--- title: "Force on 120° bend, 300 lps" subject: "Fluid Mechanics" chapter: 7 key: "semester_3/fluid_mechanics:pipe-bend-120-force-009" source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/pipe-bend-120-force-009 site: BCT Engineering --- # Force on 120° bend, 300 lps Fluid Mechanics · Chapter 7: Momentum principle and flow analysis - Asked in: 2076 Asoj, 8 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-009 - 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 vertical plane, the axis of inlet is horizontal and the centre of the outlet section is 1.5 below the centre of the inlet section, the total volume of fluid contained in the bend is 0.09 m³. Neglecting friction, calculate the magnitude and direction of the force exerted by the water on the bend by the water flowing through it at 300 lps when the inlet pressure is 130 KPa. ## Short answer Force exerted by water on the bend is 13.6 kN, at 15.7° above the horizontal. Components: 13.09 kN horizontal (inlet direction) and 3.68 kN upward. ## Answer **Given:** $d_1 = 0.3$ m, $d_2 = 0.15$ m, deflection $120^\circ$ in a vertical plane, inlet horizontal, outlet 1.5 m below the inlet, volume of fluid $= 0.09$ m³, $Q = 300$ L/s $= 0.3$ m³/s, $p_1 = 130$ kPa (gauge). Friction is neglected. ```text 1 ---> [ bend ] \ 120 deg \ v 2 (1.5 m lower) ``` Take $x$ along the inlet flow and $y$ upward. Outlet direction: $(\cos120^\circ, -\sin120^\circ) = (-0.5, -0.866)$. #### Velocities and outlet pressure $$ \begin{aligned} A_1 &= 0.070686\ \text{m}^2,\quad A_2 = 0.017671\ \text{m}^2\\ V_1 &= 4.244\ \text{m/s},\quad V_2 = 16.977\ \text{m/s} \end{aligned} $$ Bernoulli (no loss): $$ p_2 = p_1 + \gamma\left[\frac{V_1^2 - V_2^2}{2g} + 1.5\right] = 130\,000 + 9810\,(0.918 - 14.690 + 1.5) = 9620\ \text{Pa} $$ #### Forces - $p_1A_1 = 130\,000\times 0.070686 = 9189\ \text{N}$ - $p_2A_2 = 9620\times 0.017671 = 170\ \text{N}$, acting on the water along $(+0.5, +0.866)$, i.e. opposite to the outlet flow - $W = 9810\times 0.09 = 883\ \text{N}$ (downward) - $\rho Q = 300$ kg/s **Momentum equation ($R$ = force of the bend on the water):** $$ \begin{aligned} x:\ \ 9189 + 170(0.5) + R_x &= 300\,(16.977\times(-0.5) - 4.244) = -3820\\ R_x &= -13\,094\ \text{N}\\[4pt] y:\ \ 170(0.866) - 883 + R_y &= 300\,(-16.977\times 0.866) = -4411\\ R_y &= -3675\ \text{N} \end{aligned} $$ The force exerted **by the water on the bend** is $(+13\,094,\ +3675)$ N, i.e. $F_x = 13\,094$ N and $F_y = 3675$ N: $$ F = \sqrt{13\,094^2 + 3675^2} = 13\,600\ \text{N},\qquad \theta = \tan^{-1}\frac{3675}{13\,094} = 15.7^\circ $$ **Answer: the force of the water on the bend is $\approx 13.6$ kN, acting at about $15.7^\circ$ above the horizontal in the direction of the inlet flow ($F_x = 13.09$ kN, $F_y = 3.68$ kN upward).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is p2 determined (no loss)? A. p2 = p1 - gamma 1.5 B. Bernoulli: p2 = p1 + gamma[(V1^2 - V2^2)/2g + 1.5] C. p2 = p1 D. Continuity only Answer: B. Energy balance with elevation drop of 1.5 m. (Key point: Outlet pressure) 2. How is the outlet velocity resolved? A. (-0.5, -0.866) times V2 for 120 degree turn downward B. (1, 0) C. (0.5, 0.866) D. (0, -1) Answer: A. cos120 = -0.5 and downward y component is -sin120 = -0.866. (Key point: Outlet direction) 3. Which statement about p2 is correct? A. It is above p1 B. It is low (9.6 kPa), because velocity rises to 16.98 m/s C. It is negative absolute D. It equals 130 kPa Answer: B. Velocity head rises from 0.92 to 14.69 m. (Key point: Pressure at outlet) 4. What is the force of water on the bend? A. About 13.6 kN at 15.7 degrees above horizontal B. About 9.2 kN C. About 0.9 kN D. About 3.7 kN horizontal Answer: A. Fx = 13.10 kN and Fy = 3.68 kN. (Key point: Final force) ## Close variants set in other exams - [Force on 120° pipe bend, 0.23 m³/s](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Apipe-bend-120-force-0085) — 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 - [Pipe outlet minimum water level, 50 lps](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Apipe-outlet-min-water-level-50lps) — Irrigation and Drainage Engineering, chapter 7 - [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