Answer:
The temperature and relative humidity when it leaves the heating section = <u>T2 = 19° C and ∅2 = 38%</u>
Heat transfer to the air in the heating section = Qin = <u> 420 KJ/min</u>
Amount of water added = <u>0.15 KG/min</u>
Explanation:
The Property of air can be calculated at different states from the psychometric chart.
At T1 = 10° C and ∅ = 70%
h1 = 87 KJ/KG of dry air
w1 = 0.0053 kg of moist air/ kg of dry air
v1 = 0.81 m^3/kg
AT T3 = 20° C , 3 ∅ = 60%
h3 = 98 KJ/KG of dry air
w3 = 0.0087 kg of moist air/ kg of dry air
The moisture in the heating system remains the same when flowing through the heating section hence, (w1 = w2)
The mass flow rate of dry air,
m1 = V'1/V1 = 35/0.81
m1 = 43.21 kg/min
By balancing the energy in heating section we get:
mwhw + ma2h2 = mah3
(w3 -w2)hw + h2 = h3
h2 = h3 - (w3 -w2)hw @ 100 C
Hence, hw = hg @ 100 C and w2 = w1
h2 = h3 - (w3 -w2) hg @ 100 C
h2 = 98 - ( 0.0087 - 0.0053) * 2676
h2 = 33.2 KJ/KG
The exit temperature and humidity will be,
<u>T2 = 19.5° C and 2 ∅ = 37.8%</u>
(b) Calculating the transfer of heat in the heating section
Qin = ma(h2 -h1) = 43.21(33.2 - 23.5)
<u>Qin = 420 KJ/min</u>
(c) Rate at which water is added to the air in the humidifying section,
mw = ma(w3 - w2) = (43.2)(0.0087 - 0.0053)
<u>mw = 0.15 KG/min</u>
