--- title: "Force on 120° pipe bend, 0.23 m³/s" subject: "Fluid Mechanics" chapter: 7 key: "semester_3/fluid_mechanics:pipe-bend-120-force-0085" source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/pipe-bend-120-force-0085 site: BCT Engineering --- # Force on 120° pipe bend, 0.23 m³/s Fluid Mechanics · Chapter 7: Momentum principle and flow analysis - Asked in: 2078 Kartik, 9 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/pipe-bend-120-force-0085 - Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-7 ## Question The diameter of a pipe-bend is 300 mm at inlet and 150 mm at outlet and the flow is turned through 120° in a vertical plane. The axis at inlet is horizontal and the center of outlet section is 1.4 m below the center of inlet section. The total volume of fluid contained in the bend is 0.085 m³. Neglecting friction, calculate the magnitude and direction of the net force exerted on the bend by water flowing through it at 0.23 m³/s when the inlet gauge pressure is 140 kPa. Take head loss in the bend as $0.25 V^2/2g$, where V = velocity at inlet pipe. ## Short answer Net force on the bend is 13.1 kN, directed 12.7° above the horizontal. Components are 12.79 kN horizontally (inlet flow direction) and 2.88 kN upward. ## Answer **Given:** $d_1 = 0.3$ m, $d_2 = 0.15$ m, deflection $120^\circ$, $Q = 0.23$ m³/s, outlet 1.4 m below inlet, volume of fluid in bend $= 0.085$ m³, $p_1 = 140$ kPa (gauge), $h_L = 0.25V_1^2/2g$. ```text inlet 1 ---> [ bend ] (horizontal) \ 120 deg \ v outlet 2 (1.4 m lower) ``` Take $x$ along the inlet flow and $y$ upward. The outlet velocity direction is $120^\circ$ from the inlet direction, pointing downward: unit vector $(\cos120^\circ, -\sin120^\circ) = (-0.5, -0.866)$. #### Velocities and outlet pressure $$ \begin{aligned} A_1 &= 0.070686\ \text{m}^2,\quad A_2 = 0.017671\ \text{m}^2\\ V_1 &= \frac{0.23}{0.070686} = 3.254\ \text{m/s},\quad V_2 = \frac{0.23}{0.017671} = 13.015\ \text{m/s}\\ h_L &= 0.25\times\frac{3.254^2}{19.62} = 0.135\ \text{m} \end{aligned} $$ $$ \frac{p_2}{\gamma} = \frac{p_1}{\gamma} + \frac{V_1^2 - V_2^2}{2g} + 1.4 - h_L $$ $$ p_2 = 140\,000 + 9810\left(0.5396 - 8.6338 + 1.4 - 0.1349\right) = 73\,005\ \text{Pa} $$ #### Forces - $p_1A_1 = 140\,000\times 0.070686 = 9896\ \text{N}$ (in $+x$) - $p_2A_2 = 73\,005\times 0.017671 = 1290\ \text{N}$ (acts opposite to the outlet flow direction, i.e. along $(+0.5, +0.866)$) - Weight of fluid $W = 9810\times 0.085 = 834\ \text{N}$ (downward) - $\rho Q = 1000\times 0.23 = 230$ kg/s **Momentum equation (forces on fluid, $R$ = force of the bend on fluid):** $$ \begin{aligned} x:\ \ p_1A_1 + p_2A_2\,(0.5) + R_x &= \rho Q\,(V_2\cos120^\circ - V_1)\\ 9896 + 645 + R_x &= 230(-6.508 - 3.254) = -2245\\ R_x &= -12\,786\ \text{N}\\[4pt] y:\ \ p_2A_2\,(0.866) - W + R_y &= \rho Q\,(V_2\sin(-120^\circ))\\ 1117 - 834 + R_y &= 230(-11.272) = -2593\\ R_y &= -2876\ \text{N} \end{aligned} $$ The force exerted by the water **on the bend** is the opposite of $R$: $$ F_x = +12\,786\ \text{N},\qquad F_y = +2876\ \text{N} $$ $$ F = \sqrt{12\,786^2 + 2876^2} = 13\,106\ \text{N},\qquad \theta = \tan^{-1}\frac{2876}{12\,786} = 12.7^\circ $$ **Answer: net force on the bend $\approx 13.1$ kN, acting at $12.7^\circ$ above the horizontal (in the direction of the inlet flow), with $F_x = 12.79$ kN and $F_y = 2.88$ kN upward.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the outlet pressure p2 found? A. Assume p2 = 0 B. Bernoulli from inlet to outlet including elevation 1.4 m and head loss C. Assume p2 = p1 D. Use continuity only Answer: B. p2/gamma = p1/gamma + (V1^2 - V2^2)/2g + 1.4 - hL. (Key point: Outlet pressure) 2. Why is p2 much lower than p1? A. The weight of fluid pulls down B. Friction increases pressure C. The bend adds energy D. Velocity increases from 3.25 to 13.0 m/s in the smaller outlet Answer: D. Increase in velocity head of about 8 m dominates over the 1.4 m drop. (Key point: Why p2 is lower) 3. Which forces are included when applying the momentum equation to the fluid? A. Weight only B. Pressure forces at both ends, weight of fluid in the bend, and the bend reaction C. Pressure at inlet only D. Friction and viscosity only Answer: B. Both pressure forces, weight (0.085 m3 of water) and the reaction R are included. (Key point: Momentum terms) 4. What is the force of water on the bend? A. About 1.3 kN downward B. About 9.9 kN vertical C. About 2.9 kN horizontal D. About 13.1 kN at 12.7 degrees above horizontal Answer: D. Fx = 12.79 kN and Fy = 2.88 kN give 13.1 kN at 12.7 degrees. (Key point: Force on bend) ## Close variants set in other exams - [Force on 120° bend, 300 lps](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Apipe-bend-120-force-009) — Fluid Mechanics, chapter 7, asked 2× ## Related in Fluid Mechanics - [Force on bend-cum-reducer](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Abend-cum-reducer-force) — Fluid Mechanics, chapter 7 - [Resultant force on bifurcating bend](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Abifurcating-bend-resultant-force) — Fluid Mechanics, chapter 7 - [Hydrodynamic force on bend-nozzle assembly](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Avertical-bend-nozzle-hydrodynamic-force) — Fluid Mechanics, chapter 7 ## Related in other subjects - [Force on a reducing pipe bend](https://www.bctengineer.com/question.md?key=semester_4%2Ffluid_mechanics_mechanical%3Areducing-bend-force-numerical) — Fluid Mechanics, chapter 4 - [Wall friction force for developing pipe flow](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Afriction-force-parabolic-velocity-pipe) — Hydraulics, chapter 1 - [MOI of arch section about horizontal centroidal axis](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Amoi-section-horizontal-centroidal-axis-arch) — Applied Mechanics, chapter 4