. The average Specific heats for hydraulic oil that I use are:
at 100 oF = 0.450 Btu/lb-oF
at 200 oF = 0.500 Btu/lb-oF
You don't give us the temperature range you are using, so I'm not giving any more Specific heat data. If your range is between 100 and 200 oF, use the average value.
This should give you the answer I think you're looking for.
Art Montemayor
Spring, TX
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