Calculating Thermal Expansion of Water: Is the Formula Correct?

[de_Sitter]

Hello,

I'm trying to determine a way of calculating the thermal expansion of a volume of water. The formula I have come across is:

ΔV = βV₀ΛT

The general consensus seems to be that water expands roughly 4% from 20°C to 100°C, or 4.2% from 4°C to 100°C (http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html). Using β = 0.000214 as found on (http://www.engineeringtoolbox.com/cubical-expansion-coefficients-d_1262.html) the result of this formula is not around the expected 4% from 20 - 100°C. e.g. 25L gives ≈ 0.43 L increase.

Am I using the formula correctly? Some rudimentary examples use β as a constant, while others remark that β varies with temperature, but don't provide examples of using this in practice. Any suggestions would be appreciated.

Thank you.

 

[Bystander]

CRC Hndbk. tables of density of water. Definitely not constant for water.

 

[Chestermiller]

The coefficient of volume expansion varies with temperature. The equation is only supposed to apply in the vicinity of the base temperature. In your case, the coefficient is for around 20 C.

Chet

 

[de_Sitter]

Hello,

Thank you both for your replies. I'm a little unconvinced that using the coefficient of the base temperature is correct. Using 25L from 20 to 100°C with coefficient 0.000207 as per (http://www.engineeringtoolbox.com/volumetric-temperature-expansion-d_315.html) I get ΔV = 1.9715 L, which is around a 7.8% increase, rather than the expected 4%, as given on (http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html). Does anyone have an example of how they would model this increase?

 

[Chestermiller]

To do it over a larger temperature range, you would have to know the thermal expansion coefficient as a function of temperature, and you would have to integrate.

 

[nasu]

Hello,

Thank you both for your replies. I'm a little unconvinced that using the coefficient of the base temperature is correct. Using 25L from 20 to 100°C with coefficient 0.000207 as per (http://www.engineeringtoolbox.com/volumetric-temperature-expansion-d_315.html) I get ΔV = 1.9715 L, which is around a 7.8% increase, rather than the expected 4%, as given on (http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html). Does anyone have an example of how they would model this increase?

You don't get that value for that coefficient of thermal expansion assumed constant over the temperature range. What you get is around 0.4 L or about 1.6%.
But you have a table with the variation of the coefficient with temperature right there. You can see that it increases with the temperature so it will be reasonable to get 4% over the whole range.
You can do an estimate by using these values.

 

[Nidum]

There is good tabular data available for variation of density with temperature :

http://www.kayelaby.npl.co.uk/

http://www.kayelaby.npl.co.uk/general_physics/2_2/2_2_1.html

Invert to get specific volume .

Also specific volume can be found directly in some thermodynamic tables .

 

[de_Sitter]

Thank you all for your assistance. I'm a little rusty on my calculus, but Nidum's links are perfect for my purposes.

Thanks again!

 

[de_Sitter]

So just to confirm I'm doing it right Chestermiller. I've fitted a curve to the data points given in some of the tables I linked earlier, which gives β(T). I then find the definite integral of this function over the temperature range (e.g 20°-100°C), then multiply by the initial volume?

ΔV = ∫[β(T)] dT ⋅ V_o , with 20° and 100° as the limits of integration

 

[Chestermiller]

So just to confirm I'm doing it right Chestermiller. I've fitted a curve to the data points given in some of the tables I linked earlier, which gives β(T). I then find the definite integral of this function over the temperature range (e.g 20°-100°C), then multiply by the initial volume?

ΔV = ∫[β(T)] dT ⋅ V_o , with 20° and 100° as the limits of integration

Basically correct for a small temperature change, but I would do it a little differently. The exact equation is:

$$\frac{1}{V}\frac{dV}{dT}=β(T)$$

So, integrating this equation, I get:

$$V=V_0e^{\int_{T_0}^T{β(T')dT'}}$$

where T' is a dummy variable of integration. This equation will give a more accurate result.

Chet

 

[de_Sitter]

Thanks Chet, I'm not quite familiar with this dummy variable concept, but I'll do some reading. Thanks for the input.