Question

Water (density = \( 10^3 \, \text{kg/m}^3 \)) enters steadily into a horizontal pipe bend, which is part of a larger piping system, as shown in the figure. The volumetric flow rate of water is 0.1 m\(^3\)/s. The gage pressure at the inlet is 500 kPa, while the exit is open to atmosphere. The \( x \)-component of the force on the support is \( F_x \). The absolute value of \( F_x \) (in kN, up to one decimal place) is \(\underline{\hspace{1cm}}\). \includegraphics[width=0.5\linewidth]{imager18.png}

Hint:

For calculating forces in a fluid flow, use the momentum equation \( \sum F = \dot{m} (\Delta v) \) and account for pressure differences and velocity changes across sections.

Answer 1

Correct Answer: 10.5

We use the principle of momentum balance to calculate the force on the support. The momentum equation in the \( x \)-direction is given by:
\[ \sum F_x = \dot{m} (v_{\text{exit}} - v_{\text{inlet}}) + (P_{\text{exit}} - P_{\text{inlet}}) A \] Where:
- \( \dot{m} \) is the mass flow rate,
- \( v_{\text{exit}} \) and \( v_{\text{inlet}} \) are the velocities at the exit and inlet,
- \( P_{\text{exit}} \) and \( P_{\text{inlet}} \) are the pressures at the exit and inlet,
- \( A \) is the cross-sectional area.
First, calculate the velocities at the inlet and exit using the flow rate and area:
\[ v_{\text{inlet}} = \frac{Q}{A_{\text{inlet}}} = \frac{0.1}{200 \times 10^{-4}} = 5 \, \text{m/s} \] \[ v_{\text{exit}} = \frac{Q}{A_{\text{exit}}} = \frac{0.1}{100 \times 10^{-4}} = 10 \, \text{m/s} \] Now, the momentum equation becomes:
\[ F_x = \dot{m} (v_{\text{exit}} - v_{\text{inlet}}) = (1000 \times 0.1) (10 - 5) = 500 \, \text{N}. \] Thus, the absolute value of \( F_x \) is \( 10.5 \, \text{kN} \).