Let's break down and solve the problem **step by step**. --- ## **1. Problem Overview and Given Data** **Oil at 50°F** is heated in a **2-inch Schedule 40 steel pipe** (60 ft long, surface temperature 120°F). **Flow rate:** 150 gal/hr at inlet (50°F). **Find:** - **Oil temperature as it leaves the pipe** - **Average heat-transfer coefficient** **Given Table 12.5: Properties of oil** | Property | 60°F | 120°F | |---------------------------|-----------|-----------| | Specific gravity (SG) | 0.79 | 0.74 | | Thermal conductivity (k) | 0.072 | 0.074 | | Viscosity (μ) (cP) | 18 | 8 | | Specific heat (cp) | 0.75 | 0.75 | --- ## **2. Convert Units and Find Key Quantities** ### **a. Pipe and Flow Details** - **Pipe diameter (2 in):** \( D = 2 \) in = \( 2 \times 0.0833 = 0.167 \) ft - **Pipe length:** \( L = 60 \) ft - **Flow rate:** \( Q = 150 \) gal/hr = \( 150 \times 0.1337 = 20.055 \) ft³/hr \( = 20.055/3600 = 0.00557 \) ft³/s --- ### **b. Density of Oil** Assume water density = 62.4 lb/ft³. At 60°F, \( \rho = 0.79 \times 62.4 = 49.3 \) lb/ft³ --- ### **c. Mass Flow Rate (\( \dot{m} \))** \[ \dot{m} = \text{density} \times \text{volumetric flow rate} \] \[ \dot{m} = 49.3 \text{ lb/ft}^3 \times 0.00557 \text{ ft}^3/\text{s} = 0.274 \text{ lb/s} \] --- ## **3. Energy Balance (Temperature Out Calculation)** For steady flow and negligible heat loss: \[ q = \dot{m} c_p (T_{out} - T_{in}) \] But \( q \) is also given by the heat transfer from the pipe: \[ q = hA(T_s - T_{avg}) \] However, often for a pipe with constant wall temperature: \[ \frac{T_s - T_{out}}{T_s - T_{in}} = \exp \left(-\frac{hA}{\dot{m}c_p}\right) \] Where: - \( T_s \): Surface temp = 120°F - \( T_{in} \): 50°F - \( T_{out} \): To find - \( c_p \): 0.75 Btu/lb·°F - \( A = \pi D L \): Surface area --- ### **a. Calculate Surface Area \(A\)** \[ A = \pi D L = \pi \times 0.167 \times 60 = 31.5 \text{ ft}^2 \] --- ### **b. Calculate \(\dot{m} c_p\):** \[ \dot{m}c_p = 0.274 \text{ lb/s} \times 0.75 \text{ Btu/lb·°F} = 0.206 \text{ Btu/s·°F} \] --- ## **4. Find Average Heat-Transfer Coefficient (\( h \))** ### **Step 1: Find Velocity (\(V\)):** \[ V = \frac{\text{Volumetric flow rate}}{\text{Cross-sectional area}} \] \[ A_c = \frac{\pi}{4} D^2 = \frac{\pi}{4} (0.167)^2 = 0.0219 \text{ ft}^2 \] \[ V = \frac{0.00557}{0.0219} = 0.254 \text{ ft/s} \] --- ### **Step 2: Reynolds Number** Convert viscosity to lb/ft·s: \[ 18 \text{ cP} = 0.018 \text{ lb/ft·s} \] \[ \text{Re} = \frac{\rho V D}{\mu} = \frac{49.3 \times 0.254 \times 0.167}{0.018} \approx 116 \] (Laminar flow, since Re < 2100) --- ### **Step 3: Nusselt Number for Laminar Flow, Constant Wall Temp** For fully developed, laminar flow in a circular pipe, constant wall temp: \[ \text{Nu} = 3.66 \] ### **Step 4: Calculate \( h \):** \[ \text{Nu} = \frac{hD}{k} \implies h = \frac{\text{Nu} \times k}{D} \] Use \(k = 0.072\) (60°F): \[ h = \frac{3.66 \times 0.072}{0.167} = 1.58 \text{ Btu/hr·ft}^2\cdot^\circ\text{F} \] (If you need per second, divide by 3600. But usually, we keep it per hour.) --- ## **5. Solve for Outlet Temperature (\(T_{out}\))** \[ \frac{T_s - T_{out}}{T_s - T_{in}} = \exp \left(-\frac{hA}{\dot{m}c_p}\right) \] \[ \frac{120 - T_{out}}{120 - 50} = \exp \left(-\frac{1.58 \times 31.5}{0.206 \times 3600}\right) \] - \( hA = 1.58 \times 31.5 = 49.77 \) Btu/hr·°F = \( 0.0138 \) Btu/s·°F (since 1 hr = 3600 s) - \( \dot{m}c_p = 0.206 \) Btu/s·°F So, \[ \frac{hA}{\dot{m}c_p} = \frac{0.0138}{0.206} = 0.0670 \] \[ \frac{T_s - T_{out}}{70} = e^{-0.0670} = 0.935 \] \[ T_s - T_{out} = 70 \times 0.935 = 65.45 \] \[ T_{out} = T_s - 65.45 = 120 - 65.45 = 54.6^\circ F \] --- ## **6. Final Answers** ### **a. Oil temperature as it leaves the pipe:** \[ \boxed{54.6^\circ F} \] ### **b. Average heat-transfer coefficient:** \[ \boxed{1.58 \ \text{Btu/hr·ft}^2\cdot^\circ F} \] --- ## **Summary Table** | Quantity | Value | |-----------------------------|------------------------------| | Outlet Temperature (\(T_{out}\)) | \(54.6^\circ F\) | | Heat-transfer coefficient (\(h\)) | \(1.58\) Btu/hr·ft²·°F | --- ### **Let me know if you want the calculation for the temperature "after mixing," or more detail on any step!**