Adiabatic Mixing Chamber: 04-BS-10, December 2017, Q8

Question 8: Adiabatic Mixing Chamber (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, $P=200$ kPa throughout. $T_{hot}=70\ ^\circ$C, $\dot m_{hot}=1.8$ kg/s. $T_{cold}=20\ ^\circ$C (unknown flow rate). $T_{mix}=42\ ^\circ$C.

Find. (a) $\dot m_{cold}$ [kg/s]; (b) $\dot S_{gen}$ [kW/K].

Approach

Apply a steady-flow mass balance ($\dot m_{mix}=\dot m_{hot}+\dot m_{cold}$) together with an energy balance (adiabatic, no work: $\dot m_{hot}h_{hot}+\dot m_{cold}h_{cold}=\dot m_{mix}h_{mix}$) to solve directly for $\dot m_{cold}$, using compressed-liquid water properties at the stated 200 kPa (rather than the saturated-liquid approximation, since 200 kPa exceeds the saturation pressure at all three temperatures). Entropy generation follows from an entropy balance on the same adiabatic control volume.

Stream enthalpies and entropies at 200 kPa. $h_{hot}=293.20$, $h_{cold}=84.10$, $h_{mix}=176.06$ kJ/kg. $s_{hot}=0.95503$, $s_{cold}=0.29644$, $s_{mix}=0.59893$ kJ/kg·K.
Mass and energy balance ⇒ cold-water flow rate (part a). $$\dot m_{hot}h_{hot}+\dot m_{cold}h_{cold}=(\dot m_{hot}+\dot m_{cold})h_{mix}$$ $$\dot m_{cold}=\dot m_{hot}\,\frac{h_{hot}-h_{mix}}{h_{mix}-h_{cold}}=1.8\times\frac{293.20-176.06}{176.06-84.10}=1.8\times\frac{117.14}{91.96}=\boxed{2.2928\ \text{kg/s}}.$$ Total mixed flow: $\dot m_{mix}=1.8+2.2928=4.0928$ kg/s.
Entropy generation (part b). Adiabatic control volume, steady flow: the entropy balance reduces to $\dot S_{gen}=\dot m_{mix}s_{mix}-\dot m_{hot}s_{hot}-\dot m_{cold}s_{cold}$ (no heat-transfer term, since $\dot Q=0$). $$\dot S_{gen}=4.0928\times0.59893-1.8\times0.95503-2.2928\times0.29644$$ $$=2.451336-1.719057-0.679693=\boxed{0.05259\ \text{kW/K}}.$$ Positive, as required for an irreversible (finite-temperature-difference) mixing process.

Quantity	Result
(a) $\dot m_{cold}$	2.2928 kg/s
(b) $\dot S_{gen}$	0.05259 kW/K

Concept & Theory

Topic. Steady-flow adiabatic mixing of two liquid-water streams at a common pressure.

Key relations. Steady-flow mass balance: $\dot m_{mix}=\dot m_{hot}+\dot m_{cold}$. Steady-flow energy balance (adiabatic, no work, negligible ΔKE/PE): $\sum\dot m_{in}h_{in}=\sum\dot m_{out}h_{out}$. Entropy balance for the same control volume: $\dot S_{gen}=\sum\dot m_{out}s_{out}-\sum\dot m_{in}s_{in}$ (no $\dot Q/T$ term since adiabatic).

Why this works. Combining two independent streams at a fixed, common pressure into one outlet stream is a purely irreversible process (temperature and composition/enthalpy gradients relax spontaneously) even though no heat crosses the system boundary — hence $\dot S_{gen}>0$ is guaranteed by the second law despite $\dot Q=0$, and the entropy balance collapses to the simple mass-flow-weighted difference shown above.

Common pitfall. Using saturated-liquid ($h_f$ at the stream temperature) property shortcuts instead of the actual compressed-liquid state at 200 kPa — for near-atmospheric temperatures the difference is small but non-zero, and the more accurate compressed-liquid values are used throughout this solution.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 5 (Mass and Energy Analysis of Control Volumes) & Ch. 7 (Entropy).

Question 8 of 9: Adiabatic Mixing Chamber

Question 8: Adiabatic Mixing Chamber (15 marks)

Approach