Reaction Force on a Flanged Reducing Bend: 22-Mec-A6 Fluid Machinery, May 2017, Q3

Question 3: Reaction Force on a Flanged Reducing Bend (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. Steady incompressible flow of water ($\rho=1000\ \text{kg/m}^3$) through a reducing bend; the exit is a free jet at atmospheric pressure (gauge $P_2=0$); friction and gravity neglected.

Given data
Quantity	Value
Volumetric flow $Q$	15 L/s = 0.015 m³/s
Inlet diameter $D$ / exit diameter $d$	0.10 m / 0.05 m
Exit angle $\alpha$ above horizontal	30°
Exit pressure (gauge)	$P_2=0$

Find. The anchoring force ($F_x$, $F_y$, magnitude and direction) the flange must supply to hold the bend.

Figure 3: Reducing bend — horizontal inlet (diameter $D$), free-jet exit at $\alpha=30^\circ$ (diameter $d$). The flange at the inlet transmits the anchoring force.

Approach. Continuity gives the two velocities; Bernoulli (no friction, no gravity) gives the inlet gauge pressure; the steady integral momentum equation on the fluid control volume, with gauge pressures, gives the force the pipe exerts on the fluid, whose reaction is the force on the flange.

Velocities from continuity. With $A_1=\tfrac{\pi}{4}D^2=7.854\times10^{-3}\ \text{m}^2$ and $A_2=\tfrac{\pi}{4}d^2=1.963\times10^{-3}\ \text{m}^2$, $$V_1=\frac{Q}{A_1}=1.910\ \text{m/s},\qquad V_2=\frac{Q}{A_2}=7.639\ \text{m/s}.$$
Inlet pressure from Bernoulli. Neglecting gravity, $P_1+\tfrac12\rho V_1^2=P_2+\tfrac12\rho V_2^2$ with $P_2=0$ gives $$P_1=\tfrac12\rho\,(V_2^2-V_1^2)=27.4\ \text{kPa (gauge)}.$$
$x$-momentum. With $\dot m=\rho Q=15\ \text{kg/s}$ and the flange force on the fluid $(F_x,F_y)$, $$P_1A_1+F_x=\dot m\,(V_2\cos\alpha-V_1)\;\Rightarrow\;F_x=\dot m(V_2\cos\alpha-V_1)-P_1A_1=-144\ \text{N}.$$
$y$-momentum. No inlet $y$-momentum and $P_2=0$, so $$F_y=\dot m\,(V_2\sin\alpha-0)=15(7.639)(0.5)=+57.3\ \text{N}.$$ Thus the flange must exert $(F_x,F_y)=(-144,\,+57.3)\ \text{N}$ on the fluid. By Newton's third law the flow exerts the opposite force on the bend, $$\boxed{F_{x,\text{bend}}=+144\ \text{N},\quad F_{y,\text{bend}}=-57.3\ \text{N},\quad |F|=155\ \text{N}.}$$ This resultant acts at $\arctan(57.3/144)=21.7^\circ$ below the horizontal, in the direction of the incoming flow.

Question 3 — results
Quantity	Value
$V_1$ / $V_2$	1.91 / 7.64 m/s
Inlet gauge pressure $P_1$	27.4 kPa
Force on bend $F_x$ / $F_y$	+144 N / −57.3 N
Resultant $\|F\|$, direction	155 N, 21.7° below horizontal

Topic: Steady integral linear-momentum balance on a control volume (pipe bend / reducer).

Key relations: continuity $Q=V_1A_1=V_2A_2$; Bernoulli $P+\tfrac12\rho V^2=\text{const}$; momentum $\sum\mathbf F=\dot m(\mathbf V_{\text{out}}-\mathbf V_{\text{in}})$ with gauge pressures on the inlet/exit faces.

Why this works: the reaction comes from two effects — the change in fluid momentum direction/magnitude, and the pressure acting over the inlet area. Using gauge pressures automatically cancels the uniform atmospheric pressure acting all around the outside of the bend, so the anchoring force is exactly what the bolts must carry. The reducer raises the exit speed fourfold (area ratio 4), which both builds the inlet pressure and swings the momentum vector.

Common pitfall: dropping the $P_1A_1$ pressure term (it dominates $F_x$ here, exceeding the momentum flux), or confusing the force on the fluid with the force on the bend — they are equal and opposite. Also, the exit free-jet pressure is zero gauge, so no exit pressure term appears.

Source: Fox, Pritchard & McDonald, Introduction to Fluid Mechanics, 10th ed., Ch. 4 (linear momentum for a control volume).

Question 3 of 6: Reaction Force on a Flanged Reducing Bend

Question 3: Reaction Force on a Flanged Reducing Bend (20 marks)