Question

15 m3 /min of air at 104o F and 60% relative humidity is cooled to 59o F and 100% relative humidity, and then reheated to 68o F. Find the following:

a. The flow rate of dry air (BDA) in kg /min.

b. The amount of water condensed during cooling stage.

c. kW of power needed for cooling the air in the cooling stage.

d. kW of power needed for the heating step.

Answer 1

Volumetric flow rate of air V = 15 m3/min

Temperature T1 = 104 F and relative humidity R. H = 60%

At this temperature T1 = 104 F , vapor pressure of water Pv1 = 7384.4 Pa

R. H = partial pressure of water vapor /vapor pressure of water at given gas temperature

R. H = 0.6

Pw1/Pv1 = 0.6

Pw1 = 0.6*7384.4 = 4430.64 Pa

Partial pressure of water vapor Pw1 = 4430.64 Pa

Specific humidity Y1 = 0.621*(Pw1/(P-Pw1))

Y1 = 0.621* 4430.64/(101325 - 4430.64)

Y1 = 0.0283 Kg H2O/kg dry air.

a) flow rate of dry air in kg/min

T = 104 F, V = 15 m3/min

Partial pressure of dry air Pair = P - Pw1 = 101325 - 4430.64 = 96894.36 Pa

(F - 32)*5/9 + 273 = K

T1 = (104 - 32)*5/9 + 273 = 313 K

Molar flow rate of BDA(bone dry air) by using ideal gas equation,

n = PV/RT = 96894.36Pa*15m3/min / (8.314Pa.m3/mol.k)*313k = 558.51 mol/min

Mass flow rate of BDA, ma = 558.51mol/min * 18gm/mol = 10053.28 gm/min = 10.053 kg/min

mass flow rate of BDA, ma = 10.053 kg/min

b) after cooling air get 100% saturated and T2 = 59 F

T2 = (59 - 32)*5/9 + 273 = 288 K

Vapor pressure of water at this temperature Pv2 = 1705.74 Pa

R. H = 100% it means partial pressure of water vapor is equal to vapor pressure of water.

Humidity Y2 = 0.621*1705.74/(101325 - 1705.74) = 0.0106 kg H2O/kg dry air

Amount of water get condensed w = ma*(Y1 - Y2)

w = 10.053 kg BDA/min * (0.0283 - 0.0106) kgH2O/kg BDA

w = 0.178 Kg/min

amount of water condensed during cooling stage w = 0.178 kg/min

C) kW of power needed for cooling the air in cooling stage,

Q = ma*(H2 - H1)

Specific enthalpy of air,

H = 1.005*T + Y(2502 + 1.84*T) kJ/kg BDA

Where T is in C.

At inlet Temperature T1 = 313K = 40 C

Y1 = 0.0283 kg H2O/kg BDA

Specific enthalpy, H1 = 1.005*40 + 0.0283*(2502 + 1.84*40) = 113.089 kJ/kg

After cooling T2 = 288K = 15 C and Y2 = 0.0106

H2 = 1.005*15 + 0.0106*(2502 + 1.84*15) = 41.88 kJ/kg

Q = 10.053kg/min * (41.88 - 113.089) kJ/kg = - 715.86 kJ/min = -715.86 kJ/60s =- 11.931 kW

(-)sign shows cooling is there and heat liberated from the system.

D) power needed for heating system,

Q = + 11.931 kW

(+)sign shows power need for heating.