Question 5.10: Water flows steadily through the 90° reducing elbow, as show...
Water flows steadily through the 90° reducing elbow, as shown in the Fig. 5.12. At the inlet to the elbow, the pressure, velocity and cross-sectional area are p_{1}, V_{1} and A_{1}, respectively. At the outlet, the corresponding values are p_{2}, V_{2} and A_{2}, respectively. The weight of the elbow is W. The elbow discharges to the atmosphere. Determine the force required to hold the elbow in place. Neglect the non-uniformity in the velocity profile.
Choose a fixed CV (water + elbow) as shown by the dashed line in Fig. 5.12(a).
The forces acting on the control volume include those due to
- Pressure p_{1} acting on area A_{1}
- Pressure p_{2} acting on area A_{2}
- Weight of the elbow
- Weight of the water in the control volume
- Reacting force acting on control volume (force exerted by the pipe supports on the elbow).
It is extremely important to mention here that the resultant force due to uniform pressure distribution acting over a closed contour is zero. Therefore, a uniform atmospheric pressure around the CV exerts no resultant force acting on the same. Thus, any pressure difference relative to the atmospheric pressure (i.e., gauge pressure) is only capable enough of exerting any net force on the CV due to pressure. Here, such deviation occurs only over sections 1 and 2, with the corresponding gauge pressures as ( p_{1}-p_{a t m}) and ( p_{2}-p_{a t m}), respectively
Now, writing the linear momentum conservation equation for the CV results in
\sum \vec{F}=0+\int_{C S} \rho \vec{V}(\vec{V} \cdot \hat{n}) d A (5.27)
The left-hand side of Eq. (5.27)
\sum \vec{F}=\vec{F}_{\text {reaction }}+\vec{F}_{\text {pressure }}+\vec{F}_{\text {water weight }}+\vec{F}_{\text {elbow weight }}Or \sum \vec{F}=\vec{F}_{\text {reaction }}+\left(p_{1}-p_{\text {atm }}\right) A_{1} \hat{i}+\left(p_{2}-p_{\text {atm }}\right) A_{2} \hat{j}-\rho \forall_{\text {water }} g \hat{j}-W \hat{j}
(Noting that pressure is always acting inward normal to any given surface) The right-hand side of Eq. (5.27) becomes
\int_{C S} \rho \vec{V}(\vec{V} \cdot \hat{n}) d A=\rho\left(V_{1} \hat{i}\right)\left(-V_{1} A_{1}\right)+\rho\left(-V_{2} \hat{j}\right)\left(V_{2} A_{2}\right)Thus, Eq. (5.27) simplifies to
\vec{F}_{\text {reaction }}+\left(p_{1}-p_{\text {atm }}\right) A_{1} \hat{i}+\left(p_{2}-p_{\text {atm }}\right) A_{2} \hat{j}-\rho \forall_{\text {water }} g \hat{j}-W \hat{j}=\rho\left(V_{1} \hat{i}\right)\left(-V_{1} A_{1}\right)+\rho\left(-V_{2} \hat{j}\right)\left(V_{2} A_{2}\right)\vec{F}_{\text {reaction }}=-\left(p_{1}-p_{a t m}\right) A_{1} \hat{i}-\left(p_{2}-p_{a t m}\right) A_{2} \hat{j}+\rho \forall_{\text {water }} g \hat{j}+W \hat{j}+\rho\left(V_{1} \hat{i}\right)\left(-V_{1} A_{1}\right)+\rho\left(-V_{2} \hat{j}\right)\left(V_{2} A_{2}\right)
