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shaded pole evaporator fan motors

rick9890

Feb 19, 2022
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I am installing a small well insulated walkin cooler in my basement. The evaporator fan I was going to use a Bohn TL12 (1200 btu rated @ 10degree TD) has 2 208/230v shaded pole motors with a total FLA of .8 amps My understanding is that shaded pole motors along with being inefficient operate near full load amps even at no load. so even if the draw was .7 amps that would be 154 watts @ 220v
So 154w x 3.41 would be 525 btu. This would mean that the evaporator fan motors would use almost 1/2 of the cooling capacity. Could this be correct?
 

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crutschow

May 7, 2021
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This would mean that the evaporator fan motors would use almost 1/2 of the cooling capacity. Could this be correct?
I assume some of that is for the compressor motor that generates the cooling.
And depending upon the design, the fan motor heat is likely dissipated in the outside air flow, not the cooling flow.
 

rick9890

Feb 19, 2022
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Crutschow thanks for the response
Fans are in the freezer box so the heat goes into the box.
 

Bluejets

Oct 5, 2014
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The fan has nothing to do with the rated cooling output of the system.
Most of the 150 odd watts the motor draws would be converted into mechanical energy, not heat energy.
 

crutschow

May 7, 2021
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Fans are in the freezer box so the heat goes into the box.
But isn't there an outside air connection for the condenser?
Most of the 150 odd watts the motor draws would be converted into mechanical energy, not heat energy.
There's no free lunch.
That mechanical energy is converted into heat by the fan moving the air, which heats the air.
 

rick9890

Feb 19, 2022
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But isn't there an outside air connection for the condenser?
There's no free lunch.
That mechanical energy is converted into heat by the fan moving the air, which heats the air.
They are evaporator fans not condenser fans
 

Bluejets

Oct 5, 2014
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But isn't there an outside air connection for the condenser?
There's no free lunch.
That mechanical energy is converted into heat by the fan moving the air, which heats the air.

A small portion not all.
This is then absorbed by the refrigerant action and dispersed via the condensor.
Amount would be negligible compared to the cooling of the refrigerant system.
In a standard cool room this would be many kW, perhaps 20 to 30 depending on overall size.
I think the OP has his figures all wrong and panicking over nothing.
 

rick9890

Feb 19, 2022
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A small portion not all.
This is then absorbed by the refrigerant action and dispersed via the condensor.
Amount would be negligible compared to the cooling of the refrigerant system.
In a standard cool room this would be many kW, perhaps 20 to 30 depending on overall size.
I think the OP has his figures all wrong and panicking over nothing.
Evaporator is only 1200 btu. @ 10degree TD. I came up with 1290 btu load loss on my calculation for box to maintain -5 degree temperature with no new product or door openings
I have 2 1/3 hp condensers and will be using 2 evaporators I am actually going to get about 1320 btu out of each evaporator with 9 degree TD. My thinking was that should 1 unit go down the other should maintain (with defrost cycles) 0 degree box temp as long as I keep the box closed. I put an email into Heatcraft to see if their specs are a typo because the 2 motors are only 1/150hp each but if not does any one know if shaded pole motors draw at least 75% of FLA no mater what the load? as this is what I always thought.
 

Bluejets

Oct 5, 2014
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As I said , you are getting totally confused over an issue that does not exist.
 

Bluejets

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It's a user imposed problem that doesn't exist.
Any radiated heat from the fan coil motor is negligible in the fan coil output.
 

rick9890

Feb 19, 2022
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Not true if the amp draw stated by Bohn is correct. Although the amp draw is FLA Is my understanding that shaded pole motors (which are ineficient) draw at least 75% power even at no load correct?,
 
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