Optimizing Heater Size for Winter Protection of Centrifugal Water Pump

In summary, the water pump can be protected from the winter months by installing a heated enclosure around it. However, the exposed motor and platform may lose heat to the air, which may cost more than it is worth in energy. Work out the budget first before proceeding.
  • #1
steves1080
65
1
I have a centrifugal water pump that I would like to protect during the winter months. The pipes can be drained, but there are some locations that are at risk for retaining trapped water, which can be problematic whenever we actually need to run the pumps. I am proposing that a heated enclosure be placed around the pump (see attached picture). However, I fear the exposed motor and the platform on which the pump sits will suck out a lot more heat than I anticipate. How can I accurately size a heater for this enclosure and take into account any conductive heat losses from these heat sinks? The platform is steel and the pump/motor are cast iron.

Thanks
 

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  • #2
How can I accurately size a heater for this enclosure and take into account any conductive heat losses from these heat sinks?
Take measurements.
 
  • #3
Thanks, but I was looking more along the lines of relevant heat transfer equations to use for determining how much is being lost due to conduction from the platform and the adjacent motor.
 
  • #4
But you also wanted "accurate".
You can get ballpark figures by adopting a model for heat conduction ... like:

I = AΔT/R

I = heat flux
A = area of the contact
ΔT = temperature difference between the two surfaces
R = thermal resistance: which is a material property of the join.

It is the material property you'll probably have to measure.
You can look up these sorts of models in a standard engineering reference.
 
  • #5
So if I wanted the conduction losses from enclosure to the ambient air via the exposed motor, would I just use the cross-sectional diameter of the motor as the A in the surface area? There will be a 17" hole in the enclosure the accommodate the motor connected to the pump.

Thanks
 
  • #6
You would use the entire surface area of contact.

Air is a poor conductor of heat though, heat losses to the air are dominated by convection.
 
  • #7
So I guess I'm not entirely sure how to determine the conduction loss from the heated pump to the exposed motor, and then the convection loss from the exposed motor to the cold air.
 
  • #8
You do it by conducting an experiment.
It is non-trivial to calculate in detail except for simple geometries and relies on measured material properties plugged in as coefficients. What you really need to know is how accurate the calculation needs to be.

Really you just need to de-ice the pump right?
What would you normally do?
 
  • #9
The pump does not completely drain, so some sections ice up and the pump has to be thawed with a heater when we need to use it, which is problematic when we're in a time crunch. Also, we could keep running the pump continuously as to keep water flowing. The purpose of this is to keep everything warm so that if we need to run the system we can do so immediately, and also not worry about seals leaking or ice forming inside an internal cavity.
 
  • #10
So one idea is to keep the pump warm all the time so it does not ice up - but you are also going to be concerned that this may cost more than its worth in energy.

The only simple way to proceede is to over-engineer the thing.
Work out the budget first.
 
  • #11
You can get an upper bound to the heat loss at the base and to the motor by assuming that the contact areas of the pump with the base and with the motor are at the temperature that you are trying to maintain inside the enclosure. This will eliminate the need to consider the enclosure and the pump. Even with this simplification, you may still have to use finite element heat transfer analysis. Do you have to worry about heat conduction from the base to the ground? I would treat the motor as a cylinder, and estimate the convective heat transfer coefficient by considering the range of wind velocities that will be encountered in practice. I don't see anything in the figure about the geometry in the depth direction.

Chet
 
  • #12
This is nearly impossible to address without seeing the design configuration, and assumptions. I'm assuming it is around 50+ years old.
 

1. How do I calculate heat loss for a specific material?

In order to calculate heat loss for a specific material, you will need to know the material's thermal conductivity (k), thickness (t), and surface area (A). The formula for calculating heat loss is Q = k * A * (T1-T2)/t, where T1 and T2 are the temperatures on either side of the material. Using this formula, you can determine the amount of heat lost per unit time (Q) for the material in question.

2. What units should I use when calculating heat loss?

The units used for calculating heat loss will depend on the units used for thermal conductivity, thickness, and surface area. Generally, thermal conductivity is measured in watts per meter per Kelvin (W/mK), thickness is measured in meters (m), and surface area is measured in square meters (m²). When using these units, the resulting heat loss will be in watts (W).

3. Can I calculate heat loss for a multi-layered material?

Yes, you can calculate heat loss for a multi-layered material by using the formula Q = k * A * (T1-T2)/t for each layer and adding the results together. Make sure to consider the thermal conductivity and thickness of each layer when performing the calculation.

4. Is there a simplified way to calculate heat loss?

Yes, there is a simplified way to calculate heat loss known as the R-value method. This method involves dividing the thickness of the material (in inches) by its thermal conductivity (in BTU/h·ft²·°F). The resulting value is the R-value, and the higher the R-value, the better the material is at resisting heat loss.

5. How can I use the calculated heat loss to improve energy efficiency?

By calculating heat loss for a material, you can identify areas where heat is escaping and take steps to improve energy efficiency. This could include adding insulation, sealing air leaks, or choosing more efficient materials. Additionally, knowing the heat loss of a material can help with proper sizing of heating and cooling systems, which can save energy and reduce costs in the long run.

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