Question 13.2: 0.6 m³/s of gas is to be dried from a dew point of 294 K to ...
0.6 m³/s of gas is to be dried from a dew point of 294 K to a dew point of 277.5 K. How much water must be removed and what will be the volume of the gas after drying? Vapour pressure of water at 294 K = 2.5 kN/m². Vapour pressure of water at 277.5 K = 0.85 kN/m².
When the gas is cooled to 294 K, it will be saturated and Pw0 = 2.5 kN/m² .
From Section 13.2:
mass of vapour = P_{w0}M_{w}/RT = (2.5 × 18)/(8.314 × 294) = 0.0184 kg/m³ gas.
When water has been removed, the gas will be saturated at 277.5 K, and
P_{w} = 0.85 kN/m².
At this stage, mass of vapour = (0.85 × 18)/(8.314 × 277.5) = 0.0066 kg/m³ gas
Hence, water to be removed = (0.0184 – 0.0066) = 0.0118 kg/m³ gas
or: (0.0118 × 0.6) = 0.00708 kg/s
Assuming the gas flow, 0.6 m³/s, is referred to 273 K and 101.3 kN/m², 0.00708 kg/s
of water is equivalent to (0.00708/18) = 3.933 × 10^{4} kmol/s.
1 kmol of vapour occupies 22.4 m³ at STP,
and: volume of water removed = (3.933 × 10^{4} × 22.4) = 0.00881 m³/s Assuming no volume change on mixing, the gas flow after drying
= (0.60 – 0.00881) = 0.591 m³/s at STP .