Question

1) 1.7 lbm/min of air-H20 mixture is conditioned in the following manner, from a starting point of a dry bulb temperature of 100 degrees F and a relative humidity of 20%.

First, the mixture is spray-humidified adiabatically until the relative humidity is 80%.

Second, the mixture is heated until the dry-bulb temperature returns to 100 degrees F.

A) What is the net rate of heat transport required?

B) What is the rate at which water is added?

Please help asap and please show work. This is my second time posting. Thank you!

Answer 1

Mass flow rate of air – H2O mixture = m = 1.7 lbm / min = 0.012852 kg/s

DBT 1 = 100 °F = 37.7778 °C

RH1 = 20%

Post adiabatic spray humidification,

RH2 = 80%

Post Adiabatic spray humidification & heating,

DBT 3 = 100 °F = 37.7778 °C

Net rate of heat transport =?

Amount of water added =?

From Psychrometric chart,

h1 = 60.5 kJ/ kg of dry air

h2 = 76.5 kJ/kg of dry air

Net rate of heat transport = heat added = 0.012852 kg/s * (76.5 – 60.5) kJ/ kg of dry air = 0.205632kW

From Psychrometric chart,

w1 = specific humidity at point 1 = 0.00825 kg / kg of dry air

w2 = specific humidity at point 2 = 0.01480 kg / kg of dry air

Moisture removed during the process = 0.01480 kg / kg of dry air - 0.00825 kg / kg of dry air

                                                            = 0.006550 kg / kg of dry air

Rate of water addition = 0.006550 kg / kg of dry air * 0.012852 kg/s = 0.000084181 kg/s = 0.011135249298929385 lbm / minute