Calculate the change in the momentum flux of the water flowing through the plane contraction shown in Fig. P4.154 if the flow rate is 0.2 m^3/s. The slope of the two profiles is the same. The upstream profile is created by a plate containing slots of various widths.

Question

Calculate the change in the momentum flux of the water flowing through the plane contraction shown in Fig. P4.154 if the flow rate is 0.2 [latex]m^3[/latex]/s. The slope of the two profiles is the same. The upstream profile is created by a plate containing slots of various widths.

Accepted Answer

[latex]0.2 = \overline{V}_1A_ 1 = \overline{V}_ 1 ×.2×1.0. ∴\overline{V}_ 1 =1 m/s. ∴V_{1 max} = 2 m/s. ∴V_1(y) = 20(0.1− y)[/latex].

flux in =[latex] 2∫_0^{0.1} ρV^2dy=2\int_{0}^{0.1} 1000 imes 20^2 (0.1-y )^2 dy= 800 000 \frac{0.1^3}{3}= 267 N[/latex].

The slope at section 1 is −20 . ∴ [latex]V_2 (y) = −20y + A[/latex].

Continuity: [latex] A_1\overline{V} _1 = A_2\overline{V}_2. ∴\overline{V}_2 = 2\overline{V}_1 = 2 m/s[/latex].

[latex]\left. \begin{matrix} V_2(0)=A \ V_2(0.05)=A-1 \end{matrix} ight\} \begin{matrix} ∴ \overline{V} _2= A−1/2 \end{matrix}[/latex] .

[latex]2= A−\frac{1}{2}. ∴ A=2.5. ∴ V_2(y)=2.5-20y[/latex].

flux out [latex] =2∫_0^{0.05} 1000(2.5 -20 y)^2dy= 800 000\left[\frac{(y- 0.125)^3}{3} ight]_0^{0.05} =\frac{800 000}{3} [0.00153][/latex]

= [latex]408.3 N. ∴ change = 408 − 267 = \underline{141 N}[/latex] .

Question 4.154: Calculate the change in the momentum flux of the water flowi...

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