Incoming air flow of 200 m3/h at a temperature of 20?C and a humidity ratio of 0
ID: 2997013 • Letter: I
Question
Incoming air flow of 200 m3/h at a temperature of 20?C and a humidity ratio of 0.012 kgv/kga is mixed in a chamber with another incoming air flow of 500 m3/h at a temperature of 15?C and a humidity ratio of 0.008 kgv/kga. Based on the corresponding mass flow rates (not the original volumetric flow rates) determine:
a) The mass flow rate of the mixed air (i.e., the combination of the two flows) leaving the chamber in kg/s.
b) The temperature of the mixed air leaving the chamber.
Tip: part (a) you will first need to find the density or specific volume for each incoming state. (7 marks)
Explanation / Answer
Given
V1 = 200 m^3/h = 0.056 m^3/s
T1 = 20 C = 293 K
w1 = 0.012
V2 = 500 m^3/h = 0.139 m^3/s
T2 = 15 C = 288 K
w2 = 0.008
From the Pyschrometric chart at T1 = 20 C and w1 = 0.012
v1 = 0.846 m^3/kg
From the Pyschrometric chart at T2 = 15 C and w1 = 0.008
v2 = 0.824 m^3/kg
v1 = V1/m1
m1 = V1/v1 = 0.056/0.846
m1 = 0.0662 kg/s
m2 = V2/v2 = 0.139/0.824
m2 = 0.1687 kg/s
Therefore from conservation of mass
m3 = m1 + m2
m3 = 0.1687 + 0.0662
m3 = 0.2349 kg/s
For Mixing Process
m1*w1 + m2*w2 = m3*w3------(1)
m1*h1 + m2*h2 = m3*h3------(2)
In general h is related to w and T by
h = 1.005*T + w*(2500 + 1.88*T) ------_(3)
From 1
0.0662*0.012 + 0.1687*0.008 = 0.2349*w3
w3 = 0.009 kgv/kga
From 3
h1 = 1.005*T1 + w1*(2500 + 1.88*T1)
h1 = 1.005*293 + 0.012*(2500+1.88*293)
h1 = 331.08 kJ/kg
h2 = 1.005*T2 + w2*(2500 + 1.88*T2)
h2 = 1.005*288 + 0.008*(2500+1.88*288)
h2 = 313.77 kJ/kg
Using 3
m1*h1 + m2*h2 = m3*h3
0.2349*h3 = 0.0662* 331.08 + 0.1687*313.77
h3 = 318.65 kJ/kg
Using 3
h3 = 1.005*T3 + w3*(2500 + 1.88*T3)
318.65 = 1.005*T3 + 0.009*(2500 +1.88*T3)
318.65 = 1.005*T3 + 22.5 + 0.017*T3
1.022*T3 = 296.15
T3 = 289.78 K
T3 = 16.78 C