A forced jet of water at 20^◦ C with a diameter of 0.02 m and a flowrate of 0.076 m^3 /sec hits a stationary sloping flat plate, where the jet splits into two paths, as illustrated in Figure EP 5.18. The plate angle α is 60^◦ , assume that the jet and the plate lie in a horizontal plane, (x, y),

Question

A forced jet of water at [latex] 20^{\circ} C[/latex] with a diameter of 0.02 m and a flowrate of [latex] 0.076 m^{3}/sec[/latex] hits a stationary sloping flat plate, where the jet splits into two paths, as illustrated in Figure EP 5.18. The plate angle α is [latex] 60^{\circ} [/latex] , assume that the jet and the plate lie in a horizontal plane, (x, y), and assume ideal flow. (a) Determine the velocity of the flow at points 2 and 3. (b) Determine the flowrate at control faces 2 and 3. (c) Determine the reaction force acting on the plate due to the thrust force of the flow (anchoring force required to keep plate from moving).

Accepted Answer

(a) In order to determine the velocity of flow at points 2 and 3, the Bernoulli equation is applied between points 1 and 2, and between points 1 and 3 as follows:

[latex]Q: = 0.076 \frac{m^{3}}{sec} [/latex] [latex]D_{1}: = 0.02 m [/latex] [latex]A_{1}: = \frac{\pi .D^{2}_{1}}{4} = 3.142 imes 10^{-4} m^{2}[/latex]

[latex]V_{1}: = \frac{Q_{1}}{A_{1}} = 241.916 \frac{m}{s} [/latex] [latex] P_{1}: = 0 [/latex] [latex] P_{2}: = 0 [/latex]

[latex] P_{3}: = 0 [/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]V_{2}: = 1 \frac{m}{sec} [/latex] [latex]V_{3}: = 1 \frac{m}{sec} [/latex]

Given

[latex] \frac{P_{1}}{\gamma } + Z_{1} + \frac{V^{2}_{1}}{2. g} =\frac{P_{2}}{\gamma } + Z_{2} + \frac{V^{2}_{2}}{2. g}[/latex]
[latex] \frac{P_{1}}{\gamma } + Z_{1} + \frac{V^{2}_{1}}{2. g} =\frac{P_{3}}{\gamma } + Z_{3} + \frac{V^{2}_{3}}{2. g}[/latex]
[latex] \left ( \begin{matrix} V_{2} \ V_{3} \end{matrix} ight ) : = Find (V_{2}, V_{3}) = \left ( \begin{matrix} 241.916 \ 241.916 \end{matrix} ight ) \frac{m}{s} [/latex]

Thus,[latex] v_{1} = v_{2} = v_{3} [/latex].However, because the velocity of the jet hitting the plate is high, a minor head loss is indirectly modeled by a reduced velocity at control faces 2 and 3 for the flow out of the control volume. Assume a 10% reduction in the velocity at control faces 2 and 3 as follows:

[latex]V_{2}: =0.90.V_{2} = 217.724 \frac{m}{s} [/latex] [latex]V_{3}: =0.90.V_{3} = 217.724 \frac{m}{s} [/latex]

(b) In order to determine the flowrate at control faces 2 and 3, in addition to the continuity equation, [latex] Q_{1} =Q_{2} + Q_{3} [/latex], a second equation is required to solve for the two unknowns ([latex]Q_{2} [/latex]and [latex]Q_{3} [/latex]). That second equation is the integral form of the momentum equation applied along the s-axis of the sloping plate; because the friction (flow resistance) between the fluid jet and the sloping plate is assumed to be negligible (ideal flow), there is no reaction force acting on the plate along the s-axis of the sloping plate. The flow enters at control face 1 and leaves at control face 2 and control face 3. Furthermore, the viscous forces are ignored because of the assumption of ideal flow, there is no pressure force component because the jet flow is open to the at mosphere([latex] p_{1} = p_{2} = p_{3} = 0 N/m^{2} [/latex] ),and there is no gravitational force component along the s-axis because the pipe is horizontal ([latex] z_{1} = z_{2} = z_{3} = 0 [/latex]).

[latex] \sum{F_{x}} = R_{s} = 0 = ( ho Q_{2}v_{2s} + ho Q_{3}v_{3s}) - ( ho Q_{1}v_{1s})[/latex]

[latex] \alpha : = 60 deg[/latex] [latex] V_{1s}: = V_{1}. \cos (\alpha ) = 120.958 \frac{m}{s} [/latex]

[latex] V_{2s}: = V_{2} = 217.724 \frac{m}{s} [/latex] [latex] V_{3s}: = - V_{3} = - 217.724 \frac{m}{s} [/latex]

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

Given

[latex] Q_{1} = Q_{2} + Q_{3} [/latex]
[latex]0 = ( ho . Q_{2}.V_{2s} + ho . Q_{3}.V_{3s}) - ( ho . Q_{1}. V_{1s}) [/latex]

[latex]A_{2}: = \frac{ Q_{2}}{V_{2}} [/latex] [latex]A_{3}: = \frac{ Q_{3}}{V_{3}} [/latex]

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

[latex]Q_{2}: = 0.059 \frac{m^{3}}{sec} [/latex] [latex]Q_{3}: =0.017 \frac{m^{3}}{sec} [/latex] [latex]Q_{2} + Q_{3}: =0.076 \frac{m^{3}}{sec} [/latex]

[latex]A_{2}: = 2.715 imes 10^{-4} m^{2} [/latex] [latex]A_{3}: =7.757 imes 10^{-5} m^{2} [/latex]

(c) In order to determine the reaction force acting onthe plate 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 5.18. 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, there is no pressure force component because the jet flow is open to the atmosphere ([latex] p_{1} = p_{2} = p_{3} = 0 N/m^{2} [/latex] ), and there is no gravitational force component along the x- and y-axes because the pipe is horizontal ([latex] z_{1} = z_{2} = z_{3} = 0 [/latex]).

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

[latex] V_{1x}: = V_{1} = 241.916 \frac{m}{s} [/latex] [latex] V_{1y}: = 0 \frac{m}{s} [/latex]

[latex] V_{2x}: = V_{2} . \cos (\alpha ) = 108.862\frac{m}{s} [/latex] [latex] V_{2y}: = V_{2} . \sin (\alpha )= 188.554 \frac{m}{s} [/latex]

[latex] V_{3x}: =- V_{3} . \cos (\alpha ) = - 108.862\frac{m}{s} [/latex] [latex] V_{3y}: = - V_{3} . \sin (\alpha )= - 188.554 \frac{m}{s} [/latex]

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

Given

[latex] - R_{x}: =( ho .Q_{2}. V_{2x} + ho .Q_{3}. V_{3x} ) - ( ho .Q_{1}. V_{1x}) [/latex]
[latex] 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} 1.376 imes 10^{4} \ 7.945 imes 10^{3} \end{matrix} ight ) [/latex]

[latex] R: = \sqrt{R_{x}^{2} + R_{y}^{2}} = 1.589 imes 10^{4} N [/latex] [latex] heta : = atan \left(\frac{\left|R_{y} ight| }{\left|R_{x} ight|} ight) = 30 deg [/latex]

Thus, the reaction force acting on the plate, [latex]R_{x} [/latex]acts to the left, and the reaction force, [latex]R_{y}[/latex] acts upward, both as assumed, and they are illustrated in Figure EP 5.18. Note that the resultant force, R acts at an angle, θ = [latex]30^{\circ}[/latex] with the horizontal, which is normal (at an angle of [latex]90^{\circ} [/latex] ) to the plate. Such would be expected because there is no reaction force acting parallel to the plate (along the s-axis of the sloping plate), as illustrated in (b) above.

Question 5.18: A forced jet of water at 20^◦ C with a diameter of 0.02 m an...

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