Test_Data_Explanation

1 Data

This calculation is for the test data:

T₁ (^∘ C)	T₂ (^∘ C)	T₃ (^∘ C)	T₄ (^∘ C)	T₅ (^∘ C)	T₆ (^∘ C)
T_air,in^db	T_air,in^wb	T_air,out^db	T_air,out^wb	T_H₂Oⁱⁿ	T_H₂O^out
20.8	17.0	22.9	22.7	29.5	23.1

Δ P (mmH₂O)	P_atmosphere (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 = T_H₂O^out − T_air,in^wb = 23.1 − 17.0 = 6.1 ^∘ C (1)

Cooling Range = T_H₂Oⁱⁿ − T_H₂O^out = 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, P_H₂O^∘[T], given by the Antoine equation.

P_H₂O^∘[T] = 10

⎛
⎜
⎜
⎝

A −

C+T

⎞
⎟
⎟
⎠

where A = 8.07131, B = 1730.63, C = 233.426 (3)

P_H₂O^∘[T_air,in^db] = 10

⎛
⎜
⎜
⎝

A −

C+20.8

⎞
⎟
⎟
⎠

= 18.36 mmHg (4)

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

P_H₂O^∘[T_air,in^wb] = 19.30 mbar (5)

P_H₂O^∘[T_air,out^db] = 27.83 mbar (6)

P_H₂O^∘[T_air,out^wb] = 27.50 mbar (7)

3.2 Partial Pressures of inlet and outlet air

Partial Pressure of water in the air at some T, P_H₂O, given by Regnault, August, and Apjohn.

P_H₂O = P_H₂O^∘[T_air^wb] − P_atmosphere × 6.666× 10⁻⁴ ×

⎛
⎝

T_air^db − T_air^wb

⎞
⎠

(8)

P_H₂Oⁱⁿ = 19.30 − 1010.0 × 6.666× 10⁻⁴ ×

⎛
⎝

20.8 − 17.0

⎞
⎠

= 16.74 mbar (9)

P_H₂O^out = 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 T_air^db

φ =

P_H₂O

P_H₂O^∘[T_air^db]

× 100% (11)

φ_in =

16.74 mbar

24.48 mbar

× 100% = 68.38% (12)

φ_out =

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

MM_H₂O × P_H₂O

MM_air ×

⎛
⎝

P_atmosphere − P_H₂O

⎞
⎠

(14)

MM_H₂O = 18.0152 g/mol and MM_air ≈ 28.9640 g/mol (15)

ω_in =

18.0152 × 16.74

28.9640 ×

⎛
⎝

1010.0 − 16.74

⎞
⎠

= 0.01048 (16)

ω_out =

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, ṁ_air.

ṁ_air = 0.0137 ×

√

Δ P

v_{wet air,out}

= 0.0137 ×

√

Δ P

(1+ω_out)v_{dry air,out}

(18)

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

v_{dry air,out} =

R · T_air,out^db

⎛
⎝

P_atmosphere−P_H₂O^out

⎞
⎠

MM_air

(19)

v_{dry air,out} =

0.083147

L · bar

mol · K

× 295.9 K

⎛
⎝

1.01 bar − 0.02737 bar

⎞
⎠

28.9640

mol

= 0.8645

m³

kg dry air

(20)

ṁ_air = 0.0137 ×

√

16 mmH₂O

(1+0.01732)0.8645

m³

kg dry air

= 0.05844

(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:

ṁ_E = ṁ_air

⎛
⎝

ω_out − ω_in

⎞
⎠

= 0.05844

⎛
⎝

0.01732 − 0.01048

⎞
⎠

= 4.0 × 10⁻⁴

(22)

5 Energy Balance

5.1 Specific Enthalpy of Water Vapor in Unsaturated Air

Reverse engineered Antoine equation to get the dew point.

T_air^dp = −C−

log₁₀

⎛
⎜
⎜
⎜
⎜
⎜
⎝

P_H₂O^out mbar × 760

mmHg

atm

1013.25

mbar

atm

⎞
⎟
⎟
⎟
⎟
⎟
⎠

−A

(23)

T_air,out^dp = −C−

log₁₀

⎛
⎜
⎜
⎝

27.37 × 760

1013.25

⎞
⎟
⎟
⎠

−A

= 22.62 ^∘C (24)

T_air,in^dp = −C−

log₁₀

⎛
⎜
⎜
⎝

16.74 × 760

1013.25

⎞
⎟
⎟
⎠

−A

= 14.78 ^∘C (25)

Numerical fit to enthalpy data of water vapor.

h_v^o[T_air^dp] = 1.8033 × T_air^dp + 2501.7 (26)

h_v,out^o[T_air,out^dp] = 1.8033 × 22.62 + 2501.7 = 2542.49

(27)

h_v,in^o[T_air,in^dp] = 1.8033 × 14.78 + 2501.7 = 2528.36

(28)

Numerical fit to heat capacity data of water vapor.

C_p^o

⎡
⎢
⎢
⎣

T_air^db+T_air^dp

⎤
⎥
⎥
⎦

= 1.6 × 10⁻⁶

⎛
⎜
⎜
⎝

T_air^db+T_air^dp

⎞
⎟
⎟
⎠

+ 1.5408 × 10⁻⁴

⎛
⎜
⎜
⎝

T_air^db+T_air^dp

⎞
⎟
⎟
⎠

+ 1.85871 (29)

C_p^o,out

⎡
⎢
⎢
⎣

22.9+22.62

⎤
⎥
⎥
⎦

= 1.6 × 10⁻⁶ × 22.76² + 1.5408 × 10⁻⁴ × 22.76 + 1.85871 = 1.863

kg · ^∘C

(30)

C_p^o,in

⎡
⎢
⎢
⎣

20.8+14.78

⎤
⎥
⎥
⎦

= 1.6 × 10⁻⁶ × 17.79² + 1.5408 × 10⁻⁴ × 17.79 + 1.85871 = 1.862

kg · ^∘C

(31)

Specific enthalpy of water vapor.

h_v = h_v^o[T_air^dp] + C_p^o

⎡
⎢
⎢
⎣

T_air^db+T_air^dp

⎤
⎥
⎥
⎦

⎛
⎝

T_air^db − T_air^dp

⎞
⎠

(32)

h_v^out = 2542.49

+ 1.863

kg · ^∘C

⎛
⎝

22.9 − 22.62

⎞
⎠

= 2543.01

(33)

h_vⁱⁿ = 2528.36

+ 1.862

kg · ^∘C

⎛
⎝

20.8 − 14.78

⎞
⎠

= 2539.57

(34)

Specific enthalpy of wet air.

h = C_p^{dry air}T_air,out^db + ω_outh_v^out (35)

h_out = 1.005

kg · ^∘C

× 22.9 ^∘C + 0.01732 × 2543.01

= 67.06

(36)

h_in = 1.005

kg · ^∘C

× 20.8 ^∘C + 0.01048 × 2539.57

= 47.52

(37)

5.2 Specific Enthalpy of Make-up Water

h_E = C_p^waterT_air,in^db = 4180

g · ^∘C

× 20.8 ^∘C = 86940

(38)

5.3 Final Energy Balance

H_out−H_in = Q − P where P = −96 W (39)

ṁ_air

⎛
⎝

h_out − h_in

⎞
⎠

− ṁ_Eh_E = 1000 W − −96 W = 1096 W (40)

0.05844

⎛
⎜
⎜
⎝

67.06

− 47.52

⎞
⎟
⎟
⎠

− 86.94

× 4.0 × 10⁻⁴

= 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 =

∫

h_out

h_in

h^o[T_H₂O]−h

(46)

h^o[T_H₂O] = ω h_v^o[T_H₂O] + C_p^{dry air} T_H₂O (47)

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

NTU =

(h_out−h_in)

⎛
⎜
⎜
⎝

h^o[T_H₂Oⁱⁿ] − h_out

h^o[T_H₂O^out] − h_in

⎞
⎟
⎟
⎠

(48)

NTU =

(67.06−47.52)

⎛
⎜
⎜
⎝

h^o[29.5] − 67.06

h^o[23.1] − 47.52

⎞
⎟
⎟
⎠

(49)

NTU =

(67.06−47.52)

⎛
⎜
⎜
⎝

97.09 − 67.06

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:

∫

f(x) =

(b − a)

⎛
⎜
⎜
⎝

∑

i=1

⎛
⎜
⎜
⎝

i(b−a)

n+1

⎞
⎟
⎟
⎠

(51)

NTU =

(h_out−h_in)

⎛
⎜
⎜
⎜
⎜
⎜
⎝

∑

i=1

h^o

⎡
⎢
⎢
⎣

T_H₂O^out+

i(T_H₂Oⁱⁿ−T_H₂O^out)

⎤
⎥
⎥
⎦

−

⎛
⎜
⎜
⎝

h_in+

i(h_out−h_in)

⎞
⎟
⎟
⎠

⎞
⎟
⎟
⎟
⎟
⎟
⎠

(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 L^AT_EX by H^EV^EA.