Water at 20^◦ C flows in a 0.85-m-diameter horizontal pipe at a flowrate of 2.9 m^3 /sec. A regular 45^◦ threaded elbow is installed in between point 1 and 2 in order to change the direction of the flow by 45^◦ , as illustrated in Figure EP 5.15. Assume that the pipe and the elbow lie in a

Question

Water at [latex] 20^{\circ} C[/latex] flows in a 0.85-m-diameter horizontal pipe at a flowrate of [latex] 2.9 m^{3} /sec [/latex]. A regular [latex] 45^{\circ} [/latex] threaded elbow is installed in between point 1 and 2 in order to change the direction of the flow by [latex] 45^{\circ} [/latex] , as illustrated in Figure EP 5.15. Assume that the pipe and the elbow lie in a horizontal plane, (x, y). The pressure at point 1 is [latex] 3.9 × 10^{5} N/m^{2} [/latex] , and assume a minor head loss coefficient due to the elbow, k of 0.4. (a) Determine the minor head loss due to the elbow between points 1 and 2. (b) Determine the pressure at point 2. (c) Determine the reaction force acting on the elbow due to the thrust force of the flow (anchoring force required to keep the elbow from moving).

Accepted Answer

(a)–(b) The minor head loss due to the elbow is evaluated by applying the minor 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]D: = 0.85 m[/latex] [latex]A: = \frac{\pi .D^{2}}{4} = 0.567 m^{2}[/latex] [latex] k_{ elbow}: = 0.4 [/latex] [latex]Q: =2.9 \frac{m^{3}}{sec} [/latex]

[latex]V: = \frac{Q}{A }= 5.111 \frac{m}{s} [/latex] [latex]V_{1}: = V = 5.111 \frac{m}{s} [/latex] [latex]V_{2}: = V = 5.111 \frac{m}{s} [/latex]

[latex]P_{1}: = 3.9 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_{ elbow} \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} = 0.532 m[/latex] [latex]P_{2}= 3.848 imes 10^{5} \frac{N}{m^{2}} [/latex]

(c) In order to determine the reaction force acting on the elbow due to the thrust force of the flow, the integral form of the momentum equation is applied in both the x and y directions for the control volume illustrated in Figure EP 5.15. 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- 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} + R_{x} = ho Q(v_{2x} - v_{1x})[/latex]
[latex]\sum{F_{y}} = (P_{1}A_{1} )_{y} + (P_{2}A_{2} )_{y} + R_{y} = ho Q(v_{2y} - v_{1y})[/latex]

[latex]A_{1}: = A = 0.567 m^{2} [/latex] [latex]A_{2}: = A = 0.567 m^{2} [/latex] [latex]\alpha : = 45 deg [/latex]

[latex]P_{1x}: = P_{1} =3.9 imes 10^{5} \frac{N}{m^{2}} [/latex] [latex]P_{2x}: = - P_{2}. \cos (\alpha ) = - 2.721 imes 10^{5} \frac{N}{m^{2}} [/latex]

[latex]V_{1x}: = V_{1} = 5.111 \frac{m}{s} [/latex] [latex]V_{2x}: = V_{2}. \cos (\alpha ) = 3.614 \frac{m}{s} [/latex]

[latex]P_{1y}: =0\frac{N}{m^{2}} [/latex] [latex]P_{2x}: = P_{2}. \sin (\alpha ) = 2.721 imes 10^{5} \frac{N}{m^{2}} [/latex]

[latex]V_{1y}: =0 \frac{m}{s} [/latex] [latex]V_{2y}: = - V_{2}. \sin (\alpha ) = - 3.614 \frac{m}{s} [/latex]

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

Given

[latex]P_{1x}. A_{1} + P_{2x}. A_{2} - R_{x} = ho Q(v_{2x} - v_{1x})[/latex]
[latex]P_{1y}. A_{1} + P_{2y}. A_{2} - R_{y} = ho Q(v_{2y} - v_{1y})[/latex]
[latex] \left ( \begin{matrix} R_{x} \ R_{y} \end{matrix} ight ) : = Find (R_{x}, R_{y}) = \left ( \begin{matrix} 7.124 imes 10^{4} \ -1.649 imes 10^{5} \end{matrix} ight ) N[/latex]

[latex]R : = \sqrt{R_{x}^{2}R_{y}^{2}} = 1.796 imes 10^{4} N [/latex] [latex] heta : = atan \left(\frac{\left|R_{y} ight| }{\left|R_{x} ight| } ight)= 66.628 deg [/latex]

Thus, the reaction force acting on the elbow, [latex]R_{x} [/latex] acts to the left as assumed, and there action force acting on the elbow, [latex]R_{y} [/latex] acts downward (not upward as assumed), as illustrated in Figure EP 5.15.

Question 5.15: Water at 20^◦ C flows in a 0.85-m-diameter horizontal pipe a...

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