Given: Velocity of flow: \[ \vec{V} = 2x\,\hat{i} + 3y\,\hat{j} \] We are to find the **equation of the streamline** passing through the point (4, 3). ### Step 1: Streamline Equation The differential equation of a streamline is: \[ \frac{dx}{u} = \frac{dy}{v} \] where \( u = 2x \), \( v = 3y \). So, \[ \frac{dx}{2x} = \frac{dy}{3y} \] ### Step 2: Integrate \[ \frac{dx}{2x} = \frac{dy}{3y} \] Integrate both sides: \[ \int \frac{dx}{2x} = \int \frac{dy}{3y} \] \[ \frac{1}{2} \int \frac{dx}{x} = \frac{1}{3} \int \frac{dy}{y} \] \[ \frac{1}{2} \ln x = \frac{1}{3} \ln y + C \] Multiply both sides by 6: \[ 3 \ln x = 2 \ln y + 6C \] \[ 3 \ln x - 2 \ln y = 6C \] \[ \ln x^3 - \ln y^2 = 6C \] \[ \ln \left(\frac{x^3}{y^2}\right) = 6C \] Let \( K = e^{6C} \): \[ \frac{x^3}{y^2} = K \] \[ x^3 = K y^2 \] ### Step 3: Passes through (4, 3) Plug in \( x = 4, y = 3 \): \[ 4^3 = K \times 3^2 \] \[ 64 = K \times 9 \] \[ K = \frac{64}{9} \] Plug back: \[ x^3 = \left(\frac{64}{9}\right) y^2 \] Or, \[ \frac{x^3}{y^2} = \frac{64}{9} \] ### Step 4: Manipulate to Match Options Take square root: \[ x^{3/2} = \frac{8}{3} y \] Or: \[ y = \frac{3}{8} x^{3/2} \] Or, take cube root: \[ x = \left(\frac{64}{9}\right)^{1/3} y^{2/3} \] But let's check which of the given options matches. #### Option 1: \(0.72 x^{1/2} = y^{1/2}\) - Not matching. #### Option 2: \(0.72x = y^{12}\) - Not matching. #### Option 3: None of these Neither of the equations matches the derived streamline equation. Therefore, \[ \boxed{\text{None of these}} \] is the correct answer.

Given data: - Fraction of alpha ferrite phase, \(f_{\alpha} = 0.85\) - Composition of alpha ferrite, \(C_{\alpha} = 0.022\, \text{wt\% C}\) (since ferrite is nearly pure iron with very low carbon content) - Composition of cementite (or the remaining phase), \(C_{C} = 6.7\, \text{wt\% C}\) Using the lever rule for the alloy: \[ C_{alloy} = f_{\alpha} \times C_{\alpha} + (1 - f_{\alpha}) \times C_{C} \] Substitute the values: \[ C_{alloy} = 0.85 \times 0.022 + 0.15 \times 6.7 \] Calculate: \[ 0.85 \times 0.022 = 0.0187 \] \[ 0.15 \times 6.7 = 1.005 \] Add together: \[ C_{alloy} = 0.0187 + 1.005 = 1.0237\, \text{wt\% C} \] **Rounding to three decimal places:** \[ C_{alloy} \approx 1.024\, \text{wt\% C} \] **Matching with options:** - 0.689 wt% C - 0.765 wt% C - **1.023 wt% C** ← closest match - None of the above **Answer:** \[ \boxed{\text{1.023 wt\% C}} \]

Given Data: - Heat generation rate per unit volume, \( q''' = 4 \times 10^7\, \text{W/m}^3 \) - Fuel rod radius, \( R = 5\, \text{mm} = 0.005\, \text{m} \) - Convective heat transfer coefficient, \( h = 1000\, \text{W/m}^2\text{K} \) - Surface temperature of the rod, \( T_s = 125^\circ C \) --- ### Step 1: Find the **average temperature** of the fuel rod In steady state, the heat generated within the rod is transferred to the coolant via convection. The **total heat generated per unit length**: \[ Q' = q''' \times \text{Volume per unit length} = q''' \times \pi R^2 \] Calculate: \[ Q' = 4 \times 10^7 \times \pi \times (0.005)^2 \] \[ Q' = 4 \times 10^7 \times \pi \times 25 \times 10^{-6} \] \[ Q' = 4 \times 10^7 \times \pi \times 2.5 \times 10^{-5} \] \[ Q' = 4 \times 10^7 \times 7.854 \times 10^{-5} \] \[ Q' \approx 4 \times 10^7 \times 7.854 \times 10^{-5} = 3,141.6\, \text{W/m} \] --- ### Step 2: Calculate the **mean temperature difference** between the center and the surface For a **cylindrical rod with uniform heat generation**, the temperature distribution is **parabolic**: \[ T(r) = T_c - \frac{q''' }{4k} (R^2 - r^2) \] where \( T_c \) is the **center temperature** and \( k \) is the thermal conductivity of the fuel (not given). Alternatively, for the **temperature difference between center and surface**: \[ \Delta T_{center-surface} = \frac{q''' R^2}{4k} \] But since \(k\) is not given, we use the **energy balance with convection** to find the **average temperature**. --- ### Step 3: Find the **average temperature** \( T_{avg} \) The **total heat flux** from the surface (per unit length): \[ Q' = h \times P \times (T_{avg} - T_{coolant}) \] where: - \( P = 2 \pi R \) (perimeter) - \( T_{coolant} = T_{fluid} \), which is what we want to find Rearranged: \[ T_{fluid} = T_s - \frac{Q'}{h \times 2 \pi R} \] Calculate: \[ Q' \approx 3,141.6\, \text{W/m} \] \[ 2 \pi R = 2 \pi \times 0.005 = 0.031416\, \text{m} \] \[ T_{fluid} = 125^\circ C - \frac{3,141.6}{1000 \times 0.031416} \] \[ = 125 - \frac{3,141.6}{31.416} \approx 125 - 100 \] \[ T_{fluid} \approx 25^\circ C \] --- ### Step 4: Determine the **temperature distribution** Since the heat generation is **uniform** and the boundary condition is **convective cooling**, the temperature distribution inside the rod is **parabolic**. --- ### **Final answer:** - The **fluid temperature** is approximately **25°C**. - The **temperature distribution** within the fuel rod is **parabolic**. **Matching options:** **Option b:** 25°C, Parabolic temperature distribution --- ### **Answer:** \[ \boxed{\text{b. 25°C, Parabolic temperature distribution}} \]