--- title: "Resultant force on bifurcating bend" subject: "Fluid Mechanics" chapter: 7 key: "semester_3/fluid_mechanics:bifurcating-bend-resultant-force" source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/bifurcating-bend-resultant-force site: BCT Engineering --- # Resultant force on bifurcating bend Fluid Mechanics · Chapter 7: Momentum principle and flow analysis - Asked in: 2074 Asoj, 8 marks - Video: [Force Exerted by a Flowing Fluid on a Pipe Bend](https://www.youtube.com/watch?v=1RKz3P7vBXY) — TutorialsPoint - Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/bifurcating-bend-resultant-force - Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-7 ## Question Ignoring friction losses, calculate the magnitude and direction of resultant force, exerted on the bend when water discharges at the atmosphere as shown in figure below. Both nozzles discharge water with a velocity of 20 m/sec. Consider the axes of the pipe and the nozzles lie in a horizontal plane. [Figure: pipe with inlet velocity $V_1$ branching into a straight outlet $d_2$ = 12 cm ($V_2$) and a branch $d_3$ = 10 cm ($V_3$) at 60° to the axis] ## Short answer Assume d1=20 cm, compute V1=12.2 m/s, p1=125.6 kPa via Bernoulli. Momentum equations yield force components on bend: Rx=2526 N, Ry=-2721 N. Resultant force is 3.71 kN, directed 47.1° below the inlet axis. ## Answer The resultant force is found by applying the Bernoulli equation to get the inlet pressure and then the momentum equation in the $x$ and $y$ directions. The result is about **3.71 kN**, acting at about $47^\circ$ to the inlet axis. #### Assumptions The figure does not give the inlet diameter, so I take $d_1 = 20$ cm. The inlet pipe lies along the $x$-axis, nozzle 2 ($d_2 = 12$ cm) continues straight, and nozzle 3 ($d_3 = 10$ cm) leaves at $60^\circ$ to the axis. The plane is horizontal, so weight does not matter. Both nozzles discharge to the atmosphere, so $p_2 = p_3 = 0$ (gauge). $\rho = 1000\ \text{kg/m}^3$. ```text / --> V3 (d3, 60 deg) V1, p1 ------->( \ --> V2 (d2, straight) ``` #### Discharges and inlet velocity $$ \begin{aligned} A_1 &= 0.03142\ \text{m}^2,\; A_2 = 0.011310\ \text{m}^2,\; A_3 = 0.007854\ \text{m}^2 \\ Q_2 &= A_2V_2 = 0.2262\ \text{m}^3/\text{s},\quad Q_3 = A_3V_3 = 0.1571\ \text{m}^3/\text{s} \\ Q_1 &= Q_2 + Q_3 = 0.3833\ \text{m}^3/\text{s},\quad V_1 = Q_1/A_1 = 12.20\ \text{m/s} \end{aligned} $$ #### Inlet pressure (Bernoulli, no losses) $$ p_1 = \tfrac12\rho\,(V_2^2 - V_1^2) = \tfrac12(1000)(20^2 - 12.2^2) = 125\,580\ \text{Pa} = 125.6\ \text{kPa} $$ #### Momentum equation (force of bend on fluid, $F_x$, $F_y$) $$ \begin{aligned} F_x + p_1A_1 &= \rho\,(Q_2V_2 + Q_3V_3\cos 60^\circ - Q_1V_1) \\ F_x &= 1000\,(4.524 + 1.571 - 4.676) - 3945 = -2526\ \text{N} \\ F_y &= \rho Q_3V_3\sin 60^\circ = 1000(0.1571)(20)(0.866) = 2721\ \text{N} \end{aligned} $$ The force exerted **on the bend** by the water is equal and opposite: $$ R_x = 2526\ \text{N},\quad R_y = -2721\ \text{N} $$ $$ R = \sqrt{2526^2 + 2721^2} = 3713\ \text{N},\qquad \theta = \tan^{-1}\frac{2721}{2526} = 47.1^\circ $$ **Answer: $R \approx 3.71$ kN, directed $47.1^\circ$ below the inlet axis (away from the branch side), pushing the bend along the flow direction.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the inlet velocity V1 found in this problem? A. V1 equals the nozzle velocity of 20 m/s B. Q1 = Q2 + Q3 from the two nozzle discharges, then V1 = Q1/A1 C. V1 = V2 + V3 D. From the momentum equation directly Answer: B. Continuity gives the inlet discharge as the sum of the nozzle discharges. (Key point: Inlet velocity by continuity) 2. How is the inlet pressure p1 obtained? A. Bernoulli between inlet and a nozzle exit where p = 0 (gauge) B. Momentum equation in the y direction C. p1 = rho g h with the nozzle height D. p1 equals atmospheric pressure Answer: A. With no losses, p1 = 1/2 rho (V2^2 - V1^2) because the nozzles discharge to the atmosphere. (Key point: Inlet pressure) 3. Which forces appear on the left side of the x-momentum equation for the fluid in the bend? A. Only the weight of the water B. Only the pressure forces at the two nozzles C. Friction force and the weight D. Force of the bend on fluid (Fx) plus the pressure force p1A1 (nozzle pressures are zero) Answer: D. Plane is horizontal so weight is ignored, no friction, and p2 = p3 = 0, leaving Fx and p1A1. (Key point: Momentum equation terms) 4. How is the force exerted on the bend found from the force on the fluid? A. It is equal and in the same direction B. It is half of that force C. It is equal and opposite to the force of the bend on the fluid D. It is found by adding the weight Answer: C. By Newton's third law the water pushes on the bend with equal and opposite force; result about 3.71 kN. 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