Solar radiation effect on electrical motors 1

[martinrelayer]

Hi all!

We have several outdoor motors and the sun radiation makes the surface temp rise up to 50°C and more.

I was wondering how it affects on the ambient temperature derating of the motor power. And if there's any table, paper, coefficient, formulae, etc. to consider when it comes to the selection of the motor

Thanks all,
Martín.

 

[edison123]

I am sure direct sunlight on motors would increase the temp. rise but to quantify that would be problem. How about some "sun-roof" for the motors ?

 

[martinrelayer]

Yep, that would be an option.
But I was looking for some info about that... I believe that is adding to the ambient temp. but I couldn´t find any info

regards

 

[LionelHutz]

I don't really see it as a case of just adding to the ambient temperature or specifying a motor for say a 50C ambient. The motor can still radiate that heat from it's 50C surface to the cooler ambient (30C maybe?). It's more like the motor has an always on space heater installed. Remember, the amount of heat the motor can dissipate depends on the difference between it's surface and the ambient air around it. So, adding an initial temperature rise means it's capable of dissipating a bit more heat as the operation heats it up more. I hope that makes sense.

It likely would end up being some sort of intermediate value. For example, if the sun heats the motor 20C over ambient then that is equivalent to installing the motor in a location with an ambient 10C higher than it really is.

The biggest determining factor is the actual temperature rise of the motor in operation. As long as the motor isn't overheating it is OK.

 

[martinrelayer]

so it would be like trial and error and then verify it isn´t overheating?

 

[waross]

Most motors have fan cooling of some type. The motor will be cooled by the ambient air passing through it or over it.
The effect of the sunlight should be judged when the motor is running, not when it is stopped.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[hanksmith]

Could you specify a white paint?

Not an electrical answer but would help to reduce the case temperature during times of intense sun.

 

[Unotec]

Shelter it. Easy and unexpensive. It will also keep the rain away

<<A good friend will bail you out of jail, but a true friend
will be sitting beside you saying ” Damn that was fun!” - Unknown>>

 

[7anoter4]

As a speculative appreciation:
If we could consider a motor as a short cable with a very expanded diameter we could use
IEC 60287 formulae in order to appreciate the supplementary rise of copper temperature due
to sun radiation.
Let's take for copper max. temperature of 130 oC[Class B].
ambient max. temperature of 40 oC
Sun radiation=700 w/m^2.
Motor overall dia. =0.4-1.2 m [length =1m]
Then calculated the maximum rise of temp. due sun exposure= 23 oC
I suppose 15 degrees C temperature drop from copper to outside surface
then:
Max. permitted copper rise temp.=130-40-23-15= 52 [instead of 130-40=90 o C]

 

[rbulsara]

What waross said. I think sunlight has little impact on running motors and the heat is taken away by the cooling air running over it which is at ambient temp.

If a motor is rated for outdoor use, it is supposed to work in direct sunlight. Make sure that ambient temp is selected for the geographical area.

I have had motors running near full load out in the sun in abmbient temps up to 43 deg.C. The motors of course were rated for 40 deg C ambient and 50 deg rise, (or vice versa, cant recall it correctly now) for Tropical applications.

 

[7anoter4]

Correction:
Max. permitted copper rise temp.=130-40-23-15= 52 [instead of 130-40-15=75 oC as rated]
That means in order to maintain maximum 130 oC on copper one has to reduce the current
I/Irated=sqrt(52/75)= 0.83

 

[martinrelayer]

7another4 lots of cool stuff! Is it possible to post the formulae?
They're all class B rise with class F insulation, but the table I have states the power output drop with higher ambient temperatures or higher altitudes, so i started wondering this.
If I can relate this formulae to ambient temp, then I'm done!!
For example, can I add your 23ºC calculated?? (40ºC + 23ºC = 63ºC it's 80% of rated power according to the motor manual) and that matches sqrt(52/75)... but I don't know where does it come from... don't have the standard.

 

[7anoter4]

Hi martinrelayer
First of all as I already mentioned it is only an attempt to appreciate the copper temperature rise
Based on calculation for cable heating due Sun radiation.
The method is very conservative as doesn't take into consideration the cooling through ventilation.
Most of motor losses are evacuated by forced convection. Also, a great part of the Sun radiation does not
reach the windings as the ventilation eliminates it.
The cable losses-all the losses-are evacuated through the overall surface mostly by radiation and a very
small part through natural convection.
The change of surrounding air temperature and pressure are more important factors for motor current derating.
I translated the cable parameters into motor parameters so:
TC = copper conductor maximum temp.[degrees C] = 130
TA = air ambient temperature = 40
SUN = sun radiation [w/m^2] = 700
SIG = SIG=1 for surface no coated = 1
DE = overall diameter [m] = 0.4
QS = QS=1
H = 0.21 / (DE) ^ 0.6 + 3.94
KA = PI()*DE*H*therm.resist.
DTT = SUN * SIG * KA / H / PI()
DDSO = is initial value for DDSF and [=2.77]
have to change it up DDSO=DDSF
DDSF = ((TC - TA + DTT) / (1 + KA * DDSO)) ^ 0.25= 2.84
Tfive = 1 / PI() / DE / H / DDSF=0.065
TSUN = The temp.rise due to SUN[no ventilation considered] = DE * SIG * SUN * Tfive=18
dTdrop = Temperature drop from copper to surface = Supposed 15 degrees C
Toutsurf = overall surface temperature = TC-dTdrop=115
dtCU = temperature drop of copper due heating[variable] = TC-TA-TSUN-dTdrop=57
dTrated = temperature drop of copper due heating[rated] = TC-TA-dTdrop=75
Inew/Irated = Inew is the reduced value in order to maintain TC = SQRT(dtCU/dTrated)=0.87
NewTC = increased copper temperature if I=Irated = TC+TSUN=148
Newoutsurf = new outer surface temperature = NewTC-dTdrop=133

therm.resist.=dTdrop/P
P is part of losses evacuated through
exterior envelope[most of the losses are evacuated through
forced ventilation]
P=radiation from outside surface to the ambient
P=qrad*PI()*DE=564 w
qrad=krad*(((273+Toutsurf)/100)^4-((273+TA)/100)^4)
krad=3.44 w/m^2 for a varnished surface
Regards

 

[7anoter4]

I'll try to attach a small excel program. The program has to be iterative [you have to change DDSO=DDSF and also
Pradattached=Prad] up to the time where these value will be the same.

 

[martinrelayer]

Yes I agree to the fact of the ambient temp. (assumed 40ºC) and paint colour are to consider.

hey thanks a lot... but I got lost in H = 0.21 / (DE) ^ 0.6 + 3.94... where that comes from?

sorry my ignorance! but great stuff... this is certainly close to the derating of ABB motors on higher ambient temp

 

[7anoter4]

h=heat dissipation factor as per IEC 287 para 9.1.1[for exposed cable CABLE FREE IN AIR]

 

[thinker]

Many years ago I read about a rule of thumb for motor derating: for example, if the motor is designed for 40C max ambient temp. then for each 1 degree above 40 the motor should be de-rated 10%.