Question 13.12: A hydrogen stream at 300 K and atmospheric pressure has a de...
A hydrogen stream at 300 K and atmospheric pressure has a dew point of 275 K. It is to be further humidified by adding to it (through a nozzle) saturated steam at 240 kN/m² at the rate of 1 kg steam: 30 kg of hydrogen feed. What will be the temperature and humidity of the resultant stream?
At 275 K, the vapour pressure of water = 0.72 kN/m² (from Tables) and the hydrogen is saturated.
The mass of water vapour: P_{w0}M_{w}/RT = (0.72 × 18)/(8.314 × 275) = 0.00567kg/m³ and the mass of hydrogen: (P – P_{w0})M_{A}/RT = (101.3 – 0.72)2/(8.314 × 275) = 0.0880 kg/m³
Therefore the humidity at saturation, \mathscr{H}_{0} = (0.00567/0.0880) = 0.0644 kg/kg dry hydrogen and at 300 K, the humidity will be the same, \mathscr{H}_{1} = 0.0644 kg/kg.
At 240 kN/m² pressure, steam is saturated at 400 K at which temperature the latent heat is 2185 kJ/kg.
The enthalpy of the steam is therefore:
H_{2} = 4.18(400 – 273) + 2185 = 2715.9 kJ/kg
Taking the mean specific heat capacity of hydrogen as 14.6 kJ/kg K, the enthalpy in 30 kg moist hydrogen or 30/(1 + 0.0644) = 28.18 kg dry hydrogen is:
(28.18 × 14.6)(300 – 273) = 11,110 kJ
The latent heat of water at 275 K is 2490 kJ/kg and, taking the specific heat of water vapour as 2.01 kJ/kg K, the enthalpy of the water vapour is:
(28.18 × 0.0644)(4.18(275 – 273) + 2490 + 2.01(300 – 275)) = 4625 kJ
Hence the total enthalpy: H_{1} = 15,730 kJ
In mixing the two streams, 28.18 kg dry hydrogen plus (30 – 28.18) = 1.82 kg water is mixed with 1 kg steam and hence the final humidity:
\mathscr{H} = (1 + 1.82)/28.18 = 0.100 kg/kg
In the final mixture, 0.1 kg water vapour is associated with 1 kg dry hydrogen or
(0.1/18) = 0.0056 kmol water is associated with (1/2) = 0.5 kmol hydrogen, a total of 0.5056 kmol.
∴ partial pressure of water vapour = (0.0056/0.5056)101.3 = 1.11 kN/m²
Water has a vapour pressure of 1.11 kN/m² at 281 K at which the latent heat is
2477 kJ/kg. Thus if T K is the temperature of the mixture, then:
(2716 + 15730) = (28.18 × 14.6)(T – 273) + 2.82[4.18(281 – 273) + 2447 + 2.01(T – 281)]
and T = 300.5 K
It may be noted that this relatively low increase in temperature occurs because the latent heat in the steam is not recovered, as would be the case in, say, a shell and tube unit.