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Hydrodynamic force on bend-nozzle assembly

Short answer

Find velocities from Q=0.04 m³/s: V1=5.09 m/s, V2=20.37 m/s. Inlet pressure from Bernoulli with losses: p1=276 kPa. Momentum analysis gives hydrodynamic force components: opposite to jet 815 N, upward 2193 N; resultant 2.34 kN at 69.6° above horizontal.

Water flows into atmosphere through a vertical bend nozzle assembly as shown in figure below. The pipe diameter is 10 cm and nozzle exit diameter is 5 cm. The rate of flow of water is 2400 lpm. The interior volume of the assembly is 18.2 litres. The head loss in the bend is 0.5v22g0.5\frac{v^2}{2g} and in the nozzle it is 2v22g2\frac{v^2}{2g}, where V is the velocity of water in the pipe. Compute the hydrodynamic force on the system. [Figure: vertical pipe of 10 cm diameter turning through a bend to a horizontal nozzle with exit diameter 5 cm; 5 m vertical distance marked]

Given: pipe d1=0.10d_1 = 0.10 m, nozzle exit d2=0.05d_2 = 0.05 m, Q=2400Q = 2400 L/min =0.04= 0.04 m³/s, volume of assembly =18.2= 18.2 L =0.0182= 0.0182 m³, head loss in bend 0.5V2/2g0.5V^2/2g and in nozzle 2V2/2g2V^2/2g (VV = velocity in the pipe).

Assumed from the figure: water flows upward in the vertical pipe, passes a bend and leaves through a horizontal nozzle into the atmosphere, 5 m above the inlet section.

   exit ====>  (nozzle, d = 5 cm)
   |
   |  5 m
   |
   ^  inlet 1 (pipe d = 10 cm, flow up)

Velocities

A1=7.854×10−3 m2,A2=1.9635×10−3 m2V1=5.093 m/s,V2=20.372 m/s\begin{aligned} A_1 &= 7.854\times 10^{-3}\ \text{m}^2,\quad A_2 = 1.9635\times 10^{-3}\ \text{m}^2\\ V_1 &= 5.093\ \text{m/s},\quad V_2 = 20.372\ \text{m/s} \end{aligned}

Inlet pressure (Bernoulli with losses)

p1γ+V122g=0+V222g+5+(0.5+2)V122g\frac{p_1}{\gamma} + \frac{V_1^2}{2g} = 0 + \frac{V_2^2}{2g} + 5 + (0.5 + 2)\frac{V_1^2}{2g} p1γ=21.15+5+2.5(1.322)−1.322=28.13 mp1=276 009 Pa≈276 kPa (gauge)\begin{aligned} \frac{p_1}{\gamma} &= 21.15 + 5 + 2.5(1.322) - 1.322 = 28.13\ \text{m}\\ p_1 &= 276\,009\ \text{Pa} \approx 276\ \text{kPa (gauge)} \end{aligned}

(using V22/2g=21.15V_2^2/2g = 21.15 m, V12/2g=1.322V_1^2/2g = 1.322 m).

Momentum equation

  • p1A1=276 009×7.854×10−3=2168 Np_1A_1 = 276\,009\times 7.854\times 10^{-3} = 2168\ \text{N} (up, on the inlet face)
  • W=9810×0.0182=178.5 NW = 9810\times 0.0182 = 178.5\ \text{N} (down)
  • ρQ=40\rho Q = 40 kg/s. Inlet velocity =(0,+V1)= (0, +V_1); exit velocity =(+V2,0)= (+V_2, 0).

Forces on the fluid (RR = force of the assembly on the fluid):

x:  Rx=ρQ V2=40×20.372=815 Ny:  p1A1−W+Ry=ρQ (0−V1)2168−178.5+Ry=−203.7Ry=−2193 N\begin{aligned} x:\ \ R_x &= \rho Q\,V_2 = 40\times 20.372 = 815\ \text{N}\\ y:\ \ p_1A_1 - W + R_y &= \rho Q\,(0 - V_1)\\ 2168 - 178.5 + R_y &= -203.7\\ R_y &= -2193\ \text{N} \end{aligned}

The hydrodynamic force exerted by the water on the assembly is the opposite of RR:

Fx=−815 N,Fy=+2193 NF_x = -815\ \text{N},\qquad F_y = +2193\ \text{N} F=8152+21932=2339 N,θ=110.4∘ from the +x direction (the direction of the exit jet)F = \sqrt{815^2 + 2193^2} = 2339\ \text{N},\qquad \theta = 110.4^\circ\ \text{from the +x direction (the direction of the exit jet)}

Answer: the hydrodynamic force on the system ≈2.34\approx 2.34 kN, directed upward and backwards (against the jet), at 69.6∘69.6^\circ above the horizontal, with components 0.815 kN opposite to the jet and 2.19 kN upward.

✓ Answer checked by Claude Opus 5.5 (high effort)

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