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naldopr
01/07/2008, 05:05 PM
I have a chiller that water flow is rated 300 to 600 gph. my return pump is 1500 gph.
what happen if I use this pump.

should I use a separate pump

thank you!

naldopr
01/07/2008, 06:28 PM
^_^

naldopr
01/08/2008, 01:18 PM
some one ?

AZDesertRat
01/08/2008, 01:38 PM
I would venture to say two things will happen. One is it will not cool efficiently since the velocity through the cooling coils will be too high. Water needs to travel at a leisurely pace so you get heat transfer between the water and refrigerant coils. The other is you will probably reduce the pumps GPH drastically due to additional head through the chiller due to friction loss and velocity.

One option would be to install a tee in your return piping and run just a portion of the return flow through the chiller and back to a seperate chamber in yoyr sump. Another would be a dedicated smaller pump.

pvtschultz
01/08/2008, 02:36 PM
A tee would be the perfect solution so you can throttle your pump and use the extra water to feed a refugium et al.

As for the flow rate, the higher the flow rate (up to a point) the higher the heat transfer rate. Cooling power in a two-medium heat exchanger is driven by the delta T (temperature difference) and the convective heat transfer coefficient which is dependent on the Reynold's Number which is a function of the square of the fluid velocity. Basically, the higher the flow rate, the greater the Reynolds number, the greater the convective heat transfer rate (power). Unfortunately the conduction through the titanium will usually limit the rate so there is a point of diminishing returns. Also, the larger pump will add more energy (heat) to the water that will also need to be removed by the cooler. Pluse AZDesertRat is correct about the friction adding heat to the water.

The cooler will "seem" to be not working as "efficiently" because the water exiting the heat exchanger will be close to the same temperature as the water entering, but the increased mass flow rate is what increases the cooling power. If the cooling power is greater than the power going into the water from lights, pumps, and other things, the water temperature will decrease. It is a thermodynamic law (first one at that).

Power=hA(T_in-T_out): Cooling power is dependent on temperature differential and heat transfer coefficient (h in this case).