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Complete Heat Gain Analysis of Motor / PumpHelpful Member! 

Hack3398 (Mechanical)
22 Jul 10 10:56
I have reviewed previous posts and couldn't find anything quite the same as this, but feel free to direct me to previous post or let me know if it is better suited in different forum.

Background:  Was hired to design life test lab HVAC and chilled water systems for motor manufacturer.  Owner has swimming pools in test facility with 30 pumps/motors per pool around perimeter pulling water straight out and back into pool (in closed room/chamber).  Pool water is to be pumped through heat exchanger with chilled water to maintain 70 deg F pool water.  Pumps are assumed to all be nominal 5hp.  Space is to be maintained at 75deg F and 50-60% rh.  Chiller will provide chilled water to pool water heat exchanger and air handler to condition the space (w/ chilled water coil).  Owner's opinion and past experiences with their tests and equipment stated that approx 1/3 of rated hp will to to water as heat and only motor inefficiency will be rejected to space.  They do not feel all hp of motor needs to be accounted for.

Question(s): Assuming 30 motors/pumps at 5hp each (pump and motor in conditioned space), what is heat rejected to pool water and what is heat rejected to space?  Following ASHRAE calcs with driven equipment in space, total power (motor + inefficiency of motor) must be accounted for and all rejected to pool or space.  Does the friction of the water in the pool from the pumps really all convert to heat? (conservation of energy)  I'm assuming ineffiency of motor goes to space, and inefficiency of pump mostly goes to water (say 90/10 or 80/20 split), but does the remaining energy of the pumped water transfer to heat within the pool?  Any tips on how to convey this to an owner so they understand and buy off on it?

Final thought: Other half of lab is similar but instead of driving pumps the motors drive fans all contained in test chamber/room.  Does this behave the same way with the fans? (energy of moving air within room all converts to heat)

Sorry for the long post.  Long time follower, first time 'poster'.

Thanks in advance
imok2 (Mechanical)
22 Jul 10 20:27
Hack3398

Take a look at http://www.engineeringtoolbox.com/electrical-motor-heat-loss-d_898.html

Electrical Motors and Heat Loss
Heat gain from electrical motors to the surroundings
AbbyNormal (Mechanical)
22 Jul 10 20:55
Power needed to move the water in horsepower

GPM x ht of head x specific gravity / ( 3960 x impeller efficiency)

Probably if you could measure the power draw into the pumps, you could break out the heat into the water with that estimation and the reminder of the heat goes to the air.

The way we build has a far greater impact on our comfort, energy consumption and IAQ, than any HVAC system we install

AbbyNormal (Mechanical)
22 Jul 10 21:00
or even to a flow and temperature rise measurement across the pumps to get the heat into the water.

The way we build has a far greater impact on our comfort, energy consumption and IAQ, than any HVAC system we install

Helpful Member!  SAK9 (Mechanical)
22 Jul 10 22:04
The heat loss due to inefficiency of the motor gos into the surrounding air.This is straight forward.

The theoretical temperature rise of the water within the pump at steady state conditions assuming that all the heat generated remains in the fluid.


dT = H / (778*Cp*n)

dT = Temperature rise of fluid, deg-F
H = Pump head, Feet
Cp = Specific Heat of the liquid, Btu/(lb-deg F)
n = Pump efficiency, decimal value

This is the  theoretical temperature rise of the liquid within the pump at steady state conditions assuming that all the heat generated remains in the fluid.

The heat loss through the pump casing and insulated piping will be minimal(given the low values of dT you get).So allocating 100 % work out put from the pump to water is a fair assumption in my opinion.
 
SAK9 (Mechanical)
22 Jul 10 22:06
I think your client's claim can be verified by comparing the calculated and actual measured temperature values.
Hack3398 (Mechanical)
23 Jul 10 8:24
Thank you for your responses.  Unfortunately this is all new construction  they have some other testing they have done in the past, but nothing we can go out and measure before our portion of the design is completed.  I think the biggest stumbling block we are having is not the temperature rise of the water at the pump, but whether or not the energy of the moving water out of the pump really does all turn into heating of the water out in the pool(not talking about heat generated within the pump from friction/inefficiency here).

SAK9 - Thanks, yes I had used your equation (same one I found in Cameron Hyd. Data) to find temp rise in pump but still wasn't sure where the "left over" power/energy went from the moving water.  The energy of the moving water in the pool converts to heat, right?  I don't know how else to account for it in an energy balance in a closed system like this.  Is this what you mean by, "So allocating 100 % work out put from the pump to water is a fair assumption in my opinion."

Thanks again
marcoh (Mechanical)
24 Jul 10 4:11
I would assume that 5hp is transferred to the water and the motor inefficiency is rejected as heat to the plantroom.

Where the pump impeller inefficiency energy goes I am not sure, but would assume it goes into the water?

But then you need to start considering heat transfer from the pipes in/out of the plantroom.  If the water is colder than the plantroom the water in the pipe will absorb plantroom heat and it all too complicated and I would ignore for my plant sizing.
SAK9 (Mechanical)
24 Jul 10 7:49
All the pump shaft power goes into rasing the pressure energy of water.This gets expended in overcoming the friction in pipe work,valves and heat exchanger.All of the work converts into heat and is absorbed by the water and pipe work(both the pipe and water being at the same same temperature).Water having a greater specific heat as well greater mass and being in motion will absorb most of the heat  both by convection and conduction.Therefore the heat added by pump power needs to be accounted for in chiller sizing.

Similarly if you examine the computer heat load calculations you will find that all the fan power is added in as sensible heat and is used in arriving  at room air flow rates.In the olden days the motor was mounted outside the air handling unit(using a belt drive)and therefore the heat rejected by the motor was not accounted for in the cooling load calculations.
KLH (Mechanical)
27 Jul 10 13:35
See attached from Carrier's load estimating manual.
tys90 (Mechanical)
27 Jul 10 17:03
You mentioned ASHRAE calcs, why not follow them?  They specify the instantaneous heat gain to the space and to the driven fluid.  If you know the operating HP you could use that instead of nameplate HP to be a little more precise.
SAK9 (Mechanical)
28 Jul 10 3:42
KLH,
The clouded paragraph  is against the laws of Physics.Imagine a closed pipe loop with a 5 hp pump driving water through it, pump head being equal to frictional losses.According to Carrier there will no temperature rise and 5 Hp will be vanishing into thin air .The situation I have described above is very similar to churning in a centrifugal pump(pump operation with a closed discharge valve).If you can try this out somewhere,it will be a good learning experience for you.
 
tys90 (Mechanical)
28 Jul 10 11:07
Reread the paragraph.  It's clarifying where the heat is generated at.  It's saying that it's not generated by frictional losses in the pipe, it's generated by compression of the fluid at the pump.   

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