Two kg mass of air at 40 °C with 0.023 kg water vapour per kg dry air is mixed with 3 kg mass of air at 20 °C with 0.008 kg water vapour per kg dry air to produce 5 kg mass of air at 60% relative humidity at 28 °C. Assume all the streams are at normal atmospheric pressure (101.325 kPa). The saturation vapour pressure of water in kPa at 28 °C is _____.
Official Answer Key:
3.65 to 3.75
Step-by-Step Solution & Explanation
Moisture (mass) balance on humidity ratio, treating the given masses as dry-air masses (2 kg + 3 kg = 5 kg dry air):
W_mix = (2×0.023 + 3×0.008)/5 = (0.046+0.024)/5 = 0.070/5 = 0.014 kg/kg dry air.
The humidity ratio relates to saturation vapour pressure through W = 0.622×RH×P_sat/(P_atm − RH×P_sat). Solving for P_sat with W=0.014, RH=0.6, P_atm=101.325 kPa:
0.014×(101.325 − 0.6×P_sat) = 0.622×0.6×P_sat
1.4186 − 0.0084×P_sat = 0.3732×P_sat ⇒ 1.4186 = 0.3816×P_sat ⇒ P_sat = 3.72 kPa. This lies within the accepted range 3.65–3.75. (This back-calculated value is consistent with the standard steam-table saturation pressure of water near 28 °C.)
Section 7 — Dairy and Food Engineering
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