## Step-by-Step Solution and Explanation **Problem Summary:** A horizontal cylindrical pipe of length \(L = 1\) m and inner diameter \(d = 5\) mm (radius \(r = 2.5\) mm) is considered. The pipe is made of a material with thermal conductivity \(k\) (not specified in the image, but let's denote it as \(k\)). Three scenarios are given for temperature boundary conditions at the ends of the pipe. ### Given Data: - Length, \(L = 1\) m - Inner diameter, \(d = 5\) mm \(\rightarrow r = 2.5\) mm = \(.0025\) m - Cross-sectional area, \(A = 2\pi r L\) (for a thin pipe wall, but for a solid rod it would be \(\pi r^2\)) - \(T_1 = 500\) K at one end, \(T_2 = 200\) K at the other Let's solve for part (a): --- ### **a) Temperature Profile with Both Ends at Fixed Temperatures (Steady-State 1D Conduction)** For a steady-state, one-dimensional heat conduction in a rod (Fourier's Law): \[ q = -kA \frac{dT}{dz} \] Where: - \(q\) is the heat transfer rate (W) - \(k\) is the thermal conductivity (W/m·K) - \(A\) is the cross-sectional area (\(m^2\)) - \(\frac{dT}{dz}\) is the temperature gradient #### **Step 1: Set Up the Differential Equation** Since \(q\) is constant in steady state and there is no heat generation in the rod, \[ \frac{d^2T}{dz^2} = \] #### **Step 2: Integrate** \[ \frac{dT}{dz} = C_1 \] \[ T(z) = C_1 z + C_2 \] #### **Step 3: Apply Boundary Conditions** At \(z = \), \(T = 500\) K At \(z = L\), \(T = 200\) K So, \[ T() = C_2 = 500 \] \[ T(L) = C_1 L + 500 = 200 \implies C_1 = \frac{200 - 500}{L} = \frac{-300}{1} = -300 \] #### **Step 4: Write the Temperature Profile** \[ T(z) = -300z + 500 \] where \(z\) is in meters, \( \leq z \leq 1\). --- ### **Final Answer (Part a)** \[ \boxed{T(z) = -300z + 500} \] --- If you need solutions for parts (b) or (c), let me know!