Water at 20^◦ C flows in a 0.45-m-diameter pipe at a flowrate of 0.34 m^3 /sec. The pipe is fitted with a gradual pipe contraction (nozzle) with a diameter of 0.15 m in order to accelerate the flow, as illustrated in Figure EP 5.14. Assume that the pipe and the nozzle lie in a horizontal plane,

Question

Water at [latex] 20^{\circ} C[/latex] flows in a 0.45-m-diameter pipe at a flowrate of [latex]0.34 m^{3} /sec[/latex]. The pipe is fitted with a gradual pipe contraction (nozzle) with a diameter of 0.15 m in order to accelerate the flow, as illustrated in Figure EP 5.14. Assume that the pipe and the nozzle lie in a horizontal plane, (x, y). The pressure at point 1 is [latex]3 × 10^5 N/m^{2}[/latex] , and assume a minor head loss coefficient due to the nozzle, k of 0.07. (a) Determine the minor head loss due to the nozzle. (b) Determine the pressure at point 2. (c) Determine the reaction force acting on the nozzle due to the thrust force of the flow (anchoring force required to keep the nozzle from moving).

Accepted Answer

(a)–(b) Theminor headloss due tothe nozzleisevaluated byapplying theminor head loss equation, and the pressure at point 2 is determined by applying the energy equation between points 1 and 2, and as follows:

[latex] k_{ nozzle}: = 0.07 [/latex] [latex]Q: =0.34 \frac{m^{3}}{sec} [/latex] [latex]D_{1}: = 0.45 m[/latex] [latex]A_{1}: = \frac{\pi .D^{2}_{1}}{4} = 0.159 m^{2}[/latex]

[latex]D_{2}: = 0.15 m[/latex] [latex]A_{2}: = \frac{\pi .D^{2}_{2}}{4} = 0.018 m^{2}[/latex] [latex]V_{1}: = \frac{Q}{A_{1} }= 2.138 \frac{m}{s} [/latex]

[latex]V_{2}: = \frac{Q}{A_{2} }= 19.24 \frac{m}{s} [/latex] [latex]P_{1}: = 3 imes 10^{5} \frac{N}{m^{2}} [/latex] [latex]Z_{1}: = 0 m[/latex] [latex]Z_{2}: = 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]h_{fmin}: = 1 m[/latex] [latex]P_{2}: = 1 imes 10^{4} \frac{N}{m^{2}} [/latex]

Given

[latex]\frac{P_{1}}{\gamma } + Z_{1} + \frac{V^{2}_{1}}{2. g} - h_{fmin} = \frac{P_{2}}{\gamma } + Z_{2} + \frac{V^{2}_{2}}{2. g}[/latex]
[latex]h_{fmin} = k_{nozzle} \frac{V^{2}_{2}}{2. g}[/latex]
[latex]\left ( \begin{matrix} h_{fmin} \ P_{2} \end{matrix} ight ) : = Find (h_{fmin}, P_{2} )[/latex]

[latex]h_{fmin} = 1.321 m[/latex] [latex]P_{2}= 1.046 imes 10^{5} \frac{N}{m^{2}} [/latex]

(c) 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 the x direction for the control volume illustrated in Figure EP 5.14. The flow enters at control face 1 and leaves at control face 2. The viscous forces have already been accounted for indirectly through the minor head loss term, [latex]h_{f,minor}[/latex]in the energy equation, and there is no gravitational force component along the x-axis 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} + R_{x} = ho Q(v_{2x} - v_{1x})[/latex]

[latex]P_{1x}: = P_{1} =3 imes 10^{5} \frac{N}{m^{2}} [/latex] [latex]P_{2x}: = - P_{2} = - 1.046 imes 10^{5} \frac{N}{m^{2}} [/latex]

[latex]V_{1x}: = V_{1} = 2.138 \frac{m}{s} [/latex] [latex]V_{2x}: = V_{2} = 19.24 \frac{m}{s} [/latex]

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

Given

[latex]P_{1x}. A_{1} + P_{2x}. A_{2} - R_{x} = ho Q(v_{2x} - v_{1x})[/latex]
[latex]R_{x}: = Find (R_{x}) = 4.006 imes 10^{4} N [/latex]

Thus, the reaction force acting on the nozzle, [latex]R_{x}[/latex] due to the thrust force of the flow acts to the left, as assumed and illustrated in Figure EP 5.14.

Question 5.14: Water at 20^◦ C flows in a 0.45-m-diameter pipe at a flowrat...

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