## Information regarding sub zero motor performance.

## Information regarding sub zero motor performance.

(OP)

I'm getting ready to send some motors out to test at -30°C and -60°C. I know how they run in typical conditions, but where would be a good source for information regarding extreme cold effects on a motor. I know my motor runs slower, but why? I need a good source to justify what's going on. From my books and some searches on NEMA they stop caring at 0°C

/Mech engineer dealing with a 24VDC brushed motor.

Thanks!

/Mech engineer dealing with a 24VDC brushed motor.

Thanks!

## RE: Information regarding sub zero motor performance.

Small motors may experience enough increased lubrication drag to make a noticeable effect.

The field resistance of a compound motor will decrease, causing increased field current and field strength.

Where field weakening causes an increase in speed, field strengthening will cause a drop in speed.

Bill

--------------------

"Why not the best?"

Jimmy Carter

## RE: Information regarding sub zero motor performance.

## RE: Information regarding sub zero motor performance.

Regardless f the motor type, have you considered increased bearing drag with very cold lubricant?

Bill

--------------------

"Why not the best?"

Jimmy Carter

## RE: Information regarding sub zero motor performance.

The lube in the gear box is rated for temp, still working on the bearings themselves. I believe they are a COTS sealed bearing.

## RE: Information regarding sub zero motor performance.

Assume 100 Ohms.

Check the temperature co-efficient of copper and calculate the resistance at minus 60.

Express the cold resistance as a percentage of 100 Ohms.

Ballpark, you can expect that speed at minus 60 due to the drop in field resistance.

1/4 HP is not much. Despite the temperature rating the lube resistance may be causing some slowdown.

Bill

--------------------

"Why not the best?"

Jimmy Carter

## RE: Information regarding sub zero motor performance.

looks like I could expect

@ -30°F=-78%

@ -60°F=-71%

I'll find out here in a few hours! Thanks

## RE: Information regarding sub zero motor performance.

Bill

--------------------

"Why not the best?"

Jimmy Carter

## RE: Information regarding sub zero motor performance.

Muthu

www.edison.co.in

## RE: Information regarding sub zero motor performance.

Second guess; Arctic Oil and Gas.

Third guess; Quick freeze tunnels. I worked on one quick freeze tunnel where the fans were working at minus 100 F.

Bill

--------------------

"Why not the best?"

Jimmy Carter

## RE: Information regarding sub zero motor performance.

Muthu

www.edison.co.in

## RE: Information regarding sub zero motor performance.

Why did you jump use -30F/-60F in your calc when you used -30C/-60C in your op?

=======

I agree with Bill that friction could be a big factor.

I was trying to work through effects on torque speed curve (neglecting friction).

Initial conditions (subscript 0) at reference temperature:

Flux Phi0 = V/Rf0

Zero-speed torque Tz0 = K*Phi0*Vbatt / Ra0.

No-load speed Nn0 = Vbat/(K*Phi0)

Torque-speed curve is straight line between (T,N) = (Tz0,0) and (T,N) = (0,Nn0)

Due to cold, Ra and Rf both decrease by some factor (let's say 0.78).

Ra = 0.78*Ra0

Rf = 0.78*Rf0

Flux Phi = V/Rf = V/(0.78*Rf0) = Phi0/0.78

Zero-speed torque Tz = K*Phi*Vbatt / Ra = K*(Phi0/0.78)*Vbatt / (Ra0*0.78) = Tz0/0.78^2

No-load speed Nn = Vbat/(K*Phi) = Vbat / (K*Phi0/0.78) = 0.78*Nn0

So if I've done my algebra correctly, zero-speed torque increases and no-load speed decreases. In that case the direction of speed change is indeterminate depending on intersection with load curve: it would speed up if initially heavily loaded but slow down if initially lightly loaded. Repeated disclaimers: 1 - ignored friction which may be very relevant; 2 - check my algebra.

=====================================

(2B)+(2B)' ?

## RE: Information regarding sub zero motor performance.

I chose to ignore armature resistance. At light loads the armature resistance is little.

I may have been wrong to ignore armature resistance.

You tell me.

Thanks.

Bill

--------------------

"Why not the best?"

Jimmy Carter

## RE: Information regarding sub zero motor performance.

I think oversizing of motor at these lower temperatures is a standard practice in my company at extreme temperatures as your describing. It was general rule to go from 50-100& oversize on HP to -40c. below that range it was difficult to find equipment that could operate in extreme temp.

Powder coated motors was also standard. Lube for gearboxes was used that can stand range of temp. The use of air was not used on equipment if it could be avoided. If air was used then it had to dry air. Rust was also a concern so all components had to have some finish on parts, not sure what finish that was to me?

controls

I also thought the use of 24vdc was not used, so all controls had to be 120vac for devices since components could survive the shock of lower temps. Control panel was in warm area not in freezing area. If area had to have control panel in area then you have to pre warm the panel before you turned on system or interlock not to turn on until temp inside was high enough for components.

## RE: Information regarding sub zero motor performance.

To quantify it for this situation, I'd plot the two lines (T vs N) and look for the intersection. Neglecting armature resistance is fine for qualitative purposes as long as we are operating to the right (higher N) of the intersection of the two curves.

Find that intersection by solving two equations (one per curve) in two unknowns (T and N).

The general form of the torque speed line is

T = Tz - N (Tz/Nn)

The original line (ref temperature subscript 0) is:

T = Tz0 - N (Tz0/Nn0)

The cold line at -30F is

T = [Tz0/0.78^2] - N ([Tz0/0.78^2] / [0.78*Nn0])

To find the intersction, set them equal:

Tz0 - N (Tz0/Nn0) = [Tz0/0.78^2] - N ([Tz0/0.78^2] / [0.78*Nn0])

…divide through by Tz0…

1 - (N/Nn0) = 1/0.78^2 - (N/Nn0)/0.78^3

…move N/Nn0 terms to the right and constant terms to left…

(1 - 1/0.78^2) = N/nN0 (1 - 1/0.78^3)

… solve N/Nn0…

N/N0 = (1 - 1/0.78^2) / (1 - 1/0.78^3) = 0.58

T/Tz0 = 1 - N N/Nn0 = 1 - 0.58 = 0.42

So for these particular numbers, the intersection of the original line and the cold line is (N,T) = (0.58*Nn0, 0.42*Tz0). As long as speed remains above 58% of original no-load speed (equivalently Torque remains below 42% of original zero-speed torque) then we reach the correct qualitative conclusion by neglecting armature resistance. It's a very safe bet.

Let me circle back to associate a more generic solution starting with the specific solution:

N/N0 = (1 - 1/0.78^2) / (1 - 1/0.78^3)

Let's subsitute 0.78 = 1-x where x represents fractional decrease in resistance (in our specific case x=0.22, we will limit our general consideration to x<< 1)

N/N0 = (1 - 1/[1-x]^2) / (1 - 1/[1-x]^3)

N/N0 = (1 - [1-x]^<-2>) / (1 - [1-x]^<-3>)

Substitute [1-x]^p ~ 1-px for x<<1

N/N0 ~ (1 - [1--2x]) / (1 - [1--3x])

N/N0 ~ -2x / -3x = 2/3

So for any small decrease in temperature (resistance), we can neglect armature resistance in qualitative prediction as long as no-load speed doesn't decrease to 2/3 of nominal. It's still a very safe bet.

Also neglected was additional external resistance in series with the armature…if it does not change temperature as much (perhaps portion of the supply wiring is in a temperature controlled area) then the effects of motor armature resistance variation are even smaller (we could neglect armature resistance variation over an even wider range of speeds and torques).

TLDR - Bill you were right as usual. It's probably intuitive to you after many years of working with dc motors. For me, dc motors are something I'm still learning about, so I took the opportunity to explore it in my own way.

=====================================

(2B)+(2B)' ?