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An air-conditioning system operates at a total pressure of 95kPa and consists of a heating section and a humidifier that supplies wet steam (saturated water vapor) at 100°C. Air enters the heating section at 10°C and 70% relative humidity at a rate of 35 m^3/min, and it leaves the humidifying section at 20°C and 60 percent relative humidity. Determine:


(a) the temperature and relative humidity of air when it leaves the heating section.

(b) the rate of heat transfer in the heating section.

(c) the rate at which water is added to the air in the humidifying section.

Respuesta :

danishmallick97

Answer:

The temperature and relative humidity when it leaves the heating section = T2 = 19° C and ∅2 = 38%

Heat transfer to the air in the heating section = Qin = 420 KJ/min

Amount of water added =  0.15 KG/min

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,

T2 = 19.5° C and  2 ∅ = 37.8%

(b) Calculating the transfer of heat in the heating section

Qin = ma(h2 -h1) = 43.21(33.2 - 23.5)

Qin = 420 KJ/min

(c) Rate at which water is added to the air in the humidifying section,

mw = ma(w3 - w2) = (43.2)(0.0087 - 0.0053)

mw = 0.15 KG/min

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