# Homework Help: Open loop cooling system

1. Jun 22, 2017

### JaredC

1. The problem statement, all variables and given/known data
I am trying to find out the rate at which I need to pump water into a device generating a specific amount of heat to keep it at a low temperature. I have not taken thermodynamics yet, and so I am trying to use a very basic model. The water is pumped through a copper pipe.

I know the following:

- Temperature of incoming water
- length and diameter (both inner and outer) of the copper pipe
- The total heat being generated by the device

I am looking for an equation that will have these as independent variables and give me the needed flow rate of the water.

Again I am looking for a ballpark answer, so if there are some other assumptions I have to make, like uniform heating, I can get away with it.

2. Relevant equations
unkown

3. The attempt at a solution
I have tried looking around the internet but am not familular enough with the material to know what I am looking for.

Any help would be greatly appreciated, even if it's just pointing me in the right direction.

2. Jun 22, 2017

### Asymptotic

You'll want to think of it in SI units, but one equation burned into my memory is BTU/hr=GPM*500*ΔT.
'500' is the weight of water (~8.34 lb/gal under standard conditions) * 60 (converting GPM to GPH).
1 BTU is the energy required to change the temperature of 1 pound of water by 1°F.
1 watt = 3.412 BTU/hr

Perform a Google search on "heat transfer copper pipe", and explore the results.

3. Jun 23, 2017

### CWatters

The energy required to heat water is...

E=MC*deltaT

Where
m is the mass
C is the specific heat capacity
DeltaT the temperature change

So rearrange...

M=E/(c*deltaT)

dm/dt=(dE/dt)/(c*deltaT)

dm/dt is the water flow rate in kg/s
dE/dt is the power removed in Watts or Joules per second.

4. Jun 23, 2017

### JaredC

@CWatters
How can I find out the change in the water's temperature? It seems like it would also vary with the change in flow.

5. Jun 23, 2017

### CWatters

Yes it does.

Lets say you want the object to be kept below 45C. If you assume the object is in good thermal contact with the water then you don't want the water to exceed 45C. You say the incoming water temperature is known (call it 20C) so the change in water temperature must be less than the difference between these values

deltaT = 45-20=25C

The equation (in post #3) gives you the minimum flow rate required to stop the water rising more that that much.

In practice you may need a slightly higher flow rate to account for any thermal resistance between the object and the water. For example if the object dissipates 2W and the thermal resistance between the object and water is 3C/W then the max temperature of the water must be lower. Instead of 45 degrees the water must be..

45 - (2*3) = 39C

If the incoming water is still 20C then the change in water temperature must be kept down to

deltaT = 39-20 = 19C

Since deltaT is the denominator the equation shows you need a higher flow rate to achieve a lower deltaT.