Temperature rise calculations for generators 4

[Generation]

For a new project we have to supply generators with class F insulation, utilized for class B.
(So the temperature rise may only go up to temp. rise B temperature)
The only way to achieve this requirement is to select an oversized generator.

Is there a way to calculate what the amount of generator oversizing should be, in order to stay within the given B temperature rise?

 

[tgott]

DISCLAIMER: I am by no means an expert on this and am only posting this for your consideration. It may be totally off base.
Now, I'm not sure I understand your question, but for a class F totally enclosed, nonventilated motor/gen. the allowable temperature rise is 110 degrees Celcius. For a class B, the allowable temperature rise is 85 degrees Celcius. So if you supply the generator with class F insulation it will easily stay within the class B temperature limits. The above allowable temperature rise is based on a 40 degree C ambient temperature, and assumes measurement of temperature rise by determining the increase in winding resistances.
However if you can only supply class B gen's and want to oversize so they areeffectively class F, then how about this?
R = Ro(1 + (alpha times delta T)). For copper alpha = 3.9 x 10^ -3 per degree C. So R at 110 = 1.429Ro and R at 85 = 1.11815Ro. Taking ratio R110/R85 = 1.278. So maybe need to oversize by 27.8%. What do you think?

 

[electricpete]

Assume class F rise is 105C and class B is 80C above 40C.

Assume to begin with that the temperature rise is proportional to I^2.

Call the power level corresponding to 105C "full load". (not exactly... but makes the discussion easier).

80/105=76% rise would occur at sqrt(76%)=87% of the current. (i.e. generator is being limited by it's nameplate to 87% of "full load" = what it could do if allowed class F rise).

But also a portion of the losses do not vary with the square of current. Let's call alpha the portion of 105C rise that is constant.

Rise(I/I_FL)=alpha*105 + (1-alpha)*105*(I/I_FL)^2.

Pick a value of alpha and solve for value of I/I_FL which makes the rise equal to 80. That is the fraction of full load you'll be limited to (somewhere lower than 87%).

There may be some differences in treatment of ambient temperature and hot-spot in determining whether you want to select 80 and 105C or some other number as the rise.

 

[Generation]

Thanks, it seems to be very helpfull for me.
So a portion of the losses do NOT vary with the square of current.
1) Is this caused by the magnetic losses, converted to heat?
2) Is there a rule of thumb to estimate the value of "alpha", or can this be calculated as well?

 

[electricpete]

The no-load losses are core losses (hysteresis and eddy current) and windage and friction.

The load losses are I^2*R losses and stray losses.

From what I know about motors I would guess generators would be in the range of alpha between 20% and 60%. (sorry about the big range). Maybe someone else has better info?

Typically the purchase documentation should break down the losses into various categories or manufacturer should be able to provide a good estimate.

 

[jbartos]

Suggestion to tgott (Electrical)Jul 23, 2002 marked ///\\However if you can only supply class B gen's and want to oversize so they areeffectively class F, then how about this?
R = Ro(1 + (alpha times delta T)). For copper alpha = 3.9 x 10^ -3 per degree C. So R at 110 = 1.429Ro
///R110=Ro(1+0.0039x110)=1.429xRo
R85=Ro(1+0.0039x85)=1.3315xRo
So that
R110/R85=1.429/1.3315=1.073
1.073 resulting in the 7.3% oversize, rather than 1.278 resulting in oversize by 27.8%\\
and R at 85 = 1.11815Ro. Taking ratio R110/R85 = 1.278. So maybe need to oversize by 27.8%. What do you think?

///I would adhere to the following insulation system values:
Class B insulation system 130°C. If the ambient maximum temperature is 40°C then the temperature rise is 130-40=90°C
Class F insulatin system 155°C. If the ambient maximum temperature is 40°C then the temperature rise is 155-40=105°C
Then
R155/R130=1.6045/1.507=1.0647
which implies the 6.47% worst case oversize.\\\

 

[electricpete]

jbartos - In case you haven't figured it out yet, the oversizing arises from the fact that class F insulation is installed, but it's temperature rating is not used.

The copper temperature coefficient of resistance is not what creates the oversizing (margin). In fact if anything it slightly reduces the factor by which the system is oversized. Proof.... Ask yourself the question: how much higher could I increase load if I took full advantage of the class F insulation rating. You will come up with a smaller increase if you consider the temperature coefficient than if you don't.

 

[tgott]

As I originally stated, I am no expert, but now you've got my interest and I'd like to understand. I guess I don't even understand the question correctly. Whatever allowable temperature rise numbers you use, the underlying fact is that class F machines have a larger allowable temperature rise than class B machines, correct? So if you supply a class F machine, won't it be able to handle the temperature rise of a class B machine? (Maybe I'm thinking of it backwards.) Does a class F machine have more or less insulation than a class B machine? What role does the insulation play in the allowable temperature rise of a machine?

 

[electricpete]

tgott - I have no problem with your discussion whatsoever. You qualified your opinion from the beginning and you've made many valuable and intelligent contributions. My comments were not directed towards your post.

You are correct that class F insulation has a higher allowable temperature than class B.

The situation we are talking about is not a comparsion among machines. It is a discussion of one machine. That machine is is built with class F insulation but the power rating (limit) is established under the false (conservative) assumption that it has class B insulation. i.e. the class B temperature limit is applied during design of a machine which has class F insulation. More copper (for example) will have to be added to meet that conservatively-low temperature limit. When we operate it at it's rating the hottes spot should be at the class B limit which is ~25C lower than that machine could safely operate (wiht it's class F materials). It has some margin in it's design (oversized).

 

[Generation]

Hi,

Just be informed that insulation class F generator, utilized for class B, is a requirement according to the Shell DEP's and the Norwegian NORSOK standards.

The motivation is to save the insulation varnish due to the normal operating temperature (B) which is far below the max. allowed insulation temperature (F).

Thanks everybody for your good cooperation.
(I'm still looking forward for more suggestions)

 

[tgott]

light bulb! I think I get it now. No offense taken at all electricpete. Thanks for your reply. Think I'll stick to transmission planning though.

 

[jbartos]

Suggestion to electricpete (Electrical) Jul 23, 2002 marked ///\\\ (next one in the column following my previous posting:
Assume class F rise is 105C and class B is 80C above 40C.
Assume to begin with that the temperature rise is proportional to I^2.
///Generally, the temperature rise is proportional to power rise, i.e. R x I**2 in Watts
R is supposed to be considered since it will wary with the temperature rise stepdown from 105°C for insulation Class F to 90°C (or 80°C as assumed) for insulation Class B.\\\\
Call the power level corresponding to 105C "full load". (not exactly... but makes the discussion easier).

80/105=76% rise would occur at sqrt(76%)=87% of the current. (i.e. generator is being limited by it's nameplate to 87% of "full load" = what it could do if allowed class F rise).
But also a portion of the losses do not vary with the square of current. Let's call alpha the portion of 105C rise that is constant.

Rise(I/I_FL)=alpha*105 + (1-alpha)*105*(I/I_FL)^2.
///This obviously does not consider the R variation with respect to the temperature.\\Pick a value of alpha and solve for value of I/I_FL which makes the rise equal to 80. That is the fraction of full load you'll be limited to (somewhere lower than 87%).
///Where is the F insulation class generator oversize stated?
The F Class generator size must be higher than considered for the above calculation, perhaps 1/0.87 = 1.149 or 14.9% higher to be able to derated to B class generator rated power.\\

 

[jbartos]

Comment on electricpete (Electrical) Jul 24, 2002 marked ///\\jbartos - In case you haven't figured it out yet, the oversizing arises from the fact that class F insulation is installed, but it's temperature rating is not used.
///Yes. This is correct. That is why you have to in your posting increase the size of F insulation class generator power rating to be able to supply the same power as the B insulation class generator. This still appears to be due in your postings. I have concentrated on the tgott july 23, 2002 posting first since there were some errors but it captured a correct idea that the generator resistance will vary with the operating F insulation class generator downsized to B insulation class generator. Then, obviously with the upsizing the F insulation class generator power, the R of windings will be somewhat smaller.\\\

The copper temperature coefficient of resistance is not what creates the oversizing (margin).
///Yes, this was never disputed anywhere.\\ In fact if anything it slightly reduces the factor by which the system is oversized.
///This remains to be expressed mathematically.\\ Proof.... Ask yourself the question: how much higher could I increase load if I took full advantage of the class F insulation rating.
///The tgott posting on July 23, 2002 considered F insulation class generator power upsize to meet the B insulation class operational requirements based on resistance changes while the current changes were not considered. This is obviously inaccurate, since the major variable is neglected.\\ You will come up with a smaller increase if you consider the temperature coefficient than if you don't.
///This remains to been done and seen.\\

 

[electricpete]

jbartos - You are correct that the temperature coefficient of resistance should be considered along with the equations that I provided. Thank you for pointing that out.

The direction of that effect is not to increase the oversize/margin as you suggest above, but to decrease the margin between nameplate limit and how high the machine could theoretically go without exceeding insulation temperature limit.

Assuming that the same temperature applies to all the load-varying losses (rotor, stator and stray) (not entirely correct), I think an appropriate correction to my formulation would be someting like:

Solve:
80C=Rise(I/I_FL)=alpha*105 + (1-alpha)*105*(I/I_FL)^2*(0.9)

The 0.9 corresponds to 10% change which I calculated as 25C*0.4%/C. I think your 6% calculation would agree better if you reexamine your calculation "155-40=105°C"

The oversize margin will approximately be the difference between the calculated I and 1. There will not be much difference whether you compute (1-I) or (1/I - 1) since I is close to 1.

Alpha still remains a big unknown.

 

[electricpete]

I should not have included rotor losses in my list of load-varying losses. (it would be included for induction motor).

Also I'll mention that my premise that loads can be segregated into two simple classes (constant and varying with load squared) is not 100% accurate. But it's a starting point for an approximate answer.

 

[GOG]

Hi,

Since I'm always trying to keep it as simple as possible please consider the following:

1) Every generator manufacturer can inform you with the generator efficiency numbers for 25, 50, 75 and 100% load.

2) With this information it's easy to calculate the total heat losses in Watt's for each load (25, 50, 75 and 100%)

3) The generator temp. rise is proportional with the total heat losses of the generator.

4) Assume that the max. temperature rise for 100% load is 105°C for class F generator (@ ambient temp. of 40°C).

5) You can make a graph now for temperature rise versus generator load.

6) The only thing you have to do know is reading what generator load can be performed at class B (=80°C) temperature rise.

What do you think, a proper sollution?

 

[electricpete]

gog - that definitely sounds like a way to get some insight into the problem. Three thoughts:

#1 - you will have to extrapolate beyond the data since the 100% data point will correspond approx to the class B temp rise. (I labled the power associated with class F rise as 100% full load for simplicity above, but it should be the class B rise)
#2 - One weakness in both of our suggested methods would seem to be the simplicity of the heat transfer model. A watt of losses in the core does not produce affect the winding temp rise the same as a watt of losses in the winding. There are certainly some more refinements that could be made, including some discussion of winding hottest spot rise above winding average.
#3 - Along the lines of your suggestion, if the winding temperature were tabulated as a function of load, that would be an even more useful (direct) paramter to extrapolate up to the class F rise and determine associated power level.

 

[paddyB]

From the alternator industry.
This rule of thumb can be applied generally. The difference in output of a machine from Class B rating to F as an example is the square root of the temperatures. i.e. 105/80= 1.31. Square root = 1.14. 14% increase in output.

However remember that often generators are specified at Class B rise to be at peak position of efficiency curve and lower temperatures extend the insulation life.

 

[electricpete]

paddy - this would be roughly the same approach as I outlined, with alpha=0 (assumes all losses vary with current squared).