Question

The dry- and wet-bulb temperatures of atmospheric air at 95 kPa are 25 and 178C, respectively. Determine (a) the specific humidity, (b) the relative humidity, and (c) the enthalpy of the air, in kJ/kg dry air.

Answer 1

Correct question is;

The dry- and wet-bulb temperatures of atmospheric air at 95 kPa are 25 and 17 °C, respectively. Determine (a) the specific humidity, (b) the relative humidity, and (c) the enthalpy of the air, in kJ/kg dry air.

Answer:

A) w1 = 0.00967

B) Φ1 = 0.459

C) h1 = 49.75 KJ/Kg

Explanation:

We are given;

A) Atmospheric Pressure;P = 95 KPa

Dry temperature;T1 = 25 °C

Wet temperature;T2 = 17°C

From table A-4 attached and at temperature of 17°C and by interpolation, we have a saturation pressure of P_g2 = 1.938 kpa

First of all, we will calculate the specific humidity from the given pressure and saturation pressure with the formula;

w2 = (0.622 × P_g2)/(P - P_g2)

w2 = (0.622 × 1.938)/(95 - 1.938)

w2 = 0.013

Now, let's calculate specific humidity with the enthalpies at 17 °C and by interpolation. We have specific enthalpies from table A-4 as;

h_fg2 = 2460 KJ/Kg

h_g1 = 2546.5 KJ/Kg

h_f2 = 71.36 KJ/Kg

The formula for the specific humidity under these conditions is;

w1 = (c_p(T2 - T1) + w2•h_fg2)/(h_g1 - h_f2)

c_p of air has a value of 1.005 KJ/Kg.°C

Thus;

w1 = (1.005(17 - 25) + 0.013*2460)/(2546.5 - 71.36)

w1 = 0.00967

B) The relative humidity is determined from the equation;

Φ1 = (w1*p)/(0.622 + w1)p_g1

From table A-4 attached and at temperature of 25 °C, we have a saturation pressure of P_g1 = 3.1698 KPa

Φ1 = (0.00967*95)/(0.622 + 0.00967)3.1698

Φ1 = 0.459

C)For the enthalpy of air, h1 we will use the formula;

h1 = (c_p × T1) + (w1 × h_g1)

h1 = (1.005 × 25) + (0.00967 × 2546.5)

h1 = 49.75 KJ/Kg