A 5-m 5-m 3-m room shown in Fig. 14–7 contains air at 25°C and 100 kPa at a rela

Question

A 5-m 5-m 3-m room shown in Fig. 14–7 contains air at 25°C and 100
kPa at a relative humidity of 75 percent. Determine (a) the partial pressure
of dry air, (b) the specific humidity, (c) the enthalpy per unit mass of the dry
air, and (d ) the masses of the dry air and water vapor in the room.

Explanation / Answer

(a) The partial pressure of dry air can be determined from Eq. 14–2: where Thus, (b) The specific humidity of air is determined from Eq. 14–8: (c) The enthalpy of air per unit mass of dry air is determined from Eq. 14–12: The enthalpy of water vapor (2546.5 kJ/kg) could also be determined from the approximation given by Eq. 14–4: which is almost identical to the value obtained from Table A–4. (d ) Both the dry air and the water vapor fill the entire room completely. Therefore, the volume of each gas is equal to the volume of the room: The masses of the dry air and the water vapor are determined from the idealgas relation applied to each gas separately: The mass of the water vapor in the air could also be determined from Eq. 14–6: mv vma 10.01522 185.61 kg2 1.30 kg mv PvVv RvT 12.38 kPa2 175 m3 2 10.4615 kPa # m3>kg # K2 1298 K2 1.30 kg ma PaVa RaT 197.62 kPa2 175 m3 2 10.287 kPa # m3>kg # K2 1298 K2 85.61 kg Va Vv Vroom 15 m2 15 m2 13 m2 75 m3 hg @ 25°C 2500.9 1.82 1252 2546.4 kJ>kg 63.8 kJ/kg dry air 11.005 kJ>kg # °C2 125°C2 10.01522 12546.5 kJ>kg2 h ha vhv cpT vhg v 0.622Pv P Pv 10.6222 12.38 kPa2 1100 2.382 kPa 0.0152 kg H2O/kg dry air Pa 1100 2.382 kPa 97.62 kPa Pv fPg fPsat @ 25°C 10.752 13.1698 kPa2 2.38 kPa

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