# 5.6: Untitled Page 104

- Page ID
- 18237

## Chapter 5

Returning to Eq. 17 and making use of Eq. 18a leads to

*p*

*. *

*p*

H O

0 80

H O ,

( *T*

80F)

2 *, * 787 Pa

*vap*

2

2

1

(19)

and we are now ready to determine the humidity in Stream #1.

We are given that the total pressure in the system is equivalent to 755

mm Hg and this leads to a pressure given by

3

755 mm Hg

101.3 10 Pa

*p *

100 *, * 634 Pa

(20)

760 mm Hg/atm

atm

Use of this result in Eq. 14 leads to the humidity in Stream #1 given by *MW*

*p*

H O

H O

2

2 1

*(* humidity)

=

1

*p*

*MW*

*p*

*air *

H

2O

1

18.015 g H O/mol

2,795 Pa

2

(21)

28.84 g dry air/mol 100 *, * 634 Pa 2795 Pa

0 *. * 0178 g H O g dryair

2

We are given that Stream #2 is in equilibrium with Stream #3, thus the relative humidity in Stream #2 is 100% and the partial pressure of water vapor can be determined in terms of a *process equilibrium relation* as ( *p*

*p*

*p*

H O )

(

*vap *)

1 *, * 705 Pa

(22)

2

H O ,

2

2

3

H

*, vap*

2O

*T *15 *C*

This allows us to repeat the type of calculation illustrated by Eq. 21 to obtain

*MW*

*p*

H O

H O

2

2 2

(humidity)

=

2

*p*

*MW*

*p*

*air *

H

2O

2

(23)

0 *. * 01076g H O g dry air

2

Finally we return to Eq. 13 to determine the mass flow rate of water leaving the air conditioning unit according to

*Two‐Phase Systems* & *Equilibrium Stages *

189

*m*

H2O 3

kg H O

2

1000 kg dryair/h 0 01783

*. *

0 01076

*. *

(24)

kg dry air

7 0

*. * 7 kg H O h

2

EXAMPLE 5.8. Use of air to dry wet solids

In Figure 5.8a we have illustrated a co‐current air dryer. The solids entering the dryer contain 20% water on a mass basis and the mass flow rate of the wet solids entering the dryer is 1000 lbm/h. The dried solids contain 5% water on a mass basis, and the temperature of the solid stream leaving the dryer is 65 C. The complete *design* of the dryer is a complex process that requires a knowledge of the flow rate of the dry air entering the dryer. This can be determined by a macroscopic mass balance analysis.

*Figure 5.8a*. Air dryer

As the air and the wet solids pass through the dryer an equilibrium condition is approached and mass transfer of water from the wet solids to the air stream diminishes. As an example, we assume that the air leaving the dryer is in equilibrium with the solids leaving the dryer, and this allows us to determine the *maximum amount* of water that can be removed from the wet solids. For this type of approximation the dryer becomes an *equilibrium stage*.

To construct a control volume for the analysis of this system, we need only make cuts where information is *given* and *required* and these lead to the control volume shown in Figure 5.8b. We begin this problem with the species macroscopic *mole* balance for a steady‐state process in the absence

190