Question 3 of 6: Forces on the flange of a curved discharge pipe

Question 3: Forces on the flange of a curved discharge pipe (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. The $x$ and $y$ components of the force the flange must carry, together with the magnitude and direction of their resultant.

[Figure not reproduced: Figure 3 (redrawn). The flanged bend, its control volume (shaded) and the resultant load the pipe transmits to the flange bolts — 154.9 N acting 21.7° below the inlet flow direction. See the official exam paper.]

Approach. Take a control volume that cuts the pipe at the flange and encloses the whole bend; get the inlet gauge pressure from Bernoulli between the inlet section and the free jet; then apply the steady linear-momentum equation in $x$ and $y$, with the flange reaction as the only unknown.

1. Compute the two section velocities from continuity. $$A_1 = \frac{\pi D^{2}}{4} = \frac{\pi (0.100)^2}{4} = 7.854\times10^{-3}\;\text{m}^2, \qquad A_2 = \frac{\pi d^{2}}{4} = 1.963\times10^{-3}\;\text{m}^2$$ $$V_1 = \frac{Q}{A_1} = \frac{0.0150}{7.854\times10^{-3}} = 1.910\;\text{m/s}, \qquad V_2 = \frac{Q}{A_2} = \frac{0.0150}{1.963\times10^{-3}} = 7.639\;\text{m/s}$$ The velocity ratio is $(D/d)^2 = 4.00$ exactly, which is a free arithmetic check on both numbers. The mass flow is $\dot m = \rho Q = 998 \times 0.0150 = 14.97\;\text{kg/s}$.
2. Find the inlet gauge pressure from Bernoulli. With friction and elevation neglected as instructed and the exit a free jet at atmospheric pressure ($p_2 = 0$ gauge), $$p_1 + \tfrac{1}{2}\rho V_1^{2} = p_2 + \tfrac{1}{2}\rho V_2^{2} \quad\Longrightarrow\quad p_1 = \tfrac{1}{2}\rho\left(V_2^{2}-V_1^{2}\right)$$ $$p_1 = \tfrac{1}{2}(998)\left(58.36 - 3.648\right) = \boxed{27.30\;\text{kPa gauge}}$$ This pressure acts over the full inlet area and, as step 4 shows, it dominates the answer.
3. Set up the momentum balance. Let $x$ point along the inlet flow and $y$ vertically upward, and let $\mathbf{F}_R = (F_{Rx}, F_{Ry})$ be the force the flange exerts on the control volume. Only the inlet pressure force and the flange reaction act on the CV, because the outside of the bend and the exit jet are both at atmospheric pressure and gravity is neglected. For steady flow with uniform profiles, $$\sum F_x = \dot m\left(V_2\cos\alpha - V_1\right),\qquad \sum F_y = \dot m\left(V_2\sin\alpha - 0\right)$$
4. Solve the $x$ component. The pressure force at the inlet is $p_1 A_1 = 27\,302 \times 7.854\times10^{-3} = 214.4\;\text{N}$ in the $+x$ direction, so $$F_{Rx} + p_1A_1 = \dot m\left(V_2\cos 30^\circ - V_1\right)$$ $$F_{Rx} = 14.97\,(6.615 - 1.910) - 214.4 = 70.4 - 214.4 = \boxed{-144.0\;\text{N}}$$ The negative sign says the flange must pull the bend backwards, against the incoming flow. That is physically right: the jet leaving at 7.64 m/s carries away far less $x$-momentum flux (70.4 N of change) than the 214.4 N the inlet pressure pushes forward, so the bolts must absorb the difference.
5. Solve the $y$ component. No pressure force has a $y$ component, so the flange supplies the entire vertical momentum change: $$F_{Ry} = \dot m\,V_2\sin 30^\circ = 14.97 \times 7.639 \times 0.5000 = \boxed{+57.2\;\text{N}}$$
6. Assemble the resultant and reverse it onto the flange. The force of the flange on the pipe is $\mathbf{F}_R = (-144.0,\;+57.2)\;\text{N}$, of magnitude $$|\mathbf{F}_R| = \sqrt{144.0^{2}+57.2^{2}} = 154.9\;\text{N}$$ at $\arctan(57.2/144.0) = 21.7^\circ$ above the $-x$ axis, i.e. 158.3° measured from $+x$. By Newton’s third law the bend transmits the equal and opposite load to the flange: $$\boxed{\mathbf{F}_{\text{on flange}} = (+144.0\,\hat{\mathbf{i}} - 57.2\,\hat{\mathbf{j}})\;\text{N}, \qquad |\mathbf{F}| = 154.9\;\text{N} \text{ at } 21.7^\circ \text{ below } +x}$$ The bolts are therefore loaded in tension along the pipe axis and in shear vertically, which is the practical conclusion a designer needs.