22-Mec-A1 Applied Thermodynamics and Heat Transfer · May 2014
Question 8 of 8: Cross-Flow Heat Exchanger (Both Fluids Unmixed)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Examinations — 07-Mec-A1 Applied Thermodynamics and Heat Transfer, May 2014. Open-book, 3-hour paper. Part A (Thermodynamics, Q1–Q4) and Part B (Heat Transfer, Q5–Q8); a complete paper is any five questions — three from one part and two from the other, all of equal value. Full worked solutions to all eight questions are given below.
Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed., McGraw-Hill) — closed- and open-system energy balances, steam tables, gas power cycles, reciprocating compressors and vapour-compression refrigeration; Çengel & Ghajar, Heat and Mass Transfer (6th ed.) and Incropera, DeWitt, Bergman & Lavine, Fundamentals of Heat and Mass Transfer (8th ed., Wiley) — composite-wall conduction, internal-flow and cross-flow convection correlations, natural convection with radiation, and the ε–NTU heat-exchanger method. Freon-12 property data are taken from the appendix supplied with the exam; steam and air data from standard tables.
Given. Air $\dot V=285\ \text{m}^3/\text{min}=4.75\ \text{m}^3/\text{s}$ (atmospheric, inlet 5 °C), heated 5 → 40 °C; water in at 90 °C; $A=27.5\ \text{m}^2$, $U=340\ \text{W/m}^2\text{K}$; $c_{p,\text{air}}=1005$, $c_{p,\text{water}}=4184\ \text{J/kg}\cdot\text{K}$.
Find. The water exit temperature and the water mass flow rate.
Figure 6 — Cross-flow arrangement (both fluids unmixed): air is heated 5 → 40 °C while water enters at 90 °C. The water carries the smaller heat-capacity rate here, so it undergoes the larger temperature change.
Approach. Fix the duty from the air side, identify which stream has the minimum capacity rate, then use the cross-flow (both-unmixed) $\varepsilon$–NTU relation to solve simultaneously for the water capacity rate (hence flow) and its exit temperature.
Air capacity rate and duty. Inlet air density $\rho=P/RT=101.325/(0.287\cdot278.15)=1.269\ \text{kg/m}^3$, so $\dot m_\text{air}=1.269(4.75)=6.03\ \text{kg/s}$ and $C_\text{air}=6059\ \text{W/K}$. The duty is$$\dot Q=C_\text{air}(40-5)=\boxed{212\ \text{kW}}.$$
Which stream is $C_\text{min}$? If air were $C_\text{min}$, its effectiveness $\varepsilon=(40-5)/(90-5)=0.41$ would be below the value the cross-flow relation gives at $\text{NTU}=UA/C_\text{air}=1.54$ (which exceeds 0.56 for any $C_r\le1$). That is impossible, so the water is $C_\text{min}$: it carries the smaller capacity rate.
Solve the cross-flow $\varepsilon$–NTU relation. With $C_\text{min}=C_w$, $\text{NTU}=UA/C_w$, $C_r=C_w/C_\text{air}$, and effectiveness $\varepsilon=\dot Q/[C_w(90-5)]$ fixed by the duty, the both-unmixed relation$$\varepsilon=1-\exp\!\left\{\tfrac{1}{C_r}\text{NTU}^{0.22}\!\left[e^{-C_r\text{NTU}^{0.78}}-1\right]\right\}$$ is satisfied by $C_w\approx2969\ \text{W/K}$ ($\text{NTU}=3.15$, $C_r=0.49$).
Water flow and exit temperature. $$\dot m_w=\frac{C_w}{c_{p,w}}=\frac{2969}{4184}=\boxed{0.71\ \text{kg/s}}\ (\approx42.6\ \text{kg/min}),\qquad T_{w,o}=90-\frac{\dot Q}{C_w}=90-\frac{212000}{2969}=\boxed{18.6\ ^\circ\text{C}}.$$
Check
The water exit temperature (18.6 °C) falls below the air exit temperature (40 °C) — a genuine temperature cross, which a cross-flow exchanger can produce when the hot stream is $C_\text{min}$ and the NTU is high (here 3.15). The energy balance closes: $C_w(90-18.6)=2969(71.4)=212\ \text{kW}=C_\text{air}(35)$. Air properties are taken at the inlet state (5 °C, 101.3 kPa) where the volume flow is quoted.
