Let's solve the problem step by step: --- ## **Step 1: Given Data** - **Velocity along streamline AB:** \( u = [10(1 + \frac{x}{8.5})^3] \) m/s - **\( x \):** Position along streamline (in meters) - **Pressure at \( A \) (\( x = 5 \) m):** \( p_A = 5 \) kPa - **Density (\( \rho \)):** 100 kg/m³ --- ## **Step 2: Use Bernoulli's Equation** For steady, inviscid, incompressible flow along a streamline (horizontal, so elevation cancels): \[ p + \frac{1}{2} \rho u^2 = \text{constant} \] At point \( A \) (\( x = 5 \)) and any other point \( x \): \[ p(x) + \frac{1}{2} \rho u(x)^2 = p_A + \frac{1}{2} \rho u_A^2 \] \[ p(x) = p_A + \frac{1}{2} \rho (u_A^2 - u(x)^2) \] --- ## **Step 3: Substitute Known Values** - \( p_A = 5 \) kPa \( = 500 \) Pa - \( \rho = 100 \) kg/m³ - \( u_A = 10(1 + \frac{5}{8.5})^3 \) Calculate \( u_A \): \[ u_A = 10 \left(1 + \frac{5}{8.5}\right)^3 = 10 \left(1 + .5882\right)^3 = 10 \times (1.5882)^3 \approx 10 \times 4.013 = 40.13 \text{ m/s} \] --- ## **Step 4: Write the Pressure Equation** \[ p(x) = 500 + \frac{1}{2}(100)[(40.13)^2 - u(x)^2] \] \[ u(x) = 10 \left(1 + \frac{x}{8.5}\right)^3 \] So, \[ p(x) = 500 + 500 \left[ (40.13)^2 - \left(10\left(1 + \frac{x}{8.5}\right)^3\right)^2 \right] \] \[ = 500 + 500 \left[ 161.42 - 100 \left(1 + \frac{x}{8.5}\right)^6 \right] \] \[ = 500 + 500 \times 161.42 - 500 \times 100 \left(1 + \frac{x}{8.5}\right)^6 \] \[ = 500 + 805210 - 50000 \left(1 + \frac{x}{8.5}\right)^6 \] \[ = 810210 - 50000 \left(1 + \frac{x}{8.5}\right)^6 \] --- ## **Step 5: Convert to kPa** Divide by 100: \[ p(x) = 810.21 - 50 \left(1 + \frac{x}{8.5}\right)^6 \text{ kPa} \] --- ## **Step 6: Final Answer** \[ \boxed{ p(x) = 810.21 - 50 \left(1 + \frac{x}{8.5}\right)^6 \text{ kPa} } \] --- ### **Plotting** Plot \( p(x) \) for \( -2.5 \leq x \leq .5 \) m (or as specified). --- ### **Explanation** - Used **Bernoulli's equation** for incompressible, inviscid, steady flow along a streamline. - Substituted the **velocity function** and given values. - Simplified to get the **pressure as a function of \( x \)**. --- **If you want to plot, use any graphing tool with the equation above.** --- **Let me know if you want the plot or further evaluation at specific \( x \) values!**

## Step-by-Step Solution ### **Step 1: Given Data** - **Velocity along streamline AB:** \( u = 10\left(1 + \frac{x}{8.5}\right)^3 \) m/s - **Pressure at \( A \) (\( x = 5 \) m):** \( p_A = 5 \) kPa \( = 5000 \) Pa - **Density (\( \rho \)):** \( \rho = 100 \) kg/m³ ### **Step 2: Use Bernoulli's Equation** For steady, inviscid, incompressible flow along a streamline, Bernoulli's equation states: \[ p + \frac{1}{2} \rho u^2 = \text{constant} \] At point \( A \) and any other point \( x \): \[ p(x) + \frac{1}{2} \rho u(x)^2 = p_A + \frac{1}{2} \rho u_A^2 \] ### **Step 3: Calculate Velocity at \( A \)** Calculate \( u_A \): \[ u_A = 10 \left(1 + \frac{5}{8.5}\right)^3 = 10 \left(1 + 0.5882\right)^3 = 10 \times (1.5882)^3 \approx 10 \times 4.013 = 40.13 \text{ m/s} \] ### **Step 4: Write the Pressure Equation** Using Bernoulli's equation: \[ p(x) = p_A + \frac{1}{2} \rho (u_A^2 - u(x)^2) \] Substituting values: \[ p(x) = 5000 + \frac{1}{2}(100) \left((40.13)^2 - \left(10 \left(1 + \frac{x}{8.5}\right)^3\right)^2\right) \] ### **Step 5: Simplifying the Pressure Equation** Compute \( (40.13)^2 = 1610.8169 \). Thus, \[ p(x) = 5000 + 50 \left(1610.8169 - 100\left(1 + \frac{x}{8.5}\right)^6\right) \] This simplifies to: \[ p(x) = 5000 + 80540.845 - 5000 \left(1 + \frac{x}{8.5}\right)^6 \] ### **Final Pressure Equation** Combining and simplifying: \[ p(x) = 85540.845 - 5000 \left(1 + \frac{x}{8.5}\right)^6 \] ### **Convert to kPa** Divide by 1000 to convert to kPa: \[ p(x) = 85.54 - 5 \left(1 + \frac{x}{8.5}\right)^6 \text{ kPa} \] ### **Final Answer** \[ \boxed{p(x) = 85.54 - 5 \left(1 + \frac{x}{8.5}\right)^6 \text{ kPa}} \] ### **Explanation** - Used **Bernoulli's equation** for incompressible, inviscid, steady flow. - Substituted the **velocity equation** and calculated the resulting pressure function in terms of \( x \). - The final equation expresses pressure as a function of position along the streamline.