Water/Hot-Air Heat Exchanger: 04-BS-10, May 2015, Q6

Question 6: Water/Hot-Air Heat Exchanger (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. Water: $\dot m_w=60$ kg/s, inlet saturated liquid at $70$ kPa ($T_{sat}=89.93\ ^\circ$C), exit at $70$ kPa, $240\ ^\circ$C (superheated). Air: inlet $1007\ ^\circ$C, exit $457\ ^\circ$C, constant (atmospheric) pressure. Insulated, no pressure drops.

Find. (a) $\dot m_{air}$; (b) $\dot S_{gen}$.

Approach

Close the overall energy balance (insulated exchanger: heat lost by the air equals heat gained by the water) to solve for the unknown air mass flow rate, then evaluate the entropy balance across the whole exchanger — no external heat transfer, so all entropy generation appears as the net outlet-minus-inlet entropy flow of the two streams combined.

Water-side enthalpies. Saturated liquid at 70 kPa: $h_{w,in}=376.75$ kJ/kg, $s_{w,in}=1.1921$ kJ/kg·K. Superheated at 70 kPa, 240°C: $h_{w,out}=2955.69$ kJ/kg, $s_{w,out}=8.1624$ kJ/kg·K.
Air-side enthalpies (variable-cp). At 1280.15 K: $h_{a,in}=1498.80$ kJ/kg, $s^\circ_{a,in}=5.4407$ kJ/kg·K. At 730.15 K: $h_{a,out}=872.24$ kJ/kg, $s^\circ_{a,out}=4.8038$ kJ/kg·K.
Energy balance → air mass flow rate (part a). $$\dot m_{air}=\dot m_w\,\frac{h_{w,out}-h_{w,in}}{h_{a,in}-h_{a,out}} =60\times\frac{2955.69-376.75}{1498.80-872.24}=\boxed{246.96\ \text{kg/s}}.$$
Entropy generation (part b). No external heat transfer, so the exchanger's entropy balance is simply the sum of both streams' entropy changes (air entropy uses $s^\circ(T)$ alone since its pressure is unchanged, so the $-R\ln(P/P_{ref})$ offset cancels): $$\dot S_{gen}=\dot m_w(s_{w,out}-s_{w,in})+\dot m_{air}(s^\circ_{a,out}-s^\circ_{a,in})$$ $$=60\times(8.1624-1.1921)+246.96\times(4.8038-5.4407)=418.22-157.29=\boxed{260.93\ \text{kW/K}}.$$

Quantity	Result
(a) $\dot m_{air}$	246.96 kg/s
(b) $\dot S_{gen}$	260.93 kW/K

Concept & Theory

Topic. Two-stream, insulated heat exchanger with an unknown mass flow rate on one side, solved via the overall energy balance, followed by an entropy-generation check.

Key relations. Insulated HX energy balance: $\dot m_{cold}\Delta h_{cold}=\dot m_{hot}\Delta h_{hot}$. Entropy balance (no external heat transfer): $\dot S_{gen}=\sum\dot m_{out}s_{out}-\sum\dot m_{in}s_{in}$.

Why this works. Because both streams' pressures are unchanged across the exchanger (pressure drop neglected), each stream's own $-R\ln(P/P_{ref})$ offset cancels in its $\Delta s$ term, so $s^\circ(T)$ alone is sufficient for the air side — the general full-entropy correction from Q3/Q5 is only needed when a stream's own pressure actually changes.

Common pitfall. Assuming the exit water pressure differs from 70 kPa because the exit temperature (240°C) is well above the 70 kPa saturation temperature — "neglect pressure drops" means the water stays at 70 kPa throughout, simply becoming superheated rather than staying saturated.

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

Question 6 of 9: Water/Hot-Air Heat Exchanger

Question 6: Water/Hot-Air Heat Exchanger (15 marks)

Approach