Question 13.2: In a vessel at 101.3 kN/m² and 300 K, the percentage relativ......

(a) From the definition of percentage relative humidity:

P_{w}={P_{w0}}{\frac{R H}{100}}=3600\times{\bigg(}{\frac{25}{100}}{\bigg)}=900\;\mathrm{N/m^{2}} = \underline{\underline{0.9\ kN/m^{2}}}

(b) In 1 m³ of air:
mass of water vapour = {\frac{(900\times18)}{(8314\times300)}}=0.0065\ \mathrm{kg}

mass of air = {\frac{[(101.3-0.9)\times1000\times29]}{(8314\times300)}}=1.167\ \mathrm{kg}

Hence: specific volume of water vapour at 0.9 kN/m² = \left({\frac{1}{0.0065}}\right) = \underline{\underline{154\ m^{3}/kg}}

specific volume of air at 100.4 kN/m² = \left({\frac{1}{1.167}}\right) = \underline{\underline{0.857\ m^{3}/kg}}

(c) Humidity: \mathscr{H} =\left(\frac{0.0065}{1.1673}\right) = \underline{\underline{0.0056\ kg/kg}}

(Using the approximate relationship:

\mathscr{H} =\frac{(18\times900)}{(29\times101.3\times1000)}=0.0055\mathrm{~kg/kg}.)

∴ Humid volume = volume of 1 kg air + associated vapour = specific volume of air at 100.4 kN/m²

= \underline{\underline{0.857\ m^{3}/kg}}

(d) From equation 13.3:

Percentage humidity = \left(\frac{{P}-{P}_{w0}}{{P}-{P}_{w}}\right)\cdot\left(\frac{{P}_{w}}{{P}_{w0}}\right)\times\,100 = \frac{(P-P_{w0})}{(P-P_{w})}\times (percentage relative humidity)                (13.3)

percentage humidity = \frac{\left[(101.3-3.6)\times1000\right]}{\left[(101.3-0.9)\times1000\right]}\times25

= \underline{\underline{24.3\ per\ cent}}