Calculating heat added by a pump 3

[MScarn]

I am working on a project where we are sizing a chiller for a system. The system circulates water using a pump and a VFD to control the pump. The VFD will adjust the pump output at specific times. I am trying to find the heat produced by the pump.

I have gone through the calculation for the Temp Rise from the Pump (dt=Ps(1-u)/(cp*q*p). The result from this equation gives me a large temperature rise (almost 32 F). This does not seem right. I believe the issue is that my pump is not running full speed (GPM) the entire time (the VFD will cycle it back). Is there a better equation to use?

 

[1503-44]

Heat from a Constant speed pump is basically due to its inefficiency, generally around 25% of input energy at BEP. It will generate more heat as you vary flow away from BEP.

If you use a vfd, you effectively increase efficiency for all flow rates, hence do not expect a vfd controlled pump to be a great source of heat.

Sometimes in reased efficiency is a false net gain. After putting LED lights in my office, I had to turn up the temperature of the HVAC.

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[EdStainless]

Your pump efficiency will drop with lower speeds.
So a larger fraction of your power input will turn to fluid heating.
But it will be a larger fraction of a smaller amount of power.
Look at the pump curves carefully, at lower speed and off of BEP you may have pump efficiency of only 20%.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, consulting work welcomed

 

[LittleInch]

Can you post the calculation. I suspect you've got a unit error somewhere as I would normally expect less than 5C and normally less than 3.

But note that this is the sudden temperature rise as you've calculated it immediately d/s of the pump.

If your pipe is heavily insulated, the entire power of the pump comes out as frictional heat somewhere. Hence in your formula the 1 - efficiency term may not apply....

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[1503-44]

More info here
thread407-220878

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[TiCl4]

Heat input just isn't about the pump efficiency (efficiency on the curve), though. I mean, if you have pump on a 100% recirculation line the entire fluid power input will eventually turn into heat, either through pump inefficiency or through frictional losses after the fluid leaves the pump.

In a full recirculation system, the total heat input should be the motor power draw minus the power loss from motor efficiency (not pump efficiency). Most motors are usually >85% efficient (you can look up the minimum efficiency for NEMA B motors) unless you are less than 5 HP.

In your case, when the VFD turns down, the power input will drop, as will your heat input. If you have a pump curve from the manufacturer that shows pump curves at different rpms, use that data to figure out your power draw when the VFD slows the pump down. Otherwise, you can use the affinity laws to get close.

dT(liquid) =[ Q (power draw of the pump at current speed) * E (efficiency of the motor) ] / [(cP (liquid) * mdot (liquid mass flow)]

Make sure your units agree. To me, you getting a 32 F temperature rise indicates you have either mixed up your units, are not considering the drop in input power when the VFD reduces pump speed, or you have a pump that is running waaaaaay back on its curve.

 

[1503-44]

The motor is not in contact with the fluid so it will not increase fluid temperature.

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[TiCl4]

1503-44,

I didn't say the motor was heating the fluid...My point was that the entire motor power draw minus motor efficiency loss goes into the fluid either in the form of immediate heat or as flow/head. If a motor is drawing 10 HP and is a 80% efficient motor, 2HP is lost as heat in the motor, which is removed via fan cooling. The pumped fluid receives the remaining 8 HP. If the pump is running 50% efficient, then 4 HP will go to immediate heating of the fluid and the other 4 HP will go towards flow/head. However, that flow/head will all eventually be reconverted to heat via frictional losses, so the fluid will eventually receive all 8 HP as heat. Thus the equation above.

 

[LittleInch]

I recall a very similar debate about the heating effect of fans within the cool air ducts and that was the end result - all the power you put into a flow which is either a closed circuit or ends with the fluid having no kinetic or static energy relative to the pump or fan turns to heat.

and then if the motor is inside the air duct it's total power not shaft power.

but not 32F rise in a water system unless you're doing something seriously wrong...

In most pipelines or piping systems you only normally see the temperature rise after the pump as the fluid then cools down to ambient temperature. But if you insulate it then that doesn't happen.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[1503-44]

OK, that's clear way of saying it. However there is a tiny bit of a problem with "so the fluid will eventually receive all 8 HP as heat". If you said energy instead of heat, I'd agree. The fluid receives 4HP of immediate sensible heat, but no more. The other 4HP, carried out of the control volume are pressure, its potential energy, and velocity, its kinetic energy, which are not in the form of "heat". It is energy which might be converted into heat, if subjected to friction, but being outside the control volume, that is not for us to know, but we might see at least some of that as friction on the pipe wall directly increasing the pipe temperature and conducted out to surroundings without ever have been received as heat in the fluid. How much goes where depends on turbulence, viscosity, conductivity, convection, heat capacities, geometry and insulation, but fortunately, that's outside our control volume.

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[TiCl4]

1503-44,

This is a chilled system. Unless this system is in the frozen tundras of the north, the ambient temperature will be higher than the pumped fluid. Thus any heat generated from friction will serve to increase the fluid temperature, as heat is transferred INTO the pipe from the surroundings. Even if the ambient conditions were colder than the pumped fluid, the fact that frictional losses generate heat is still there. In that case, however, you’d balance the heat generated friction with losses to the surroundings. It still doesn’t change the underlying equations.

Secondly, you can’t say “we can’t predict because it’s not in our control volume”. We CAN predict because of the known conditions, by considering the entire loop as the system, not just the pump inlet/outlet conditions. There is no change in potential or kinetic energy in a recirculation loop. There is only a change in pressure. Any power that goes into developing pressure WILL be lost as heat. There is no other thermodynamic option.

Why do I say heat? There is no other option for the head/flow energy to go in a recirc system. At steady state on a recirc loop, the fluid input power WILL equal the frictional losses.

 

[MScarn]

I really appreciate the responses.

TiCl4; your equation is slightly different than the one I was using. If I use yours with the below variables;
Pump power draw @ flow = 9.35 hp
Efficiency @ flow = 74.89%
Cp Specific Heat (water) = 1 btu/lb f
Liquid mass flow = 150 GPM

Using your equation i get; 0.047 deg F.

Was it my equation that was off? Originally I was using dt= (power draw*(1-efficiency) / (specific heat*flow*density)

 

[TiCl4]

Mscarn,
It looks like you haven't converted your units properly (You've used horsepower for power and gallons per minute as a mass flow rate, for one), and don't understand what I meant by discussing motor efficiency vs pump efficiency. Here's my calc for total extra load imposed on the cooling system by the pump:

Brake HP = 9.35 HP (assuming 85% efficient NEMA motor)
Q = Brake HP * Motor Efficiency = 9.35 * 0.85 = shaft HP = 7.95 HP = 337 BTU/min

E: Efficiency @ Flow = I believe these values on the pump curves are only the pump efficiency values, not a combination of pump and motor efficiency. In this case, only the motor efficiency matters, as all the shaft horsepower is converted to heat eventually.

Cp = 1

mdot = 150 * 8.34 = 1,251 lb/min.

Q = mdot * Cp * dT
dT = Q / (mdot * Cp)

dT = 337 BTU/min / (1,251 lb/min * 1 BTU/lb*F) = 0.27 F.

I mean no insult, but you don't seem to understand brake vs shaft vs hydraulic horsepower, nor the associated efficiencies. I base this on the fact that you used 74.89% as the efficiency value, which I can only assume is the efficiency point on the pump curve. Please correct me if I'm wrong here. If not, I suggest you read up on the terms above.


Edit: Probably a more important factor than temperature rise is total heat input for chiller design. Chillers are often designed on a tons basis, and 337 BTU/min is about 1.7 tons. I don't know the size of your system, but this pump alone will require about 1.7 tons of chilling by itself.

 

[1503-44]

I think we both understand whats going on, so I'm happy to leave it at that.

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I take no responsibility for any damages or injuries of any kind that may result.

 

[MScarn]

TiCl4,

Thanks again for breaking this down for me. Yes, I need to familiarize myself with the values and their applied intentions.

The efficiency I used came from a pump sheet we have, 74.89% is what the Efficiency is listed as, based on the 150 gpm flow of our system.

 

[Tony_M]

MScarn hi

If you are using a VFD then presumably you are controlling the delta pressure. If you know the delta pressure across the pump the delta temperature is independent of the flowrate:

Wh = Q ΔP
Wi = Wh / e
Wi = m cp ΔT
m = Q ρ
Q ΔP / e = Q ρ cp ΔT
ΔP / e = ρ cp ΔT
ΔT = ΔP / (e ρ cp)

Wh = Hydraulic Power
Wi = Pump Input Power
Q = Flowrate
ΔP = Delta Pressure Across Pump
e = Pump efficiency
m = mass flowrate
cp = heat capacity
ΔT = Delta Temperture
ρ = density

I have estimated your delta pressure from the horsepower - about 470 kPa (SI units) which gives ΔT = 0.15 C or 0.27 F which agrees with
TiCl4 above