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Force on bend-cum-reducer

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.

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).

Apply Bernoulli between the two sections to find p2p_2, then apply the momentum equation in the xx and yy directions on the water in the bend.

Assumptions

The bend lies in a vertical plane. Inlet flow is along +x+x (horizontal). "120° bend" means the flow is turned through 120∘120^\circ, so the outlet velocity makes 120∘120^\circ with the inlet direction and points upward (V2x=V2cos⁡120∘V_{2x} = V_2\cos120^\circ, V2y=V2sin⁡120∘V_{2y} = V_2\sin120^\circ). W=1500W = 1500 N acts downward. ρ=1000 kg/m3\rho = 1000\ \text{kg/m}^3.

            ^ y        2 (outlet, 0.4 m higher)
            |         /
            |        /  120 deg from inlet direction
   1 -----> +-------+--------> x

Velocities and pressures

A1=0.07069 m2,  A2=0.03142 m2V1=0.30/0.07069=4.244 m/s,V2=0.30/0.03142=9.549 m/sp2=p1+ρg(z1−z2)+12ρ(V12−V22)=210 000−3924+12(1000)(18.01−91.19)=169 488 Pa\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: p1A1=14 844p_1A_1 = 14\,844 N, p2A2=5325p_2A_2 = 5325 N. Momentum flux: ρQV1=1273\rho QV_1 = 1273 N, ρQV2=2865\rho QV_2 = 2865 N.

Momentum equations (forces by bend on fluid)

x:  Fx+p1A1−p2A2cos⁡120∘=ρQ (V2cos⁡120∘−V1)Fx=1000(0.3)(−4.775−4.244)−14 844−2662=−20 212 Ny:  Fy−W−p2A2sin⁡120∘=ρQ V2sin⁡120∘Fy=2481+1500+4611=8592 N\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=20 2122+85922=21 962 N,θ=180∘−tan⁡−1859220 212=157.0∘ from the inlet directionF = \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: Fx=−20.2F_x = -20.2 kN (opposite to inlet flow), Fy=+8.59F_y = +8.59 kN (upward); resultant ≈22.0\approx 22.0 kN at 157∘157^\circ to the +x+x axis. The force on the bend is equal and opposite.

✓ Answer checked by Claude Opus 5.5 (high effort)

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