Adiabatic Mixing of Two Moist-Air Streams: 04-BS-10, December 2013, Q7

Question 7: Adiabatic Mixing of Two Moist-Air Streams (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Steady-flow adiabatic mixing chamber, $P=1$ atm throughout. Stream 1: $T_1=42\ ^\circ$C, $\phi_1=30\%$. Stream 2 (unknown): $T_2=?$, $\phi_2=?$. Mixed exit (stream 3): $T_3=29\ ^\circ$C, $\phi_3=40\%$, total mass flow $\dot m_3=2$ kg/s. Equal (dry-air-basis) mass flow rates entering: $\dot m_{a1}=\dot m_{a2}=1$ kg/s.

Stream	$T$	$\phi$	$\dot m_a$
1 (in)	42°C	30%	1 kg/s
2 (in, unknown)	?	?	1 kg/s
3 (out, mixed)	29°C	40%	2 kg/s

Find. (a) $\phi_2$; (b) $T_2$.

Approach

An adiabatic mixing chamber with no work conserves dry-air mass, water mass, and energy. With equal entering dry-air flow rates the water-mass and energy balances reduce to simple averaging identities ($w_1+w_2=2w_3$, $h_1+h_2=2h_3$) that pin down the humidity ratio and moist-air enthalpy of the unknown stream; its temperature and relative humidity are then recovered by inverting the psychrometric state equations.

Humidity ratio and enthalpy of the two known streams. Using the ASHRAE humid-air property functions (equivalent to Table A-... psychrometric chart data) at $P=101.325$ kPa: $$w_1=0.01557\ \text{kg/kg}_{da},\quad h_1=82.41\ \text{kJ/kg}_{da}\quad(42\ ^\circ\text{C},\ 30\%\text{RH}).$$ $$w_3=0.01005\ \text{kg/kg}_{da},\quad h_3=54.83\ \text{kJ/kg}_{da}\quad(29\ ^\circ\text{C},\ 40\%\text{RH}).$$
Water-mass and energy balances ⇒ state of stream 2. With $\dot m_{a1}=\dot m_{a2}=\dot m_{a3}/2$, both balances reduce to a simple average condition solved for stream 2: $$\dot m_{a1}w_1+\dot m_{a2}w_2=\dot m_{a3}w_3\ \Rightarrow\ w_2=2w_3-w_1=2(0.01005)-0.01557=0.00452\ \text{kg/kg}_{da}.$$ $$\dot m_{a1}h_1+\dot m_{a2}h_2=\dot m_{a3}h_3\ \Rightarrow\ h_2=2h_3-h_1=2(54.83)-82.41=27.25\ \text{kJ/kg}_{da}.$$
Recover $T_2$ and $\phi_2$ from $(w_2,h_2)$. Inverting the psychrometric state functions at $P=101.325$ kPa with $w=w_2$, $h=h_2$: $$T_2=\boxed{15.73\ ^\circ\text{C}}\ \ (\text{part b}),\qquad \phi_2=\boxed{40.74\%}\ \ (\text{part a}).$$

Quantity	Result
(a) $\phi_2$ (stream 2 relative humidity)	40.74%
(b) $T_2$ (stream 2 temperature)	15.73°C

Concept & Theory

Topic. Adiabatic mixing of two moist-air streams (psychrometrics), a steady-flow device with three simultaneous conservation laws.

Key relations. Dry-air mass balance $\dot m_{a1}+\dot m_{a2}=\dot m_{a3}$; water balance $\dot m_{a1}w_1+\dot m_{a2}w_2=\dot m_{a3}w_3$; energy balance (no heat, no work, negligible KE/PE) $\dot m_{a1}h_1+\dot m_{a2}h_2=\dot m_{a3}h_3$, all on a per-kg-dry-air basis.

Why this works. Because psychrometric properties are conventionally referenced per unit mass of dry air (the dry-air mass is conserved exactly, unlike total moist-air mass, which changes if condensation occurs), writing all three balances on a dry-air basis avoids ever needing to know whether any moisture condenses during mixing — the state of stream 3 already reflects the mixed equilibrium result.

Common pitfall. Averaging temperatures directly ($T_2\ne 2T_3-T_1$) instead of averaging the humidity ratio and enthalpy — moist-air enthalpy is not a linear function of temperature alone (it also depends on $w$), so only $w$ and $h$ obey the simple linear mixing rule; $T$ and $\phi$ must be recovered afterward from the psychrometric state equations.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 14 (Gas–Vapor Mixtures and Air-Conditioning).

Question 7 of 9: Adiabatic Mixing of Two Moist-Air Streams

Question 7: Adiabatic Mixing of Two Moist-Air Streams (15 marks)

Approach