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Posted
Does someone have handy the formula (along with all constant parameters) for calculating thermal growth ?

I'll really appreciate any help. Kind of an emergency occured during an alignment procedure...

David
 
Posts: 994 | Location: Texas | Registered: 22 February 2005Reply With QuoteEdit or Delete MessageReport This Post
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You'll get a lot of responses, but here's my 2 cents worth...

Measure the temperature in at least five places, evenly spaced,from the base, to the centerline of the shaft. This will give you an average of the temperature. Let's say this average is 125 degrees. Subtract the ambient temperature. If it's 75, you end up with a 50 rise in temperature, cold to hot.

Next, measure the distance from the base to the centerline of the shaft. Let's say for example - 36".

You have to know the material's coefficient for thermal expansion. Steel and cast iron are close - 0.0000065 is more than close enough.

Now the math.

(36")x(50 degrees)x(0.0000065) = 0.012" thermal rise.

One more thing. Most machines only get hot toward the parts that are doing the work. for example, if a Feedwater Pump is pumping 125 degree water, the pumpcase, bearings, and shaft are pretty warm, but most of the base is about room temperature. In that case, only measure the distance of what is getting hot.

If we used the example above, and found out that only 8" worth of pump is getting above ambient temperature, then we would use 8" instead of 36", which would change our result:

(8)(50)(0.0000065) = 0.0026"

I hope this helps. Thermal growth on most machines is usually way overfigured.

Stan Riddle
 
Posts: 168 | Location: North Carolina | Registered: 21 February 2005Reply With QuoteEdit or Delete MessageReport This Post
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Stan is right as usual.

A few other ways to say that 0.0000065 inch per inch per deg F thermal coefficient of expansion that may (or may not be) easier to carry around in your head:

1 - 6.5 microinch per inch per deg F.

2 - Thermal growth in mils will be temperature change in F times growing height in inches divided by 150. i.e. 30" growing 10F would be 30x10/150 = 2 mils.
 
Posts: 3119 | Location: Texas Gulf Coast | Registered: 20 February 2005Reply With QuoteEdit or Delete MessageReport This Post
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This file has some different things in it. Pretty complex, but read the instruction page. It has some coefficients of expansion for different materials and a 'mini growth' calculation page.


Patrick


Excel SpreadsheetThermal_Growth_Master_311.xls (128 Kb, 64 downloads)
 
Posts: 382 | Location: NJ | Registered: 19 April 2004Reply With QuoteEdit or Delete MessageReport This Post
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Thermal Growth (TG) Formula

TG=TLC

T= delta T (Deg F)
L= measurement from base to Center Line (inches)
C= growth Coeff. from Table

AL = 0.0126
Bronze= 0.01
Cast Iron= 0.0059
Mild steel= 0.0063
Stainless steel= 0.0074
 
Posts: 95 | Location: Tennessee | Registered: 21 February 2005Reply With QuoteEdit or Delete MessageReport This Post
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Spencer,

Do these factors give you the growth in mils?

Thanks,

Danny


Danny
 
Posts: 1630 | Location: Midlothian, VA, US | Registered: 22 February 2005Reply With QuoteEdit or Delete MessageReport This Post
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Thanks a lot to all !!!!!
That is real Forum in action !!!!!

David
 
Posts: 994 | Location: Texas | Registered: 22 February 2005Reply With QuoteEdit or Delete MessageReport This Post
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Hatfield, where did the coefficient numbers you have come from? I just wanted to make sure my info. was OK. I got the numbers from "Machinery's Handbook", and the coefficients are different from yours by 3 decimal places.
 
Posts: 168 | Location: North Carolina | Registered: 21 February 2005Reply With QuoteEdit or Delete MessageReport This Post
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For completeness it should be clarified that the units of TG from that formula are mils.

Expressing L in inches and TG in mils creates the factor of 1000.
 
Posts: 3119 | Location: Texas Gulf Coast | Registered: 20 February 2005Reply With QuoteEdit or Delete MessageReport This Post
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Got it from LUDECA and the growth is in mills.
 
Posts: 95 | Location: Tennessee | Registered: 21 February 2005Reply With QuoteEdit or Delete MessageReport This Post
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