Question 6 of 8: Oil Heated by Hot-Gas Crossflow Over a Tube — Wall-Temperature Check

Question 6: Oil Heated by Hot-Gas Crossflow Over a Tube — Wall-Temperature Check

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. Oil: $\dot{m}=0.25$ kg/s, tube $D=50$ mm, $L=6$ m, $T_{m,i}=23\,{}^{\circ}\text{C}$. Hot gas (air properties): crossflow at $V=10$ m/s, $T_\infty=300\,{}^{\circ}\text{C}$, uniform along the tube's length. Wall-temperature ceiling $T_t=100\,{}^{\circ}\text{C}$.

Find. Whether the tube-wall temperature stays below $100\,{}^{\circ}\text{C}$ everywhere along the 6 m length.

Approach. Get the external convection coefficient from the Churchill– Bernstein cross-flow-cylinder correlation (gas properties as air); get the internal coefficient from the oil's Reynolds number and the appropriate internal-flow Nusselt correlation; combine them into an overall $U$ (thin wall) to march the oil's mean temperature along the tube via the constant-$T_\infty$ exponential relation, then back out the LOCAL wall temperature at each axial station from the two convective resistances in series.

1. External (gas-side) convection — Churchill–Bernstein, evaluated at the film temperature. Air properties at $T_f\approx165\,{}^{\circ}\text{C}$ give $\text{Re}_D\approx1.28\times10^4$, and $$\overline{\text{Nu}}_D=0.3+\frac{0.62\,\text{Re}_D^{1/2}\Pr^{1/3}}{[1+(0.4/\Pr)^{2/3}]^{1/4}}\left[1+\left(\frac{\text{Re}_D}{282{,}000}\right)^{5/8}\right]^{4/5}$$ $$h_o\approx 49.2\text{ W/m}^2\text{K}$$
2. Internal (oil-side) convection. Engine-oil properties at the oil's mean bulk temperature give a very small Reynolds number, $$\text{Re}_D=\frac{4\dot{m}}{\pi D\mu}\approx 12 \quad(\text{deeply laminar, }\text{Re}_D\ll2300)$$ so the constant-surface-temperature laminar limit applies, $\text{Nu}_D=3.66$, giving $$h_i\approx 10.6\text{ W/m}^2\text{K}$$
3. Overall coefficient (thin wall, $A_i\approx A_o$). $$U=\left(\frac{1}{h_i}+\frac{1}{h_o}\right)^{-1}=\left(\frac{1}{10.6}+\frac{1}{49.2}\right)^{-1}\approx8.7\text{ W/m}^2\text{K}$$
4. Oil mean-temperature rise along the tube (external fluid at constant $T_\infty$ behaves like the classic "constant surface temperature" internal-flow case, with $U$ in place of a single film coefficient): $$\frac{T_\infty-T_{m,o}}{T_\infty-T_{m,i}}=\exp\!\left(-\frac{UA_s}{\dot{m}c_p}\right),\qquad A_s=\pi DL=0.942\text{ m}^2$$ $$\boxed{T_{m,o}\approx 27.8\,{}^{\circ}\text{C}}\qquad(q=\dot{m}c_p(T_{m,o}-T_{m,i})\approx2.26\text{ kW})$$
5. Local wall temperature. At any station, the same heat flux crosses both films: $h_o(T_\infty-T_s)=h_i(T_s-T_m)$, giving $$T_s=\frac{h_o T_\infty + h_i T_m}{h_i+h_o}$$ Because $h_o\gg h_i$, $T_s$ sits MUCH closer to the hot-gas temperature than to the oil's own bulk temperature, everywhere along the tube: $$T_s(x{=}0)=\frac{(49.2)(300)+(10.6)(23)}{10.6+49.2}=250.8\,{}^{\circ}\text{C}$$ $$T_s(x{=}L)=\frac{(49.2)(300)+(10.6)(27.8)}{10.6+49.2}=\boxed{251.7\,{}^{\circ}\text{C}}$$

Since $h_o$ (gas-side, ~49 W/m²K) is nearly five times $h_i$ (oil-side, laminar, ~11 W/m²K), the internal film carries almost all of the temperature drop, and the wall sits close to the 300°C gas temperature everywhere — not just near the outlet. Yes, there is a serious problem: the wall runs at roughly $250\,{}^{\circ}\text{C}$ along the entire 6 m length, more than $150\,{}^{\circ}\text{C}$ above the 100°C decomposition limit, even though the oil itself barely warms (23°C to 27.8°C) because its own convective resistance is what is throttling the heat transfer INTO the oil, not out of the gas.