Why does a constant speed pump to variable speed pump result in electrical energy savings and an equivalent heating energy increase?

asked 2016-01-08 10:17:46 -0500

Steven W
43 ●5

updated 2018-06-13 08:31:39 -0500

NOTE: I've saved files here

I'm modeling the change from a constant speed pump to a variable speed pump for a hot water loop with a standard natural gas boiler in OpenStudio. The measure is applied as expected. However, the results are not as expected. The electrical savings are as expected at ~3000 kWh, but I did not expect an energy equivalent increase in natural gas usage of 10 Million BTU (the kWh plus the efficiency of the boiler). Looking deeper, I was able to see that the output power is directly applied as an increase in heat to the water.

Looking deeper into the Energy Plus code, it appears this functionality is on purpose. This page has the source code and has a comment on line 1726 that says "We assume that all of the heat ends up in the fluid eventually since this is a closed loop". I understand there will be an increase in heating energy, but my expectation is that the heat gain is a fraction of the pumping. I assumed more heat would is lost to the surroundings in the case of a constant volume pump.

So, is EnergyPlus's assumption correct that most or all of the pumping energy ends up in the fluid eventually?

To say this another way, if you have an electric boiler, would a VFD on a hot water pump result in an increase in energy usage and not save a building owner anything?

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answered 2016-01-08 15:01:30 -0500

aaron

888 ●1 ●7 https://github.com/aar...

updated 2016-01-08 15:24:48 -0500

Yes, all of the mechanical energy of the pump does end up as heat in the fluid. There's a really good article in HPAC Engineering by Gerald Williams which explains pump/fan heat called "Fan Heat and Pump Heat: Sources and Significance." You might be able to find a pdf copy floating around on the internet.

You can think of hot water pumps as in-line electrical pre/post heaters for the boilers. By reducing the speed of the pumps, you're reducing the "supplemental heat" from the pumps, and this must be overcome by the boilers. There are still good reasons for implementing variable hot water flow systems:

Electricity can be on the order of 3 times more expensive than natural gas (per Btu). So even though you're total energy consumption has risen, you're still going to save money. Natural gas is also usually perceived as greener than electricity due to the inefficiencies in the electric grid. This is the same reason people tend to opt for natural gas boilers over electric boilers even though electric boilers are more efficient on a Btu basis. Using the previous analogy, you can think of variable flow hot water systems as removing an in-line supplemental electric heater.
The delta-T in a variable flow system will tend to be greater than those in a constant volume. If you have condensing boilers, this allows for lower return temperatures which increases the efficiency of the boiler. Larger delta-Ts also will result in lower environmental/radiant losses in the piping because the average loop temperature will be lower.

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answered 2016-01-08 10:58:32 -0500

rraustad

13599 ●72 ●12 http://www.fsec.ucf.edu/

updated 2016-01-08 11:38:02 -0500

The shaft power (impeller) winds up as heat to the fluid. If you have specified a motor heat to fluid fraction greater than 0, additional motor heat will be added to the fluid proportional to that fraction.

! This adds the pump heat based on User input for the pump
! We assume that all of the heat ends up in the fluid eventually since this is a closed loop
ShaftPower = Power * PumpEquip(PumpNum)%MotorEffic
PumpHeattoFluid = ShaftPower + (Power - ShaftPower)*PumpEquip(PumpNum)%FracMotorLossToFluid

So there are 2 fluid heating components. One is imparted by the impeller to increase fluid pressure, the other is the motor electrical losses that may or may not heat the fluid.

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Comments

Thanks for responding and expanding on what I wrote! My model has a FracMotorLossToFluid of 0, so the shaft power heats the fluid. When changing from a constant speed pump to a variable speed pump, and not changing the motor efficiency, all the energy savings from a VFD result in a heating penalty. My question is, should all electrical savings from a VFD should result in a heating penalty? I've never seen the heating penalty mentioned or discussed before.

Steven W ( 2016-01-08 11:21:14 -0500 )edit

What did you use for the pump coefficients:

Pump:VariableSpeed,
  0,                       !- Coefficient 1 of the Part Load Performance Curve
  1,                       !- Coefficient 2 of the Part Load Performance Curve
  0,                       !- Coefficient 3 of the Part Load Performance Curve
  0,                       !- Coefficient 4 of the Part Load Performance Curve