mech 430: psychrometrics 5 - queen's universitymy.me.queensu.ca/courses/mech4301/430 psychro 5...

MECH 430: psychrometrics 5

Humidification

Steam Injection Liquid Injection

See Ex. 12.14 in text. Alternative solution recommended with Psychrometric Chart.


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Evaporative Cooling


Suitable for cooling in hot, relatively dry climates

Low moist air enters and readily “soaks” up liquid via evaporation.

This evaporation accomplished by Q extracted from moist air intake.

Results in a decrease in the temperature of the moist air

Liquid either in soaked pad or sprayed in air stream.


Mass and Energy Balances

Assuming 0cvW and Q with surroundings = 0.

2 2 2 2 1 1 1 1( ) ( ) ( )a g f a gh h h h h Eq. 30.1

where,

All of the water is assumed to evaporate into air stream

2 1( ) fh = Energy carried in with the injected liquid

water. Typically, this is much less than the enthalpy of moist air in or out.

Thus mixH constant and we can represent process as a

constant mixH process. Since mix

a

H

m and WBT lines coincide, it is

also a constant WBT process.

In reality, the process deviates from a line of constant mix

a

H

m to

a slightly higher mix

a

H

m.


Example Text Problem 12.86 Air at 35oC, 1 bar and 10% enters an evaporative cooler

(operating at steady state) at 50 m3/min. Liquid water at 20oC enters the cooler and fully evaporates. Moist air leaves the

cooler at 25oC, 1 bar. Assume Q with surroundings = 0.

Find ,w exitm

wm at T3 = 20oC

T1 = 35oC T2 = 25oC P1 = 1 bar P2 = 1 bar

2 ? 10%

3

1( ) 50min

mAV


Solution

1 2a a am m m

2 1w v vm m m

Since v

a

m

m , then 2 1

w

a

m

m

Therefore, to find wm , we need: 1 2, , am

To find 1 :

11

1

0.622 v

v

P

P P

where

1 1 1

1

1

(0.1)(0.05628)

0.005628

v g

v

v

P P

P

P bar

1

1

(0.622)(0.005628)

(1 0.005628)

0.00352 ( ) / ( )kg v kg a


To find 2 :

Recall equation 30.1:

2 2 2 2 1 3 1 1 1( ) ( ) ( )a g f a gh h h h h

Isolating 2w :

1 2 1 1 3

2

2 3

2

2

( ) ( )

(308.23 298.18) 0.00352(2565.3 83.96)

2547.2 83.96

( )0.00763

( )

a a g f

g f

h h h h

h h

kg v

kg a


To find am :

1 1

1

3 5 2

( )

(50 / min)(0.994373 10 / )

8314 28.97 (308 )

56.25 ( ) / min

aa

a

a

a

AV Pm

RT

M

m x N mm

N m kgK

kmol K kmol

m kg a

Thus

2 1( )

56.25 ( ) / min 0.00763 0.00352

0.231 / min

w a

w

w

m m

m kg a

m kg


2( ) (25 )o

g gP T P C

To find 2 :

22

2

2

2

0.622

(0.00763)(1 )

0.00763 0.622

0.01212

v

v

v

PP

barP

P bar

22

2

2

2

0.01212

0.03169

0.382 or 38.2%

v

g

P

P

bar

bar


Cooling Towers


Used, for example, when:

Sufficient cooling unattainable from natural sources e.g., lakes, rivers.

Environmental concerns limit Tmax of cooling water returned to environment.

Can be operated by Natural or Forced Convection Configurations:

Counterflow

Cross-flow

Usually neglect FANW and Q with surroundings as these are

usually much smaller than other quantities considered in energy balances on cooling towers.


Analysis Mass Rate Balances

3 4a am m (dry air)

1 5 3 2 4v vm m m m m (water)

Since 1 2m m :

then 5 4 3v vm m m

Since 4 4v am m and 3 3v am m :

then 5 4 3( )am m Eq. 30.2


Energy Rate Balance

1 1 3 3 3 5 5 2 2 4 4 40 ( ) ( )w a a v v w w a a v vm h m h m h m h m h m h m h

Let each v gh h at given T

Let each w fh h at given T

with

1 2m m , 5 4 3( )am m , 3 3v am m , 4 4v am m

We can re-write the above equations as:

1 1 2

4 4 4 3 3 3 4 3 5

( )

( ) ( ) ( )

f f

a

a g a g f

m h hm

h h h h h

Eq. 30.3


Example Cooling water leaves the condenser of the power plant and enters a cooling tower at a rate of 100 kg/s. The water is cooled to 22oC in the cooling tower by air which enters the

tower at 1 atm, 20oC, 60% and leaves saturated at 30oC.

Neglect the power input to the fan. Find: ) )a makeup watera m b m

4

4

30

100%

oT C

1

1

35

100 /

oT C

m kg s

2

2

22

100 /

oT C

m kg s

5

5

22

?

oT C

m

3

3

3

1

60%

20

?

o

a

P atm

T C

m


Solution To solve, evaluate Eq. 30.3 and then 30.2 2 approaches:

1) Tables 2) Psychrometric Chart

* Review text Ex. 12-17 to familiarize yourself with tabular approach. In this example, we will use a psychrometric chart.

1

1

2

5 2 5 2

100 /

(35 ) 146.68 /

(22 ) 92.33 /

(in this example since )

o

f f

o

f f

f f

m kg s

h h C kJ kg

h h C kJ kg

h h T T

Mixture enthalpies, 4 3, from psychrometric chart:

4

3

4 4 4

4

3 3 3

3

0.0273

0.0087

( ) 100.0 / ( )

( ) 42.2 / ( )

mixa v

a

mixa v

a

Hh h kJ kg a

m

Hh h kJ kg a

m


2T 4T

3

4w

3

mix

a

H

m

4

mix

a

H

m


Evaluate Eq. 30.3

1 1 2

4 4 4 3 3 3 4 3 5

( )

( ) ( ) ( )

(100 / )(146.68 92.33) /

(100 42.2) / (0.0273 0.0087)(92.33) /

96.9 /

f f

a

a g a g f

a

a

m h hm

h h h h h

kg s kJ kgm

kJ kg kJ kg

m kg s

Evaluate Eq. 30.2

5 4 3

5

5

( )

(96.9 / )(0.0273 0.0087)

1.80 /

am m

m kg s

m kg s

Note: , 100 /in waterm kg s

, 100 1.80 98.2 /returned waterm kg s

That is, only 1.8% of condensate lost from cooling tower.


What if the cooling effect was provided by circulating cooling water coming in at 5oC and leaving at 30oC? ?cwm

( (35 ) (22 )) ( (30 ) (5 ))

(146.68 92.33)100 /

(125.79 20.98)

51.86 /

o o o o

cond f f cw f f

cw

cw

m h C h C m h C h C

m kg s

m kg s

Recall for a cooling tower, 1.80 /makeup waterm kg s

35oC 35oC

22oC 5oC

100 /condm kg s ?cwm

1

2

3

4

mech 430: psychrometrics 5 - queen's universitymy.me.queensu.ca/courses/mech4301/430 psychro 5...

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