TITLE.PM5

PSYCHROMETRICS 469

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(ii)Heat added to air per minute :

Enthalpy, h 2 = cptdb + Whvapour

= 1.005 × 30 + 0.0109 × [hg + 1.88 (tdb – tdp)]

= 1.005 × 30 + 0.0109 [2556.3 + 1.88 (30 – 15.5)]

= 58.31 kJ/kg of dry air

Mass of dry air in 90 m^3 of air supplied

ma =

pV

RT

ppV

RT

=()tv−

= −××

×+

(.. )

()

1 0132 0 01755 10 90

287 273 20

5
= 106.5 kg/min.
Amount of heat added per minute
= 106.5 (h 2 – h 1 ) = 106.5 (58.31 – 47.85) ~−1114 kJ. (Ans.)
Example 10.7. Sensible cooling : 40 m^3 of air at 35°C DBT and 50% R.H. is cooled to
25 °C DBT maintaining its specific humidity constant. Determine :
(i)Relative humidity (R.H.) of cooled air ;
(ii)Heat removed from air.
Solution. For air at 35°C DBT and 50% R.H. :
pvs = 0.0563 bar (At 35ºC, from steam tables)
φ = p
p

v

vs

∴ pv = φ × pvs = 0.5 × 0.0563 = 0.02815 bar

W 1 =

622 622 02815

0132 02815

p

pp

v

tv−

= ×

− = 0.0177 kg/kg of dry air

h 1 = cp tdb 1 + W 1 [hg 1 + 1.88 (tdb 1 – tdp 1 ]

tdp 1 ~− 23ºC (corresponding to 0.02815 bar)

∴ h 1 = 1.005 × 35 + 0.0177 [2565.3 + 1.88 (35 – 23)] = 80.98 kJ/kg of dry air

For air at 25°C DBT :

(i)R.H. of cooled air :

Since the specific humidity remains constant the vapour pressure in the air remains constant.

φ =

p

p

v

vs

=0 02815

0 0317

.

.

= 0.888 or 88.8%

i.e., Relative humidity of the cooled air = 88.8%. (Ans.)
(ii)Heat removed from air :
h 2 = cp tdb 2 + W 2 [hg 2 + 1.88 (tdb 2 – tdp 2 )]
= 1.005 × 25 + 0.0177 [2547.2 + 1.88 (25 – 23)]
= 70.27 kJ/kg of dry air.
To find mass of dry air (ma), using the relation :
pava = maRaTa

∴ ma = pv

RT

aa

aa

= −××

×+

(.. )

()

1 0132 0 02815 10 40

287 273 35

5

= 44.57 kg

∴ Heat removed from 40 m^3 of air

= ma (h 1 – h 2 ) = 44.57 (80.98 – 70.27) = 477.3 kJ. (Ans.)

Q WW

ttdp dp pv

12 0 0177

2123

==

==°

L

NM

O

QP

.kg/kgof dryair

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