Heat developed in circulating fluid

[sb00]

Hi guys,
I have a vessel with say 100m3 of water at room temp. I have to keep it recirculating with a centrifugal pump of 1MW capacity. I am concerned abt the temp rise of the fluid. The temp rise is only due to the friction losses, rite?
So, i should calculate as follows:
Electrical power (1MW)=mass flow rate x pressure developed + friction loss (f).
Now, f/(mass flow rate x specific heat) should be the rise in temperature, rite?
Is it the correct way or am i wrong, please advise.
Thanks in advance.

 

[25362]

Since you are speaking of an hypothetical case, I bring to you an hypothetical question: barring pipe friction and the pump's inherent loss of efficiency, what's the difference between the proposed external circulation and an internal circulation brought about by a mixer expending the same amount of energy?

 

[TD2K]

Look at thread407-72665

Typically, the temperature rise across a pump is based on the pump's inefficiency. That energy goes into heating the liquid and generally results in a fairly small temperature rise except for high head, low efficiency pumps. The equations will be in any pump book.

However, if you are essentially pumping in a loop, you first put the energy into the fluid due to the pump's inefficiency. The remaining energy goes into velocity and head energy which in this case, is just recirculated to the suction tank. Since you have to have conservation of energy, your frictional losses also show up as heat.

As an example, I have a hot tub at home with about a 5 Hp motor. During the summer, I can't run the circulation pump constantly as the hot tub reaches the high temperature trip and shuts down.

 

[22082002]

E. Ludwig's book has answer for your question..
This will solve the problem.

 

[saxon]

sb00, Once again, this is the classic example of the First Law of Thermodynamics at work, following the equation:
Sum dQ = Sum dW.

That is, "when a system undergoes a thermodynanic cycle the net heat supplied to a system from its surrondings is equal to the net work done by the system on its surrondings." Also, any energy input into a system that is not converted to work by the system is converted to heat.

So in a pump recirc. system, every watt of energy that is supplied to a system that is not utilised as work will be converted to heat. A general rule of thumb for these types of systems is:
1Kwh = 3600 Kj (this assumes no work is done on the environment)

Hope this helps.
saxon

 

[sb00]

thnx guys...

 

[MarkkraM]

As said earlier, the inefficiency of the pump will add heat energy to the fluid. Do calculations of the expected heat losses through the surface of your vessel at various water temperatures until you match the energy input from the pump. You may find that the water will not accumulate much temperature after all. There may also be some losses from the process pipe going to and from the pump.

I unfortunately have a similar case but with an insulated vessel containing hexene. Due to a change in process we have long periods where we need to kick back 99% of our pump flow back into this vessel. After about a day the temperature accumulates around 15degC, where it has reached an equilibrium of heat loss through piping/vessel and heat gain through pump. This may become unacceptable in summer so I've got to come up with a (inexpensive) solution before then! We can't afford to install a cooler!

 

[moltenmetal]

Note saxon's comment is that all energy expended by the pump motor which does not result in work done ON THE SURROUNDINGS, will ultimately end up as heat in your closed loop system. The only thing which is doing work on the surroundings is the fan which cools the motor- a tiny fraction of the motor's energy input. Therefore, every watt which is not lost to driving the fan, or to the motor's internal inefficiency (i.e. the resistance and eddy current losses in the motor itself), ends up in your fluid.

This will help you calculate the heat duty. Then you have to do heat transfer calculations if you want to know the temperature rise in your tank versus time.