--- title: "Adiabatic mixing of two air streams" subject: "Applied Thermodynamics" chapter: 4 key: "semester_3/applied_thermodynamics:mixing-of-air-streams-numerical" source: https://www.bctengineer.com/q/semester_3/applied_thermodynamics/mixing-of-air-streams-numerical site: BCT Engineering --- # Adiabatic mixing of two air streams Applied Thermodynamics · Chapter 4: Air-Conditioning - Asked in: year not given, 6 marks - Video: [Adiabatic Mixing of Two Airstreams](https://www.youtube.com/watch?v=xKI7ENKzVDo) — Mitchell Paulus - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/applied_thermodynamics/mixing-of-air-streams-numerical - Whole chapter: https://www.bctengineer.com/md/semester_3/applied_thermodynamics/chapter-4 ## Question 1500 m³/h of fresh air at 38 °C and 40% relative humidity is mixed with 4500 m³/h of recirculated room air at 25 °C and 50% relative humidity. Calculate the dry bulb temperature, specific humidity, enthalpy and relative humidity of the mixture. Take $p = 101.325$ kPa; saturation pressures are 6.624 kPa at 38 °C and 3.169 kPa at 25 °C. ## Short answer From volumetric flows and specific volumes, dry air mass flows are 0.4604 kg/s and 1.4570 kg/s. Mass-weighted averages give humidity 0.0115 kg/kg, enthalpy 57.7 kJ/kg; solving for temperature yields 28.1 °C, with relative humidity about 48%. ## Answer #### State of each stream | Stream | $t$ (°C) | $\phi$ | $\omega$ (kg/kg) | $v$ (m³/kg) | $h$ (kJ/kg) | |---|---|---|---|---|---| | Fresh (f) | 38 | 0.40 | 0.01670 | 0.9050 | 81.15 | | Recirculated (r) | 25 | 0.50 | 0.00988 | 0.8579 | 50.30 | (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 %. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Before mixing, each stream's flow must be converted to: A. Volume flow at the mixture temperature B. Mass flow of dry air, using each stream's own specific volume C. Total moist air mass D. Relative humidity flow Answer: B. Balances are on dry air mass, so m = V/v for each stream: 0.4604 kg/s fresh and 1.4570 kg/s recirculated. (Key point: Mass flow basis) 2. The mixture specific humidity and enthalpy are found by: A. Simple average of the two values B. Mass-flow-weighted averages from mass and energy balances C. Volume-flow-weighted average of RH D. Taking the larger stream's values Answer: B. omega_m = (m_f omega_f + m_r omega_r)/m_m, and likewise for h, since both are conserved. (Key point: Mixture omega and h) 3. How is the mixture dry bulb temperature obtained precisely? A. Average the two RH values B. From h_m = 1.005 t + omega_m(2501 + 1.88 t) solved for t, about 28.2 °C C. t_m = 38 - 25 D. Equal to the recirculated air temperature Answer: B. Using the mixture h and omega, solving the enthalpy equation gives t_m = 28.2 °C, close to the mass-weighted average. (Key point: Mixture DBT) 4. Why can't RH of the mixture be averaged directly? A. RH is not conserved; it is found from p_v,m and p_s at t_m (about 48%) B. RH is always 100% after mixing C. RH equals omega D. RH is the mass average of 40% and 50% Answer: A. RH depends nonlinearly on temperature, so compute p_v,m from omega_m and divide by p_s at t_m: 1.843/3.816 = 0.483. 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