TITLE.PM5

PSYCHROMETRICS 475

dharm

\M-therm\Th10-2.pm5

D B T

2

1

3

55%

100% R. H.

3Cº

W

(b)

Fig. 10.17

(i)Mass of spray water required

At 22ºC

From steam tables : pvs = 0.0264 bar

φ 3 =

p

P

v p

vs

3 vs

3

3

0 0264

=

.

= 0.55

∴ pv 3 = 0.55 × 0.0264 = 0.01452 bar

∴ W 3 = 0 622 0 01452

1 0132 0 01452

..

(.. )

×

−

= 0.00904 kg/kg of dry air

At 3ºC

From steam tables : pvs = 0.0076 bar

φ 1 =

p

p

v

vs

1

1

= 1.00

∴ ppvvs 11 = = 0.0076 bar

W 1 =

622 0076

0132 0076

×

− = 0.0047 kg/kg of dry air

W 3 – W 1 = 0.00904 – 0.0047 = 0.00434 kg/kg of dry air

v RT

a p p

a

(^3) a 3 v 3
3
55
287 273 22
1 0132 10
287 295
0 9987 10
==×+
−×
= ×
×
()
(. ).
= 0.847 m^3 /kg of dry air
Spray water =
0 00434
0 847
.
.
= 0.005124 kg moisture/m^3. (Ans.)
(ii)Temperature to which the air must be heated tdb 2 :
Now h 2 + (W 3 – W 2 ) h 4 = h 3