February 21, 2009

Heat Capacity with Dissolved Solids

You are working on some design problem where a concentrated salt solution is under consideration and you are looking for its heat capacity. In such situation, will you use it to be equal to water Or it will be significantly different.

Yes, the answer is that it will be significantly different than water depending on its concentration. The deviation goes higher & higher if concentration increases.

The discussed equations are very useful for heat transfer calculations for slurry systems as well as solid handling systems.

So how to calculate it? Find out the easy way to calculate it.

A quick estimation method was proposed by Dimoplon in 1972. The proposed expression is:

Cpsoln = W1 x Cps + W2 x Cpw

Where Cpsoln = Specific heat of soluion mix.
Cps = Specific heat of solids
Cpw = Specific heat of water
W1 = Wt% of solids
W2 = Wt% of water (Usually 1 - W1)

The equation is valid for a given T. So if T changes you need to change the values accordingly.

Following chart / table is very important for above equation.



Example
Calculate the heat capacity of a 20-wt% Na2CO3 solution at 150 °F

Solution
Step-1
Look up the heat capacity of this solid from table. If it is not available, apply Kopp's Rule, which says

Cp(Na2CO3) = 2 x Cp(Na) + 1 x Cp(C) + 3 x Cp(O)

From Table, we read the values at 150 °F (339 K). Notice that the heat capacity for oxygen is given as O2 (it's natural form). This value must be divided by two to get the heat capacity for one atom of oxygen.

So,
Cp(Na) = 28.5612 KJ/Mole/K
Cp(C) = 11.6364 KJ/Mole/K
Cp(O) = 14.0611 KJ/Mole/K

So by Kopp's Rule

Cp (Na2CO3) = 2 x 28.5612 + 11.6364 + 3 x 14.0611
= 110.9421 KJ/Mole/K

Step-2
Now since our Dimoplon equation uses only weight basis, we need to divide this figure by molecular weight of compound. So,

Cp (Na2CO3) = 110.9421 / 105.9 = 1.0476 KJ/Kg/K

Now if you convert to Kcal then it becomes = 1.0476 x 0.23886 = 0.25 Kcal/Kg/K

Step-3
Now note down the heat capacity of water at 150 F which is 0.9975 Kcal/Kg/K.

Step4
Now finally apply Dimoplon rule as,
W1 = 0.2 (20%)
W2 = 0.8 (80%)
Cpw = 0.9975
Cps = 0.25

Hence,
Cps = 0.2 x 0.25 + 0.8 x 0.9975
Cps = 0.848 Kcal/Kg / K

Result
The literature data for this system is reported as 0.850, SO there is a variation of only 0.2%. Thus Dimoplon rule gives a very good estimate of specific heat of solutions with dissolved solids.

The equations are useful for slurry systems where impacts are significant.
Also it can be used for identification of specific heats of solids, where solid handling systems are involved with heat transfer.

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3 comments:

Lukman Nulhakiem said...

That's very good suggestion. However, I am always facing similiar problem. Simply because there is minimun physical properties source available.

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