Let's break down **Question 2** step by step: --- ## **Given Data** - Thickness of wall: \( 2L = 86\, \text{mm} \implies L = 43\, \text{mm} = .043\, \text{m} \) - Uniform heat generation: \( \dot{q} = 100\, \text{W}/\text{mm}^3 = 1 \times 10^9\, \text{W}/\text{m}^3 \) - Convective cooling at \( x = \pm 40\, \text{mm} = \pm .04\, \text{m} \) - Ambient temperature: \( T_\infty = 30^\circ \text{C} \) - Temperature distribution in wall: \( T(x) = a(L^2 - x^2) + b \), where \( a = 16\,^\circ\text{C}/\text{m}^2 \), \( b = 46^\circ\text{C} \) --- # **(a) Thermal Conductivity of the Wall (\(k\))** ### **Step 1: General Heat Equation for Steady State with Internal Generation** \[ \frac{d^2 T}{dx^2} + \frac{\dot{q}}{k} = \] ### **Step 2: Compare with Given Temperature Profile** Given: \[ T(x) = a(L^2 - x^2) + b = aL^2 - a x^2 + b \] Take the second derivative: \[ \frac{dT}{dx} = -2a x \\ \frac{d^2 T}{dx^2} = -2a \] ### **Step 3: Plug into Heat Equation** \[ -2a + \frac{\dot{q}}{k} = \implies k = \frac{\dot{q}}{2a} \] ### **Step 4: Substitute Values** - \( \dot{q} = 1 \times 10^9\, \text{W}/\text{m}^3 \) - \( a = 16\,^\circ\text{C}/\text{m}^2 \) \[ k = \frac{1 \times 10^9}{2 \times 16} = \frac{1 \times 10^9}{32} \] \[ k = 31,250,000\, \text{W}/(\text{m}\cdot ^\circ\text{C}) \] --- ### **Final Answer for (a):** \[ \boxed{ k = 3.13 \times 10^7\, \text{W}/(\text{m}\cdot \text{C}) } \] --- # **(b) Value of the Convection Heat Transfer Coefficient (\( h \))** ### **Step 1: Boundary Condition at \( x = .04\, \text{m} \)** At the surface (\( x = L_s = .04\, \text{m} \)), energy balance: \[ - k \left. \frac{dT}{dx} \right|_{x=L_s} = h [T(L_s) - T_\infty] \] ### **Step 2: Compute Needed Values** #### 1. **\(\frac{dT}{dx}\) at \(x = .04\)** \[ \frac{dT}{dx} = -2a x \Rightarrow \left. \frac{dT}{dx} \right|_{x=.04} = -2 \times 16 \times .04 = -1.28\, ^\circ\text{C}/\text{m} \] #### 2. **\(T(x)\) at \(x = .04\)** \[ T(.04) = 16[(.043)^2 - (.04)^2] + 46 \] \[ (.043)^2 = .001849,\quad (.04)^2 = .0016 \] \[ T(.04) = 16(.001849 - .0016) + 46 = 16(.000249) + 46 = .003984 + 46 \approx 46.004^\circ\text{C} \] \[ T(.04) - T_\infty = 46.004 - 30 = 16.004^\circ\text{C} \] ### **Step 3: Solve for \( h \)** \[ - k \left( -1.28 \right) = h (16.004) \] \[ k \times 1.28 = h \times 16.004 \] \[ h = \frac{k \times 1.28}{16.004} \] Recall \( k = 3.13 \times 10^7 \): \[ h = \frac{3.13 \times 10^7 \times 1.28}{16.004} \] \[ h = \frac{4.0064 \times 10^7}{16.004} \] \[ h \approx 2.50 \times 10^6\, \text{W}/(\text{m}^2 \cdot ^\circ\text{C}) \] --- ### **Final Answer for (b):** \[ \boxed{ h = 2.50 \times 10^6\, \text{W}/(\text{m}^2 \cdot \text{C}) } \] --- ## **Summary** - (a) \( k = 3.13 \times 10^7\, \text{W}/(\text{m}\cdot \text{C}) \) - (b) \( h = 2.50 \times 10^6\, \text{W}/(\text{m}^2 \cdot \text{C}) \) Let me know if you need further clarification on any step!