correcting measured temperature rise 2

electricpete · Jun 5, 2002

I was looking at a group of motors that seem to be running much hotter than expected by comparison to factory data.

Then I realized that the factory calcs seemed way off base.

It's a 1250hp motor (4kv, 1190rpm).
They measured a temperature rise of 61.1C when running at 1003 horsepower.

Based on the above they estimated the temperature rise if the motor were running at full horsepower would be 69.4C. (I estimate way higher... what do you come up with?).

Here was their logic:
Losses=HP*0.746*(1-eff)/eff

They have "measured" efficiency values of 0.94 at 80% load (1000hp) and 0.945 at 100% load (1250hp)

Losses at 1003hp=HP*0.746*(1-eff)/eff=1003*0.746*(1-.94)/.94=47.76kw

Losses at 1250hp = HP*0.746*(1-eff)/eff= 1250*0.746*(1-.945)/.945=54.27kw.

Since temperature rise is proportional to losses:
rise(1250)=rise(1003)*losses(1250)/losses(1003)=61.1*54.27/47.76=69.42

The above is their calculation.

Here's the way I look at it (the way we did it when I worked on transformers)
TL=NLL+LL
TL= total losses
NLL = No load losses (constant with load)
LL = load losses (vary with approx square of load).

TL(1250)=TL(1003)*[alpha + (1-alpha)*(1250/1003)^2]
Rise(1250)=Rise(1003)*[alpha + (1-alpha)*(1250/1003)^2]
where alpha is the fraction of total losses which are no-load losses.

If alpha=0.2 => Rise = 88C
If alpha=0.4 => Rise = 81C
If alpha=0.6 => Rise = 75C
If alpha=0.75 => Rise = 69.5C

I have to get awfully high alpha to predict their result. Maybe there is a small problem with my definition of alpha as written.... it will be the fraction of NLL at 80%, not 100%. I'll see if that makes a difference.

In the meantime any comments?

cbarn24050 · Jun 5, 2002

Hi, in the second set of calculations you are assuming that the efficiency is constant regardless of load. I'm not so sure that the transformer model is the same as a motor.

motorman · Jun 5, 2002

Let me see now. You are actually running at full load so this is a test right ?

I estimate that at 1250HP the rise is 91.2 C. But others would guess 84.8 C at the lowest possible

How'd I do ?

electricpete · Jun 5, 2002

cbarn - I'm not sure where you're coming from. In the first calculation there is a slight difference in measured effieicney 0.940 vs 0.945. In the 2nd calc I'm only assuming that the losses follow a model TL=NL+LL where LL varies with load^2.

motorman - your answer falls fairly close in line with what we see in the field (what we see is 80C rise at approx 70% load..... But I don't want to get sidetracked into analysing the field data yet, I want to really understand whether the factory calc is bogus). I'm glad you came to similar conclusion as myself. Although I'm having a hard time believeing a motor manufacturer could be so far off in calcs....

In my formula above alpha represents the fraction of NL/TL at 80% load (1000hp).

If I compute beta=fraction of NL/TL at full load I get around 65%. That still sounds high to me (but not as far out as 75%).

electricpete · Jun 6, 2002

Here's where the 65 came from at the end of my previous post:

In my formula above alpha represents the fraction of NL/TL at 80% load (1000hp).

I am more interested in beta which I define to represent the fraction of NL/TL at 100% load (1250hp).

Before:
Rise(1250)=Rise(1003)*[alpha + (1-alpha)*(1250/1003)^2]

Similar formulation using beta:
Rise(1003)=Rise(1250)*[beta + (1-beta)*(1003/1250)^2]

Rearrange to solve for Rise(1250):
Rise(1250)=Rise(1003)/[beta + (1-beta)*(1003/1250)^2]

Plug in Rise(1003)=61.1 and beta varying values:

beta Rise(1250)
0.2 85.4
0.3 81.4
0.4 77.7
0.5 74.3
0.6 71.3
0.65 69.8
0.7 68.4

Their prediction (69C) implies that beta (the fraction of losses which are nonload varying at full load... ie NLL/TLL at full load) to be approx 0.65. If actual beta is lower, then predicted rise should be higher.

Losses are:
core losses
friction and windage
stator losses
rotor losses
stray losses.

The first two are approx const with load and the last three vary approx as load^2. I have a hard time believing that the first two make up 65% of the total. (If they make up less than 65% of total, then the predicted temperature rise should be higher).

cbarn24050 · Jun 6, 2002

Hi, now I'm awake I've had another look. If you have a hot point hear a cold point the heat flows and creates a thermal gradient,similar to a voltage gradient down a resistor chain.The energy flow would equate to the current, and the temperature to the voltage.To calculate the temp at any point along the gradient you need to know the thermal resistance from core to case and from case to ambient.Your calcs do not mention the ambient temp which must be included.

electricpete · Jun 6, 2002

cbarn - I am talking about temperature "rise" which I define as winding tempearutre (by RTD) minus ambient (so I have already accounted for ambient.

My assumption is that rise is proportional to losses.
You are right that there is a temperature coefficient involved in the proportionality constant, but if I'm estimating the ratio of two rises (based on ratio of losses)osses), I need not concern myself with explicity compluting that temperature coefficient.

electricpete · Jun 6, 2002

Here's a simpler way of asking the equivalent question.

Is it reasonable that the efficiency at 100% load is a full 0.5% higher than at 80% load?

(typically the curve of efficiency vs power is constant or decreasing as it approaches 100%.... peak efficiency somewhere at or below 100%).

electricpete · Jun 6, 2002

My understanding of efficiency curves vs load for motors. For small motors they often show incrasing efficiency vs load in the range of 100% load. For large motors they show constant or decreasing efficiency vs load in the range of 100% load.

electricpete · Jun 6, 2002

I have located some other large motor curves which show similar behavior of increase in efficiency up to 0.8% from 80% to 100%.

I believe this implies that no load losses form >65% of total full load losses.

That is not in agreement with typical breakdowns of losses shown from various sources, but it is apparenlty possible for certain motors (even though it conflicts with my gut feel).

When cbarn suggested that it is improper to compare motors to transformers, I believe he was right from the standpont that the fraction beta can be much larger for motors.

cbarn24050 · Jun 6, 2002

Are you saying that you don't believe the manafacturers measured data on the motor efficiency?

electricpete · Jun 6, 2002

Up until my recent post I was skeptical of the measured data on two aspects;

#1 - I was thinking about the possibility of measurement error (a 1% error in efficiency measurement equates to 20% error is losses at 95%).

#2 - I was thinking about the possibility that the manufacturer fudged the efficiency data in order to force the calc'd rise below required 80C. (worth considering given that these motor has a 80C rise at 70% load in the field .... but these are 20-year old motors in dirty environement and we surely have some clogging in the air ducts).

All of the above based on my intuition and understanding that no-load losses were certainly less than half of the total. Apparently that was flawed assumption on my part.

Also there were some very gross assumptions on my part that all of the losses contribute equally to winding temperature rise... ignores the fact that friction/windage don't have as big effect on winding rise as do stator I^2R.

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