# 5.5: Untitled Page 103

- Page ID
- 18236

## Chapter 5

humidity precisely defined by Eq. 5‐36. We start by examining the ratio of mass flow rates (water to dry air) given by

**v ** **n ** *dA*

H2O

( *m*

)

H O

*A*

1

1

2

(7)

( *m*

)

*air * 1

**v ** **n ** *dA*

*air*

*A* 1

If the velocity, **v ** **n **, is *uniform* across the entrance, or the species densities,

and

, are *uniform* across the entrance, we can follow the H O

*air*

2

development in Sec. 4.5.1 to conclude that

**v ** **n ** *dA*

H2O

( *m*

)

*Q*

H O 1

*A*

H O 1

1

H O 1

1

2

2

2

(8)

( *m*

)

*Q*

*air * 1

**v ** **n ** *dA*

*air * 1

1

*air * 1

*air*

*A* 1

In addition, if we accept the ratio of the *area average species densities* as our measure of the humidity we obtain

( *m*

)

H O 1

H O

2

2

1

humidity

(9)

1

( *m*

)

*air * 1

*air * 1

This suggests a particular strategy for solving this mass balance problem.

Dividing Eq. 5 by the mass flow rate of dry air in Stream #1 leads to ( *m*

)

( *m*

)

( *m*

)

H O 1

H O 2

H O

2

2

2

3

0

(10)

( *m*

)

( *m*

)

( *m*

)

*air * 1

*air * 1

*air * 1

and on the basis of the mass balance for “dry air” given by Eq. 4, we can express this result in the form

( *m*

)

( *m*

)

( *m*

)

H O 1

H O 2

H O

2

2

2

3

0

(11)

( *m*

)

( *m*

)

( *m*

)

*air * 1

*air * 2

*air * 1

Use of equations of the form of Eq. 9 leads to a “humidity balance”

equation given by

*Two‐Phase Systems* & *Equilibrium Stages* 187

( *m*

)

H

humidity humidity

O

2

3

(12)

1

2

( *m*

)

*air * 1

Our objective in this example is to determine how much liquid water is removed from the air, and Eq. 12 provides this information in the form

( *m*

)

( *m* )

humidity

humidity

(13)

H O 3

*air * 1

2

1

2

Here it becomes clear that the solution to this problem requires that we determine the humidity in Streams #1 and #2. This motivates us to make use of Eq. 5‐37 that we list here as

*MW*

*p*

H O

H O

2

2

point humidity =

(14)

*MW*

*p * *p*

*air *

H2O

In addition, we are given information about the percent relative humidity defined by Eq. 5‐38 and listed here as

*p* H O

2

*%* relative humidity =

100

(15)

*p* H O, *vap*

2

In Stream #1 the percent relative humidity is 80% and this provides

*p* H2O1

*%* relative humidity =

100 80

(16)

1

*p* H O, *vap*

2

1

which allows us to express the *partial pressure* of water vapor in Stream #1

as

*p*

*. *

*p*

H

0 80

(17)

O

O *vap*

2

H ,2

1

1

The *vapor pressure* of water, in both Stream #1 and Stream #2, can be determined by Antoine’s equation (see Appendix A3) that provides

*p*

*p*

H O ,

H O ,

( *T*

80F)

3 484

*. *

kPa

*vap*

*vap*

(18a)

2

2

1

*p*

*p*

( *T * 15C) 1 705

*. *

kPa

H O , *vap*

*,vap*

(18b)

2

H2O

2