22-Mec-A1 Applied Thermodynamics and Heat Transfer · May 2017
Question 6 of 8: Oil in a Tube Heated by Cross-Flow Gas — Wall-Temperature Check
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, filling of evacuated vessels, air-standard dual and gas-turbine (turbojet) 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) — radial composite-cylinder conduction with convection, external cross-flow over a cylinder, internal-flow temperature decay, radiation between concentric spheres with a shield, and the ε–NTU cross-flow heat-exchanger method. Ammonia and ideal-gas air properties are read from the tables appended to the examination paper; the turbojet uses cold-air-standard constant specific heats.
Paper format: National Examination 16-Mec-A1, May 2017, 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 questions from one part and two from the other). All eight questions are solved in full below.
Question 6: Oil in a Tube Heated by Cross-Flow Gas — Wall-Temperature Check (equal value)
Given. The tube, oil and gas data above, with the 100 °C wall limit. Find. whether the wall temperature exceeds 100 °C anywhere (i.e. is there a problem).
Figure 6 — Oil is heated inside the tube while hot gas sweeps across it. Because the external gas film is a better conductor to the wall than the sluggish oil film inside, the wall floats close to the gas temperature.
Approach. Compute the external cross-flow coefficient (Churchill–Bernstein) and the internal laminar coefficient (thermal-entry correlation), combine them through the thin wall to find the oil outlet temperature, then evaluate the wall temperature — which is highest at the oil exit — and compare with the 100 °C limit.
External gas-side coefficient. $Re_D=\dfrac{VD}{\nu}=\dfrac{10(0.05)}{43.6\times10^{-6}}=1.15\times10^{4}$. Churchill–Bernstein gives $Nu\approx56.9$, so $h_o=\dfrac{Nu\,k}{D}=\dfrac{56.9(0.0430)}{0.05}\approx48.9$ W/m²·K.
Internal oil-side coefficient. $Re_D=\dfrac{4\dot m}{\pi D\mu}=8.2$ (deeply laminar; $\mu=\rho\nu=0.0779$). With $Pr\approx1148$, the flow is thermally developing over the whole tube; the entry-length correlation gives $Nu\approx6.7$, so $h_i=\dfrac{6.7(0.141)}{0.05}\approx18.8$ W/m²·K.
Overall coefficient and oil outlet temperature. Thin wall: $\dfrac1U=\dfrac1{h_i}+\dfrac1{h_o}\Rightarrow U=13.6$ W/m²·K; $A=\pi DL=0.942$ m². The single-stream decay against the constant gas gives
$$T_{m,o}=T_\infty-(T_\infty-T_{m,i})e^{-UA/\dot m c_p}=300-277\,e^{-12.8/51.95}=83.5\ \text{°C}$$
Wall temperature (worst case = oil exit). Equating the two films, $T_s=\dfrac{h_oT_\infty+h_iT_{m}}{h_o+h_i}$. At the outlet ($T_m=83.5$ °C):
$$T_{s,max}=\frac{48.9(300)+18.8(83.5)}{48.9+18.8}$$
$T_{s,max}\approx240$ °C — even at the inlet $T_s\approx223$ °C
Verdict. The wall runs at roughly 223–240 °C over the entire length, far above the 100 °C limit.
Yes — there is a serious problem: the oil will overheat and decompose.
Check — property sensitivity and the physical reason.
Oil viscosity is strongly temperature-dependent, so $h_i$ (and $T_{m,o}$) shift with the assumed mean oil temperature; but the conclusion is robust — because $h_o>h_i$, the thin wall sits far closer to the 300 °C gas than to the oil. Even doubling $h_i$ leaves $T_s>200$ °C. The remedy would be to lower the gas temperature, reduce the gas velocity/coefficient, or interpose a wall resistance so the oil-side controls.
