Water flows at mass flow rate \(\dot{m}\) through a \(90^{\circ}\) vertically oriented elbow of elbow radius \(R\) (to the centerline) and inner pipe diameter \(D\) as sketched. The outlet is exposed to the atmosphere. The pressure at the inlet must obviously be higher than atmospheric in order to push the water through the elbow and to raise the elevation of the water. The irreversible head loss through the elbow is \(h_{L}\). Assume that the kinetic energy flux correction factor \(\alpha\) is not unity, but is the same at the inlet and outlet of the elbow \(\left(\alpha_{1}=\alpha_{2}\right)\). Assume that the same thing applies to the momentum flux correction factor \(\beta\) (i.e., \(\beta_{1}=\beta_{2}\) ).

(a) Using the head form of the energy equation, derive an expression for the gage pressure \(P_{\text {gage, } 1}\) at the center of the inlet as a function of the other variables as needed.

(b) Plug in these numbers and solve for \(P_{\text {gage, } 1}: ho=998.0 \mathrm{~kg} / \mathrm{m}^{3}\), \(D=10.0 \mathrm{~cm}, R=35.0 \mathrm{~cm}, h_{L}=0.259 \mathrm{~m}\) (of equivalent water column height), \(\alpha_{1}=\alpha_{2}=1.05, \beta_{1}=\beta_{2}=1.03\), and \(\dot{m}=25.0 \mathrm{~kg} / \mathrm{s}\). Use \(g=9.807 \mathrm{~m} / \mathrm{s}^{2}\) for consistency. Your answer should lie between 5 and \(6 \mathrm{kPa}\).

(c) Neglecting the weight of the elbow itself and the weight of the water in the elbow, calculate the \(x\) and \(z\) components of the anchoring force required to hold the elbow in place. Your final answer for the anchoring force should be given as a vector, \(\vec{F}=F_{x} \vec{i}+F_{z} \vec{k}\). Your answer for \(F_{x}\) should lie between -120 and \(-140 \mathrm{~N}\), and your answer for \(F_{z}\) should lie between 80 and \(90 \mathrm{~N}\).

(d) Repeat Part (c) without neglecting the weight of the water in the elbow. Is it reasonable to neglect the weight of the water in this problem?

FIGURE P6-92

Transcribed Image Text:

In, 1 m R anchoring Out, 2 Inner dia. D Fx.anchoring