Let's solve the problem step by step and find the **correct surface temperature of the tube**. --- ## **Given Data** - **Mass flow rate** (\(\dot{m}\)): 42 g/s = 0.042 kg/s - **Tube inner diameter** (\(D\)): 25 mm = 0.025 m - **Tube length** (\(L\)): 3 m - **Heat supplied** (\(\dot{Q}\)): 1400 W - **Inlet temperature of propane** (\(T_{in}\)): \(-50^\circ\)C - **Minimum allowable tube surface temp (ASME code)**: \(-30^\circ\)C - **Properties at \(-40^\circ\)C:** - \(c_p = 2258\) J/kg·K - \(k = 0.1281\) W/m·K - \(\mu = 1.926 \times 10^{-4}\) kg/m·s - \(Pr = 3.395\) --- ## **Step 1: Energy Balance** Assume all the heat goes into raising the temperature of the fluid: \[ \dot{Q} = \dot{m} c_p (T_{out} - T_{in}) \] But the problem asks for the surface temperature of the tube, which is maintained isothermal, not the fluid outlet temperature. Since the surface is isothermal and heat is being supplied, we use the **heat transfer equation**: \[ \dot{Q} = h A_s (T_s - T_{m}) \] Where: - \(T_s\) = Surface temperature of the tube - \(T_m\) = Mean temperature of the fluid (assumed close to inlet since fluid is saturated liquid) - \(A_s = \pi D L\) = Surface area --- ## **Step 2: Calculate Surface Area** \[ A_s = \pi D L = \pi \times 0.025 \times 3 = 0.2356~\text{m}^2 \] --- ## **Step 3: Find the Heat Transfer Coefficient \(h\)** First, calculate **Reynolds number** (\(Re\)): \[ \text{Mass flow rate},~ \dot{m} = \rho u A \] \[ A = \frac{\pi D^2}{4} = \frac{\pi \times (0.025)^2}{4} = 4.91 \times 10^{-4}~\text{m}^2 \] \[ u = \frac{\dot{m}}{\rho A} \] Assume \(\rho\) of saturated liquid propane at \(-40^\circ\)C is about **610 kg/m³** (from tables). \[ u = \frac{0.042}{610 \times 4.91 \times 10^{-4}} = \frac{0.042}{0.2995} = 0.140~\text{m/s} \] \[ Re = \frac{\rho u D}{\mu} = \frac{610 \times 0.140 \times 0.025}{1.926 \times 10^{-4}} \] \[ Re = \frac{2.135}{1.926 \times 10^{-4}} \approx 11088 \] This is **turbulent flow**. --- ### **Nusselt Number (Dittus-Boelter Equation for turbulent flow):** \[ Nu = 0.023 Re^{0.8} Pr^{0.4} \] \[ Nu = 0.023 \times (11088)^{0.8} \times (3.395)^{0.4} \] \[ (11088)^{0.8} \approx 1582 \] \[ (3.395)^{0.4} \approx 1.589 \] \[ Nu = 0.023 \times 1582 \times 1.589 \approx 57.9 \] --- \[ h = \frac{Nu \cdot k}{D} = \frac{57.9 \times 0.1281}{0.025} = \frac{7.424}{0.025} = 297.0~\text{W/m}^2\cdot\text{K} \] --- ## **Step 4: Calculate Surface Temperature** \[ \dot{Q} = h A_s (T_s - T_{m}) \] \[ T_s = T_{m} + \frac{\dot{Q}}{h A_s} \] Assume \(T_m = T_{in} = -50^\circ\)C (since the temperature rise is small and fluid is saturated liquid). \[ T_s = -50 + \frac{1400}{297 \times 0.2356} \] \[ 297 \times 0.2356 = 69.98 \] \[ \frac{1400}{69.98} = 20.0 \] \[ T_s = -50 + 20.0 = \boxed{-30.0^\circ\text{C}} \] --- ## **Step 5: Correct Answer** - **The red-circled answer (\(-34.069^\circ\)C) is incorrect.** - **The correct surface temperature is \(-30.0^\circ\)C.** --- ### **Summary Table** | Quantity | Value | |-----------------------|--------------------| | Surface area (\(A_s\))| 0.2356 m² | | Heat transfer coeff. \(h\) | 297 W/m²·K | | Surface temp. (\(T_s\)) | **-30.0°C** | --- ### **Conclusion** - The surface temperature of the tube is **\(-30.0^\circ\)C**. - This is exactly the minimum allowed by the ASME code for this material. - The tube **is suitable** under these conditions. - \(-40^\circ\)C **is** an appropriate temperature to use for fluid properties (since it's close to the mean fluid temperature). --- **If you have more questions or need clarification, just ask!**