TITLE.PM5

PSYCHROMETRICS 481

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\M-therm\Th10-2.pm5

l Locate the point ‘2’ where the above two lines intersect.

12 Cº 20 Cº 24.5 Cº

W = 6.8

1

W = 8.6

3

W(gm / kgof dry air)

(^12)
3
4
60% RH
h(kJ/kg)
D B T
h = 29.3 1
h = 42.3 2
Fig. 10.22
From the psychrometric chart :
h 1 = 29.3 kJ/kg, h 2 = h 3 = 42.3 kJ/kg
tdb 2 = 24.5ºC, vs 1 = 0.817 m^3 /kg
The mass of air circulated per minute,
ma =
030 60
0817
.
.
×
= 22.03 kg/min.
(i)Heating capacity of the heating coil
= ma(h 2 – h 1 ) = 22.03 (42.3 – 29.3) = 286.4 kJ/min.
= 4.77 kJ/s or 4.77 kW. (Ans.)
The by-pass factor (BF) of heating coil is given by :
BF =
tt
tt
db db
db db
42
41
−
−
0.4 =
t
t
db
db
4
4
24 5
12
−
−
.
∴ 0.4 (tdb 4 – 12) = tdb 4 – 24.5
i.e., tdb 4 (coil surface temperature) = 32.8ºC. (Ans.)
(ii)The capacity of the humidifier
=mW Wa()^31 −
1000 × 60 kg/h =
22 03 8 6 6 8
1000
.(. .)−
× 60 = 2.379 kg/h. (Ans.)