Water at 20^◦ C flows in a 0.8-m-diameter pipe at a flowrate of 1.6 m^3 /sec. A double nozzle is installed downstream of point 1 in order to divide the flow in two, as illustrated in Figure EP 5.5. Assume that the pipe and the nozzle lie in a horizontal plane, (x, y). The diameter of the nozzle at

Question

Water at [latex] 20^{\circ} C [/latex] flows in a 0.8-m-diameter pipe at a flowrate of [latex]1.6 m^{3} /sec[/latex]. A double nozzle is installed downstream of point 1 in order to divide the flow in two, as illustrated in Figure EP 5.5. Assume that the pipe and the nozzle lie in a horizontal plane, (x, y). The diameter of the nozzle at point 2 is 0.45 m and is at an angle of [latex] 20^{\circ} C [/latex] with the x-axis, and the diameter of the nozzle at point 3 is 0.35 m and is at an angle of [latex] 25^{\circ} C [/latex] with the x-axis. The flows at points 2 and 3 are open to the atmosphere. Assume that the flow is ideal and thus ignore any minor head loss due to the nozzle. (a) Determine the pressure at point 1. (b) Determine the reaction force acting on the nozzle due to the thrust force of the flow (anchoring force required to keep nozzle from moving).

Accepted Answer

(a) In order to determine the pressure at point 1, the energy equation is applied between points 1 and 2 and between points 1 and 3. Furthermore, in order to determine the velocity of flow in the pipe and the double nozzle, the continuity equation is applied as follows:

[latex]D_{1}: = 0.8 m [/latex] [latex]A_{1}: = \frac{\pi .D^{2}_{1}}{4} = 0.503 m^{2} [/latex] [latex]Q_{1}: = 1.6 \frac{m^{3}}{sec} [/latex]

[latex]V_{1}: = \frac{Q_{1}}{A_{1}} : = 3.183 \frac{m}{s} [/latex] [latex]D_{2}: = 0.45 m [/latex] [latex]A_{2}: = \frac{\pi .D^{2}_{2}}{4} = 0.159 m^{2} [/latex]

[latex]D_{3}: = 0.35 m [/latex] [latex]A_{3}: = \frac{\pi .D^{2}_{3}}{4} = 0.096 m^{2} [/latex] [latex]P_{2}: = 0 \frac{N}{m^{2}} [/latex]

[latex]P_{3}: = 0 \frac{N}{m^{2}} [/latex] [latex]Z_{1}: = 0 m [/latex] [latex]Z_{2}: = 0 m [/latex] [latex]Z_{3}: = 0 m [/latex]

[latex] ho : = 998 \frac{kg}{m^{3}} [/latex] [latex]g: = 9.81 \frac{m}{sec^{2}} [/latex] [latex]\gamma : = ho .g = 9.79 imes 10^{3} \frac{kg}{m^{2}.s^{2}} [/latex]

Guess value: [latex]P_{1}: = 1 imes 10^{4} \frac{N}{m^{2}} [/latex] [latex]V_{2}: = 1 \frac{m}{sec} [/latex] [latex]V_{3}: = 1 \frac{m}{sec} [/latex]

[latex]Q_{2}: = 1 \frac{m^{3}}{sec} [/latex] [latex]Q_{3}: = 1 \frac{m^{3}}{sec} [/latex]

Given

[latex]\frac{P_{1}}{\gamma } + Z_{1} + \frac{V^{2}_{1} }{2g} = \frac{P_{2}}{\gamma } + Z_{2} + \frac{V^{2}_{2} }{2g} [/latex] [latex]\frac{P_{1}}{\gamma } + Z_{1} + \frac{V^{2}_{1} }{2g} = \frac{P_{3}}{\gamma } + Z_{3} + \frac{V^{2}_{3} }{2g} [/latex]

[latex]Q_{1} = Q_{2}+ Q_{3} [/latex] [latex]Q_{2} = V_{2}+ A_{2} [/latex] [latex]Q_{3} = V_{3}+ A_{3} [/latex]

[latex] \left ( \begin{matrix} V_{2} \ V_{3} \ Q_{2} \ Q_{3} \ P_{1} \end{matrix} ight ) : = Find (V_{2},V_{3}, Q_{2},Q_{3} , P_{1} )[/latex]

[latex]V_{2} = 6.268 \frac{m}{s} [/latex] [latex]V_{3} = 6.268 \frac{m}{s} [/latex] [latex]Q_{2} = 0.997 \frac{m^{3}}{s} [/latex] [latex]Q_{3} = 0.603 \frac{m^{3}}{s} [/latex]

[latex] P_{1} = 1.455 imes 10^{4} \frac{N}{m^{2}} [/latex]

(b) In order to determine the reaction force acting on the nozzle due to the thrust force of the flow, the integral form of the momentum equation is applied in both the x and ydirections forthe control volume illustrated inFigure EP5.5.The flow enters at control face 1 and leaves at control face 2 and control face 3. The viscous forces are ignored because of the assumption of ideal flow, and there is no gravitational force component along the x- and y-axes because the pipe is horizontal [latex] (z_{1} = z_{2} = 0)[/latex].

[latex] \sum{F_{x}} = (P_{1}A_{1})_{x} + (P_{2}A_{2})_{x} + (P_{3}A_{3})_{x} + R_{x} = ( ho Q_{2} v_{2x} +( ho Q_{3} v_{3x}) - ( ho Q_{1} v_{1x}) [/latex]
[latex] \sum{F_{y}} = (P_{1}A_{1})_{y} + (P_{2}A_{2})_{y} +(P_{3}A_{3})_{y} + R_{y} = ( ho Q_{2} v_{2y} +( ho Q_{3} v_{3y}) - ( ho Q_{1} v_{1y}) [/latex]

[latex] P_{1x}: = P_{1} = 1.455 imes 10^{4} \frac{N}{m^{2}} [/latex] [latex] P_{2x}: = - P_{2}. \cos (20 deg) = 0 \frac{N}{m^{2}} [/latex]
[latex] P_{3x}: = - P_{3}. \cos (25 deg) = 0 \frac{N}{m^{2}} [/latex] [latex] V_{1x}: = V_{1} = 3.183 \frac{m}{s} [/latex]
[latex] V_{2x}: = V_{2}. \cos (20 deg) = 5.89 \frac{m}{s} [/latex] [latex] V_{3x}: = V_{3}. \cos (25 deg) = 5.681 \frac{m}{s} [/latex]

[latex] P_{1y}: = 0 \frac{N}{m^{2}} [/latex] [latex] P_{2y}: = -P_{2} . \sin (20 deg) = 0 \frac{N}{m^{2}} [/latex]
[latex] P_{3y}: = P_{3} . \cos (25 deg) = 0 \frac{N}{m^{2}} [/latex] [latex] V_{1y}: = 0 \frac{m}{s} [/latex]

[latex] V_{2y}: = V_{2}. \sin (20 deg) = 2.144 \frac{m}{s} [/latex] [latex] V_{3y}: = - V_{3}. \sin (25 deg) = -2.649 \frac{m}{s} [/latex]

Guess value: [latex] R_{x}: = 1000 N [/latex] [latex] R_{y}: = 1000 N [/latex]

[latex]P_{1x}. A_{1} + P_{2x}. A_{2} + P_{3x}. A_{3} - R_{x}= ( ho .Q_{2}. V_{2x} + ho .Q_{3} .V_{3x} ) - ( ho . Q_{1}. V_{1x}) [/latex]
[latex]P_{1y}. A_{1} + P_{2y}. A_{2} + P_{3y}. A_{3} + R_{y}= ( ho .Q_{2}. V_{2y} + ho .Q_{3} .V_{3y} ) - ( ho . Q_{1}. V_{1y}) [/latex]

[latex] \left ( \begin{matrix} R_{x} \ R_{y} \end{matrix} ight ) : = Find (R_{x}, R_{y} ) = \left ( \begin{matrix} 3.117 imes 10^{3} \538.595 \end{matrix} ight ) N [/latex]

[latex] R: = \sqrt{R_{x}^{2} + R_{y}^{2} } = 3.163 imes 10^{3} N[/latex]
[latex] heta : = atan \left(\frac{\left| R_{y} ight| }{\left| R_{x} ight|} ight) = 9.804 deg [/latex]

Thus, the reaction force acting on the nozzle, [latex]R_{x} [/latex]acts to the left as assumed, and the reaction force, [latex]R_{y} [/latex]acts upwards as assumed, as illustrated in Figure EP 5.5.

Question 5.5: Water at 20^◦ C flows in a 0.8-m-diameter pipe at a flowrate...

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