Chemical Engineering February 2012 - 29

TABLE 1.
Water
Temperature, t
Water
Vapor
Saturation
Pressure,
pws
°C K
33
kPa
306.15 5.0343
33.5 306.65 5.1774
34
307.15 5.3239
34.5 307.65 5.4740
35
308.15 5.6278
35.5 308.65 5.7853
36
37
38
39
40
41
42
43
309.15 5.9466
310.15 6.2810
311.15 6.6315
312.15 6.9987
313.15 7.3835
314.15 7.7863
315.15 8.2080
316.15 8.6492
Partial
Press.
of H2O
Vapor,
pw
Humidity
Ratio, W
Enthalpy
of Film, h'
Enthalpy
of Air, ha
Enthalpy
Difference,
(h'-ha)
kPa
kg water/
kg dry air
5.0343 0.0326
5.1774 0.0335
5.3239 0.0345
5.4740 0.0356
5.6278 0.0366
5.7853 0.0377
5.9466 0.0388
6.2810 0.0412
6.6315 0.0436
6.9987 0.0462
7.3835 0.0489
7.7863 0.0518
8.2080 0.0549
8.6492 0.0581
1/(h'-
ha)
∆t
NTU
∑NTU Cumulative
Cooling
Range
kJ/kg
kJ/kg kJ/kg kg/kJ
116.5686 94.6668 21.9018 0.0457
119.5982 97.9585 21.6397 0.0462 0.5 0.096
122.6986 101.2503 21.4483 0.0466 0.5 0.097
125.8718 104.5420 21.3298 0.0469 0.5 0.098
129.1197 107.8338 21.2859 0.0470 0.5 0.098
132.4444 111.1255 21.3189 0.0469 0.5 0.098
135.8480 114.4173 21.4307 0.0467 0.5 0.098
142.9006 121.0008 21.8999 0.0457 1.0 0.193
150.2958 127.5843 22.7116 0.0440 1.0 0.187
158.0530 134.1678 23.8852 0.0419 1.0 0.180
166.1928 140.7513 25.4415 0.0393 1.0 0.170
174.7371 147.3348 27.4023 0.0365 1.0 0.158
183.7094 153.9183 29.7911 0.0336 1.0 0.146
193.1348 160.5018 32.6330 0.0306 1.0 0.134
∑NTU 1.75334674
Capacity (F):
3,000 m3/h
Wet bulb temperature (twb): 29°C
Relative humidity ()
Cooling water inlet (t2):
Cooling water outlet (t1):
Altitude (Z):
92%
43°C
33°C
10 m
Step I. This step involves heat load
calculations as follows:
1. Range = (t2 - t1) = 43 - 33 = 10°C
2. Approach = (t1 - twb) = 33 - 29 =
4°C
3. Heat load, Q = mCp(t2 - t1)
= 998.13  F  Range
= 998.13  3,000  10
= 29,943,900 kcal/h
Step II. This step involves total psychometric
calculations as follows:
1. Barometric pressure (p) at the given
altitude (Z) is calculated by using
the following equation:
(3)
For an altitute of 10 m, this becomes
p = 101.2 kPa
2. Assume a dry bulb temperature of
say, tdb = 32°C
3. Calculate water vapor saturation
pressure (pws) at the assumed tdb for
the temperature range of 0 to 200°C
using the equation:
(4)
Where:
C1 = -5.8002206  103
C2 = 1.3914993  100
C3 = - 4.8640239  10-2
C4 = 4.1764768  10-5
C5 =-1.4452093  10-8
C6 =6.5459673  100
and T represents the dry bulb temperature
in Kelvin. This results in the
value:
pws = 4.7585 kPa
4. The partial pressure of water (pw)
at given relative humidity is found
using the following equation:
(5)
pw = 4.3779 kPa
5. The partial pressure (pws) is again
calculated using Equation 4. This
time T represents the wet bulb
temperature in Kelvin, which calculates
to:
pws = 4.0083 kPa
6. Using pws calculated in Step 5 we
recalculate twb using the Carrier
equation:
(6)
which gives the result:
twb = 37.7°C
7. This step is an iterative process,
whereby the assumed value of tdb in
Step 2 is varied in such a way that
the calculated twb in Step 6 equals
the actual (real) twb.
8. After a number of iterations, the
calculated tdb value converges to
30.12°C.
Step III. This step involves the calculation
of the inlet air enthalpy (h1)
as follows:
1. The humidity ratio (W) for dry air is
calculated using the following equation:
(7)
W
= 0.02515 kg water/kg dry air
2. The specific volume (v) for dry air
is calculated using the following
equation:
(8)
v = 0.89511 m3/kg, dry air
3. Calculate the enthalpy of inlet air
(h1) using the following equation:
(9)
h1 = 94.750 kJ/kg
4. Calculate the humidity ratio at saturation
(Ws) for wet air using same
Equation 7. Here we now use pws:
Ws = 0.02743 kg water/kg moist air
CHEMICAL ENGINEERING WWW.CHE.COM FEBRUARY 2012 27
0.096 0.5
0.193 1
0.291 1.5
0.389 2
0.487 2.5
0.585 3
0.778 4
0.965 5
1.145 6
1.314 7
1.473 8
1.619 9
1.753 10
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Chemical Engineering February 2012

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