Respuesta :
Answer: Third stream’s temperature is [tex]33.2^{o}C[/tex] and its relative humidity is 50%.
Explanation:
According to the Psychrometric chart.
For the first stream, we have the following.
[tex]T_{1} = 40^{o}C [/tex], R.H = 40%
V = 3L/s , W = 19 gm of moisture/kg of dry air
[tex]h_{1}[/tex] = 89 kj/kg , [tex]v_{1} = 0.913 m^{3}/kg[/tex]
For the second stream, we have the following.
[tex]T_{2} = 15^{o}C[/tex]
R.H = 80%
V = 1 L/s
W = 8.5 gm of moisture/kg of dry air
[tex]h_{2}[/tex] = 36.5 kj/kg
[tex]v_{2} = 0.828 m^{3}/kg[/tex]
Now,
[tex]ma_{1} = \frac{V}{v_{1}}[/tex]
= [tex]\frac{3}{0.913}[/tex]
= 3.286 kg/s
[tex]ma_{2} = \frac{V}{v_{2}}[/tex]
= [tex]\frac{1}{0.828}[/tex]
= 1.2077 kg/s
We will calculate the value of [tex]ma_{3}[/tex] as follows.
[tex]ma_{3} = ma_{1} + ma_{2}[/tex]
= 3.286 kg/s + 1.2077 kg/s
= 4.5 kg/s
Now, heat balance will be as follows.
[tex]ma_{1}h_{1} + ma_{2}h_{2} = ma_{3}h_{3}[/tex]
[tex]h_{3}[/tex] = 74.7855 kj/kg
Hence, the moisture balance will be calculated as follows.
[tex]ma_{1}W_{1} + ma_{2}W_{2} = ma_{3}W_{3}[/tex]
[tex]W_{3}[/tex] = 16.155
Now, again using the psychrometric chart we will find the value of temperature and relative humidity as follows.
Values of [tex]T_{3}[/tex] and [tex]R.H_{3}[/tex] against [tex]W_{3}[/tex] and [tex]h_{3}[/tex] are:
[tex]T_{3} = 33.2^{o}C[/tex]
R.H = 50%
Thus, we can conclude that third stream’s temperature is [tex]33.2^{o}C[/tex] and its relative humidity is 50%.
