## Step-by-Step Solution Let's address each part independently as requested, focusing on deriving the temperature profile equation for a horizontal pipe under steady-state conditions. --- ### Given Data - Pipe radius: \( R \) (in meters) - Pipe length: \( L \) (in meters) - Pipe wall thickness: \( 5 \) mm \( = .005 \) m --- ### Part (a): Pipe Between Two Walls at Different Temperatures **Problem:** One end at \( 500\,\text{K} \), the other at \( 200\,\text{K} \). **Assumptions:** - Steady-state, 1D heat conduction along the length (\( x \)-axis) - No heat generation in the wall - Constant thermal conductivity \( k \) **Governing Equation:** 1D heat conduction equation without generation: \[ \frac{d^2 T}{dx^2} = \] **Solution:** Integrate twice: \[ \frac{dT}{dx} = C_1 \] \[ T(x) = C_1 x + C_2 \] Apply boundary conditions: - At \( x = \): \( T() = 500 \) - At \( x = L \): \( T(L) = 200 \) Plug in the boundary conditions: \[ T() = C_1 \cdot + C_2 = 500 \implies C_2 = 500 \] \[ T(L) = C_1 L + 500 = 200 \implies C_1 = \frac{200 - 500}{L} = -\frac{300}{L} \] **Final Answer (a):** \[ T(x) = 500 - \frac{300}{L}x \] where \( x \) is the distance from the \( 500\,K \) end, \( \leq x \leq L \). --- ### Part (b): Pipe with Fluid Flowing Inside at 500 K **Problem:** Fluid inside at constant \( 500\,K \). **Assumptions:** - Inner surface maintained at \( 500\,K \) (isothermal boundary) - Outer surface exposed to air (temperature not specified, so assume heat loss or insulation) - If the pipe wall is thin and the fluid dominates, wall temperature equals fluid temperature Since only the fluid temperature is given and no external temperature or heat flux is specified, the temperature profile in the wall will be uniform: **Final Answer (b):** \[ T(x) = 500\,K \] for all \( x \). --- ### Part (c): One End at 500 K, Other End with Internal Heat Generation **Problem:** - One end at \( 500\,K \) - Other end connected to a reactor with first-order exothermic reaction (\( q_{generated} = 200\,W/L \)) **Assumptions:** - Steady-state, 1D conduction with uniform internal heat generation - Let \( q \) be the heat generated per unit volume (\( W/m^3 \)), but here it is given per unit length (\( W/L \)), so convert as needed **Governing Equation:** 1D conduction with generation: \[ \frac{d^2 T}{dx^2} + \frac{q}{k A} = \] where \( A \) is the cross-sectional area. Given \( q_{generated} = 200\,W/L \), so per unit length, not per volume. For a pipe of length \( L \), the total heat generated is \( Q = 200 \times L \) W. But for the differential equation, we need \( q \) per unit volume: \[ q = \frac{q_{generated}}{A} \] where \( A = \pi (R_{outer}^2 - R_{inner}^2) \) and thickness \( = .005 \) m, so \( R_{outer} = R \), \( R_{inner} = R - .005 \). Integrate twice: \[ \frac{d^2 T}{dx^2} = -\frac{q}{k} \] \[ \frac{dT}{dx} = -\frac{q}{k}x + C_1 \] \[ T(x) = -\frac{q}{2k}x^2 + C_1 x + C_2 \] Apply boundary conditions: - At \( x = \): \( T() = 500\,K \implies C_2 = 500 \) - At \( x = L \): (if the reactor is at some temperature or has some heat flux, but if not specified, you can leave in terms of \( T(L) \)) If the end at \( x=L \) is insulated or specified at temperature \( T_L \), use that; otherwise, the generic solution: **Final Answer (c):** \[ T(x) = -\frac{q}{2k}x^2 + C_1 x + 500 \] If the end at \( x=L \) is also held at a temperature \( T_L \), then: \[ T(L) = -\frac{q}{2k}L^2 + C_1 L + 500 = T_L \] \[ C_1 = \frac{T_L - 500 + \frac{q}{2k}L^2}{L} \] So, the full temperature profile: \[ T(x) = -\frac{q}{2k}x^2 + \left[\frac{T_L - 500 + \frac{q}{2k}L^2}{L}\right] x + 500 \] where \( q = \frac{200}{A} \), \( A = \pi (R^2 - (R-.005)^2) \). --- ### **Summary Table** | Part | Equation for \( T(x) \) | |------|------------------------| | (a) | \( T(x) = 500 - \frac{300}{L}x \) | | (b) | \( T(x) = 500 \) | | (c) | \( T(x) = -\frac{q}{2k}x^2 + C_1 x + 500 \) (with \( q = \frac{200}{A} \)), \( C_1 \) depends on \( T_L \) | This approach covers the general steady-state heat profiles for each scenario described.