Adiabatic mixing of two air streams

State of each stream

(using $\omega = 0.622\,p_v/(p-p_v)$, $v = 0.287\,T(1 + 1.608\omega)/p$ and $h = 1.005t + \omega(2501 + 1.88t)$.)

Mass flow of dry air

$$\begin{aligned} \dot m_f &= \frac{1500}{0.9050 \times 3600} = 0.4604\ \text{kg/s}\\ \dot m_r &= \frac{4500}{0.8579 \times 3600} = 1.4570\ \text{kg/s}\\ \dot m_m &= 1.9174\ \text{kg/s} \end{aligned}$$

Mixture (mass and energy balances)

$$\omega_m = \frac{\dot m_f\omega_f + \dot m_r\omega_r}{\dot m_m} = 0.01152\ \text{kg/kg}$$ $$h_m = \frac{\dot m_f h_f + \dot m_r h_r}{\dot m_m} = 57.71\ \text{kJ/kg}$$

Dry bulb temperature from $h_m = 1.005\,t_m + \omega_m(2501 + 1.88\,t_m)$:

$$t_m = \frac{57.71 - 0.01152 \times 2501}{1.005 + 1.88 \times 0.01152} = 28.1\ ^\circ\text{C}$$

(The mass-weighted average temperature, 28.1 °C, agrees closely.) The vapour pressure is $p_{v,m} = \omega_m p/(0.622 + \omega_m) = 1.842$ kPa and the saturation pressure at $t_m$ is about 3.80 kPa, so $\phi_m = 0.48$.

Answer: $t_m \approx 28.1$ °C; $\omega_m \approx 0.0115$ kg/kg; $h_m \approx 57.7$ kJ/kg; RH about 48 %.