Adiabatic Mixing Chamber — Hot and Cold Water Streams: 04-BS-10, December 2014, Q7

Question 7: Adiabatic Mixing Chamber — Hot and Cold Water 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. Hot stream: $T=70\ ^\circ$C, $\dot m_{hot}=1.8$ kg/s. Cold stream: $T=20\ ^\circ$C. Mixture exit: $T=42\ ^\circ$C. All streams at $P=200$ kPa. Adiabatic mixing chamber, steady flow.

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

Approach

Close the steady-flow mass and energy balances across the adiabatic mixing chamber to solve for the unknown cold-water flow rate, then evaluate the entropy balance (no heat-transfer term, since the chamber is adiabatic) to get the generation rate directly from the three streams' entropies.

Stream properties at 200 kPa. Hot (70°C): $h_{hot}=293.20$ kJ/kg, $s_{hot}=0.95503$ kJ/kg·K. Cold (20°C): $h_{cold}=84.10$ kJ/kg, $s_{cold}=0.29644$ kJ/kg·K. Mixture (42°C): $h_{mix}=176.06$ kJ/kg, $s_{mix}=0.59893$ kJ/kg·K. (Five decimals are carried on the entropies deliberately: the generation term in part (b) is a difference of three products that nearly cancel, so four-decimal entropies would move the answer in its third significant figure.)
Cold-water mass flow rate (part a). Mass balance $\dot m_{mix}=\dot m_{hot}+\dot m_{cold}$ combined with the energy balance $\dot m_{hot}h_{hot}+\dot m_{cold}h_{cold}=\dot m_{mix}h_{mix}$ gives $$\dot m_{cold}=\dot m_{hot}\,\frac{h_{hot}-h_{mix}}{h_{mix}-h_{cold}}=1.8\times\frac{293.20-176.06}{176.06-84.10}=\boxed{2.2928\ \text{kg/s}}.$$
Entropy generation rate (part b). Adiabatic, steady flow, no work: $$\dot S_{gen}=\dot m_{mix}s_{mix}-\dot m_{hot}s_{hot}-\dot m_{cold}s_{cold}=4.09283\times0.59893-1.8\times0.95503-2.29283\times0.29644=\boxed{0.05259\ \text{kJ/K}\cdot\text{s}}.$$

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

Concept & Theory

Topic. Steady-flow adiabatic mixing of two liquid-water streams at different temperatures.

Key relations. Steady-flow mass balance $\sum\dot m_{in}=\dot m_{out}$; energy balance $\sum\dot m_{in}h_{in}=\dot m_{out}h_{out}$ (adiabatic, no work); entropy balance $\dot S_{gen}=\dot m_{out}s_{out}-\sum\dot m_{in}s_{in}$ (adiabatic).

Why this works. Mixing two streams at different temperatures is inherently irreversible even though no heat crosses the chamber boundary — the entropy increase comes entirely from the internal irreversibility of two different-temperature streams equilibrating, which is exactly what the entropy balance's positive $\dot S_{gen}$ term captures.

Common pitfall. Assuming $\dot S_{gen}=0$ because the process is adiabatic — adiabatic only rules out the heat-transfer entropy term $\dot Q/T$; it says nothing about the generation term, which must still be positive for this genuinely irreversible mixing process.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 7 (Entropy).

Question 7 of 9: Adiabatic Mixing Chamber — Hot and Cold Water Streams

Question 7: Adiabatic Mixing Chamber — Hot and Cold Water Streams (15 marks)

Approach