Question

5) In a steamy sauna, the wall is colder than the rest of the sauna. The sauna starts off at a relative humidity of 0.60 and steam is continously added to the sauna. If the wall stays at a constant temperature of 15 celcius, what is the water vapor partial pressure (kPa) next to the wall when condensation forms? 6) For atmospheric air at dry bulb temperature of 30 celcius and wet bulb temperature of 25 celcius, what is the humidity ratio (g vapor/ kg dry air)? 7) For atmospheric air at dry bulb temperature of 30 celcius and wet bulb temperature of 25 celcius, what is the intensive enthalpy (kJ/ kg dry air)? 8) For atmospheric air at dry bulb temperature of 30 celcius and wet bulb temperature of 25 celcius, what is the dew point temperature (celcius)? 9) For atmospheric air at dry bulb temperature of 30 celcius and wet bulb temperature of 25 celcius, what is the relative humidity (%)? 10) During a simple cooling process, atmospheric air with humidity ratio of 4.5 g vapor/ kg dry air is cooled from 30 celcius to 20 celcius. What is the relative humidity (%) of the cooled atmospheric air? 11) During a simple cooling process, atmospheric air with humidity ratio of 4.5 g vapor/kg dry air is cooled from 30 celcius to 20 celcius. what is the wet bulb temperature (celcius) of the cooled atmospheric air? 12) During a cooling process with dehumidification, is the humidity ratio LARGER or SMALLER or SAME after the process? 13) During a cooling process with dehumidification, is the mass of dry air LARGER or SMALLER or SAME fter the process?

Answer 1

The temperature of wall is 15 degree C & condensation takes place as wall Therefore P_V = saturation pressure corresponding to 15 degree C = [P_v = 1.705 kpa] DBT = t = 30 degree C & WBT = t� = 25 degree C & p = p_atm = 1.01325 times 10^5 pa = 101.325 Kpa We knoe that p_v = p�_v - 1.8p (t - t�)/2700 P�_V = P_sat corresponding to WbT i.e 25 degree C = 3.169 kpa = P_V = 3.169 - 1.8 times 1.01325 times 10^5 (30 - 25)/2700 times 10^3 = 3.169 - 0.3378 = [p_v = 2.8312 kpa] w = 0.622 p_v/p - p_v = 0.622 times 2.8312/101.325 - 2.8312 kg vapour/kg of dry air \r\n = w = 17.897 times 10^-3 kg vapour/kg of dry air = [w = 17.879 g vapor/kg of dry air] h = 1.005 t + w[2500 + 1.8 t] = (1.005 times 30) + 17.879 times 10^-3 (2500 + 1.8 times 30) kJ/Kg = [h = 75.813