Question

a) Define relative humidity (2 marks) b) Describe the relation between relative humidity, the capacity of air to hold moisture a ambient temperature. (2 marks) c) Moist air at 22°C and a wet bulb temperature of 9°C enters a steam spray humidifier. The mass flow of the dry air is 90 kg/min. Saturated water vapor at 110 °C is injected into the mixture at a rate of 52 kg/hr. There is no heat transfer with the surroundings, and the pressure is constant throughout at 1 bar. Determine at the exit the humidity ratio ii) temperature in C (16 marks) [20 marks)

Answer 1

Relative Humidity: Amount of water vapoor present in moist air expressed as a percentage of the amount of water vapour needed for saturation of moist air at the same temperature. b Relative mu Pv = Humitity RH mus Pus mu Mass of water vapon hold in air mus Maximum mass of water vapour hold in out. sjlis SEO sum bo Exit Conservation of mass. n v/ of water vapour Humidifier Maw, + Wr - Wz maz. katht Entry W, Humidity ratio at entry Humidity ratio at Exit Amount of water added to those by the humidifier. و کیا د دیا

At Entry (DBT), = 22^C / goc (WBT), = hi - from psychometric chart. 26.812 kJ/kg W, = 0.0018433 Kg/kg of dry air Wr = 52 kg Ihr = 52 kg (sec. 60x60 M go kg lsee. ma = maz = go kg 1 min GO = W2 X 90 gox 0.0018433 + GoxGo 60 60 52 W2X W2 = 0.09844 kg 1 kg of dry air W₂ = 0.01147 kg 1 kg of day air

E: At P= 1 ban T = 110°C hw= 2696.008 kulkg Apply energy balance May hit Wvx hw= Marx hz. (ma, =mar) 1 hit hw= he ma. 26.812 t 52 3600X90 60 * 2696.008 = tha hy = 52.774 kJ/kg h = Cptz +(2500 +1-88 tz) Wz 52.774 - 1.005.xtz+ 0.01147*(2500 +1.88+2) 52.774.1.02656 tz +28.675. Itz = 23.475'c Dry bulb temperature at exist ta = 23.475'c from psychrometric chart, wet bulb tempr. WBT = 18.5°C