For a steady incompressible flow of water through the reducing elbow in Fig. 10.9, the entrance area A1 is 30 cm^2 and the exit area A2 is 1/2A1. The mean velocity v1 entering the elbow is 5 m/s with an inlet pressure of 5 Pa and outlet pressure equal to atmospheric. Find the total force required

Question

For a steady incompressible flow of water through the reducing elbow in Fig. 10.9, the entrance area [latex]A_{1}[/latex] is 30 [latex]cm^{2}[/latex] and the exit area [latex]A_{2}[/latex] is [latex]\frac{1}{2}A_{1}.[/latex] The mean velocity [latex]\overline{ u }_{1}[/latex] entering the elbow is 5 m/s with an inlet pressure of 5 Pa and outlet pressure equal to atmospheric. Find the total force required to hold the bend in place. Control Volume and Semi-empirical Methods FIGURE 10.9 Academic experiment to determine the support reactions due to a fluid flowing through an elbow. Shown, too, is a possible control volume that ignores possible effects of the fluid-induced shear stresses.

Accepted Answer

We are given the following:

[latex] A_{1}=30 cm^{2}=0.003 m^{2},[/latex]

[latex] A_{2}=\frac{1}{2}A_{1}=15 cm^{2}=0.0015 m^{2}[/latex]

[latex] \overline{ u }_{1}=5 m/s,[/latex]

[latex] P_{1}=5Pa=5N/m^{2},[/latex]

[latex] P_{2}=o(gauge),[/latex]

[latex] ho _{H_{2}O}=1000 kg/m^{3}[/latex]

We first need to solve for the velocity [latex] u_{2}[/latex] leaving the elbow. From the mass balance equation for a steady, incompressible flow,

[latex] 0=\int_{A_{1}}^{}{ u _{1}\cdot n_{1}dA}+\int_{A_{2}}^{}{ u _{2}\cdot n_{2}dA},[/latex]

where we have for [latex]CS_{1}, u _{1}= u _{1}\left(\hat{i} ight)[/latex] and [latex]n_{1}=-\hat{i},[/latex] and for [latex]CS_{2}, u _{2}= u _{2}\left(\hat{j} ight)[/latex] and [latex]n_{2}=\hat{j}. [/latex] Hence, mass balance requires

[latex] 0=\int_{A_{1}}^{}{ u _{1}}\left(\hat{i}\cdot -\hat{i} ight)dA+\int_{A_{2}}^{}{ u _{2}}\left(\hat{j}\cdot \hat{j} ight)dA.[/latex]

Using mean velocities, we have

[latex] 0=-\overline{ u }_{1}\int_{A_{1}}^{}{dA}+\overline{ u }_{2}\int_{A_{2}}^{}{dA} ightarrow \overline{ u }_{1}A_{1}=\overline{ u }_{2}A_{2}[/latex]

or, with [latex]A_{2}=A_{1}/2,[/latex]

[latex] \overline{ u }_{1}A_{1}=\overline{ u }_{2}\left(\frac{1}{2}A_{1} ight) ightarrow \overline{ u }_{2}=2\overline{ u }_{1}.[/latex]

From Newton’s third law we know that for every action, there is an equal and opposite reaction. To determine the total force required to hold the elbow in place, we need to use the balance of linear momentum equation for a steady, incompressible flow. Summing the forces for this CV, we obtain

[latex] p_{1}A_{1}\left(\hat{i} ight)+p_{2}A_{2}\left(-\hat{j} ight)+R= ho \int_{A_{1}}^{}{ u _{1}\left(\hat{i} ight) }\left[ u _{1}\left(\hat{i}\cdot -\hat{i} ight) ight]dA[/latex]

[latex] + ho \int_{A_{2}}^{}{ u _{2}\left(\hat{j} ight) }\left[ u _{2}\left(\hat{j}\cdot \hat{j} ight) ight]dA,[/latex]

where R is the reaction force, or with [latex]p_{2}=p_{atm}=0[/latex] (gauge), integration yields

[latex] p_{1}A_{1}\hat{i}+R=- ho \overline{ u }^{2}_{1}A_{1}\hat{i}+ ho \overline{ u }^{2}_{2}A_{2}\hat{j}.[/latex]

Collecting terms, we have

[latex] R=-(p_{1}A_{1}+ ho \overline{ u }^{2}_{1}A_{1} )\hat{i}+( ho \overline{ u }^{2}_{2}A_{2})\hat{j}.[/latex]

By substituting the numerical values into this equation, we obtain

[latex] R=-\left[\left(\frac{5N}{m^{2}} ight)(0.003 m^{2})+\left(\frac{1000 kg}{m^{3}} ight)(5m/s)^{2}(0.003 m^{2}) ight]\hat{i}[/latex]

[latex] +\left[\left(\frac{1000 kg}{m^{3}} ight)(10 m/s)^{2}(0.0015 m^{2}) ight]\hat{j}=+(75 N)\hat{i}+(150 N)\hat{j}.[/latex]

This force, which acts on the fluid is equal and opposite that exerted by the flow on the support. We could now solve the solid mechanics problem and Control Volume and Semi-empirical Methods determine the reactions at the base of the support and the stresses due to axial and bending loads.

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