Respuesta :
A) The specific humidity of the air with the given parameters is;
w1 = 0.00967
B) The relative humidity of the air with the given parameters is;
Φ1 = 0.459
C) The enthalpy of the air in KJ/kg dry air with the given parameters is;
h1 = 49.75 KJ/Kg
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.
We are given;
Atmospheric Pressure;P = 95 KPa
Dry temperature;T1 = 25 °C
Wet temperature;T2 = 17°C
A) 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
Read more at; https://brainly.com/question/13826648
