249326327-Humidification.pdf

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 1°C or 1°F =

+

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 ( − )+ − = = −

= +

CHEMICAL ENGINEERING SERIES 4 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 = 32°F 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 ℎ – 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

Humidity

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 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.

(

=

Enthalpy Balance: =

(

)

−

−

Humidity

Heat Balance:

) t1

t2

Dry Bulb Temperature

3. ADIABATIC HUMIDIFICATION W t2 H2 h2

Liquid L2 T2

Gas W t1 H1 h1

L1 T1

CHEMICAL ENGINEERING SERIES 7 HUMIDIFICATION HT = Hi = Hw = constant

t1

Humidity

H1

H2

t2 T = ti = t w = constant

H2

t1

H1

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 ℎ – 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

t2

Humidity

T1 H1

HT1

H2

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

HT1

H2

Humidity

t2

HT2

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 40°C. The cooled water leaves the tower at 30°C. 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 25°C and has a relative humidity of 80%. The air leaves the tower at 34°C 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 80% RH

Water, L1 T1 = 30 C

Properties of Air: For the inlet air: = 100

At 25°C, from steam table, = 0.4609 (80)(0.4609 ) = = 0.3687 100 =

( − ) (18)(0.3687 ) = (28.84)(14.7 − 0.3687)

= 0.0161

. .

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

CHEMICAL ENGINEERING SERIES 11 HUMIDIFICATION

.

= 13.5 +

.

0.0161

.

18

= 13.8507

〈77 + 460〉°

∙

0.7302

∙°

1

.

For the outlet air: = 100

At 34°C, from steam table, = 0.7739 (98)(0.7739 ) = = 0.7584 100 =

( −

)

=

(18)(0.7584 ) (28.84)(14.7 − 0.7584)

= 0.0340

. .

Solve for water absorbed by the air, = ( − ) = (0.0340 − 0.0161) = 55.8659 ⟶ 1 Consider water balance: = +ℓ At 40°C,

= 61.94

= 2,500

= 992.20

992.2

ℎ

+ 0.2 + 0.5 ℓ= 100

= 2,480,504.076

ℎ

ℓ=

= 50

ℎ

2,480,504.076

992.2

= 49,610

= 49,610 − 17,363.53 = 32,246.47

= 17,363.53

ℎ

ℎ

ℎ

ℎ

Substitute to equation 1 = 55.8659 32,246.47

ℎ

= 1,801,478.771

ℎ

For the volume of air = 1,801,478.771

= ,

,

.

. ℎ

13.8507

2.2 .

. .

0.028317

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 23.9°C 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 18.3°C. 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 80% RH

Air, W 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 ℎ ℎ For the properties of air inside the room = 100

At 23.9°C, from steam table, = 0.4303 (80)(0.4303 ) = = 0.3442 100 =

( − ) (18)(0.3442 ) = (28.84)(14.7 − 0.3442)

= 0.0150

. .

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

CHEMICAL ENGINEERING SERIES 13 HUMIDIFICATION

.

= 13.5 +

.

0.0150

.

= 13.8256 = 0.245

= 1,440

18

〈75.02 + 460〉°

∙

0.7302

∙°

1

.

∙°

= 1.0258

∙ 1

ℎ

= 1,671.90

13.8256

2.2

.

.

0.028317

.

ℎ

Assume that the air leaving the AC unit is saturated At 18.3°C, = = 0.3051 =

( − ) (18)(0.3051 ) = (28.84)(14.7 − 0.3051)

= 0.0132

. .

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

= 1.0174 ∙° ∙ Solve for water condensed: = ( − ) . (0.0150 − 0.0132) = 1,671.90 ℎ = .

.

( )

Consider heat balance: = ̅ ( − )− 1.0258 + 1.0174 ̅ = = 1.0216 2 ∙ At 18.3°C, λ = 1,056.83 BTU/lb = 2,453.05 kJ/kg =

1,671.90

= −16,948.55 = .

ℎ ℎ

1.0216 1ℎ 3,600

(18.3 − 23.9)

∙ 1 1

− 3.01

ℎ

2,453.05

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 40°C. The cooled water accumulated at the basin below the tower has a temperature of 25°C. Ambient air at 25°C and 50% RH is induced into the tower and leaves at the top at 35°C 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 t1 = 25 C 50% RH

Water, L1 T1 = 25 C

Properties of Air: For the inlet air: = 100

At 25°C, from steam table, = 0.4609 (50)(0.4609 ) = = 0.2304 100

CHEMICAL ENGINEERING SERIES 15 HUMIDIFICATION

=

( − ) (18)(0.2304 ) = (28.84)(14.7 − 0.2304)

= 0.0099

. .

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

= 13.5 +

.

0.0099

.

= 13.7157

= 0.243 =

18

∙°

1

.

∙°

( −

)+ = 32° ,

=

〈77 + 460〉°

∙

0.7302

0.243

∙°

= 1075.4

(77 − 32)°

+ 0.0099

1,075.4

= 21.5815

For the outlet air: = 100

At 35°C (95°F), from steam table,

=

= 0.8162

=

( − ) (18)(0.8162 ) = (28.84)(14.7 − 0.8162) = 0.256

=

( −

0.256

. .

∙° )+ = 32° ,

=

= 0.0367

∙°

= 1075.4

(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 40°C (104°F), = 1.0

= 61.94

992.2

(2,182.84

) 1

=

= 992.2

∙°

= 2,182.84

(104 − 77)°

(55.5952 − 21.5815) = 1,732.73

= 992.20

.

.

= = (1,732.73 =

. ) 13.7157

.

0.028317 .

( )

Substitute W to equation 1 1,732.73 = 37.3134 = 46.4372 = 46.4372 = .

1 61.94

0.028317

( )

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 160°F. If the air has a dew point of 68°F 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 68°F, and dry bulb 160°F = 0.015

.

= 90°

Specific volume of dry air = 15.6 . = 15.6 + 0.015 . . 18 = 15.9773

〈160 + 460〉°

∙

0.7302

∙°

. .

= 100

= 6.2589

15.9773 .

= 100

0.015

15.9773 ℎ

= 100

. 0.08 100

= 0.0939 = 0.08

= 0.0939 + 0.08 = 0.1739 =

0.1739 6.2589

.

= 0.0278

.

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

=

%

1

CHEMICAL ENGINEERING SERIES 18 HUMIDIFICATION

PROBLEM # 05: It is desired to condition saturated atmospheric air at 70°F with entrained 0.0008 lb water per cu ft air to hot air at 200°F dry bulb and 115°F 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% Humidity

COND 2: HEATER

HUMIDIFIER

COND 1: Saturated 70 F 0.0008 lb water per cu ft air

HEATER

COND 4: Dry Bulb: 200 F 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 ,

=

,

= 0.0533

,

=

90

( −

.

)

CHEMICAL ENGINEERING SERIES 19 HUMIDIFICATION 0.0533 = 18

.

28.84

.

.

+ 0.0533

(14.7

) .

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 70°F and air is saturated, = 0.3632 ,

=

,

=

( −

) (0.3632

18 .

28.84 ,

.

= 0.0158 =

,

)

(14.7 − 0.3632)

.

= 0.0158

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

Volume of air entering the heater = . + 0.0158 = 13.25

= 13.60

+

.

∙

10.731

.

(70 + 460)°

∙° (14.7

18

)

.

4. Final humidity of air leaving the heater (with entrained 0.0008 lb water per cu ft air) = 0.0008 = 0.01088 = 0.0158

.

+ 0.01088

13.60

.

. .

= 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 = 9.9327 9.9327 = 1,255 = 0.0080

+ 6.3648 1,255 (1,255 − 9.9327) .

5. Consider heat balance =− ( − )=− . (1 + 0.009)

400

= − 855 =

.

.

(

−

) 0.242

. ∙° . (1 + 0.0075)

(

− 56) 0.241

. ∙°

(

− 82)

°

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 55°F will be allowed to heat up to 110°F. 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/h·ft3·°F. 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

COKEPACKED HUMIDIFIER

Cooling water 110 F

Air Dry Bulb: 130 F

1. Properties of air entering the humidifier Vapor pressure at 130°F = 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 ℎ 17.5 = 6,857.1428

.

ℎ

3. Consider heat transfer equation ( − )= ∆ (130 − 110) − (65 − 55) ∆ = 130 − 110 65 − 55 ∆ = 14.427 ° 6,857.1428 =

ℎ 250

.

0.243

ℎ∙

∙°

. ∙°

(130 − 65)°

(14.427° )

= 30.0294

4. Solve for height of the humidifier . ℎ (30.0294 ) = 1,200 ℎ∙ 6,857.1428 = . 5. Solve for diameter = =

4 ) 4(30.0294 (5.25 )

= .

6. Cooling water requirement = 1,150 = ,

.

ℎ∙

4

(2.70

)

.

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/h·ft3·°F 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/h·ft2 a) b) c) d)

What shall be the height and diameter of the cooling tower? To what temperature is the air cooled in the tower? What is the temperature of the outlet water? Calculate the

Source: CHE Board Exam July 1951 SOLUTION:

Condition 2 Condition 3 HEATER 120 ft x 40 ft x 70 ft

Air 100% saturation

Air Dry Bulb: 70 F 50% saturation

Air in the room to be recharged every 6 min

COKE-PACKED DEHUMIDIFIER

Air Dry Bulb: 95 F 70% saturation

Condition 1

Cooling water 45 F

CHEMICAL ENGINEERING SERIES 26 HUMIDIFICATION

1. Volume of air required in the room (120 )(40 )(70 ) = 1ℎ 6 60 = 3,360,000

ℎ

2. Properties of air entering the room (condition 3) Vapor pressure at 70°F = 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 . ∙ (70 + 460)° 0.0079 10.731 . . ∙° = 13.40 + . (14.7 ) 18 = 13.57

= 0.241

.

. ∙°

3. Mass of dry air required = 3,360,000 = 247,605.0111

. ℎ

13.57 . ℎ

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 95°F = 0.8162 =

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

.

Since air is 70% saturated, %

=

100

0.0367 =

. 100

(70) = 0.0257

.

From the psychrometric chart, using the specific volume dry air vs temperature line . = 13.90 . ∙ (95 + 460)° 0.0257 10.731 . . ∙° = 13.90 + . (14.7 ) 18 = 14.48

= 0.255

.

. ∙°

CHEMICAL ENGINEERING SERIES 28 HUMIDIFICATION

6. For the tower diameter (1 +

=

) .

247,605.0111

(1 + 0.0257)

ℎ

=

1,200

.

ℎ∙

= 211.6404 = = =

4 4(211.6404

)

.

7. Cooling water required = 1,150 = (211.6404

) 1,150

= 243,386.46

ℎ

8. Mass water evaporated in the dehumidifier ℰ= ( − ) . (0.0257 − 0.0079) ℰ = 247,605.0111 ℎ ℰ = 4,407.3692

.

ℎ

9. Heat balance around the dehumidifier ̅ ( − ) + ℰ(− ) = + 0.255 + 0.241 ̅ = = = 0.248 2 2 . ∙° . (50.79 − 95)° = 247,605.0111 0.248 ℎ . ∙° +

4,407.3692

= −7,364,535.656

−1,055

ℎ

ℎ

=− = − −7,364,535.656 =

(

−

)

ℎ

= 7,364,535.656

ℎ

CHEMICAL ENGINEERING SERIES 29 HUMIDIFICATION 7,364,535.656 =

.

ℎ

= 243,386.46

°

ℎ

1

∙°

( )

10. Consider heat transfer equation ̅ ( − )= ∆ (95 − 75.26) − (50.79 − 45) ∆ = 95 − 75.26 50.79 − 45 ∆ = 11.3737 ° . 247,605.0111 0.248 ℎ . ∙° = 250

ℎ∙

∙°

= 954.7504

11. Solve for height of the humidifier = 954.7504 = 211.6404 = .

(95 − 50.79)°

(11.3737° )

(

− 45)