# 5.5: Untitled Page 103

## 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 H2O1

% 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

188