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liquid thermal expansion calc - help neededHelpful Member! 

heaterguy (Mechanical)
10 Jan 05 15:56
I have density data that shows this liquid is 55.9 lb/ft3 at 200'F and 53.7 lb/ft3 at 300'F. What is the thermal expansion coefficient (beta)?

I have tried inverting the density adn calculating (0.0186-0.0179)/100 = 0.000007 but that number seems too small compared to similar fluids. What is the right way to do this?

TYIA,

Heater Guy
Dean78au (Chemical)
10 Jan 05 23:26
The equation for thermal expansion of any substance is given by

B = 1/V ( dV/ T)

where B = thermal expansion coefficient, V = Volume, dV = change in Volume, and T = change in temperature.

Your method looks right.  Have you checked the units?
Helpful Member!  quark (Mechanical)
10 Jan 05 23:38
...because it should be [(0.0186-0.0179)/0.0179]/100 = 0.00039

beta = (dV/V0)/dT (PS: I am using d instead of delta and this is not a differentiation)

quark (Mechanical)
11 Jan 05 6:13
Dean,

It seems we have posted at the same time.

25362 (Chemical)
11 Jan 05 8:54

I hope heaterguy is now satisfied with the cubic thermal expansion value estimated by quark in 1/oF. When using oC the result should be multiplied by 1.8 .
heaterguy (Mechanical)
11 Jan 05 18:03
Thank you. Your responses were sent to my spam. Sorry it took so long to reply. Quark answered my question.
sailoday28 (Mechanical)
13 Jan 05 12:12
heaterguy:
The information you provided is not enough to calculate the term BETTA.  Unless, the pressure at the two conditions is the same.
25362 (Chemical)
15 Jan 05 1:57

In theory sailoday28 is right. The definition of the cubic thermal expansion β, sometimes called alpha, is:

β = (1/V)(∂ V/∂ T)P

However, the isothermal compressibility values  

κ = -(1/V)(∂ V/∂ P)T

are quite small when the liquids are well below their normal boiling points, usually lower than 2×10-4/atm, to sensibly affect the definition given by quark.
sailoday28 (Mechanical)
15 Jan 05 11:43
25362 (Chemical)
Can you explain in layman terms--How did you display the partials?
Thanks
sailoday28 (Mechanical)
15 Jan 05 11:45
heaterguy
Knowing the liquid might give more of a handle on determining Betta.
25362 (Chemical)
15 Jan 05 12:30
To sailoday28. It's easy: via the Process TGML at the end of your message, look down for TGML Character Entity Reference. There you"ll find the answer to your query.
25362 (Chemical)
16 Feb 05 7:22
Another useful expression for β would be:

ρ /Δt)(1/ρave)

which in this case would result in:

β = [(55.9-53.7)/100][1/(55.9+53.7/2] = 0.0004/oF
heaterguy (Mechanical)
16 Feb 05 8:41
25362,

I was using either of the densities, but it makes more sense to use the average of the two. Thanks.

heaterguy

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