Question

Moist air at a pressure of 100 kPa is compressed to 500 kPa and then cooled to ${35}^{\circ }C$ in an after cooler. The air at the entry to the after cooler is unsaturated and becomes just saturated at the exit of the after cooler. The saturation pressure of water at ${35}^{\circ }C$ is 5.628 kPa. The partial pressure of water vapour (in kPa) in the moist air entering the compressor is closest to

• A. 0.57
• B. 1.13
• C. 2.26
• D. 4.52

Answer 1

The correct option is B 1.13

Method I:

Process 1 - 2: Isentropic compression

Process 2 - 3: Cooling at constant pressure

$p_1=100kPa$

$p_2=p_3=500kPa$

$T_3={35}^{\circ }C$




Air at point 1 and 2 is unsturated and air at point 3 is saturated. Saturated partial vapour pressure at point 3,

$p_{s3}=5.628kPa$

Specific humidity at point 1,2 and 3 is same because no addition or removal water during process 1 - 2 and 2 -3

$i.e.,{\omega }_1={\omega }_3$

$\frac{0.622p_{v1}}{p_1-p_{v1}}=\frac{0.622p_{s3}}{p_3-p_{s3}}$

or $\frac{p_{v1}}{p_1-p_{v1}}=\frac{p_{s3}}{p_3-p_{s3}}$

$\frac{p_{v1}}{100-p_{v1}}=\frac{5.628}{500-5.628}=0.01138$

or $p_{v1}=0.01138(100-p_{v1})$

$=1.138-0.01138p_{v1}$

or
$1.01138p_{v1}=1.138$

or $p_{v1}=\frac{1.38}{1.01138}=1.125kPa$


Method II:



Let mixture of air water as ideal gas. At exit of intercooler, the mixture will be saturated.

Also, $\varphi =1$

$p_v=p_{vs}=5.628\text{kPa}$

Absolute humidity at exit of intercooler,
${\omega }_3=\frac{0.622p_v}{p-p_v}$

$=0.622\times \frac{5.628}{500-5.628}$

$=7.08\times {10}^{-3}$ kg/kg of dry air

Specific humidity or absolute humidity at point 1, 2 and 3 is same, because whole process does not involve any moisture content addition or removal,

$\therefore {\omega }_3={\omega }_1$

$7.08\times {10}^{-3}=0.622\times \frac{p_v}{100-p_v},p_v=1.13\text{kPa}$