A. Lecture Note 1

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EKC316 - SEPARATION PROCESHUMIDIFICATION

DR. AZAM T MOHD DIN

SCHOOL OF CHEMICAL ENGINEERIN

UNIVERSITI SAINS MALAYSIA

Semester II, 2014/2015



TEXT BOOKS



From left to right: a reinforced concrete tower, a wood

tower, and a hyperbolic tower built with a steel framework

and wood cladding

w



HUMIDITY

• The amount of water vapor in air.

How does relative

humidity affects people?

0% humidity 100% humidity

Psychrometry

Moist air



London, 2006-2012



Penang, 2006-2012



TERMINOLOGIES & DEFINITIONS

•Temperature : dry-bulb, wet-bulb

•Humidity : Relative, Absolute, Percent

•Enthalpy



Dry-bulb Temperature

• It is true temperature of air measured (or,

non-condensable and condensable mixture

thermometer whose bulb is dry.



Wet-bulb Temperature

• It is the steady-state temperature attaine

small amount of evaporating water in a m

such that the sensible heat transferred frair to the liquid is equal to the laten

required for evaporation.



Dew point

• A temperature at which a vapor-gas mixture must be cconstant humidity) to become saturated.

• The dew point of a saturated gas equals the gas temper

• If a vapor-gas mixture is gradually cooled at a constant

the temperature at which it just becomes saturated is aits dew point.



Relative humidity

• It is the ratio of partial pressure of water

(pA ) in air at a given temperature to the

pressure of water (pvA ) at the same tempera

% x p

piditylative hum

v A

A 100Re

Relative humidity does not ‘exp

moisture content of a gas, but g

of saturation’ of the gas at a gi



Absolute humidity (simply humidity)

• It is the direct measurement of moisture con

a gas. The mass of water vapor per unit m

dry gas is called absolute humidity, Y ’

.

Eq. 1



Percent humidity or percent saturation

• It is the relation between absolute humidity to that of saturation humidity at the same

and pressure.

• where, Y’ is absolute humidity of sample of air and Y’s is humidity at same temperature

if saturated with water vapor.

Eq. 2

Eq. 3



Humid volume

•The humid volume, vH, is defined as the vol

unit mass of dry air with accompanying

vapor at a given temperature and pressure.Assuming idea

T G is gas tempEq. 5



Humid Heat

• The humid heat, cH, is the heat energy required to

temperature of unit mass of dry air with the accom

water vapor by one (1) degree.

• At ordinary T & P, the heat capacity of dry air =

kJ/kg.K and that of water vapor as 1.88 kJ/kg.K

air)(K)drykJ/(kg88.1005.1 'Y c H E



Enthalpy

• The enthalpy of a vapor-gas mixture, H’ is the

the relative enthalpies of gas and vapor conte

• Enthalpy = Latent heat + Sensible heat)('' 00 T T cY H G H



•At 0º C, λ 0 = 2500 kj/kg

kJ/kg))('88.1005.1('2500' 0T T Y Y H G



Sensible heat

• When an object is heated, its temperature rises as heat is added. The increase in heat is called sensib

when heat is removed from an object and its temperature falls, the heat removed is also called sensibl

causes a change in temperature in an object is called sensible heat.

Latent heat

• All pure substances in nature are able to change their state. Solids can become liquids (ice to water

become gases (water to vapor) but changes such as these require the addition or removal of heat. The

these changes is called latent heat.

• Latent heat however, does not affect the temperature of a substance - for example, water remains at 10

The heat added to keep the water boiling is latent heat. Heat that causes a change of state with no chan

is called latent heat.

• Appreciating this difference is fundamental to understanding why refrigerant is used in cooling systemwhy the terms 'total capacity' (sensible & latent heat) and 'sensible capacity' are used to define a unit's

During the cooling cycling, condensation forms within the unit due to the removal of latent heat from

capacity is the capacity required to lower the temperature and latent capacity is the capacity to remove

the air.

Ref: https://www.spaceair.co.uk/faqs/what-is-the-difference-between-sensible-and-latent-heat



Adiabatic Saturation Temperature, Ts

• The adiabatic saturation temperature is a thermodynamic p

moist air that is defined as the temperature that the air stre

achieve if it were allowed to become saturated adiabati

adiabatic saturation temperature is computed by equ

enthalpy of moist air at a given temperature and relative h

the enthalpy of a saturated air-water mixture at the

saturation temperature (ASHRAE, 1996).



The process of adiabatic saturation of air



• an adiabatic saturator is a device in which air flows through an infinite

containing water.

• As the air comes in contact with water in the duct, there will be heat and mbetween water and air.

• If the duct is infinitely long, then at the exit, there would exist perfec

between air and water at steady state. Air at the exit would be fully satur

temperature is equal to that of water temperature.

• The device is adiabatic as the walls of the chamber are thermally insulatedcontinue the process, makeup water has to be provided to compensate for t

water evaporated into the air. The temperature of the make-up water is

that it is the same as that in the duct.



Psychrometric Charts

The psychrometric chart characterizes the interdependences of seven properties of w

mixture:

• dry-bulb temperature

• wet-bulb temperature,

• relative humidity,

• absolute humidity,

• dew point,

• enthalpy and

• specific volume



• If any two of these quantities are known, t

other five quantities can be readily obtain

from the Psychrometric chart.



• If TG is the dry-bulb temperature of air and Y’ is its humidity,

denoted by point ‘a’. It falls on the constant humidity line, A%

• The adiabatic saturation line through ‘a’ is ‘ab’.

• ‘c’ point indicates its humidity, Y’.

• The adiabatic saturation temperature, Ts is obtained by dr

vertical line through ‘b’. For air-water system, wet-bulb temp

is practically same as Ts.



• The humidity of the adiabatically saturated air is given by the poin

• The dew point Td is given by the point ‘d’ that can be reached by m

horizontally from the point ‘a’ to 100% humidity line and then movvertically down to the temperature axis.

• The humid volume of saturated air at TG corresponds to the point ‘

of dry air at TG is given by point ‘g’.

•The point ‘m’ gives the humid volume if the humidity is Y’ and it is rinterpolation between ‘g’ and ‘f’.

• Enthalpy of a sample of air can also be obtained from humidity cha



IN-CLASS EXERCISE 1

Determine the following psychrometric properties of a moist air sample having

temperature 27ºC and a humidity of 0.015 kg/kg dry air using the pyschrometric

the vapour pressure equation for water:

a) Relative humidity b) Dew point

c) Adiabatic saturation temperature d) Wet bulb temperaturee) Enthalpy f) Humid volume

g) Humid heat



QUIZ1

Determine the following psychrometric properties of a moist air sample having

temperature 37.5ºC and a humidity of 0.02 kg/kg dry air using the pyschrometric

the vapour pressure equation for water:

a) Relative humidity b) Dew point

c) Adiabatic saturation temperature d) Wet bulb temperaturee) Enthalpy f) Humid volume

g) Humid heat



DEFINITIONS

•Humidification involves the transfer of water fliquid phase into a gaseous mixture of air and

vapor (Geankoplis, 2003).

• The process of increasing the moisture contenis called humidification (Dutta, 2007)



What is dehumidification?



AIR-WATER CONTACTING APPS

• Water cooling – air-water contacting is done mostlypurpose of cooling the warm water before it can be reus

• Humidification of gas for drying of solids under c

condition.

• Dehumidification and cooling of gas in air conditioning.

• Gas cooling



HEATING WITHOUT HUMIDIFICATION

1 2

1

2

#1h

#

2h

1

21 2

# #

2 1 H aQ m h h

• Constant humidity ra

• Relative humidity

decreases• Outlet relative humid

may be too dry to be

comfortable



HEATING WITH HUMIDIFICATION H Q

1 2

water

1T 1'T

1

1'

#

1h

#

1'h

1

1'

1

1'

2

#

2h

2 2

# #2 1 2

H

a

Qh h

m

#

1 H a water w

Q m h m h

1 2

2

w water w

water w w

a a a

m m m

m m m

m m m

2T



• A cooling tower is a special type of heat exchanger in which the warm

the air are brought in direct contact for ‘evaporative cooling’.

• It provides a very good contact of air and water in terms of the contact

mass transfer co-efficient of water vapor while keeping air pressure drop

• Enthalpy of air is lower than enthalpy of water. Sensible heat and latent h

transfer take place from water drop to surrounding air.

• Thus, cooling is accomplished by sensible heat transfer from water to a

evaporation of a small portion of water.

Cooling Tower Construction & Operation



General cooling Tower



• The hot water which is coming from heat exchanger is sprayed at the top of t

tower.

• Air enters through the louvers at the two opposite walls of the cooling tower.

• During cooling process of water, around 2% water is evaporated. Make watcompensate the water loss due to evaporation.

• Blowdown is there to drain a part of water containing solid deposit.

• The exit cold water from the cooling tower is used in the heat exchanger or o

operation.



FACTORS GOVERN THE OPERATION OF COOTOWER

• The dry-bulb and wet – bulb temperatures of air

• Temperature of warm water

• The efficiency of contact between air and water in terms of volume transfer coefficie

• Contact time between air and water

• The uniformity of the distribution of the phases within the tower

• Air pressure drop

• Desired temperature of cooled water



Atmospheric Cooling Tower

• It is a big rectangular chamber with two opposite ‘louvered’ walls.

• Tower is packed with a suitable ‘tower fill’.

• Atmospheric air enters the tower through louvers driven by its own velocity.

• Direction and velocity of wind greatly influence its performance.



Cooling tower production: https://www.youtube.com/watch?v=qgfQXo6SI4U

N l d f li h l



• Natural draft cooling tower has a large re

concrete shell of hyperbolic shape (also

‘hyperbolic tower’).• Natural flow of air occurs through the tower; he

called natural draft



Factors responsible for creating natural dra

(a) A rise in temperature and humidity of air in the column reduces

(b) Wind velocity at the tower bottom

• Fan is used to enhance the air flow rate in fan assisted natural dr

• The typical diameter of tower is 100 m, heigh of 150 m and cap5,00,000 gallon/minute.



https://www.youtube.com/watch?v=ggg3C87UVCY



Why hyperbolic shape?

(i) More packing materials can be placed at the bottom

(ii) The entering air gets smoothly directed towards the

(iii) Greater structural strength and stability

https://www.youtube.com/watch?v=xKzenFW0ZIg



Mechanical Draft Towers

• Forced draft

• Induced draft

• Fans are used to move air through the tower in mechanical d

cooling towers.



• FORDED DRAFT: It has one or more fans located at the tower bottom to push air into tow

• Advantages:

(a) A part of the velocity head of air thrown by the blower is converted to pressure head

into the tower. It makes energy efficient than induced draft.

(b) Less susceptible to vibrations as fans are installed near the ground.

• Disadvantages:

(a) Air flow through the packing may not be uniform

(b) Some of the warm and humid air may be recirculated back. Recirculation rate becom

wind velocity is high. It is not popular except for small capacities.



Forced draft cooling tower



• Induced draft towers: One or more fans are installed at the top of the tower. Depending on

and flow pattern, induced draft towers are of two types, cross-flow and counter flow towers.

• Major advantages of countercurrent induced draft cooling tower

(a) Relatively dry air contacts the coldest water at the bottom of the cooling tower

(b) Humid air is in contact with the warm water and hence maximum average driving forc

both heat and mass transfer.

• Disadvantage of induced draft towers compared to forced draft towers

(a)It consumes more horse power.

(b) Cross-flow induced draft cooling tower requires less motor horse power than countercurr

draft cooling towers.

C fl i d d d f li



Cross-flow induced draft cooling tower

• Cross-flow induced draft cooling tower supplies horiz

air flow along the packed height and requires less m

horse power than the counter-flow type.

• Additional ‘cells’ may be added to raise the capacity



Structural Components & MOC

• The shell, the framework and casing walls – wooden, concrete, steel , glass fibre reinf

walls.

• The tower fills/packings – splash type, film type (counterflow), plastic, wood

• Louvers – air passage, glass fibre, wood

•Drift eliminators - plastic

• Fans – propeller, centrifugal

• Water distribution – gravity distribution, spray



Drift Eliminator

• In every cooling tower there is a loss of water to the environment due to the evapo

process.

• This evaporation is usually in the form of pure water vapor and presents no

environment.

• Drift, however, is the undesirable loss of liquid water to the environment via small dro



, , q

become entrained in the leaving air stream.

• There, water droplets carry with them chemicals and minerals, thus impacting the surr

environment.

• Drift eliminators are designed to capture large



water droplets caught in the cooling tower air

stream.

• The eliminators prevent the water droplets and

mist from escaping the cooling tower.Eliminators do this by causing the droplets to

change direction and lose velocity at impact

on the blade walls and fall back into the

tower.

• Efficient drift eliminators will keep drift losses

to less than .001% of the re-circulating water

flow rate.

http://www towercomponentsinc com/operation-drift-eliminator php



Tower Problems

• Scale – inorganic minerals CA2CO3 etc.

• Fouling – waterborne contaminants

• Microbial growth – bacteria, algae, fungi, legio

• Corrosion



Consequences

• Energy losses

• Reduced heat transfer efficiency

• Increased corrosion and pitting

• Loss of tower efficiency

• Wood decay and loss of structural integrity of the co

tower



Cooling Range & Approach

• Cooling range is the difference in the temperature of thewater and the outlet cooled water.

• If hot water is cooled from 40ºC to 30ºC, the range is 10ºC.

• Approach is the temperature difference between



• Approach is the temperature difference between

being produced and the “power source” that cre

product.• In the case of cooling tower, the “product” is co

leaving the tower and ambient wet bulb is the driv

that creates the cold water.

• If a cooling tower produces 85°F cold water w

ambient wet bulb is 78°F, then the cooling tower ap

7°F.



• If a small ‘approach’ is targeted, the height of

increases rapidly.

• Theoretically, the approach is zero if a tower has a

packed height.



• What happened if the designed wet bulb temperature is higher than the actual wet

temperature?



Design of Cooling Tower

We need to determine:

• The tower cross-section required to take the given loa

warm water.

• The height of packing required to achieved the desir

cooling of water.

Basic assumptions for the design of cooling towe



Basic assumptions for the design of cooling towe

follows:

(i) the rate of vaporization of water is much less thanof water input to the tower (about 1% loss of feed w

(ii) evaporative or adiabatic cooling of water occu

tower



Enthalpy balance diagram of cooling tower

• Let, L is the constant water flow rate (kg/m2s) and Gs



rate (kg dry air/m2s). Across a differential thickness d

bed, temperature of water is decreased by dT L and the

enthalpy of air is increased by dH’.

• Hence, change in enthalpy of water = L.cWL.dT L and,

• Change in enthalpy of air = Gs.dH’

• Differential enthalpy balance over dz is L.cWL.dT L = Gs.

Eq. 9

• Enthalpy balance over envelope I:



py p

• LcWL(TL - TL1) = Gs(H’ - H’1) This is the operating lin

air-water contact.

• Enthalpy balance over envelope II:

• LcWL(TL2 - TL1) = Gs(H’2 - H’1)

• The equilibrium curve for air-water system on TL-H’ p

the plot of enthalpy of saturated air versus liquid temp

at equilibrium.

Eq. 11

Eq. 12

• Rate of transfer of water vapor to air in the differen



p

volume is

• The decrease in temperature of air for sensible heat

to water is

Eq. 13

Eq. 14

• Differentiation of Equation 7and multiplication with Gs gives



Eq. 15

• The height ( z) of the packing in the cooling tower is ob



by

• Number of gas-enthalpy transfer units

Eq

Eq. 17

• Height of gas-enthalpy transfer units



g g py

• Height of gas-enthalpy transfer unit

• Hence, height of cooling tower (packing section), z

Eq

Eq

E

I Cl E l



In Class Example

A cooling tower is to be designed to cool water from 45ºC t

countercurrent contact with air of dry bulb temperature 30ºC an

temperature of 25ºC. The water rate is 5500 kg/m2.h and the air

times the minimum. Determine the tower height if the individual

mass transfer coefficient (kY /ā ) is 5743.5 kg/m3h. The volumetric wa

transfer coefficient is given by hLā=0.059L0.51Gs, in Kcal/m3hK, where L

mass flow rates of water and air (dry basis).

Antoine Equation: ln PV A(bar)=11.96481-3984.923/(T-39.724)

S l ti



Solutions

• TG1 = 30, TW = TS = 25, used a psychrometric chart to

• From chart, Y’1 = 0.019

• Now, used equation 6 & 7 to calculate H’1

• TL1 = 30, TL2 = 45



Plot the equilibrium line



• Locate point Q(TL1,H’1 ) (Lower terminal of operating line) at Q(30, 78.7) on TL-H’ p

• Draw a tangent to the equilibrium line through Q Slope of the tangent is 8 44



Draw a tangent to the equilibrium line through Q. Slope of the tangent is 8.44.



• H’2 = 180 kJ/kg

• locate point P (TL2, H’2) (Upper terminal of the operating line) at P (45, 180) on TL-



p ( , ) ( pp p g ) ( , )

• Randomly choose one upper point within graph (x2, y2).

• Choose another one point at lower part of the graph (x1, y1) with x1 is left unknown



Choose another one point at lower part of the graph (x1, y1) with x1 is left unknown

• Use the following formula to guess x1.

• Solved for x1 and draw the 1st tie line. Subsequently, replicate the line by maintainin

slope to get a set of tie lines.

230

250

Equilibrium Line



50

70

90

110

130

150

170

190

210

230

22 27 32 37 42 47

H ' ( k J / k g d r y a i r )

Temperature (Celcius)

Q point

P Point

Operating Line

Tie line

Tangent

• A set of tie lines of this slope is drawn from several points on the operating line. Thes

meet the equilibrium line at (TLi,H’i ). Hence, the points (H’, H’i ) are obtained.



T L 30 32.5 35 37.5 40 42.5 45

H' 78.7 96.4 112.8 130.3 148.8 165.3 184

T Li 28.7 31.4 33.9 36.6 39 41.4 43.7

H'i 93.0 107.5 123.2 139.9 158.4 177 198.8

1/(H'-H'i) 0.070 0.090 0.096 0.104 0.104 0.085 0.068

• Calculate the area under curve



0.000

0.020

0.040

0.060

0.080

0.100

0.120

78.0 98.0 118.0 138.0 158.0 178.0

1 ( H ' - H ' i )

H'

Simpson’s rule, five point quadrature

• NtG =Area under the curve= (184-78.7)×0.088=9.27



• HtG = Gs/k’Ya = 3410/5743.5 = 0.59 m

• z = HtGNtG = 0.59 x 9.27 = 5.47 m

Blowdown



• During the cooling process of hot water in cooling tower, around 2% w

evaporates.

• In the long run, it increases the solid content in the circulating water. So

particles also come from the environment and mix with circulating wate

• The solid content of the cooled water must be kept under a certain limi

scaling or fouling on the heat exchange equipment.

• A part of the circulating water is drained from the bottom of the coolindiscard the deposited solids from the cooling tower blowdown.

• The losses due to blowdown, evaporation, drift and leakage are comp

adding make-up water.

• Water balance in cooling tower M = B + D + E

• Solid balance MC1 = (B + D)C2 + (E)(0)



( ) ( )( )

• The water vaporized (E) does not have any solids in it, and the TDS in t

blowdown and in the drift is the same as that in the circulating water.

• (B + D + E)C1 = (B + D) C2

• Where r = C2/C1

• Once the blowdown rate B is known, the makeup rate M may be calcul

M – Makeu

B – Blowdo

D – Rate of

and leakag

C1 = dissol

makeup waC2 = dissol

circulating w

E = (water flow rat

IN CLASS EXAMPLE



An induced draft crossflow tower is rated to cool 15000

water from 40ºC to 29ºC. The total solid concentration

exceed 900 ppm. The TDS of the makeup water is 300 p

0.1% of the water is lost by drift from the tower and leaka

circulation system. Calculate the blowdown and makeup rate

Solutions



• The range = 40 – 29 = 11ºC = 19.8 ºF

• E = (15000)(19.8)(0.0008) = 237.6 gpm

• D = 0.1% x 15000 = 15 gpm

• r = C2/C1 = 900/300 = 3

• B = [(237.6 – 15(3-1)]/(3-1) = 104 gpm

• M = B + D + E = 104 + 15 + 237.6 = 356.6 gpm

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