Question

A hydrodynamically and thermally fully-developed fluid of 1 kg/s flows in a uniformly heated pipe (diameter 0.1 m, length 40 m). A constant heat flux of 15000 W/m$^2$ is applied. The fluid enters at 200$^\circ$C. Given: Re = 85000, Pr = 5, $k = 0.08$ W/m·K, $c_p = 2600$ J/kg·K. Correlation: \[ \text{Nu} = 0.023\, \text{Re}^{0.8} \text{Pr}^{0.4} \] Find the pipe surface temperature at the exit (nearest integer).

Hint:

For fully developed flow with constant heat flux, the wall-bulk temperature difference is constant: \[ T_s - T_b = \frac{q''}{h}. \] Compute fluid temperature rise first, then apply this relation at exit.

Answer 1

Correct Answer: 317

Step 1: Compute Nusselt number.
\[ \text{Nu} = 0.023 (85000)^{0.8} (5)^{0.4} \] Compute terms: \[ 85000^{0.8} \approx 8370, \qquad 5^{0.4} \approx 2.03 \] Thus: \[ \text{Nu} = 0.023 \times 8370 \times 2.03 \approx 391 \] Step 2: Compute convective heat-transfer coefficient.
\[ h = \frac{\text{Nu}\, k}{D} = \frac{391 \times 0.08}{0.1} = 312.8\ \text{W/m}^2\text{K} \] Step 3: Temperature rise of fluid at exit.
Total heat added: \[ Q = q'' A_s = 15000 \times (\pi D L) \] \[ A_s = \pi(0.1)(40) = 12.566\ \text{m}^2 \] \[ Q = 15000 \times 12.566 \approx 188490\ \text{W} \] Temperature rise: \[ \Delta T_b = \frac{Q}{\dot{m} c_p} = \frac{188490}{1 \times 2600} \approx 72.5^{\circ}C \] Exit bulk temperature: \[ T_{b,2} = 200 + 72.5 = 272.5^{\circ}C \] Step 4: Wall temperature for fully developed heating with constant heat flux.
For fully developed flow: \[ T_s - T_b = \frac{q''}{h} \] At exit: \[ T_s = T_{b,2} + \frac{15000}{312.8} \] \[ \frac{15000}{312.8} \approx 48.0^{\circ}C \] Thus: \[ T_s = 272.5 + 48.0 = 320.5^{\circ}C \] Rounded to nearest integer: \[ \boxed{321^{\circ}C} \]