Cooling Towers

Cooling Towers

MEP 460 Heat Exchanger design

King Abdulaziz University

Mechanical Engineering Department

Nov. 2018

1

Cooling towers 0-What is a cooling tower

1-Introduction

2-Review of psychrometery 3-Evaporative cooling

4-Cooling tower applications

5-Anaology between heat and mass transfer 6-Basic simplified analysis of cooling tower 7-CTC (NTU) calculations

Approximate methods By integration

8-Cooling tower effectiveness 9-Cooling tower fill packing

10-Estimating the design liquid/air flow ratio

2

0-What is a cooling tower

Cooling tower is a heat rejection heat exchanger

Where heat is rejected to atmospheric air

₃

0-What is a cooling tower

4

0-What is a cooling tower

5

Cooling towers

6

Cooling towers

7

1-Introduction

Wet cooling towers and Dry cooling towers

Cooling towers types

Can be forced or induced cooling towers

Counter flow cooling towers Cross flow cooling towers

8

1-Introduction

9

T

_wi

Hot water in

Cold water out

Air in

Air out

Packing material

Fill

𝑇

_𝑎𝑖

, 𝑇

_𝑎𝑖^∗

𝑊

_𝑖

𝑊

_𝑜

Range R= 𝑇 _𝑤𝑖 − 𝑇 _𝑤𝑜

Approach 𝑨 = 𝑇

_𝑤𝑜

− 𝑇

_𝑎𝑖^∗

Range and approach of a cooling tower

T

_wo

1-Introduction

𝑇

_𝑎𝑖^∗

= wet bulb temperature of entering air

Range

10

2-Review of psychrometery moist air thermodynamic properties

Dry, wet bulb, and dew point temperatures (T, T

^*

, T

_dew

) Humidity ratio W

Relative humidity 𝜙 Air enthalpy, h

Saturated air enthalpy, h

_s

Humidification efficiency, 𝜂 _𝐻

11

2-Review of psychrometery moist air

W

T

_a

𝑻 _𝒂 ^∗

Saturation line

Dry bulb temp.

Humidity ration

12

2-Review of psychrometery moist air

Moist air ( dry and saturated ) properties

13 p s t

W W

,

 

Degree of saturation ℎ = ℎ_𝑑𝑎 + 𝜇 ℎ_𝑠 − ℎ_𝑑𝑎

2-Review of psychrometery moist air

Water properties

14

3-Evaporative cooling Desert cooler

15

يوارحصلا فَيكملا

3-Evaporative cooling

16

An evaporative cooler (also swamp cooler, desert cooler and wet air cooler) is a device that cools air through the evaporation of water. Evaporative

cooling differs from typical air conditioning systems, which use vapor-compression or absorption

refrigeration cycles.

3-Evaporative cooling

17

18

Evaporative cooling in

Mina

^.

Cooling the Hajji’s tents

19

Huge number of tents in

Madina provided with fans and water sprays

20

Using water sprays (Evaporative cooling in ARAFAT ^{and Mina)}

21

Evaporative cooling

1 2

𝜂 _𝐻 = 𝜙 ₂ − 𝜙 ₁ 𝜙 _𝑠 − 𝜙 ₁

Humidification efficiency

22

4-Cooling tower applications

Power plant heat rejection

Refrigeration and air conditioning heat rejection (HVAC)

Oil refineries

Petrochemical industries

Cooling tower used mainly as a heat rejection heat exchanger

23

ሶ 𝑄 = ℎ

_𝑐

𝐴 (𝑇

_𝑠

− 𝑇

_∞

)

ሶ

𝑚 = ℎ

_𝑚

𝐴𝜌(𝑊

_𝑠

− 𝑊

_∞

)

ሶ

𝑚 = 𝑚

𝑠 𝑚

²

𝑘𝑔 𝑚

³

𝑘𝑔

𝑘𝑔 = 𝑘𝑔 𝑠

ሶ

𝑚 = ℎ

_𝑚

𝐴(𝜌

_𝑠

− 𝜌

_∞

)

ሶ

𝑚 = ℎ_𝑚𝐴 𝐶_𝑠 − 𝐶_∞

Heat transfer Mass transfer

h_m is the mass transfer coefficient [m/s]

h

_c

is heat transfer

coefficient [W/(m

²

.K)

ℎ

_𝑑

= 𝜌ℎ

_𝑚

h

_d

is the mass conductance coefficient

5-Anaology between heat and mass transfer

Driving force for heat transfer is the

temperature difference

Driving force for mas transfer is the concentration difference

h

_d

=[kg/(m

²

.s]

24

𝑃𝑟 = 𝜈

𝛼 = 𝑚𝑜𝑚𝑒𝑛𝑢𝑚 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑣𝑖𝑡𝑦 𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑣𝑖𝑡𝑦 𝑆𝑐 = 𝜈

𝐷_𝐴𝐵 = 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑣𝑖𝑡𝑦 𝑚𝑎𝑠𝑠 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑣𝑖𝑡𝑦 𝐿𝑒 = 𝛼

𝐷_𝐴𝐵 = 𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑣𝑖𝑡𝑦

𝑚𝑎𝑠𝑠 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑣𝑖𝑡𝑦 𝐿𝑒 = 𝑆𝑐 Pr Prandtl number

Schmidt number

Lewis number

Nusselt number

𝑁𝑢 = ℎ_𝑐𝐿 𝑘 Sherwood Number

𝑆ℎ = ℎ_𝑚𝐿 𝐷_𝐴𝐵 Stanton Number

𝑆𝑡 = 𝑁𝑢

𝑅𝑒 𝑃𝑟 = ℎ_𝑐 𝜌𝑉𝐶𝑝

Dimensionless numbers

mass transfer Stanton

number 𝑆𝑡_𝑚 = 𝑆ℎ

𝑅𝑒 𝑆𝑐 = ℎ_𝑚 𝑉 Coburn modulus j 𝐽_𝐻 = St 𝑃𝑟^{2 3Τ}

5-Anaology between heat and mass transfer

25

𝜈 = 𝜇 𝜌 𝛼 = 𝑘

𝜌 𝐶_𝑝

𝑓

2 = 𝑆𝑡_𝑚𝑆𝑐^{2 3Τ} 𝑓

2 = 𝑆𝑡𝑃𝑟^{2 3Τ} Pr≠1, Sc≠1

Relation between momentum transfer and heat transfer

Relation between momentum transfer and mass transfer

𝑁𝑢 = ℎ_𝑐𝐿 𝑘 𝑆ℎ = ℎ_𝑚𝐿

𝐷_𝐴𝐵 𝑆𝑐 = 𝜈

𝐷_𝐴𝐵

𝐿𝑒 = 𝛼

𝐷_𝐴𝐵 = 𝑆𝑐 𝑃𝑟 𝑃𝑟 = 𝜈

𝛼

ℎ_𝑐

𝜌𝐶_𝑝𝑉𝑃𝑟^{2 3Τ} = ℎ_𝑚

𝑉 𝑆𝑐^{2 3Τ} 𝑓

2 = 𝑆𝑡𝑃𝑟^{2 3Τ} = 𝑆𝑡_𝑚𝑆𝑐^{2 3Τ}

ℎ_𝑐

ℎ_𝑚 = 𝜌𝐶_𝑝 𝑆𝑐 𝑃𝑟

Τ 2 3

= 𝜌𝐶_𝑝 𝐿𝑒^{2 3Τ}

For Le1 Good for air

ℎ_𝑐

ℎ_𝑚 = 𝜌𝐶_𝑝 or

ℎ_𝑐 = 𝜌𝐶_𝑝ℎ_𝑚 = ℎ_𝑑𝐶_𝑝

5-Anaology between heat and mass transfer

𝑆𝑡 = ℎ_𝑐 𝜌𝐶_𝑝𝑉 𝑆𝑡_𝑚 = ℎ_𝑚

𝑉

26

Mass and energy balance for a small control volume of volume dV

6-Analysis of counter current cooling towers

Objective of a cooling tower is to cool hot water from T

_wi

to T

_wo

by rejecting heat to air.

27

𝑑 ሶ 𝑚

_𝑤

= ሶ 𝑚

_𝑎

𝑑𝑊 = ℎ

_𝑑

𝑎

_𝑣

𝑑𝑉(𝑊

_𝑠

− 𝑊

_𝑎

)

( ሶ 𝑚

_𝑤

+ ሶ 𝑚

_𝑎

𝑑𝑊) ℎ

_𝑓

+ 𝑑ℎ

_𝑓

+ ሶ 𝑚

_𝑎

ℎ

_𝑎

= ሶ 𝑚

_𝑤

ℎ

_𝑓

+ ሶ 𝑚

_𝑎

(ℎ

_𝑎

+ 𝑑ℎ

_𝑎

)

ሶ

𝑚

_𝑤

𝑑ℎ

_𝑓

+ ሶ 𝑚

_𝑎

𝑑𝑊ℎ

_𝑓

+ ሶ 𝑚

_𝑎

𝑑𝑊𝑑ℎ

_𝑓

= ሶ 𝑚

_𝑎

𝑑ℎ

_𝑎

Mass balance for water

Energy balance

Re-arranging terms

6-Analysis of counter current cooling towers

28

a

_v

=surface area per unit volume=m

²

/m

³

𝑑 ሶ 𝑚

_𝑤

ሶ

𝑚

_𝑤

𝑑ℎ

_𝑓

= ℎ

_𝑐

𝑎

_𝑣

𝑑𝑉 𝑇

_𝑠

− 𝑇

_𝑎

+ ℎ

_𝑑

𝑎

_𝑣

𝑑𝑉 𝑊

_𝑠

− 𝑊

_𝑎

ℎ

_𝑓𝑔

ሶ

𝑚

_𝑤

𝑑ℎ

_𝑓

+ ሶ 𝑚

_𝑎

𝑑𝑊ℎ

_𝑓

+ ሶ 𝑚

_𝑎

𝑑𝑊𝑑ℎ

_𝑓

= ሶ 𝑚

_𝑎

𝑑ℎ

_𝑎

ሶ

𝑚

_𝑎

𝑑ℎ

_𝑎

= ℎ

_𝑐

𝑎

_𝑣

𝑑𝑉 𝑇

_𝑠

− 𝑇

_𝑎

+ ℎ

_𝑑

𝑎

_𝑣

𝑑𝑉 𝑊

_𝑠

− 𝑊

_𝑎

ℎ

_𝑓𝑔

+ ℎ

_𝑑

𝑎

_𝑣

𝑑𝑉 𝑊

_𝑠

− 𝑊

_𝑎

ℎ

_𝑓

ℎ

_𝑐

= ℎ

_𝑑

𝐶

_𝑝𝑎

ሶ

𝑚

_𝑎

𝑑ℎ

_𝑎

= 𝐶

_𝑝𝑎

ℎ

_𝑑

𝑎

_𝑣

𝑑𝑉 𝑇

_𝑠

− 𝑇

_𝑎

+ ℎ

_𝑑

𝑎

_𝑣

𝑑𝑉 𝑊

_𝑠

− 𝑊

_𝑎

ℎ

_𝑔

ሶ

𝑚

_𝑎

𝑑ℎ

_𝑎

= ℎ

_𝑑

𝑎

_𝑣

𝑑𝑉 𝐶

_𝑝𝑎

𝑇

_𝑠

− 𝑇

_𝑎

+ 𝑊

_𝑠

− 𝑊

_𝑎

ℎ

_𝑔

Heat lost by water due to heat and mass transfer i.e.

Can be neglected

Using heat mass analogy and

Assuming Le=1

Substitute in the above equation to get the energy balance becomes

Energy balance

hc heat transfer coeff [W/(m2.K)]

hd mass transfer

conductance [kg/(m2.s]

29

ℎ

_𝑎

= 𝐶

_𝑝𝑎

𝑇

_𝑎

+ 𝑊

_𝑎

(ℎ

_𝑔0

+ 𝐶

_𝑝𝑣

𝑇

_𝑎

) ℎ

_𝑠

= 𝐶

_𝑝𝑎

𝑇

_𝑠

+ 𝑊

_𝑠

(ℎ

_𝑔0

+ 𝐶

_𝑝𝑣

𝑇

_𝑠

)

ℎ

_𝑠

− ℎ

_𝑎

= 𝐶

_𝑝𝑎

𝑇

_𝑠

− 𝑇

_𝑎

+ 𝑊

_𝑠

− 𝑊

_𝑎

ℎ

_𝑔0

+ 𝐶

_𝑝𝑣

(𝑊

_𝑠

𝑇

_𝑠

− 𝑊

_𝑎

𝑇

_𝑎

)

ሶ

𝑚

_𝑎

𝑑ℎ

_𝑎

= ℎ

_𝑑

𝑎

_𝑣

𝑑𝑉(ℎ

_𝑠

− ℎ

_𝑎

)

ሶ

𝑚

_𝑤

𝑑ℎ

_𝑤

= ሶ 𝑚

_𝑤

𝐶

_𝑝𝑤

𝑑𝑇

_𝑤

= ሶ 𝑚

_𝑎

𝑑ℎ

_𝑎

ሶ

𝑚

_𝑤

𝐶

_𝑝𝑤

𝑑𝑇

_𝑤

= ℎ

_𝑑

𝑎

_𝑣

𝑑𝑉 ℎ

_𝑠

− ℎ

_𝑎

න ^𝐶

^𝑝𝑤

^𝑑𝑇

^𝑤

ℎ

_𝑠

− ℎ

_𝑎

= න ^ℎ

^𝑑

^𝑎

^𝑣

^𝑑𝑉

ሶ

𝑚

_𝑤

= ^ℎ

^𝑑

^𝑎

^𝑣

^𝑉

ሶ

𝑚

_𝑤

= 𝐶𝑇𝐶

Enthalpy of moist air Saturated air

Small term

Assuming negligible difference between h

_g

and h

_g0

The difference

CTC or NTU can be found by several methods 𝑁𝑇𝑈 = ℎ

_𝑑

𝑎

_𝑣

𝑉

ሶ

𝑚

_𝑎

= 𝐶𝑇𝐶 𝑚 ሶ

_𝑤

ሶ

𝑚

_𝑎

30

Calculation of CTC or NTU for a cooling tower

1-Using Nomograph (Perry’s Chemical Engineering HB) 2-Approximate methods

a- ( Fraas Design of heat exchanger Book) b-Chebyshev integration

3-Trapizodial rule or similar integration procedure

31

From Perry’s Chemical Engineers’ Hand book

1-Using Nomograph (Perry’s Chemical Engineering HB)

32

Cooling tower characteristic=CTC

𝐶𝑇𝐶 = ℎ

_𝑑

𝑎

_𝑣

𝑉

ሶ

𝑚

_𝑤

= 𝐾𝑎𝑉

𝐿

33

1- At the given range and cold

temperature located the first point say A.

2-At the air inlet wet bulb temperature locate the second point say point B.

3-Connect the two point A, and by a straight line (call it line 1)

4-On the L/G scale locate point C 5-Draw a line parallel to the first line (line1 ) and passing through point C. call it line 2

6-Line 2 will intersect KaV/L at a point.

Read the value of this CTC (cooling tower characteristics)

Using the nomograph to approximately find the cooling tower characteristics (KaV/L)

A B

C

Hot water temp =100 F

Cold water temp. 80 F

Wet bulb temp. 70 F

L/G=1.

ℎ_𝑑𝑎_𝑣𝑉

ሶ

𝑚_𝑎 = 𝐾𝑎_𝑣𝑉

ሶ

𝑚_𝑎

34

𝛿ℎ = ℎ

_𝑠𝑤𝑖

+ ℎ

_𝑠𝑤𝑜

− 2ℎ

_𝑠𝑤𝑚

4

Δ𝐻

_𝑜

= ℎ

_𝑠𝑤𝑖

− ℎ

_𝑎𝑜

Δ𝐻

_𝑖

= ℎ

_𝑠𝑤𝑜

− ℎ

_𝑎𝑖

Δ𝐻

_𝑚

= (Δ𝐻

_𝑜

− Δ𝐻

_𝑖

) 𝑙𝑛 Δ𝐻

_𝑜

− 𝛿ℎ

Δ𝐻

_𝑖

− 𝛿ℎ 𝑡

_𝑚

= 𝑡

_𝑤𝑖

+ 𝑡

_𝑤𝑜

2

ሶ

𝑚

_𝑎

Δℎ

_𝑎

= ሶ 𝑚

_𝑤

Δℎ

_𝑤

𝐶𝑇𝐶 = 𝐶

_𝑝𝑤

Δ𝑡

_𝑤

Δ𝐻

_𝑚

1-Crude approximation for CTC

h

_sw

is the saturated air enthalpy at t

_w

h_sw= Saturated air at water temperature

35

Ref. Fraas, A.P., Heat exchanger design, 2^nd edition, John Wiely, 1989

Example on finding the CTC

Hot water in

Cold water out

Air in

Air out

Packing material

Fill

𝑇_𝑎𝑖, 𝑇_𝑎𝑖^∗

𝑊

_𝑖

𝑊

_𝑜 T_wi=50 C

T_wo=29 𝑇_𝑎𝑖^∗ = 24 𝐶

Given:

T_wi=50 C T_wo=29 C 𝑇_𝑎𝑖^∗ = 24 𝐶

ሶ

𝑚_𝑤

ሶ

𝑚_𝑎 = 𝐿_𝑤

𝐺_𝑎 = 1.25 Find

CTC or NTU

36

න 𝐶

_𝑝𝑤

𝑑𝑇

_𝑤

ℎ

_𝑠

− ℎ

_𝑎

= න ℎ

_𝑑

𝑎

_𝑣

𝑑𝑉

ሶ

𝑚

_𝑤

= ℎ

_𝑑

𝑎

_𝑣

𝑉

ሶ

𝑚

_𝑤

= 𝐶𝑇𝐶

The cooling Tower characteristic CTC is given by

t_wi t_wo h_swi h_swo h_ai t_wm h_sm Δ𝐻_𝑜 Δ𝐻_𝑖 𝛿ℎ Δ𝐻_𝑚

50 29 275.35 94.87 72.39 39.5 162.59 93.08 22.49 11.26 35.55

ሶ

𝑚_𝑎𝑑ℎ_𝑎 = ሶ𝑚_𝑤𝐶𝑝_𝑤𝑑𝑡_𝑤 𝐺_𝑎 Δℎ_𝑎 = 𝐿_𝑤𝐶𝑝_𝑤Δ_𝑤

ሶ

𝑚_𝑤 = 𝐿_𝑤 𝑚ሶ _𝑎 = 𝐺_𝑎

𝐶𝑇𝐶 = 𝐶

_𝑝𝑤

Δ𝑡

_𝑤

Δ𝐻

_𝑚

Δ𝐻

_𝑚

= (Δ𝐻

_𝑜

− Δ𝐻

_𝑖

) 𝑙𝑛 Δ𝐻

_𝑜

− 𝛿ℎ

Δ𝐻

_𝑖

− 𝛿ℎ 𝛿ℎ = ℎ

_𝑠𝑤𝑖

+ ℎ

_𝑠𝑤𝑜

− 2ℎ

_𝑠𝑤𝑚

4

Δ𝐻

_𝑜

= ℎ

_𝑠𝑤𝑖

− ℎ

_𝑎𝑜

Δ𝐻

_𝑖

= ℎ

_𝑠𝑤𝑜

− ℎ

_𝑎𝑖

𝐶𝑇𝐶 = 𝐶𝑝_𝑤 ∗ (𝑡_𝑤𝑖 − 𝑡_𝑤𝑜) Δ𝐻_𝑚

𝐶𝑇𝐶 = 4.186 ∗ (50 − 29)

35.55 = 2.47

37

2- Using Chebyshev integration method

න

𝑡_𝑤𝑜

𝑡_𝑤𝑖𝐶_𝑝𝑤𝑑𝑡_𝑤

ℎ_𝑠 − ℎ_𝑎 = 𝐶_𝑝𝑤 𝑡_𝑤𝑖 − 𝑡_𝑤𝑜 1

ℎ_𝑠 − ℎ_{𝑎 𝑎𝑣𝑔} න

𝑡_𝑤𝑜

𝑡_𝑤𝑖 𝑑𝑡_𝑤

ℎ_𝑠 − ℎ_𝑎 = 𝐶_𝑝𝑤 4

1 ℎ_𝑠 − ℎ_𝑎 ቚ

0.1 + 1 ℎ_𝑠 − ℎ_𝑎 ቚ

0.4 + 1 ℎ_𝑠 − ℎ_𝑎 ቚ

0.6 + 1 ℎ_𝑠 − ℎ_𝑎 ቚ

0.9

1

ℎ_𝑠 − ℎ_{𝑎 𝑎𝑣𝑔} = 1 4

1

ℎ_𝑠 − ℎ_𝑎 𝑒𝑣𝑎𝑙𝑢𝑎𝑡𝑒𝑑 𝑎𝑡 0.1, 0.4, 0.6 𝑎𝑛𝑑 0.9 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑎𝑛𝑔𝑒

1

ℎ_𝑠 − ℎ_{𝑎 𝑎𝑣𝑔} = 1 4

1

ℎ_𝑠 − ℎ_{𝑎 0.1} + 1

ℎ_𝑠 − ℎ_{𝑎 0.4} + 1

ℎ_𝑠 − ℎ_{𝑎 0.6} + 1

ℎ_𝑠 − ℎ_{𝑎 0.9}

න ^𝐶

^𝑝𝑤

^𝑑𝑇

^𝑤

ℎ

_𝑠

− ℎ

_𝑎

= න ^ℎ

^𝑑

^𝑎

^𝑣

^𝑑𝑉

ሶ

𝑚

_𝑤

= ^ℎ

^𝑑

^𝑎

^𝑣

^𝑉

ሶ

𝑚

_𝑤

= 𝐶𝑇𝐶

h_s is the enthalpy of air at saturation

38

Given:

t_wi=50 C t_wo=29 C 𝑇_𝑎𝑖^∗ = 24 °𝐶

ሶ

𝑚_𝑤

ሶ

𝑚_𝑎 = 𝐿_𝑤

𝐺_𝑎 = 1.25

t_wi t_wo L_w /G_g 𝑇_𝑎^∗ h_ai

50 29 1.25 24 72.4

Range=t_wi-t_wo=50-29=21 C

0.1 Δ𝑡_𝑤 0.4 Δ𝑡_𝑤 0.6 Δ𝑡_𝑤 0.9 Δ𝑡_𝑤

2.1 8.4 12.6 18.9

t_wi 31.1 37.4 41.6 47.9

h_si 105.6 145.8 180.0 246.8

h_ai 83.21 116.2 138.1 171.1

h_si-h_ai 22.41 29.62 41.89 75.69

ℎ_𝑑𝑎_𝑣𝑉

ሶ

𝑚_𝑤 = 𝐶𝑝_𝑤 𝑡_𝑤𝑖 − 𝑡_𝑤𝑜 4

1

22.41 + 1

29.62 + 1

41.89+ 1

75.69 = 2.538

Example on Chebyshev integration method

ሶ

𝑚_𝑎Δℎ_𝑎 = ሶ𝑚_𝑤𝐶_𝑝𝑤Δ𝑡_𝑤

39

1

ℎ_𝑠 − ℎ_{𝑎 𝑎𝑣𝑔} = 1 4

1

ℎ_𝑠 − ℎ_{𝑎 0.1} + 1

ℎ_𝑠 − ℎ_{𝑎 0.4} + 1

ℎ_𝑠 − ℎ_{𝑎 0.6} + 1

ℎ_𝑠 − ℎ_{𝑎 0.9}

𝐶_𝑝𝑤 = 4.186 𝑘𝐽/𝑘𝑔

3-Integration of the equation with Merkel Approximation

න 𝐶

_𝑝𝑤

𝑑𝑡

_𝑤

ℎ

_𝑠

− ℎ

_𝑎

= ℎ

_𝑑

𝑎

_𝑣

ሶ

𝑚

_𝑤

𝑉 = 𝐶𝑇𝐶

ሶ

𝑚

_𝑎

𝑑ℎ

_𝑎

= ሶ 𝑚

_𝑤

𝑑ℎ

_𝑓

= ሶ 𝑚

_𝑤

𝐶

_𝑝𝑤

𝑑𝑡

_𝑤

By starting from the bottom of the tower with t

_wi

(the exit

temperature of water and assuming a small dt

_w

(temperature difference for the water). The inlet air state is known. From Eq. 2 one can get dh

_a

and the term on left hand side of

equation 1 above can be found

1

2

40

y0

y1

y2

y3

y4

Δ𝑥

𝐴𝑟𝑒𝑎 = 𝑦₀ + 𝑦₁

2 Δ𝑥 +𝑦₁ + 𝑦₂

2 Δx + y₂ + y₃

2 Δ𝑥 + 𝑦₃ + 𝑦₄ 2 Δ𝑥 𝐴𝑟𝑒𝑎 = Δ𝑥

2 𝑦₀ + 2𝑦₁ + 2𝑦₂ + 2𝑦₃ + 𝑦₄

The Trapezoidal Rule

x y

𝐴𝑟𝑒𝑎 = න

𝑎 𝑏

𝑦 𝑥 𝑑𝑥

41

න

𝑎 𝑏

𝑓 𝑥 = 𝑑𝑥 = 𝑏 − 𝑎

𝑛 𝑦₀ + 2𝑦₁ + 2𝑦₂ + ⋯ . 2𝑦_𝑛−1 + 𝑦_𝑛

The Trapezoidal Rule

42

8-Cooling tower Effectiveness

Temperature potential

(T

_w

-T

_a

) Enthalpy potential

(h

_sw

-h

_a

)

43

Direct contact heat exchanger

Sensible heat exchangers

𝑇_ℎ𝑖

𝑇_𝑐𝑖 𝑇_ℎ𝑜

𝑇_𝑐𝑜 Δ𝑇 = 𝑇_ℎ − 𝑇_𝑐

𝑑𝑞 = 𝐶_𝑐𝑑𝑇_𝑐 𝑑𝑞 = 𝐶_ℎ𝑑𝑇_ℎ 𝑑𝑞 = 𝑈𝑑𝐴 𝑇_ℎ − 𝑇_𝑐

𝑑Δ𝑇 = 𝑑𝑇_ℎ − 𝑑𝑇_𝑐 𝑑𝑇_𝑐 = 𝑑𝑞

𝐶_𝑐 = 𝑈𝑑𝐴

𝐶_𝑐 𝑇_ℎ − 𝑇_𝑐 𝑑𝑇_ℎ = 𝑑𝑞

𝐶_ℎ = 𝑈𝑑𝐴

𝐶_ℎ 𝑇_ℎ − 𝑇_𝑐

Δ𝑇

𝜖 = 1 − 𝑒

^{−𝑁𝑇𝑈 1−𝐶𝑟}

1 − 𝐶

_𝑟

𝑒

^{−𝑁𝑇𝑈 1−𝐶𝑟}

8-Cooling tower Effectiveness

44

𝑑Δ𝑇

Δ𝑇 = 𝑈𝑑𝐴 1

𝐶_ℎ − 1 𝐶_𝑐 ln Δ𝑇₂

Δ𝑇₁ = 𝑈𝐴 1

𝐶_ℎ + 1 𝐶_𝑐

Counter flow sensible heat

exchanger

𝑁𝑇𝑈 = ℎ_𝑑𝑎_𝑣𝑉

ሶ

𝑚_𝑎 = 𝐶𝑇𝐶 ሶ Τ

𝑚_𝑎 𝑚ሶ _𝑤

𝑚^∗ = 𝑚ሶ _𝑎𝐶_𝑠

ሶ

𝑚_𝑤 𝐶_𝑝𝑤 𝑑ℎ_𝑎 = 𝑁𝑇𝑈𝑑𝑉

𝑉 (ℎ_𝑠 − ℎ_𝑎)

ሶ

𝑚_𝑤𝐶_𝑝𝑤𝑑𝑇_𝑤 = ሶ𝑚_𝑎𝑑ℎ_𝑎 𝐶_𝑠 = 𝑑ℎ_𝑠

𝑑𝑇_𝑤

ሶ

𝑚_𝑤𝐶_𝑝𝑤𝑑ℎ_𝑠

𝐶_𝑠 = ሶ𝑚_𝑎𝑑ℎ_𝑎 = ሶ𝑚_𝑎𝑁𝑇𝑈𝑑𝑉

𝑉 (ℎ_𝑠 − ℎ_𝑎) 𝑑ℎ_𝑠 = 𝑚ሶ _𝑎𝐶_𝑠

ሶ

𝑚_𝑤𝐶_𝑝𝑤 𝑁𝑇𝑈 𝑑𝑉

𝑉 ℎ_𝑠 − ℎ_𝑎

𝑑ℎ_𝑠 = 𝑚^∗ 𝑁𝑇𝑈𝑑𝑉

𝑉 ℎ_𝑠 − ℎ_𝑎

ℎ_𝑠𝑖

ℎ_𝑠𝑜 ℎ_𝑎𝑖

ℎ_𝑎𝑜

8-Cooling tower Effectiveness

ሶ

𝑚_𝑎𝑑ℎ_𝑎 = ℎ_𝑑𝑎_𝑣𝑑𝑉 ℎ_𝑠 − ℎ_𝑎

45

𝑑𝑇_ℎ = 𝑑𝑞

𝐶_ℎ = 𝑁𝑇𝑈 𝐶_𝑚𝑖𝑛𝑑𝐴

𝐶_ℎ𝐴 𝑇_ℎ − 𝑇_𝑐 For Sensible heat exchangers

𝜖 = ℎ_𝑎𝑜 − ℎ_𝑎𝑖

ℎ_𝑠𝑖 − ℎ_𝑎𝑖 = 𝑞 𝑞_𝑚𝑎𝑥

ℎ_𝑎𝑜 = 𝜖 ℎ_𝑠𝑖 − ℎ_𝑎𝑖 + ℎ_𝑎𝑖 𝑑ℎ_𝑠 = 𝑚^∗ 𝑁𝑇𝑈𝑑𝑉

𝑉 ℎ_𝑠 − ℎ_𝑎

𝜖 = 1 − 𝑒^{−𝑁𝑇𝑈 1−𝑚∗} 1 − 𝑚^∗𝑒^{−𝑁𝑇𝑈 1−𝑚∗}

8-Cooling tower Effectiveness

Sensible heat exchanger

𝑚^∗ = 𝑚ሶ _𝑎𝐶_𝑠

ሶ

𝑚_𝑤 𝐶_𝑝𝑤 𝐶_𝑠 = 𝑑ℎ_𝑠 𝑑𝑇_𝑤

𝜖 = 1 − 𝑒^{−𝑁𝑇𝑈 1−𝐶𝑟} 1 − 𝐶_𝑟𝑒^{−𝑁𝑇𝑈 1−𝐶𝑟}

46

𝑑𝑇_ℎ = 𝑑𝑞

𝐶_ℎ = 𝑁𝑇𝑈 𝐶_𝑚𝑖𝑛𝑑𝐴

𝐶_ℎ𝐴 𝑇_ℎ − 𝑇_𝑐

Assuming fixed effective value for W_s such as W_swe and integrate the above equation

𝑊_𝑎𝑜 = 𝑊_𝑠𝑤𝑒 + 𝑊_𝑎𝑖 − 𝑊_𝑠𝑤𝑒 𝑒^{−𝑁𝑇𝑈}

8-Cooling tower Effectiveness

Also the change in humidity ratio is given by

W_swe is effective humidity ratio at effective saturated enthalpy h_swe

47

𝑑 ሶ 𝑚

_𝑤

= ሶ 𝑚

_𝑎

𝑑𝑊

_𝑎

= ℎ

_𝑑

𝑎

_𝑣

𝑑𝑉(𝑊

_𝑠

− 𝑊

_𝑎

)

𝑑𝑊_𝑎 = −ℎ_𝑑𝑎_𝑣𝑑𝑉

ሶ

𝑚_𝑎 (𝑊_𝑎 − 𝑊_𝑠) 𝑑𝑊_𝑎

𝑊_𝑎 − 𝑊_𝑠𝑤𝑒 = −ℎ_𝑑𝑎_𝑣𝑑𝑉

ሶ

𝑚_𝑎

𝑊_𝑎𝑜 − 𝑊_𝑠𝑤𝑒

𝑊_𝑎𝑖 − 𝑊_𝑠𝑤𝑒 = 𝑒^{−𝑁𝑇𝑈}

48

ℎ_𝑠𝑤𝑒 = ℎ_𝑎𝑖 + ℎ_𝑎𝑜 − ℎ_𝑎𝑖 1 − 𝑒^{−𝑁𝑇𝑈} 𝑑ℎ_𝑎 = −ℎ_𝑑𝑎_𝑣𝑑𝑉

ሶ

𝑚_𝑎 (ℎ_𝑎 − ℎ_𝑠)

At this value of saturated air enthalpy find the effective W_swe and then use the equation below to find the mass of water leaving the tower

From the enthalpy (energy) balance

ሶ

𝑚_𝑤𝑜 = ሶ𝑚_𝑤𝑖 − ሶ𝑚_𝑎(𝑊_𝑎𝑜 − 𝑊_𝑎𝑖)

Amount of water evaporated= 𝑚ሶ _𝑎 (𝑊_𝑎𝑜 − 𝑊_𝑎𝑖)

8-Cooling tower Effectiveness

𝑑ℎ_𝑎

ℎ_𝑎 − ℎ_𝑠𝑤𝑒 = −ℎ_𝑑𝑎_𝑣𝑑𝑉

ሶ

𝑚_𝑎

ℎ_𝑎𝑜 − ℎ_𝑠𝑤𝑒

ℎ_𝑎𝑖 − ℎ_𝑠𝑤𝑒 = 𝑒^{−𝑁𝑇𝑈} Integrate

or

ሶ

𝑚_𝑤𝑜 = ሶ𝑚_𝑤𝑖 − ሶ𝑚_𝑎(𝑊_𝑎𝑜 − 𝑊_𝑎𝑖)

or

Amount of water evaporated= 𝑚ሶ _𝑎 (𝑊_𝑎𝑜 − 𝑊_𝑎𝑖) ℎ_𝑠𝑤𝑒 = ℎ_𝑎𝑖 + ℎ_𝑎𝑜 − ℎ_𝑎𝑖

1 − 𝑒^{−𝑁𝑇𝑈}

use phsychromery chart or moist air tables to find the corresponding W_se 𝑊_𝑎0 = 𝑊_𝑠𝑒 + 𝑊_𝑎𝑖 − 𝑊_𝑠𝑒 𝑒^{−𝑁𝑇𝑈} Then

8-Cooling tower Effectiveness

49

Given:

t_wi=50 C t_a=40 C 𝑡_𝑎𝑖^∗ = 24 𝐶 CTC=2.54

ሶ

𝑚_𝑤

ሶ

𝑚_𝑎 = 𝐿_𝑤

𝐺_𝑎 = 1.25

CTC 𝑚ሶ _𝑤Τ𝑚ሶ _𝑎 NTU T_wi 𝑇_𝑎𝑖^∗ h_swi h_ai W_ai

2.54 1.25 3.175 50 24 275.35 72.22 0.0121

4

𝐶_𝑆 = 𝑑ℎ_𝑠𝑤

𝑑𝑡_𝑤 = ℎ_𝑠𝑤 𝑡_𝑤 = 50 − ℎ_𝑠𝑤(𝑡_𝑤 = 49)

1 = (275.35 − 261.8)

1 = 13.55

𝑚^∗ = 𝑚ሶ _𝑎𝐶_𝑠

ሶ

𝑚_𝑤𝐶𝑝_𝑤 = 13.55

1.25 ∗ 4.186 = 2.59

𝑁𝑇𝑈 1 − 𝑚^∗ = 2.032 ∗ 1 − 2.59 = −5.05

𝜖 = 1 − 𝑒^{−𝑁𝑇𝑈 1−𝑚∗}

1 − 𝑚^∗𝑒^{−𝑁𝑇𝑈 1−𝑚∗} = 0.38

ℎ_𝑎𝑜 = ℎ_𝑎𝑖 + 𝜖 ℎ_𝑠𝑤𝑖 − ℎ_𝑎𝑖 =72.22+0.38(275.35-72.22)=148.7 Example on cooling tower effectiveness

and finding the state of air and water leaving the tower

50

ℎ_𝑠𝑤𝑒 = ℎ_𝑎𝑖 + ℎ_𝑎𝑜 − ℎ_𝑎𝑖 1 − 𝑒^{−𝑁𝑇𝑈} ℎ_𝑠𝑤𝑒 = 160.3 𝑘𝐽/𝑘𝑔 𝑊_𝑠𝑤𝑒 = 0.0469𝑘𝑔_𝑤

𝑘𝑔_𝑎

𝑊_𝑎0 = 𝑊_𝑠𝑤𝑒 + 𝑊_𝑎𝑖 − 𝑊_𝑠𝑤𝑒 𝑒^{−𝑁𝑇𝑈}

𝑊_𝑎𝑜 = 0.04235 𝑘𝑔_𝑤 𝑘𝑔_𝑎

ሶ

𝑚_𝑎Δℎ_𝑎 = ሶ𝑚_𝑤𝐶𝑝_𝑤Δ𝑡_𝑤 Δ𝑡_𝑤 = ℎ_𝑎𝑜 − ℎ_𝑎𝑖

ሶ Τ

𝑚_w 𝑚ሶ _a 𝐶_𝑝𝑤 = 14.62

𝑡_𝑤𝑜 = 𝑡_𝑤𝑖 − Δ𝑡_𝑤 = 50 − 14.62 = 35.38 From moist air

thermodynamics tables Find W_swe at h_swe

51

Another iteration with t_wo=35.38

𝐶_𝑠 = 𝑑ℎ_𝑠

𝑑𝑡_𝑤 = ℎ_𝑠 50 − ℎ_𝑠(35.38)

14.62 = 274.3 − 131.6

14.62 = 9.759

Cs 𝑚^∗  h_ao h_swe W_swe W_ao t_wo

9.759 1.865 0.489 171 185.9 0.0556 0.04989 31.1

52

9-Cooling tower Fill packing

53

Fill types

54

Fill types

55

Ref.: Process heat transfer, Hewitt et al Triangular slats

Rectangular slats

ℎ_𝑑𝑎_𝑣𝑉

ሶ

𝑚_𝑤 = 0.32 𝑚ሶ _𝑤

ሶ

𝑚_𝑎

−0.45

ℎ_𝑑𝑎_𝑣𝑉

ሶ

𝑚_𝑤 = 0.28𝐻 𝑚ሶ _𝑤

ሶ

𝑚_𝑎

−0.52

ℎ_𝑑𝑎_𝑣𝑉

ሶ

𝑚_𝑤 = 𝑒𝐻 𝑚ሶ _𝑤

ሶ

𝑚_𝑎

−𝑛

General form for the CTC as a function of L/G and packing height

Fill types

Corrugated asbestos sheets

ℎ_𝑑𝑎_𝑣𝑉

ሶ

𝑚_𝑤 = 0.72𝐻 𝑚ሶ _𝑤

ሶ

𝑚_𝑎

−061

56

න 𝐶_𝑝𝑤𝑑𝑡_𝑤

ℎ_𝑠 − ℎ_𝑎 = ℎ_𝑑𝑎_𝑣

ሶ

𝑚_𝑤 𝑉 = 𝐶𝑇𝐶

Merkel Integral

Packing material

ℎ_𝑑𝑎_𝑣𝑉

ሶ

𝑚_𝑤 = 𝑒𝐻 𝑚ሶ _𝑤

ሶ

𝑚_𝑎

−𝑛

10-Estimating the design flow ratio 𝒎 ሶ _𝒘 Τ 𝒎 ሶ _𝒂

57

58

Cooling water design

Variation of Cooling tower characteristics with the ratio L/G,

tower height from heat/mass transfer side and from packing

materials side.

References

1-Braun, J. E. Methodologies for the Design and Control of Central Cooling Plants."

Ph.D. Thesis, University of Wisconsin, 1988

2-Braun J. E., S. A. Klein, JW Mitchell, Effectiveness models for cooling towers and cooling coils

3-D. Stevens, J. E. Braun, SA Klein, An effectiveness model of liquid-desiccant system heat/mass exchangers

4-Kroger air cooled condenser and cooling towers 5-Perry’s chemical engineering handbook, 8

^th

edition 6-CRC Thermal Engineering Hand book

7-Fraas, A.P., Heat exchanger design, 2

^nd

edition, John Wiely, 1989

59