Psychrometric Processes Questions and Answers

Answer: C

The condition of air coming out of the cooling coil is the apparatus dew point (ADP) on the psychrometric chart, which is the temperature of the coil surface when the air is cooled below its dew point. The wet bulb and dry bulb temperatures are properties of the air, not specific to the cooling coil's outlet, and 'dry point' is not a term used on the chart.

Answer: C

The outlet humidity is about 0.021 kg/kg dry air.

Adiabatic cooling by spraying water keeps the enthalpy of the moist air essentially constant, with \(h = 1.005\,t + W(2501.7 + 1.88\,t)\) in kJ per kg dry air:

\[ \begin{aligned} h_{1} &= 1.005(70) + 0.015(2501.7 + 131.6) \\ &= 70.35 + 39.50 = 109.85 \\ 109.85 &= 1.005(55) + W_{2}(2501.7 + 103.4) \\ W_{2} &= \frac{109.85 - 55.28}{2605.1} = 0.0209 \end{aligned} \]

The air gains about 0.006 kg of water per kg dry air as it cools by 15 °C. 0.017 and 0.019 add too little moisture to balance the sensible heat given up, and 0.023 adds too much.

Answer: C

The correct matching is 1-R, 2-S, 3-P, 4-Q.

Dew point temperature depends only on the moisture content (humidity ratio) of the air, so it moves with moisture, while the dry bulb temperature follows heating or cooling.

1. Sensible cooling removes heat at constant moisture, so dry bulb falls and dew point stays constant: R.
2. Cooling and dehumidification lowers the temperature and condenses moisture out, so both fall: S.
3. Chemical dehumidification by an adsorbent removes moisture and releases the heat of adsorption, so dew point falls while dry bulb rises: P.
4. Heating and humidification adds both heat and moisture, so both rise: Q.

The first option swaps the pairings for chemical dehumidification and heating with humidification. The second and fourth options pair sensible cooling with a rising dry bulb temperature, which is impossible for a cooling process.

Answer: A

The absolute humidity is about 0.026 kg water per kg dry air.

The partial pressure of the vapour is the relative humidity times the saturation pressure, and the humidity ratio follows from the molar mass ratio:

\[ \begin{aligned} p_{v} &= 0.20 \times 20 = 4\ \text{kPa} \\ H &= \frac{18.02}{28.97} \cdot \frac{p_{v}}{P - p_{v}} \\ &= 0.622 \times \frac{4}{101.325 - 4} \\ &= 0.622 \times 0.0411 = 0.0256 \end{aligned} \]

That rounds to 0.026. 0.041 is the pressure ratio \(p_{v}/(P - p_{v})\) without the molar-mass factor 0.622. 0.004 and 0.032 do not come from the correct vapour pressure of 4 kPa; using the saturation pressure of 20 kPa instead would give a much larger 0.153.