Your equation looks very much like some equations used in chilled water systems but SG is replaced with gpm in that version. The 500 is a constant to correct for units. Where is this formula from?
I've seen heat transfer as:
Q = m Cp dT. Looking at the units, you have BTU/hr = lb/hr * BTU/lb deg F * deg F so everything is dimensionally consistent. You use any units you want, for example you can replaced deg F with deg C but you need to add a conversion factor.
yOyO, "our formula" is not the standard heat transfer expression:
Q= AUdT
U is the total heat trasnfer coefficent and will contain contribution from convection and conduction
Area NEEDS to be included. Youyr constant may include unit convers and a specific are (or it may of course be your unit for Q thats wrong and perhaps is BTU(sq ft/h.)The heat transfer will not depend directly on theCp (heat capacity) - but temperature increase will. As far as i recall Nussels number is somewhat dependant on Cp for the convection part. I would have to check my books to be sure.
I would recommend that you approach somebody who actually know heat transfer
"The heat transfer will not depend directly on theCp (heat capacity)"
A decomposition of the heat transfer equation for results in m_dot*c[sub]p[/sub]*delta_T which says that the heat transfer is directly proportional to the mass flow, the heat capacity of the mass flow, and the temperature change of the mass flow, See Lienhard's "A Heat Transfer Textbook" equation 3.2 (
OK maybe i didnt put it right, but i maintain my point of view. It is true that you can determine the amount of heat exchanged by measuring the mass flow and knowing the dT and the Cp.
However, if you are using the same type of fluid - how would you determine the change in heat flow aring from using e.g. a different tube material (with a different heat transfer coefficient)? Apperently using the mid term of 3.2 you couldnt. You would - of course be able to measure a change in dT - but you Cp would be the same!
So i may have said it incorrectly/misleading but i still maintain that you can easily change the heat exchange rate - without changing either the mass flow rate or the Cp. However, its true that you can always determine your heat exchange rate by doing an energy balance including the Cp of the fluid.
I don't disagree with the notion that there are multiple factors that affect heat transfer; obviously, area or length is one of them. From a system's perspective, all factors are fair game, including the fluid, though.
I think they have all answered it. the 500 is just a term to convert volumetric flow of a liquid (water) using gallons per minute to Btu/h. Heat transfer by a 'fluid' is related mass flow and specific heat. It also includes the specific heat 'units' so to speak.
In US terms and for WATER at 60 degrees F.. 60 min/hr x 62.4 lb/cf/7.48 gal/cf x 1.0 Btu/(lb-deg F) = 500.5 (min/hr)(Btu/(deg F-gal). Multiple by a Gal/min and a deg F you should get Btu/h left...Unless I hosed up this basic math exercise which is highly possible....
As TD2K pointed, using a coherent units system, you do not need to consider any numerical constant. But if you mix units from two or more dimensional systems, their scale/equivalence needs those numerical elements. So 500 has to be address to some particular formula, I mean it's far from to be any rule.
By the way, as an example, i saw written there two formules (in the integrated form) Q=M.Cp.ΔT, which is the energy [Joule] (or time-rate scale of energy [Joule/sec = Watt] = power. Q means the energy associated to a certain fluid in motion in a tube, for instance and Q=A.U.ΔT(log), when that fluid in motion = flow reaches or passes through an heat-exchanger and transfer its energy. So, we can couple the two formules in the following balance at the heat-exchanger in this way;
M.Cp.ΔT = A.U.ΔT(log). This balance is highly useful for parametric and iteractive thermal calculations, with no need to have any numerical values, when using the same units system.
Foxstar has given the best answer.
For a heat transfer formula you will need to equal both formulas as Foxstar mentioned, but with a specific example maybe we can give you an idea how to do it.
NOTE: the formulas to calculate U are different for different heat transfer situations, so a specific problem will be good for you to understand it or for us to give you some guidelines.