Given: $t_{d}=22^{\circ} \mathrm{C}, p_{b}=760 \mathrm{~mm}$ of $\mathrm{Hg}=0.755 \mathrm{mof} \mathrm{Hg}, R H=30 \%$, $$ p_{b}=\rho g h=13.6 \times 1000 \times 9.81 \times 0.760=101.4 \mathrm{kPa} $$ i) Vapour pressure: Finding the partial pressure of saturated air at same DBT: $$ \begin{array}{c} p_{v s}=p_{s a t}=0.02642 \text { bar }(\text { from steam table }) \\ t_{d} \quad \text { also, } \quad R H=\frac{p_{v}}{p_{v s}} \\ p_{v}=0.7926 k P a \end{array} $$ ii) Humidity ratio: $$ \begin{array}{c} \omega=0.622 \times \frac{p_{v}}{p_{b}-p_{v}}=0.622 \times \frac{0.7926}{101.4-0.7926} \\ \omega=4.9 \times 10^{-3} \frac{\mathrm{kg} \text { of } \text { water vapour }}{\mathrm{kg} \text { of dry air }} \end{array} $$ iii) Vapour density: $$ \begin{array}{c} \rho_{v}=\frac{\omega \times\left(p_{b}-p_{v}\right)}{0.287 \times\left(t_{d}+273\right)} \\ =\frac{4.9 \times 10^{-3} \times(101.4-0.7926)}{0.287 \times(22+273)} \\ \rho_{v}=5.823 \times 10^{-3} \frac{\mathrm{kg} \text { of } w . v .}{m^{3} \text { of air }} \end{array} $$ iv) Enthalpy $$ \begin{array}{c} h=C_{p a} \cdot t_{d}+\omega\left[\begin{array}{ll} h & g \\ a t p_{v} & +C_{p v} \end{array}\left(t_{d}-T_{s a t}\right)\right] \\ C_{p a}=1.005 \frac{k J}{k g K} \text { and } C_{p v}=1.88 \frac{k J}{k g K} \end{array} $$ Now, from steam tables interpolating between $0.0061$ bar and $0.01$ bar for $h$ $g$ at $0.007926 \mathrm{bar}$ $$ \begin{array}{c} h \underset{a t p_{v}}{g}=2507.593 \frac{\mathrm{kJ}}{\mathrm{kg}} \ h=1.005 \times 22+4.933 \times 10^{-3}[2507.593+1.88(22-3.269)] \ h=34.653 \frac{\mathrm{kJ}}{\mathrm{kg} \text { of } d r y \text { air }} \end{array} $$