Calculate Diffusion Coefficient in Liquids

The order of magnitude of diffusion coefficients in liquids is 10^-9 m2/s with the bulk of data in the range between 0.5 x 10^-9 m2/s and 5 x 10^-9 m2/s. Note that these data also hold for gases dissolved in liquids.

Equation of Chen and Othmer for gases is used for

• Its Simplicity
• Sufficiently Accurate
• Availability of Input Data

Correlation for liquids is that of Wilke and Chang (AIChE J. 1 (1955), 264):

The equation is very simple to use & is given below.

D(1,2) = 7.4 x 10 ^ -12 x (T / mu) x (C x M2)^0.5 x (V1)^-0.6

Where

D(1,2) = is the diffusion coeff of solute 1 in Solvent 2 (M2/s)
T = Temperature in K
mu = dynamic viscosity of solvent 2 (mPa s)
M2 = Mol Wt of solvent 2
V1 = Molar volume of solute 1 at NBP (Cm3/mol)
C = association factor of solvent 2 as below

C = 2.6 for water
C = 1.9 for methanol
C = 1.5 for Ethanol
C = 1.0 for non polar solvents.

I hope it is useful for many of you.

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Heat capacities of dissolved solids & Organic solutions - Quickest Method

In my previous article on the topic, I suggested a method called dimoplon rule for estimating the heat capacity for solution with dissolved solids.

The method was giving a detailed estimate with accurate prediction for such values of specific heat capacities.

However, I am also giving a very quick method for such evaluations with some approximation. The results from this method are accurate enough for preliminary estimates and are not too far from experimental values compared to previous method.

The Quick method is known as Vosseller method proposed by him in 1973.

Cpsolution = 1 - 0.7 x W1
Where W1 is Wt% of solids in solution.

So for our example of Sodium Carbonate solution of 20% in previous post will result in as below.

For 20% solids W1 = 0.2
So specific heat of solution shall be 1 - 0.7 x 0.2 = 0.86

Now compare this with literature data & previous detailed method.

Literature data = 0.85
Dimoplon method = 0.848
Vosseller method = 0.86

So error is still only 1.2% but the method is quite quick for initial understanding of the system & calculations.

Similarly Vosseller also proposed this kind of equation for organic solutions with water where the equation is

Cpsolution = 1 - 0.45 X W1
Here W1 is the Wt % of Organics in water. So for 20% Ethanol mixture the heat capacity can be assumed as = 1 - 0.45 x 0.2 = 0.91 which is quite close to experimental value.

Hope it will be useful for many of you in different industries.

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Reverse Osmosis

Background:
Diffusion is the movement of molecules from a region of high concentration to a region of lower concentration. Osmosis is a special case of diffusion in which the molecules are of water and the concentration gradient occurs across a semipermeable membrane.

Diffusion and osmosis are thermodynamically favorable and will continue till equilibrium is reached. Osmosis can be slowed, stopped or even reversed if sufficient pressure is applied to the membrane from the ‘concentrated’ side of the membrane.


Introduction:
Reverse Osmosis is the movement of water molecules across the semipermeable membrane against the concentration gradient, from a region of low concentration to a region of higher concentration, by applying a pressure in excess of the osmotic pressure.



The membranes used for reverse osmosis have a dense barrier layer in the polymer matrix where most separation occurs. The semipermeable membrane allows the passage of water but not ions (Na+, Ca+2, Cl-1) or larger molecules (e.g. Glucose, urea, bacteria)



Uses:

- One of the methods used for desalinating water
- Purification of ethanol ( here water is present as an undesirable impurity)
- Commercial and residential water filtration


Industrial Applications:
Reverse Osmosis systems can be used to treat boiler feed water, industrial wastewater, process water and more. A few of the major uses are :

• BOILER FEED WATER TREATMENT - RO is used to reduce the solids content of water prior to feeding into boilers for the power generation and other industries.


• PHARMACEUTICALS - RO is an approved treatment process for the production of United States Pharmacopeia(USP) grade water for pharmaceutical applications.


• FOOD & BEVERAGE - Water used to process food products and to produce beverages is often treated by a RO system.


• SEMICONDUCTORS - RO is an accepted component of a treatment process to produce ultrapure water in the semiconductor industry.


• METAL FINISHING - RO systems have been successfully applied to a variety of metal finishing operations including several type of copper, nickel and zinc electroplating; nickel acetate seal and black dye.

Cost Benefits of RO:
RO is increasingly being adopted by power producers as a treatment method for purifying boiler feed water, makeup water and in zero-liquid discharge applications. The injection of high-purity water produced by RO technology into a gas turbine can improve operating efficiency and increase energy output by 10 percent or more.

There are other cost benefits as well. For example, the prices of acid and caustic solutions continue to rise while the prices of RO units and membrane elements continue to decrease. The primary cost for operating RO systems is electricity, and since these systems consume very little energy, operating costs are relatively low.

Operating efficiency differs between ion exchange beds and RO systems. Cation and anion resin beds must be regenerated once they reach a set exchange capacity. Their efficiency is related directly to the amount of dissolved solids that pass through the system. Conversely, the operating cost for RO does not vary with the level of dissolved solids in the feed water since the operating cost is based on flow rate.

An RO system does not require significant downtime with the exception of quarterly or semi-annual routine maintenance. And RO systems are highly automated, requiring minimal operator interaction. By contrast, during regeneration, which can take up to twelve hours, ion exchange equipment cannot be used and the plant is forced to stop water production.

With such advantages, expect to see continued growth in the use of RO technology in the industrial sector, particularly for power generation applications.

By Associate Writer - Ms. Nidhi Garg

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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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