Page 218 - Chemical process engineering design and economics

Compressors, Pumps, and Turbines 201

mechanical friction in the seals and bearings. This lost work is accounted for by
a mechanical efficiency. Thus, the fan work,

2
1 f ACv ) API
W F = —— | ——— + —— (5.6)
T!F L 2g c p J

is the work delivered to the shaft of the fan.
where r| F = r| M r| H, and W F
Because power is the rate of doing work, the fan shaft power - frequently
called brake power - is calculated from Equation 5.7.

(5.7)
P F = mW F
Example 5.1 Calculation of Fan Power______________________

A fan will pneumatically convey 1360 kg/h (2300 Ib/h) of a powdered resin
from a storage bin to a mixer [11]. The duct diameter is 15.2 cm (6 in). Assume
an electrical-motor efficiency of 95%. If the air flow rate needed to convey the
3
resin is 1670 m /h (983 ftVmin) at 300 K (540 °R) and 1.013 bar (14.7 psia).
The pressure drop in the duct system is 0.0893 bar (1.29 psi), what is the re-
quired fan power?
The fan shaft work is calculated from Equation 5.6. From Table 5.2, it is
seen that a radial fan is acceptable for the conveying system. From Table 5.3 a
conservative value for the radial fan efficiency of 65% is selected. The air velocity
is

V m 3 4 1 h
v = —— = 1670 — ——————— ———— = 25.56 m/s (83.9 ft/s)
2
A h 7i(0.152) m 2 3600 s
From the ideal gas law, the air density,

PM 1.013 bar IxlO 5 N 1 kgmol-K 1 29 kg
p = -
2
RT 1 1 m -bar 8314 N-m 300 K 1 kgmol
3
3
= 1.179 kg/m (0.0736 lb/ft )
Because SI units are used, gc is not needed in Equation 5.6. From Equation
5.6, the fan work,

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