Water at 20^◦ C flows in a 0.65-m-diameter pipe at a flowrate of 0.98 m^3 /sec. The pipe is fitted with sudden pipe contraction with a diameter of 0.35 m in order to accelerate the flow, as illustrated in Figure EP 5.13. Assume that the pipe and the contraction lie in a horizontal plane, (x,y).The

Question

Water at [latex] 20^{\circ} C[/latex] flows in a 0.65-m-diameter pipe at a flowrate of [latex] 0.98 m^{3} /sec [/latex]. The pipe is fitted with sudden pipe contraction with a diameter of 0.35 m in order to accelerate the flow, as illustrated in Figure EP 5.13. Assume that the pipe and the contraction lie in a horizontal plane, (x,y).The pressure at point 2 is [latex]3.56 × 10^{4} N/m^{2}[/latex] ,and assume aminor head loss coefficient due to the sudden pipe contraction, k of 0.25. (a) Determine the minor head loss due to the contraction. (b) Determine the pressure at point1. (c) Determine the reaction force acting on the pipe contraction due to the thrust force of the flow (anchoring force required to keep the pipe contraction from moving).

Accepted Answer

(a)–(b) The minor head loss due to the pipe contraction is evaluated by applying the minor head loss equation, and the pressure at point 1 is determined by applying the energy equation between points 1 and 2, and as follows:

[latex] k_{ contrc}: = 0.25 [/latex] [latex]Q: = 0.98 \frac{m^{3}}{sec} [/latex] [latex]D_{1}: = 0.65 m[/latex] [latex]A_{1}: = \frac{\pi .D^{2}_{1}}{4} = 0.332 m^{2}[/latex]

[latex]D_{2}: = 0.35 m[/latex] [latex]A_{2}: = \frac{\pi .D^{2}_{2}}{4} = 0.096 m^{2}[/latex] [latex]V_{1}: = \frac{Q}{A_{1} }= 2.953 \frac{m}{s} [/latex]

[latex]V_{2}: = \frac{Q}{A_{2} }= 10.186 \frac{m}{s} [/latex] [latex]P_{2}: = 3.56 imes 10^{4} \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_{1}: = 6 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_{contrc} \frac{V^{2}_{2}}{2. g}[/latex]
[latex]\left ( \begin{matrix} h_{fmin} \ P_{1} \end{matrix} ight ) : = Find (h_{fmin}, P_{1} )[/latex]

[latex]h_{fmin} = 1.322 m[/latex] [latex]P_{1}= 9.596 imes 10^{4} \frac{N}{m^{2}} [/latex]

(c) In order to determine the reaction force acting on the pipe contraction 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 5.13. 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} = 9.596 imes 10^{4} \frac{N}{m^{2}} [/latex] [latex]P_{2x}: = - P_{2} = - 3.56 imes 10^{4} \frac{N}{m^{2}} [/latex]

[latex]v_{1x}: = V_{1} = 2.953 \frac{m}{s} [/latex] [latex]v_{2x}: = V_{2} = 10.186 \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}) = 2.134 imes 10^{4} N [/latex]

Thus, the reaction force acting on the pipe contraction,[latex] R_{x} [/latex] due to the thrust force of the flow acts to the left as assumed and illustrated in Figure EP 5.13.

Question 5.13: Water at 20^◦ C flows in a 0.65-m-diameter pipe at a flowrat...

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