0.08 m³/s of air at 305 K and 60% humidity is to be cooled to 275 K. Calculate, using a psychrometric chart, the amount of heat to be removed for each 10 deg K interval of the cooling process. What total mass of moisture will be deposited? What is the humid heat of the air at the beginning and end

Question

0.08 m³/s of air at 305 K and 60% humidity is to be cooled to 275 K. Calculate, using a psychrometric chart, the amount of heat to be removed for each 10 deg K interval of the cooling process. What total mass of moisture will be deposited? What is the humid heat of the air at the beginning and end of the process?

Accepted Answer

At 305 K and 60% humidity, from Fig. 13.4, the wet-bulb temperature is 299 K and [latex]\mathscr{H}[/latex] = 0.018 kg/kg. Thus, as the air is cooled, the per cent humidity will increase until saturation occurs at 299 K and the problem is then one of cooling saturated vapour from 299 K to 275 K.
Considering the cooling in 10 deg K increments, the following data are obtained from Fig. 13.4:

At 305 K: the specific volume of dry air = 0.861 m³/kg

the saturated volume = 0.908 m³/kg
and hence the specific volume at 60% humidity = [0.861 + (0.908 - 0.861)60/100]
= 0.889 m³/kg

Thus: mass flow of moist air = (0.08/0.889) = 0.090 kg/s

Thus the flowrate of dry air = 0.090/(1 + 0.018) = 0.0884 kg/s.

From Fig. 13.4, specific heat of dry air (at [latex]\mathscr{H}[/latex] = 0) = 0.995 kJ/kg K.

∴ enthalpy of moist air = (0.0884 × 0.995)(299 - 273) + (0.018 × 0.0884)
× [4.18(299 - 273) + 2435] + 0.090 × 1.032(305 - 299) = 6.89 kW

At 295 K: Enthalpy of moist air = (0.0884 × 0.995)(295 - 273) + (0.017 × 0.0884)
× [4.18(295 - 273) + 2445] = 5.75 kW

At 285 K: Enthalpy of moist air = (0.0884 × 0.995)(285 - 273) + (0.009 × 0.0884)
× [4.18(285 - 273) + 2468] = 3.06 kW

At 275 K: Enthalpy of moist air = (0.0884 × 0.995)(275 - 273) + (0.0045 × 0.0884)
× [4.18(275 - 273) + 2491] = 1.17 kW

and hence in cooling from 305 to 295 K, heat to be removed = (6.89 - 5.75) = 1.14 kW

in cooling from 295 to 285 K, heat to be removed = (5.75 - 3.06) = 2.69 kW

in cooling from 285 to 275 K, heat to be removed = (3.06 - 1.17) = 1.89 kW

The mass of water condensed = 0.0884(0.018 - 0.0045) = 0.0012 kg/s.

The humid heats at the beginning and end of the process are:

1.082 and 1.001 kJ/kg K respectively.

Question 13.11: 0.08 m³/s of air at 305 K and 60% humidity is to be cooled t...

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