$t_d=34℃, p_b=740 mm \ of \ Hg=0.74 m \ of \ Hg$

$p_b=ρ.g.h=13.6×1000×9.81×0.74=98.73 kPa$

Finding the partial pressure of saturated air at same dry bulb temperature

$p_vs=p_{sat_{34℃}}=0.05318 bar=5.32 kPa(from \ steam \ table)$

Also,

$p_v=p_{sat_{WBT=20℃}}-\frac{(p_b-p_{sat_{WBT}}=20℃)(DBT-WBT)}{(1547-1.44*WBT)}$

$p_v=2.982 kPa-\frac{(98.73-2.982)(34-24)}{(1547-1.44*24)}$
=2.3489 kPa

DPT

We know, DPT or $t_dp=T_{sat_{p_v}}=T_{sat_{0.023489 bar}}$=20.015℃ (from steam table by interpolation)

Relative Humidity

$R.H.=\frac{p_v}{p_vs} =\frac{0.023489}{0.05318}$=44.17%

Specific Humidity

$ω=0.622×\frac{p_v}{p_b-p_v}$

$ω=0.622×\frac{2.3489}{98.73-2.3489}$

=$15.16×10^-3 \frac{kg \ of \ w.v.}{kg \ of \ dry \ air}$

Degree of Saturation

$ρ_v=\frac{ω×(p_b-p_v)}{0.287×(t_d+273)}$

$ρ_v=16.58×10^{-3} \frac{kg \ of \ w.v}{m^3 \ of \ air}$

Enthalpy

$h=C_pa.t_d+ω[h_{g_at \ p_v}+C_pv (t_d-T_{sat_{p_v )}} )]$

C_pa=1.005 kJ/kgK and C_pv=1.88 kJ/kgK

Now, from steam tables interpolating between 0.02 bar and 0.025 bar for $h_{g_{at
0.023489 bar}}$

$h_{g_{at p_v}}=2538.21 \frac{kJ}{kg}$

$h=1.005×34+15.16×10^-3 [2538.21+1.88(34-20.015)]$

$h=73.05 \frac{kJ}{kg \ of \ dry \ air}$