liquid thermal expansion calc - help needed 1

[heaterguy]

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]

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?

 

[quark]

...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]

Dean,

It seems we have posted at the same time.

 

[25362]

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]

Thank you. Your responses were sent to my spam. Sorry it took so long to reply. Quark answered my question.

 

[sailoday28]

heaterguy:
The information you provided is not enough to calculate the term BETTA. Unless, the pressure at the two conditions is the same.

 

[25362]

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]

25362 (Chemical)
Can you explain in layman terms--How did you display the partials?
Thanks

 

[sailoday28]

heaterguy
Knowing the liquid might give more of a handle on determining Betta.

 

[25362]

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]

Another useful expression for [β] would be:

([Δ][ρ] /[Δ]t)(1/[ρ]_ave)​

which in this case would result in:

[β] = [(55.9-53.7)/100][1/(55.9+53.7/2] = 0.0004/^oF​

 

[heaterguy]

25362,

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

heaterguy