22-Mec-A1 Applied Thermodynamics and Heat Transfer · May 2016
Question 8 of 8: Cross-Flow Tube-Bank Heat Exchanger (ε–NTU)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed., McGraw-Hill) — closed- and open-system energy balances, steam tables, vapour and gas power cycles, 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) — cylindrical 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, air and engine-oil data from standard tables.
Paper format: National Examination 07-Mec-A1, 3 hours, open book. Part A — Thermodynamics (Q1–4); Part B — Heat Transfer (Q5–8). Each answer carries equal value; a complete paper is any five (three from one part, two from the other). All eight questions are solved in full below.
Figure 8 — Cross-flow tube bank: cold water runs inside the 40 tubes (into the page); hot air sweeps across them along the duct. Both streams are unmixed. The exchanger area is small relative to the large capacity rates, so the temperature changes are slight.
Given. A cross-flow bank of 40 tubes (1 cm dia) in a 1 m × 1 m duct: water in at 18 °C/3 m/s inside, air in at 130 °C/105 kPa/12 m/s across, U = 80 W/m²·°C. Find. Both outlet temperatures and the heat-transfer rate.
Approach. Compute both capacity rates from the stated velocities, form NTU on the tube surface area, and use the cross-flow (both-unmixed) effectiveness relation to get the duty and the two outlet temperatures.
Water capacity rate. $\dot m_w=\rho_w(n\tfrac{\pi}{4}D^2)V=998.7(40\cdot7.85\times10^{-5})(3)=9.41\ \text{kg/s}$; $C_w=\dot m_w c_{p,w}=3.93\times10^{4}$ W/K.
Surface area and NTU. $A_s=n\pi DL=40\pi(0.01)(1)=1.257\ \text{m}^2$, so $$\text{NTU}=\frac{UA_s}{C_\text{min}}=\frac{80(1.257)}{1.10\times10^{4}}=0.0091.$$
Effectiveness (cross-flow, both unmixed). $\varepsilon=1-\exp\!\big\{\tfrac{1}{C_r}\text{NTU}^{0.22}[e^{-C_r\text{NTU}^{0.78}}-1]\big\}=0.00907$ (essentially $\varepsilon\approx\text{NTU}$ at this tiny NTU).
Duty and outlet temperatures. $$\dot Q=\varepsilon C_\text{min}(T_{a,i}-T_{w,i})=0.00907(1.10\times10^{4})(112)=\boxed{11.2\ \text{kW}}.$$ Then $T_{a,o}=130-\dot Q/C_a=\boxed{129.0\,{}^\circ\text{C}}$ and $T_{w,o}=18+\dot Q/C_w=\boxed{18.3\,{}^\circ\text{C}}.$
Check
Energy balances close: $\dot Q=C_a(130-129.0)=C_w(18.3-18)=11.2$ kW. The striking result — barely 1 °C off the air and 0.3 °C on the water — is correct: with only 1.26 m² of area against ~10⁴ W/K capacity rates, NTU ≈ 0.009, so this exchanger is grossly undersized for any meaningful temperature change. The tube length is taken as the 1 m duct span, since the tubes must cross the flow for a cross-flow arrangement.
