Question

The temperature of air entering an adiabatic saturator is ${42}^{\circ }$ and that of the air leaving is ${30}^{\circ }$C.
The humidity ratio (kg/kg d.a) of the entering air is
Use the following relation:

$P_v{P}_w-\frac{(P-P_w)(T_{db}-T_{wb})\left(1.8\right)}{2800-1.3(1.8T_{db}+32)}$

where P = total air pressure (kPa)
$P_v$ = partial pressure of water vapour (kPa)
$P_w$ = saturation pressure of water vapour corresponding to wet bulb temperature (kPa)
$T_{db}$ = dry-bulb temperature ${(}^{{}^{\circ }C})$
$T_{wb}$ = Wet -bulb temperature ${(}^{{}^{\circ }C})$

Use the following data:

T${(}^{{}^{\circ }C})$	30	34	38	42
P(kPa)	4.246	4.753	6.624	8.20

• A. 0.04723
• B. 0.01396
• C. 0.03154
• D. 0.02198

Answer 1

The correct option is D 0.02198

Temperature at the exit of an adiabatic saturator is WBT.

So,
DET = ${42}^{\circ }C,WBT={30}^{\circ }C,P_s=8.20kPa,P_w=4.246kPa$

Using the relation
$P_v=P_w-\frac{(P-P_w)(T_{db}-T_{wb})\left(1.8\right)}{2800-1.3(1.8T_{db}+32)}$

$P_v=4.246-\frac{(101.325-4.246)(42-30)\left(1.8\right)}{2800-13(1.8\times 42+32)}$

$P_v=3.458kPa$

Humidity ratio, $\omega =0.622\frac{P_v}{P-P_v}=\frac{0.622\times 3.458}{101.325-3.458}=0.02198kg/kgd.a$