## 1  Data

This calculation is for the test data:

 T1 (∘ C) T2 (∘ C) T3 (∘ C) T4 (∘ C) T5 (∘ C) T6 (∘ C) Tair,indb Tair,inwb Tair,outdb Tair,outwb TH2Oin TH2Oout 20.8 17.0 22.9 22.7 29.5 23.1

 Δ P (mmH2O) Patmosphere (bar) 16 1.01

## 2  Cooling Tower Performance Calculations

Approach is how close the outlet water temperature gets to the wet bulb temperature of the incoming air. Cooling Range is the difference betwen inlet and outlet water temperatures.

 Approach = TH2Oout − Tair,inwb = 23.1 − 17.0 = 6.1 ∘ C     (1)
 Cooling Range = TH2Oin − TH2Oout = 29.5 − 23.1 = 6.4 ∘ C     (2)

## 3  Humidity Calculations

### 3.1  Vapor Pressures at the 4 "air" temperatures

Vapor Pressure of water in the air at some T, PH2O[T], given by the Antoine equation.

PH2O[T] = 10

A −
 B C+T

where A = 8.07131, B = 1730.63, C = 233.426     (3)
PH2O[Tair,indb] = 10

A −
 B C+20.8

= 18.36 mmHg     (4)

18.36 mmHg × (1 atm/760 mmHg) (1013.25 mbar/1 atm) = 24.48 mbar

 PH2O∘[Tair,inwb] = 19.30 mbar     (5)
 PH2O∘[Tair,outdb] = 27.83 mbar     (6)
 PH2O∘[Tair,outwb] = 27.50 mbar     (7)

### 3.2  Partial Pressures of inlet and outlet air

Partial Pressure of water in the air at some T, PH2O, given by Regnault, August, and Apjohn.

 PH2O = PH2O∘[Tairwb] − Patmosphere × 6.666× 10−4 × ⎛ ⎝ Tairdb − Tairwb ⎞ ⎠ (8)
 PH2Oin = 19.30 − 1010.0 × 6.666× 10−4 × ⎛ ⎝ 20.8 − 17.0 ⎞ ⎠ = 16.74 mbar     (9)
 PH2Oout = 27.50 − 1010.0 × 6.666× 10−4 × ⎛ ⎝ 22.9 − 22.7 ⎞ ⎠ = 27.37 mbar     (10)

### 3.3  Relative Humidity

Relative humidity, φ, is the ratio of the partial pressure to the vapor pressure evaluated at Tairdb

φ =
 PH2O PH2O∘[Tairdb]
× 100%     (11)
φin =
 16.74 mbar 24.48 mbar
× 100% = 68.38%     (12)
φout =
 27.37 mbar 27.83 mbar
× 100% = 98.35%     (13)

### 3.4  Absolute Humidity

Absolute humidity, ω, is the ratio of the mass of water vapor to the mass of dry air. See McCabe, chapter 19 Humidification Operations, equation 19.1.

ω =
 mass of water vapor mass of dry air
=
MMH2O × PH2O
 MMair × ⎛ ⎝ Patmosphere − PH2O ⎞ ⎠
(14)
 MMH2O = 18.0152 g/mol and MMair ≈ 28.9640 g/mol     (15)
ωin =
18.0152 × 16.74
 28.9640 × ⎛ ⎝ 1010.0 − 16.74 ⎞ ⎠
= 0.01048     (16)
ωout =
18.0152 × 27.37
 28.9640 × ⎛ ⎝ 1010.0 − 27.37 ⎞ ⎠
= 0.01732     (17)

## 4  Mass Balance

### 4.1  Mass Flow Rate of Air

Dry air mass flow rate, ṁair.

air = 0.0137 ×
 √
 Δ P vwet air,out
= 0.0137 ×
 √
 Δ P (1+ωout)vdry air,out
(18)

Rearranging the ideal gas law to solve for the specifc volume with respect to the mass of dry air present:

vdry air,out =
R · Tair,outdb
 ⎛ ⎝ Patmosphere−PH2Oout ⎞ ⎠ MMair
(19)
vdry air,out =
0.083147
 L · bar mol · K
× 295.9  K

1.01  bar − 0.02737  bar
28.9640
 g mol
= 0.8645
 L g
= 0.8645
 m3 kg dry air
(20)
air = 0.0137 ×
 √
16  mmH2O
(1+0.01732)0.8645
 m3 kg dry air
= 0.05844
 kg s
(21)

Mass flow rate of water that would be added from the make-up tank (if it existed) is equal to the mass flow rate of water leaving the tower minus entering the tower:

E = ṁair
ωout − ωin
= 0.05844
 kg s

0.01732 − 0.01048
= 4.0 × 10−4
 kg s
(22)

## 5  Energy Balance

### 5.1  Specific Enthalpy of Water Vapor in Unsaturated Air

Reverse engineered Antoine equation to get the dew point.

Tairdp = −C
B
log10

PH2Oout  mbar × 760
 mmHg atm
1013.25
 mbar atm

A
(23)
Tair,outdp = −C
B
log10

 27.37 × 760 1013.25

A
= 22.62 C     (24)
Tair,indp = −C
B
log10

 16.74 × 760 1013.25

A
= 14.78 C     (25)

Numerical fit to enthalpy data of water vapor.

 hvo[Tairdp] = 1.8033 × Tairdp + 2501.7     (26)
hv,outo[Tair,outdp] = 1.8033 × 22.62 + 2501.7 = 2542.49
 kJ kg
(27)
hv,ino[Tair,indp] = 1.8033 × 14.78 + 2501.7 = 2528.36
 kJ kg
(28)

Numerical fit to heat capacity data of water vapor.

Cpo

 Tairdb+Tairdp 2

= 1.6 × 10−6

 Tairdb+Tairdp 2

 2
+ 1.5408 × 10−4

 Tairdb+Tairdp 2

+ 1.85871     (29)
Cpo,out

 22.9+22.62 2

= 1.6 × 10−6 × 22.762 + 1.5408 × 10−4 × 22.76 + 1.85871 = 1.863
 kJ kg · ∘C
(30)
Cpo,in

 20.8+14.78 2

= 1.6 × 10−6 × 17.792 + 1.5408 × 10−4 × 17.79 + 1.85871 = 1.862
 kJ kg · ∘C
(31)

Specific enthalpy of water vapor.

hv = hvo[Tairdp] +  Cpo

 Tairdb+Tairdp 2

Tairdb − Tairdp
(32)
hvout = 2542.49
 kJ kg
+ 1.863
 kJ kg · ∘C

22.9 − 22.62
= 2543.01
 kJ kg
(33)
hvin = 2528.36
 kJ kg
+ 1.862
 kJ kg · ∘C

20.8 − 14.78
= 2539.57
 kJ kg
(34)

Specific enthalpy of wet air.

 h = Cpdry airTair,outdb + ωouthvout     (35)
hout = 1.005
 kJ kg · ∘C
× 22.9 C + 0.01732 × 2543.01
 kJ kg
= 67.06
 kJ kg
(36)
hin = 1.005
 kJ kg · ∘C
× 20.8 C + 0.01048 × 2539.57
 kJ kg
= 47.52
 kJ kg
(37)

### 5.2  Specific Enthalpy of Make-up Water

hE = CpwaterTair,indb = 4180
 J g · ∘C
× 20.8 C = 86940
 J kg
(38)

### 5.3  Final Energy Balance

 Hout−Hin = Q − P  where  P = −96 W     (39)
 ṁair ⎛ ⎝ hout − hin ⎞ ⎠ − ṁEhE = 1000  W − −96  W = 1096  W     (40)
0.05844
 kg s

67.06
 kJ kg
− 47.52
 kJ kg

− 86.94
 kJ kg
× 4.0 × 10−4
 kg s
= 1096  W     (41)
 1.142 kW − 0.03478 kW = 1096  W     (42)
 1.107  kW = 1.096  kW     (43)
 1.107−1.096 1.096
× 100% = 1.00%  error     (44)

Or since we actually have no make-up water source unlike the experiment that derived the above data, we would report:

 1.142−1.096 1.096
× 100% = 4.20%  error     (45)

## 6  Number of Transfer Units, NTU

NTU =
 hout hin

 dh ho[TH2O]−h
(46)
 ho[TH2O] = ω hvo[TH2O] + Cpdry air TH2O     (47)

This can be numerically solved by several integration methods like the trapezoidal rule:

NTU =
 (hout−hin) 2

 1 ho[TH2Oin] − hout
+
 1 ho[TH2Oout] − hin

(48)
NTU =
 (67.06−47.52) 2

 1 ho[29.5] − 67.06
+
 1 ho[23.1] − 47.52

(49)
NTU =
 (67.06−47.52) 2

 1 97.09 − 67.06
+
 1 68.60 − 47.52

= 0.789     (50)

The trapezoidal rule assumes that the path between the two function points is linear, which it is not. In such a case, we might use the Chebyshev integration technique:

 b a
f(x) =
 (b − a) n

 n ∑ i=1
f

a+
 i(b−a) n+1

(51)
NTU =
 (hout−hin) 4

 4 ∑ i=1

1
ho

TH2Oout+
 i(TH2Oin−TH2Oout) 5

hin+
 i(hout−hin) 5

(52)

However, for this problem, this results in NTU = 0.794 which doesn’t really justify its use given its complexity.

This document was translated from LATEX by HEVEA.