1  Data

This calculation is for the test data:

T1 ( C)T2 ( C)T3 ( C)T4 ( C)T5 ( C)T6 ( C)
Tair,indbTair,inwbTair,outdbTair,outwbTH2OinTH2Oout
20.817.022.922.729.523.1


Δ P (mmH2O)Patmosphere (bar)
161.01

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)
φin = 
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

PatmospherePH2Oout
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[Tairindp] = 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

HoutHin = Q − P  where  P = −96 W     (39)
air 
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 = 
(houthin)
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(ba)
n+1






    (51)
NTU = 
(houthin)
4
 





4
i=1
 
1
ho


TH2Oout+
i(TH2OinTH2Oout)
5






hin+
i(houthin)
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.