COOLING TOWER

By. Engr. Yuri G. Melliza

Hot water 1

m1

t1

h1

Cold water 2

t2, h2, m2

t3, h3, m3

Make-up water 3

B Air Out

hB,WB, ma

A Air in

hA,WA, ma

Fan

Catch Basin

Cooling Tower A Cooling tower is a wind braced enclosure or shell

usually made of wood, concrete or metal with fillings on

the inside to aid water exposure. The water to be cooled

is pumped into a distributing header at the top of the

tower from which it drops in sprays to the filling. The

water spreads out in the filling thus exposing new water

surfaces to the air circulating through the tower. The

cooled water drops to the bottom of the tower called the

catch basin. The air

circulating through the tower becomes partially saturated

with moisture by evaporating some amount of water.

This evaporation is mostly what cools the water.

Fundamental Equations:

1. Actual Cooling Range (ACR)

ACR = t1 - t2

2. Cooling Tower Approach (A)

A = t2 - tWA

3. Theoretical Cooling Range (TCR)

TCR = t1 – tWA

4. Cooling Tower Efficiency

100% xtt

tte

WA1

21

5. Vapor Pressure

PV = PW - PA(td - tW)

where: A = 6.66 x 10-4 (For tW of equal or greater

than 0C.

A = 5.94 x 10-4 (For tW of less than 0C)

6. Specific Humidity or Humidity Ratio

kgda

kgm

PvP

Pv 0.622W

7. Relative Humidity

% 100 xPd

PvΦ

8. Enthalpy

h = 1.0045td + W(2501.3 + 1.86td) KJ/kgda

9. Specific Volume

kgda

m

Pv)(P

273)0.287(t 3

d

υ

10. Degree of Saturation

PvP

PdPΦμ

11. By moisture balance in the tower:

a) With make up water, m1 = m2

m3 = ma(WB - WA) kg/sec

b) Without make up water available, m1 m2:

m1 - m2 = ma (WB - WA) kg/sec

12. By Energy Balance in the tower

a. Considering make up water

KJ/kg

m

)hW(W)h(hmhmh

kg/sec )hW(W)h(h

)h(hmm

1

3ABABa112

3ABAB

211a

b. Without considering make up water

KJ/kg )W(Wmm

)h(hmhmh

kg/sec )hW(W)h(h

)h(hmm

ABa1

ABa11

2

2ABAB

211

a

13. Driving Pressure

KPa

1000

ρ-ρgHΔP io

d

14. Mass Flow rate of air and vapor mixture

m = ma(1+W) kg/sec

m = ma + mv

15. Cooling water flow rate related to Brake

Power of an Engine

L/hr t-t

Power Brake904.3m

21w

where:

m1 - mass flow rate of water entering tower in

kg/sec

m2 - mass flow rate of cooled water in kg/sec

m3 - make up water in kg/sec

h1 - enthalpy of hot water in KJ/kg

h2 - enthalpy of cooled water in KJ/kg

h3 - enthalpy of make up water in KJ/kg

hA - enthalpy of air entering tower in KJ/kgda

hB - enthalpy of air leaving tower in KJ/kgda

WA - humidity ratio of air entering tower in kgm/kgda

WB - humidity ratio of air leaving tower in kgm/kgda

ma - mass flow rate of dry air in kg/sec

td - dry bulb temperature in C

tw - wet bulb temperature in C

t1 - temperature of hot water, C

t2 - temperature of cooled water, C

t3 - temperature of make up water, C

H - tower height, meters

o - density of outside air and vapor mixture,

kg/m3

i - density of inside air and vapor mixture,

taken at exit of the fill, kg/m3

DRYER

Dryer - is an equipment used in removing moisture or solvents from a wet

material or product.

Hygroscopic Substance - a substance that can contained bound moisture

and is variable in moisture content which they posses at different times.

Weight of Moisture - amount of moisture present in the product at the start

or at the end of the drying operation.

Bone Dry Weight - it is the final constant weight reached by a hygroscopic

material when it is completely dried out. It is the weight of the product without

the presence of moisture.

Gross Weight - it is the sum of the bone-dry weight of the product and the

weight of moisture.

Moisture Content - it is the amount of moisture expressed as a percentage

of the gross weight or the bone dry weight of the product.

A) Wet Basis - is the moisture content of the product in percent

of the gross weight.

B) Dry Basis 0r Regain - it is the moisture content of the product

in percent of the bone dry weight.

Continuous Drying - is that type of drying operation in which the material

to be dried is fed to and discharge from the dryer continuously.

Batch Drying - is that type of drying operation in which the material to be

dried is done in batches at definite interval of time.

CLASSIFICATION OF DRYERS

1. Direct Dryers - conduction heat transfer

2. Indirect Dryers - convection heat transfer

3. Infra-red Dryers - radiation heat transfer

PRODUCT SYMBOLS

1. GW = BDW + M

2. Xm = [M/GW] x 100% (wet basis)

3. Xm = [M/BDW] x 100% (dry basis or regain)

where: GW - gross weight

BDW - bone dry weight

M - weight of moisture

Xm - moisture content

HEAT REQUIREMENT BY THE PRODUCT

Q = Q1 + Q2 + Q3 + Q4

Q1 = (BDW)Cp(tB - tA) kg/hr

Q2 = MBCpw(tB - tA) kg/hr

Q2 = MB(hfB - hfA) kg/hr

Q3 = (MA - MB)(hvB - hfA) kg/hr = MR(hvB - hfA)

Q4 = heat loss

Q1 - sensible heat of product, KJ/hr

Q2 - sensible heat of moisture remaining in the product, KJ/hr

Q3 - heat required to evaporate and superheat moisture removed from

the product in KJ/hr

Q4 - heat losses, KJ/hr

A,B - conditions at the start or at the end of drying operation

t - temperature in C

hf - enthalpy of water at saturated liquid, KJ/kg

hv - enthalpy of vapor, KJ/kg

Cp - specific heat of the product, KJ/kg-C or KJkg-K

Cpw - specific heat of water, KJ/kg-C or KJ/kg-K

GWA GWB

MA

MB

MR(moisture removed)

Condition A Condition B

BDW

BDW (weight of product without moisture)

It is desired to designed a drying plant to have a capacity of 680 kg/hr of

product 3.5% moisture content from a wet feed containing 42% moisture.

Fresh air at 27C with 40% RH will be preheated to 93C before entering

the dryer and will leave the dryer with the same temperature but with a

60% RH. Find:

a) the amount of air to dryer in m3/sec ( 0.25)

b) the heat supplied to the preheater in KW (16)

At 27 C DB and 40% RH At 93 C and W = .0089 kgm/kgda

W = .0089 kgm/kgda h = 117.22 KJ/kgda

h = 49.8 KJ/kgda = 1.05 m3/kgda

At 93 C and 60% RH

W = 0.54 kgm/kgda

h = 1538.94 KJ/kgda

0 Fresh air 1 heated air 2 exhaust air

A GWA B GWB

Air Preheater

Dryer m m

Q

GW = BDW + M

GW = BDW + Xm(GW)

(GW)X M

)XGW(1BDW

)X(1

BDWGW

m

m

m

Given:

GWB = 680 kg/hr

XmB = 0.035 ; XmA = 0.42

W0 = 0.0089 ; h0 = 49.8

W1 = 0.0089 ; h1 = 117.22 ; v1 = 1.05

W2 = 0.54 ;h2 = 1538.94

W0 = W1

W2

h2

h1

h0

0 1

2 MB = 23.8 kg/hr

BDW = 656.2 kg/hr

GWA = 1131.4 kg/hr

MA = 475.2 kg/hr

By moisture balance on dryer

mW1 + MA = mW2 + MB

12

BA

WW

MMm

m = 850 kg/hr

Qa1 = 850(1.05) = 892.43 m3/hr

Qa1 = 0.25 m3/sec

By energy balance in the

preheater:

Q = m(h1 - h0)

Q = 16 KW

Raw cotton has been stored in a warehouse at 29C and 50% relative

humidity, with a regain of 6.6%. (a) the cotton goes through a mill and

passes through the weaving room kept at 31C and 70% relative humidity

with a regain of 8.1%. What is the moisture in 200 kg of cotton? (b) for

200 kg of cotton from the warehouse, how many kilograms should appear

in the woven cloth, neglecting lintage and thread

losses? ANSWER: a) 12.4 kg ; b) 202.8 kg

GW = BDW + M

M = Xm(BDW)

BDW = GW/(1+Xm)

Given:

XmA = 0.066 ; XmB = 0.081

GWA = 200 kg

BDW = 187.61 kg

MA = 12.4 kg

MB = 15.2 kg

GWB = 202.8 kg

A 10 kg sample from a batch of material under test is found to have a BDW

of 8.5 kg. This material is processed and is then found to have a regain

(dry basis moisture content) of 20%. How much weight of product appears for

each kilogram of original material. (1.02 kg/kg)

Given:

GWA = 10 kg ; BDW = 8.5 kg ; XmB = 0.20 (dry basis)

M = GW - BDW

MA = 1.5 kg

MB = XmB(BDW)

MB = 1.7 kg

GW = BDW + M

GWB = 10.2 kg

GWB/GWA = 1.02

A rotary dryer is fired with bunker oil of 41 870 KJ/kg HHV is to produce

20 metric tons per hour of dried sand with 0.5% moisture from a wet feed

containing 7% moisture, specific heat of sand is 0.879 KJ/kg-C, temperature

of wet feed is 30C and temperature of dried product is 115C. Calculate

the L/hr of bunker oil consumed if the specific gravity of bunker oil is 0.90

and dryer efficiency of 60%.

hf at 30C = 125.79 KJ/kg

hg at 101.325 KPa and 115C = 2706.12 KJ/kg

Wet Feed (sand)

Dried sand

Given:

GWB = 20,000kg/hr; XmB = 0.005; XmA = 0.07

HHV =41,870 KJ/kg; Cp = 0.879 KJ/kg-C; tA = 30C ; tB = 115C

S = 0.90; e = 60%

GW = BDW + M ; GW = BDW/(1-Xm)

M = Xm(GW)

BDW = GW - M

MB = 100 kg/hr ; e = Q/mf(HHV)

BDW = 19,900 kg/hr mf = 204.2 kg/sec

GWA = 21,398 kg/hr ; MA = 1498 kg/hr df = 900 kg/m3

MA - MB = MR ; MR = 1398 kg/hr Vf = 0.227 m3

Q1 = BDW(Cp)(tB - t=A) = 413 KW Vf = 227 Liters

Q2 = MB(Cpw)(tB - tA) = 10 KW

Q3 = MR(hg -hf) = 1002 KW

Q4 = 0

Q = Q1+ Q2+ Q3 +Q4 = 1425 KW