--- title: "Force on bend-cum-reducer" subject: "Fluid Mechanics" chapter: 7 key: "semester_3/fluid_mechanics:bend-cum-reducer-force" source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/bend-cum-reducer-force site: BCT Engineering --- # Force on bend-cum-reducer Fluid Mechanics · Chapter 7: Momentum principle and flow analysis - Asked in: 2072 Chaitra, 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/bend-cum-reducer-force - Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-7 ## Question A 120° bend-cum reducer has 300 mm diameter at inlet and 200 mm diameter at the outlet end. When the bend-cum reducer carries 0.30 m³/s of water, pressure at section 1 (inlet) is 210 kN/m². Assume no energy losses in the bend and determine the components of force exerted by the bend on the flow. Assume the weight of the bend plus water in it to be 1500 N. Assume section 2 (outlet) to be 0.40 m above sections 1 (inlet). ## Short answer Apply Bernoulli: p2=169.5 kPa. Momentum analysis including 1500 N weight gives force of bend on fluid: Fx=-20.2 kN, Fy=+8.59 kN. Resultant 22.0 kN at 157° to inlet direction. Force on bend is equal and opposite. ## Answer Apply Bernoulli between the two sections to find $p_2$, then apply the momentum equation in the $x$ and $y$ directions on the water in the bend. #### Assumptions The bend lies in a vertical plane. Inlet flow is along $+x$ (horizontal). "120° bend" means the flow is turned through $120^\circ$, so the outlet velocity makes $120^\circ$ with the inlet direction and points upward ($V_{2x} = V_2\cos120^\circ$, $V_{2y} = V_2\sin120^\circ$). $W = 1500$ N acts downward. $\rho = 1000\ \text{kg/m}^3$. ```text ^ y 2 (outlet, 0.4 m higher) | / | / 120 deg from inlet direction 1 -----> +-------+--------> x ``` #### Velocities and pressures $$ \begin{aligned} A_1 &= 0.07069\ \text{m}^2,\; A_2 = 0.03142\ \text{m}^2 \\ V_1 &= 0.30/0.07069 = 4.244\ \text{m/s},\quad V_2 = 0.30/0.03142 = 9.549\ \text{m/s} \\ p_2 &= p_1 + \rho g(z_1 - z_2) + \tfrac12\rho(V_1^2 - V_2^2) \\ &= 210\,000 - 3924 + \tfrac12(1000)(18.01 - 91.19) = 169\,488\ \text{Pa} \end{aligned} $$ Forces: $p_1A_1 = 14\,844$ N, $p_2A_2 = 5325$ N. Momentum flux: $\rho QV_1 = 1273$ N, $\rho QV_2 = 2865$ N. #### Momentum equations (forces by bend on fluid) $$ \begin{aligned} x:\;& F_x + p_1A_1 - p_2A_2\cos120^\circ = \rho Q\,(V_2\cos120^\circ - V_1) \\ & F_x = 1000(0.3)(-4.775 - 4.244) - 14\,844 - 2662 = -20\,212\ \text{N} \\[4pt] y:\;& F_y - W - p_2A_2\sin120^\circ = \rho Q\,V_2\sin120^\circ \\ & F_y = 2481 + 1500 + 4611 = 8592\ \text{N} \end{aligned} $$ #### Resultant $$ F = \sqrt{20\,212^2 + 8592^2} = 21\,962\ \text{N},\qquad \theta = 180^\circ - \tan^{-1}\frac{8592}{20\,212} = 157.0^\circ \text{ from the inlet direction} $$ **Answer: $F_x = -20.2$ kN (opposite to inlet flow), $F_y = +8.59$ kN (upward); resultant $\approx 22.0$ kN at $157^\circ$ to the $+x$ axis.** The force on the bend is equal and opposite. ## 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. Continuity equation alone B. Bernoulli between sections 1 and 2 including the 0.4 m height rise C. p2 = p1 since there are no losses D. Momentum equation in x only Answer: B. No losses, so p2 = p1 + rho g (z1 - z2) + 1/2 rho (V1^2 - V2^2) = about 169.5 kPa. (Key point: Outlet pressure) 2. Why is V2 larger than V1? A. The outlet area is smaller, so by continuity V = Q/A is higher B. The outlet is higher up C. The pressure is higher at the outlet D. The bend adds energy Answer: A. V1 = 4.24 m/s and V2 = 9.55 m/s from Q = AV with A2 < A1. (Key point: Velocities) 3. How does the 1500 N weight enter the y-momentum equation? A. It is ignored since the bend is in a vertical plane B. It adds to the upward pressure force C. It is multiplied by cos120 D. It acts downward, so Fy must also support it (Fy = ... + W) Answer: D. Bend lies in a vertical plane; weight acts downward and is balanced partly by Fy. (Key point: Weight in y equation) 4. What do Fx and Fy tell us about the force of the bend on the flow? A. Fx = +20.2 kN and Fy = -8.59 kN B. Fx = 0 and Fy = 8.59 kN C. Fx = -20.2 kN (opposite to inlet flow) and Fy = +8.59 kN (upward) D. Both are directed along the inlet flow Answer: C. The bend must push back against the flow in x and up in y; resultant is about 22 kN. 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