CHEMICAL ENGINEERING SERIES

HUMIDIFICATION
Compilation of Lectures and Solved Problems
 CHEMICAL ENGINEERING SERIES 2
HUMIDIFICATION

DEFINITION OF TERMS

HUMIDIFICATION

Involves the transfer of material between a pure liquid phase and a fixed gas that is nearly
insoluble in the liquid

HUMIDITY,

The mass of vapor carried by a unit mass of vapor-free gas; it depends only on the partial
pressure of the vapor in the mixture when the total pressure is fixed

= = = =
( ) ( )

SATURATED GAS

A gas in which the vapor is in equilibrium with the liquid at the gas temperature; the partial
pressure of vapor in saturated gas equals the vapor pressure of the liquid at the gas
temperature

=
( )

RELATIVE HUMIDITY,

Defined as the ratio of the partial pressure of the vapor to the vapor pressure of the liquid at
the gas temperature; it is usually expressed on a percentage basis; 100% relative humidity
means saturated gas and 0% relative humidity means vapor-free gas

PERCENT HUMIDITY,

Ratio of the actual humidity, H, to the saturation humidity, HS, at the gas temperature, also
on a percentage basis; percent humidity is less than the relative humidity

= =

 CHEMICAL ENGINEERING SERIES 3

HUMIDIFICATION

HUMID HEAT,

Heat energy necessary to increase the temperature of 1 g or 1 lb of gas plus whatever vapor
it may contain by 1C or 1F

= +

HUMID VOLUME,

Total volume of a unit mass of vapor-free gas plus whatever vapor it may contain at 1 atm
and the gas temperature

.
= + /

= + /

DEW POINT

Temperature to which a gas-vapor mixture must be cooled (at constant humidity) to

become saturated; the dew point of a saturated gas phase equals the gas temperature

TOTAL ENTHALPY, i

Enthalpy of a unit mass of gas plus whatever vapor it may contain; to calculate h, two
reference states must be chosen, one for the gas and one for the vapour

= ( )+ + ( )
= ( )+

ADIABATIC SATURATION TEMPERATURE,

Temperature of the gas that would be attained if the gas were saturated in an adiabatic
process

( )+ =
+
= =

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HUMIDIFICATION

DRY BULB TEMPERATURE,

Actual gas temperature

WET BULB TEMPERATURE,

Temperature obtained when the heat required to vaporize a small amount of liquid (water
for air-water system) into a large volume of gas (air for air-water system) exactly equals the
sensible heat transferred from the gas to the liquid.

HUMIDITY CHART

A convenient diagram showing the properties of mixtures of a permanent gas and a

condensable vapor

THEORY OF WET BULB TEMPERATURE

The rate of heat transfer from the gas to the liquid may be equated to the product of the rate
of vaporization and the sum of the latent heat of evaporation at temperature and the
sensible heat of the vapour; neglecting radiation
= + ( )
The rate of heat transfer may be expressed as:
= ( )
The rate of mass transfer may be expressed as:
= ( )
( )

= =

 CHEMICAL ENGINEERING SERIES 5
HUMIDIFICATION

Notations:
- mass of vapor
- mass of vapor-free gas
- mole fraction of vapor
mole fraction of vapor-free gas
molecular weight of vapor
molecular weight of vapor free gas
partial pressure of vapor
partial pressure of vapor-free gas
total pressure
saturation humidity
vapour pressure at the gas temperature
specific heat of gas
specific heat of vapor
gas temperature, in K or R
datum temperature for both gas and vapor ( T0 = 32F for air-water
problem)
latent heat of the liquid at T0
rate of sensible heat transfer to liquid
molal rate of vaporization
latent heat of the liquid at Tw
surface area of liquid
heat transfer coefficient between gas and surface of the liquid
temperature at the interface
mass transfer coefficient, mole per unit area per unit mole fraction
mole fraction of vapour at the interface
mole fraction of vapour in air-stream
(1 ) - one way diffusion factor
Schmidt Number
Prandtl Number

HUMIDIFICATION PROCESSES

1. SENSIBLE COOLING
During this process, the moisture content of air
remains constant but its temperature decreases as
it flows over a cooling coil. For moisture content to
remain constant the surface of the cooling coil
Humidity

should be dry and its surface temperature should

be greater than the dew point temperature of air. If
the cooling coil is 100% effective, then the exit
temperature of air will be equal to the coil

t2 t1
Dry Bulb Temperature
 CHEMICAL ENGINEERING SERIES 6
HUMIDIFICATION

temperature. However, in practice, the exit air temperature will be higher than the
cooling coil temperature. Below shows the sensible cooling process O-A on a
psychrometric chart

Heat Balance:
= ( )

Enthalpy Balance:
= ( )

where:
entering air temperature
leaving air temperature
entering air enthalpy
leaving air enthalpy

2. SENSIBLE HEATING

During this process, the moisture content of

air remains constant and its temperature
increases as it flows over a heating coil.

Heat Balance:

Humidity
= ( )

Enthalpy Balance:
= ( )
t1 t2
Dry Bulb Temperature
3. ADIABATIC HUMIDIFICATION

W
Liquid
t2
H2 L2
h2 T2

Gas
W L1
t1 T1
H1
h1
 CHEMICAL ENGINEERING SERIES 7
HUMIDIFICATION

t1 HT = Hi = Hw = constant

H2
t2

Humidity
H1 T = ti = t w = constant
H2
H1

t1 t2
Dry Bulb Temperature

Material Balance:

= ( )

Heat Balance:

= ( )

= ( ) ( )+ ( )

= ( ) ( )+

= ( ) ( )+

( )= ( ) ( )+

Enthalpy Balance:

+ =

Mass Transfer Equation (only gas phase involved)

( )= ( )

Heat Transfer Equation (only gas phase involved)

( )= ( )

Where:
saturation humidity at gas wet bulb temperature
gas phase heat transfer coefficient
 CHEMICAL ENGINEERING SERIES 8
HUMIDIFICATION

total contact volume

- mass transfer coefficient or enthalpy transfer coefficient

4. ADIABATIC DEHUMIDIFICATION

t1 H1

T1 t2 HT1 H2

Humidity
H1

T2 HT2
H2

t1 t2
Dry Bulb Temperature

Material Balance:

= ( )

Heat Balance:

( )= ( )+ ( )

Enthalpy Balance:

( )= ( )

Mass Transfer Equation (only gas phase involved)

( )= ( )

Heat Transfer Equation (only gas phase involved)

( )= ( )

5. WATER COOLING

When warm liquid is brought into contact with unsaturated gas, part of the liquid
evaporates and the liquid temperature drops.

Approach difference of the water discharge temperature with that of the wet bulb
temperature

Range change in water temperature of inlet to exit

 CHEMICAL ENGINEERING SERIES 9
HUMIDIFICATION

T2 HT2

H2
t2 HT1

Humidity
H1

t1
H1
H2
T1

t1 t2
Dry Bulb Temperature

Material Balance:

= ( )

Heat Balance:

( )= ( )+ ( )

Enthalpy Balance:

( )= ( )

Mass Transfer Equation (only gas phase involved)

( )= ( )

Heat Transfer Equation (only gas phase involved)

( )= ( )

pressure of vapour in saturated gas equals the vapour pressure of the liquid at the
gas temperature

=
( )
 CHEMICAL ENGINEERING SERIES 10
HUMIDIFICATION

PROBLEM # 01:

A cooling tower of a centralized air conditioning system handles 2,500 cu.m/h of water
which enters the tower at 40C. The cooled water leaves the tower at 30C. The drift loss is
0.2% while the blowdown is 0.5%. The water make-up is 50 cu.m/h. The air blown through
the tower enters at 25C and has a relative humidity of 80%. The air leaves the tower at
34C with a relative humidity of 98%. Calculate the volume of air, in cu.m/h at ambient
conditions that the forced draft fan of the cooling tower handles.

Source: CHE Board Exam May 1989

SOLUTION:

Water, L2
2,500 m3/h
T 2 = 40 C
AIR, W
t2 = 34 C
98% RH

Make-up
Water, M
50 m3/h

Drift Loss = 0.2%

Blowdown = 0.5%

AIR, W
t1 = 25 C Water, L1
T1 = 30 C
80% RH

Properties of Air:
For the inlet air:
= 100

At 25C, from steam table, = 0.4609

(80)(0.4609 )
= = 0.3687
100

=
( )
(18)(0.3687 )
= = 0.0161
(28.84)(14.7 0.3687) . .

From figure 19.2 (Unit Operations 7th edition by McCabe and Smith)
Specific volume of dry air at 25C (77F) is 13.5 ft3/lb d.a
 CHEMICAL ENGINEERING SERIES 11
HUMIDIFICATION

.
= 13.5
.
77 + 460
+ 0.0161 0.7302
. 18 1

= 13.8507
.
For the outlet air:
= 100

At 34C, from steam table, = 0.7739

(98)(0.7739 )
= = 0.7584
100

(18)(0.7584 )
= = = 0.0340
( ) (28.84)(14.7 0.7584) . .

Solve for water absorbed by the air,

= ( )
= (0.0340 0.0161)
= 55.8659 1

Consider water balance:

= +
At 40C, = 61.94 = 992.20

= 2,500 992.2 = 2,480,504.076

= +
0.2 + 0.5
= 2,480,504.076 = 17,363.53
100
= 50 992.2 = 49,610

= 49,610 17,363.53 = 32,246.47

Substitute to equation 1
= 55.8659 32,246.47 = 1,801,478.771

For the volume of air
. 2.2 . 0.028317
= 1,801,478.771 13.8507
. .

= , , .
 CHEMICAL ENGINEERING SERIES 12
HUMIDIFICATION

PROBLEM # 02:

In a plant laboratory having a floor area of 100 m2 and a ceiling height of 3 m, the
temperature and relative humidity are kept at 23.9C and 80%, respectively. The closed
loop air conditioning (AC) unit installed for the purpose has an air handling capacity to
change the air in the room of which 80% is void space, every ten minutes. The air leaving
the condenser of the AC unit has a temperature of 18.3C. Calculate:

a) The duty of the AC units in kW

b) Quantity of condensate which has to be drained from the AC unit, in kg/h

Source: CHE Board Exam May 1988

SOLUTION:

Air, W
t1 = 23.9 C Air, W
80% RH t2 = 18.3 C
AIR CONDITIONING UNIT

Condensate, w
t1 = 18.3 C

Amount of air required for the room

= (100 3 )(0.80) = 240
240 60
= = 1,440
10

For the properties of air inside the room

= 100

At 23.9C, from steam table, = 0.4303

(80)(0.4303 )
= = 0.3442
100

=
( )
(18)(0.3442 )
= = 0.0150
(28.84)(14.7 0.3442) . .

From figure 19.2 (Unit Operations 7th edition by McCabe and Smith)
Specific volume of dry air at 23.9C (75.02F) is 13.5 ft3/lb d.a
 CHEMICAL ENGINEERING SERIES 13
HUMIDIFICATION

.
= 13.5
.
75.02 + 460
+ 0.0150 0.7302
. 18 1

= 13.8256
.
= 0.245 = 1.0258

1
= 1,440
2.2 . 0.028317
13.8256
. .

= 1,671.90

Assume that the air leaving the AC unit is saturated

At 18.3C, = = 0.3051
=
( )
(18)(0.3051 )
= = 0.0132
(28.84)(14.7 0.3051) . .

From figure 19.2 (Unit Operations 7th edition by McCabe and Smith)
= 0.243 = 1.0174

Solve for water condensed:
= ( )
.
= 1,671.90 (0.0150 0.0132)
.
= . ( )

Consider heat balance:

= ( )
1.0258 + 1.0174
= = 1.0216
2
At 18.3C, = 1,056.83 BTU/lb = 2,453.05 kJ/kg
= 1,671.90 1.0216 (18.3 23.9) 3.01 2,453.05

1 1
= 16,948.55
3,600 1
= .
 CHEMICAL ENGINEERING SERIES 14
HUMIDIFICATION

PROBLEM # 03:

At an oil refinery in Batangas, cooling water for the condensers and coolers in the plant is
provided by a closed-loop cooling water system. From the plant, used cooling water is sent
to a cooling tower to reduce its temperature. Make-up water is added before the cooled
water is circulated back to the plant. In the cooling tower, the used water enters the top at
an average bulk temperature of 40C. The cooled water accumulated at the basin below the
tower has a temperature of 25C. Ambient air at 25C and 50% RH is induced into the
tower and leaves at the top at 35C fully saturated. Heat losses to the surroundings may be
assumed to be negligible. For every cu.m of cooling water used in the plant, calculate:

c) the volume of ambient air, in cu.m, that is induced into the cooling tower
d) the quantity of make-up water, in liters, that has to be added to the system

Source: CHE Board Exam November 1987

SOLUTION:

Basis: 1 m3 of cooling water

Water, L2
T2 = 40 C
AIR, W
t2 = 35 C
saturated

Make-up
Water, M

AIR, W
Water, L1
t1 = 25 C
T1 = 25 C
50% RH

Properties of Air:
For the inlet air:
= 100

At 25C, from steam table, = 0.4609

(50)(0.4609 )
= = 0.2304
100
 CHEMICAL ENGINEERING SERIES 15
HUMIDIFICATION

=
( )
(18)(0.2304 )
= = 0.0099
(28.84)(14.7 0.2304) . .

From figure 19.2 (Unit Operations 7th edition by McCabe and Smith)
Specific volume of dry air at 25C (77F) is 13.5 ft3/lb d.a

.
= 13.5
.
77 + 460
+ 0.0099 0.7302
. 18 1

= 13.7157
.

= 0.243

= ( )+

= 32 , = 1075.4

= 0.243 (77 32) + 0.0099 1,075.4

= 21.5815

For the outlet air:

= 100

At 35C (95F), from steam table, = = 0.8162

=
( )
(18)(0.8162 )
= = 0.0367
(28.84)(14.7 0.8162) . .

= 0.256

= ( )+

= 32 , = 1075.4

= 0.256 (95 32) + 0.0367 1,075.4

 CHEMICAL ENGINEERING SERIES 16
HUMIDIFICATION

= 55.5952

For the water absorbed by the air,

= ( )
= (0.0367 0.0099)
= 37.3134 1

Consider enthalpy balance:

( )= ( )

At 40C (104F), = 61.94 = 992.20

= 1.0 992.2 = 992.2 = 2,182.84

(2,182.84 ) 1 (104 77)

=
(55.5952 21.5815)
.
= 1,732.73 .

=
0.028317
= (1,732.73 . ) 13.7157
.
= . ( )

Substitute W to equation 1
1,732.73
=
37.3134
= 46.4372
1 0.028317
= 46.4372
61.94

= . ( )
 CHEMICAL ENGINEERING SERIES 17
HUMIDIFICATION

PROBLEM # 04:

The semiconductor plant at the Food Terminal Export Zone, an adiabatic dryer is used
where air enters at 160F. If the air has a dew point of 68F and it picked up 0.08 lb water
per 100 cu. ft, how saturated is the air coming out of the dryer?

Source: CHE Board Exam May 1986

SOLUTION:
0.08 lb water per
100 cu ft air
Air, W
t1 = 160 F
tdp = 68 F Air, W
ADIABATIC DRYER

Properties of Air:
For the inlet air:

From figure 19.2 (Unit Operations 7th edition by McCabe and Smith)
At dew point of 68F, and dry bulb 160F
= 0.015
.
= 90

Specific volume of dry air = 15.6

. 160 + 460
= 15.6 + 0.015 0.7302
. . 18 1

= 15.9773
.
.
= 100 = 6.2589
15.9773
. 0.015
= 100 = 0.0939
15.9773 .
0.08
= 100 = 0.08
100
= 0.0939 + 0.08 = 0.1739

0.1739
= = 0.0278
6.2589 . .

For adiabatic dryer, wet bulb remains constant, from the psychrometric chart
At 90F wet bulb and 0.0278 humidity

% = %
 CHEMICAL ENGINEERING SERIES 18
HUMIDIFICATION

PROBLEM # 05:

It is desired to condition saturated atmospheric air at 70F with entrained 0.0008 lb water
per cu ft air to hot air at 200F dry bulb and 115F wet bulb temperatures. The air is passed
thru a heater, then thru an adiabatic humidifier, then thru a reheater. The air, as it leaves
the adiabatic humidifier, has a humidity of 90%. Calculate the temperature of the air as it
leaves the (a) heater, and (b) adiabatic humidifier.

Source: CHE Board Exam May 1984

SOLUTION:

COND 3:
90%
COND 2: Humidity
HEATER HUMIDIFIER HEATER

COND 1:
Saturated
70 F COND 4:
0.0008 lb water Dry Bulb: 200 F
per cu ft air Wet Bulb: 115 F

1. Properties of air after the re-heater (condition 4)

From the psychrometric chart (figure 19.2 McCabe and Smith)
= 0.048
.
= 16.5
.

2. Properties of air after the adiabatic humidifier (condition 3)

= = 0.048

% = 100
,

(100) 0.048
, =
90
, = 0.0533
.

, =
( )
 CHEMICAL ENGINEERING SERIES 19
HUMIDIFICATION

.
0.0533 28.84 (14.7 )
. .
=
.
18 + 0.0533 28.84
. .
= 1.1566

Therefore dry bulb temperature of the air leaving the adiabatic humidifier (from steam
table)
= . ( )

Wet bulb temperature of air leaving the humidifier

= 105

3. Properties of air as it enters the heater (condition 1)

At 70F and air is saturated,
= 0.3632

, =
( )
18 (0.3632 )
, =
.
28.84 (14.7 0.3632)
.

, = 0.0158
.
= , = 0.0158
.
Using the specific volume dry air vs temperature line
.
= 13.25
.

Volume of air entering the heater

= . +

0.0158 10.731 (70 + 460)
.
= 13.25 +
. 18 (14.7 )

= 13.60
.

4. Final humidity of air leaving the heater (with entrained 0.0008 lb water per cu ft air)
= 0.0008 13.60
.
= 0.01088
.
= 0.0158 + 0.01088 = 0.02668
. . .
 CHEMICAL ENGINEERING SERIES 20
HUMIDIFICATION

5. Temperature of air leaving the heater

= = 0.02668
.
From the psychrometric chart, assume same wet bulb temperature with air leaving the
humidifier (since adiabatic conditions)

= ( )
 CHEMICAL ENGINEERING SERIES 21
HUMIDIFICATION

PROBLEM # 06:

If 400 lb of air at a dry bulb temperature of 56 F and wet bulb temperature of 50 F are
mixed with 855 lb of air at a dry bulb temperature of 82 F and a wet bulb temperature of
60 F. What will be the dry bulb and wet bulb temperature of the mixture?.

Source: CHE Board Exam May 1981

SOLUTION:

A = 400 lb
Dry Bulb: 56 F
Wet Bulb: 50 F

MIXER Mixture

B = 855 lb
Dry Bulb: 82 F
Wet Bulb: 60 F

1. Properties of air B
From the psychrometric chart (figure 19.2 McCabe and Smith)
= 0.0075
.
= 13.5
.
= 0.241
.

2. Properties of air A
From the psychrometric chart (figure 19.2 McCabe and Smith)
= 0.009
.
= 13
.
= 0.242
.
 CHEMICAL ENGINEERING SERIES 22
HUMIDIFICATION

3. Over-all material balance

+ =
= 400 + 855
= 1,255

4. Consider water balance

+ =
0.009 .
= 400
. (1 + 0.009)
= 3.5679

0.0075 .
= 855
. (1 + 0.0075)
= 6.3648

= 3.5679 + 6.3648
= 9.9327
9.9327 1,255
=
1,255 (1,255 9.9327) .
= 0.0080
.

5. Consider heat balance

=
( )= ( )
.
400 0.242 ( 56)
(1 + 0.009) .
.
= 855 0.241 ( 82)
(1 + 0.0075) .

= .

From the psychrometric chart, given the humidity and dry bulb of the mixture
=
 CHEMICAL ENGINEERING SERIES 23
HUMIDIFICATION

PROBLEM # 07:

A coke packed humidifier is to be designed to cool 2,000 cfm of saturated air from 130 to 65
F at barometric pressure. Cooling water at 55F will be allowed to heat up to 110F. Gas
velocity will be 1,200 lb of dry air per sq ft of total cross section. Water velocity is 1,150
lb/h per sq ft of total cross section. Over-all coefficient of sensible heat transfer from air to
water = 250 BTU/hft3F. Calculate the height and diameter of cooling tower required and
the amount of cooling water needed per hour.

Source: CHE Board Exam October 1977

SOLUTION:

Dry Bulb: 65 F Cooling water

55 F

COKE-
PACKED
HUMIDIFIER

Cooling water
Air 110 F
Dry Bulb: 130 F

1. Properties of air entering the humidifier

Vapor pressure at 130F

= 2.225

=
( )

Since air is saturated,

= = 2.225

(18)(2.225)
= = 0.1113
(28.84)(14.7 2.225) .

From the psychrometric chart, using the saturated volume vs temperature line

= 17.5
.
 CHEMICAL ENGINEERING SERIES 24
HUMIDIFICATION

= 0.243
.

2. Mass of dry air entering the humidifier

60 .
= 2,000
17.5
= 6,857.1428 .

3. Consider heat transfer equation

( )=
(130 110) (65 55)
=
130 110
65 55
= 14.427

6,857.1428 . 0.243 (130 65)

.
=
250 (14.427 )

= 30.0294

4. Solve for height of the humidifier

.
= 1,200 (30.0294 )
6,857.1428 .
= .

5. Solve for diameter

=
4
4(30.0294 )
=
(5.25 )
= .

6. Cooling water requirement

= 1,150 (2.70 )
4
= , .
 CHEMICAL ENGINEERING SERIES 25
HUMIDIFICATION

PROBLEM # 08:

It is desired to air condition the enclosed assembly hall of a local university. The hall
measures 120 ft x 40 ft x 70 ft. This is to be charged every 6 minutes and it is also to be
maintained at 70 F and a relative saturation of 50%. At the warmest period of the year, the
outside air is 95 F and 70% relative saturation. It is planned to cool and dehumidify this
air to the desired humidity by the use of a coke-packed tower. The air will leave the tower
saturated and it is to be reheated to the desired temperature before being blown to the hall.
City water for available for cooling is 45 F.

DATA:

1) The over-all coefficient of a sensible heat transfer from air to water, UGa = 250
BTU/hft3F
2) lb inlet water per sq ft cross section, L/S = 1,150
3) assume latent heat of vaporization = 1,055 BTU/lb
4) mass velocity of air = 1,200 lb/hft2

a) What shall be the height and diameter of the cooling tower?

b) To what temperature is the air cooled in the tower?
c) What is the temperature of the outlet water?
d) Calculate the

Source: CHE Board Exam July 1951

SOLUTION:

Condition
2

Condition Air Cooling water

3 100% 45 F
saturation
HEATER
120 ft x 40 ft x 70 ft Air
Dry Bulb: 70 F
50% saturation

Air in the room to COKE-PACKED

be recharged DEHUMIDIFIER
every 6 min

Air
Dry Bulb: 95 F Condition
70% saturation 1
 CHEMICAL ENGINEERING SERIES 26
HUMIDIFICATION

1. Volume of air required in the room

(120 )(40 )(70 )
=
1
6
60
= 3,360,000

2. Properties of air entering the room (condition 3)

Vapor pressure at 70F
= 0.3632

=
( )
(18)(0.3632)
= = 0.0158
(28.84)(14.7 0.3632) .

Since air is 50% saturated,

% = 100

0.0158 (50)
.
= = 0.0079
100 .

From the psychrometric chart, using the specific volume dry air vs temperature line
.
= 13.40
.

. 0.0079 10.731 (70 + 460)
.
= 13.40 +
. 18 (14.7 )

= 13.57
.

= 0.241
.

3. Mass of dry air required

.
= 3,360,000
13.57
.
= 247,605.0111

4. Properties of air entering the heater (leaving the tower)

 CHEMICAL ENGINEERING SERIES 27
HUMIDIFICATION

= = 0.0079
.

Since air is saturated

= = 0.0079
.
=
( )
(0.0079)(28.84)(14.7)
=
18 + (0.0079)(28.84)
= 0.1837

From steam table,

= . ( )

5. Properties of air entering the humidifier

Vapor pressure at 95F
= 0.8162

=
( )
(18)(0.8162)
= = 0.0367
(28.84)(14.7 0.8162) .

Since air is 70% saturated,

% = 100

0.0367 (70)
.
= = 0.0257
100 .

From the psychrometric chart, using the specific volume dry air vs temperature line
.
= 13.90
.

. 0.0257 10.731 (95 + 460)
.
= 13.90 +
. 18 (14.7 )

= 14.48
.

= 0.255
.
 CHEMICAL ENGINEERING SERIES 28
HUMIDIFICATION

6. For the tower diameter

= (1 + )
.
247,605.0111 (1 + 0.0257)
.
=
1,200

= 211.6404
=
4
4(211.6404 )
=

= .

7. Cooling water required

= 1,150

= (211.6404 ) 1,150

= 243,386.46

8. Mass water evaporated in the dehumidifier

= ( )
.
= 247,605.0111 (0.0257 0.0079)
.
= 4,407.3692

9. Heat balance around the dehumidifier

= ( ) + ( )
+ 0.255 + 0.241
= = = 0.248
2 2 .
.
= 247,605.0111 0.248 (50.79 95)
.
+ 4,407.3692 1,055

= 7,364,535.656

= 7,364,535.656 = 7,364,535.656

= ( )
 CHEMICAL ENGINEERING SERIES 29
HUMIDIFICATION

7,364,535.656 = 243,386.46 1 ( 45)

= . ( )

10. Consider heat transfer equation

( )=
(95 75.26) (50.79 45)
=
95 75.26
50.79 45
= 11.3737
.
247,605.0111 0.248 (95 50.79)
.
=
250 (11.3737 )

= 954.7504

11. Solve for height of the humidifier

=
954.7504
=
211.6404
= .