# 5.23: Untitled Page 120

## Chapter 5

hydroxide leaves in the vapor stream, that none accumulates in the boiler, and that the temperature of the liquid entering and leaving the boiler is a constant.

Assume that the solid calcium hydroxide leaving the boiler is in equilibrium with the dissolved carbon dioxide, i.e., the boiler is an equilibrium stage. The solubility is often expressed as;

Equilibrium relation:

solubility = S  g of Ca(OH) g of H O

2

2

however, a more precise description can be constructed.

In this problem you are asked to develop a general solution for the mass fraction of the suspended solid in the liquid stream leaving the boiler in terms of

 and S. For  = 0.50, 0.21, and 0.075, determine the mass fraction of suspended

solid when S .

3

2 5  10 .

Figure 5.29. Precipitation of calcium hydroxide in a boiler 5‐30. In problem 5‐29 an equilibrium relation between solid calcium hydroxide and dissolved calcium hydroxide was given by

Equilibrium relation:

solubility = S  g of Ca(OH) g of H O

(1)

2

2

In Sec. 5.3.1 we expressed the general gas/liquid equilibrium relation for species A in terms of the chemical potential and for a solid/liquid system we would express Eq. 5‐24 as

Equilibrium relation:

( )

 ( )

(2)

A solid

A liquid

Two‐Phase Systems & Equilibrium Stages 221

For the process considered in Problem 5‐29, we assume that the solid phase is pure calcium hydroxide so that Eq. 2 takes the form

Equilibrium relation:

(O )

 ( )

(3)

A solid

A liquid

The description of phase equilibrium phenomena in terms of the chemical potential will be the subject of a subsequent course in thermodynamics; however, at this point one can illustrate how Eqs. 1 and 3 are related.

The development begins with a general representation for the chemical potential at some fixed temperature and pressure. This is given by ( )

(

F T , p, x , x , etc.)

(3)

A liquid

A

B

where x is the mole fraction of species A (calcium hydroxide) in the liquid A

phase. A Taylor series expansion about x  0 leads to (see Problem 5‐31) A

2

F

F

2

( )

F

( x ) 

( x )  .....

(4)

A liquid

x 0

A

2

A

A

x

A x

x

0

A

A

x 0

A

The first term in this expansion is zero and when the mole fraction of species A is small compared to one, x  1 , we can make use of a linear form of Eq. 4 given A

by

F

( )

( x )

A liquid

A

x

(5)

A x 0

A

In this problem you are asked to use Eq. 3 along with Eq. 5 and the approximation

c

A

x

,

when c  c

(6)

A

A

B

cB

to derive Eq. 1. Here c represents the molar concentration of calcium A

hydroxide and c represents the molar concentration of water. In terms of B

species A and species B, it will be convenient to express the solubility in the form Equilibrium relation:

solubility = S  

(7)

A

B

and note that this can be related to the molar form by use of c MW   and A

A

A

c MW   .

B

B

B

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