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?

Question

Accepted Answer

At 275 K, the vapour pressure of water = 0.72 kN/m² (from Tables) and the hydrogen is saturated.

The mass of water vapour: [latex]P_{w0}M_{w}[/latex]/RT = (0.72 × 18)/(8.314 × 275) = 0.00567kg/m³ and the mass of hydrogen: ([latex]P - P_{w0})M_{A}[/latex]/RT = (101.3 - 0.72)2/(8.314 × 275) = 0.0880 kg/m³

Therefore the humidity at saturation, [latex]\mathscr{H}_{0}[/latex] = (0.00567/0.0880) = 0.0644 kg/kg dry hydrogen and at 300 K, the humidity will be the same, [latex]\mathscr{H}_{1}[/latex] = 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:

[latex]H_{2}[/latex] = 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: [latex]H_{1}[/latex] = 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:

[latex]\mathscr{H}[/latex] = (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.

Question 13.12: A hydrogen stream at 300 K and atmospheric pressure has a de...

Related Answered Questions

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer:

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. ...

Verified Answer: