Question 7.12: Water is pumped steadily through a 90° reducing elbow welded...

We are asked for the structural force applied to the elbow by the pipe to which it is welded. This force holds the elbow in place. From experience we know that this external force must balance the forces of the water inside and air outside acting on the elbow, as well as the weight of the elbow (Figure 7.21A). However, we are told to neglect gravity in this problem. Although we would normally write a force balance on the elbow at this point, we will instead go directly to a momentum balance on the mixed CV shown in Figure 7.21B. Our CV selection is guided by the knowledge that a mixed control volume works best for an external force. We have placed a decal surface through the weld where the external force acts, and sections of control surface at the inlet and the exit of the elbow. The last section of control surface is a decal surface placed on the outside of the elbow where the force applied by the air occurs. A mass balance shows that \dot M = ρA₁V₁ = ρA₂V₂.

Since we anticipate momentum transport terms in both the y and z directions and the flow is steady, we will use the vector form of the steady flow momentum balance as given by Eq. 7.19a:

\int_{CS}^{}{ (ρu)(u • n)dS=F_B +F_S }

and neglect the body force. We will evaluate the momentum transport at the inlet and exit, surface forces due to the pressure at these locations, the force applied by the air to the decal surface on the outside of the elbow, and the external force F_E applied by the pipe at the decal surface at the weld. On the inlet we have  u = V₁j, n = − j, and u • n = −V₁. By inspection, the momentum transport vector there is −ρV^2_1 A₁j. On the exit, u = −V₂k, n = −k, and u • n = V₂, and the momentum transport is −ρV^2_2 A₂k. The net momentum transport is (−ρV^2_1 A₁j−ρV^2_2 A₂k).

The surface force on the inlet and exit is due to the pressure at each location. These terms are (−p₁(−j)A₁) + (−p₂(−k)A₂) = (p₁A₁j + p₂A₂k). The force of the air on the outside of the elbow can be included by using gage pressure in these terms. This assumes the wall of the elbow is thin, which is not always the case. If the wall is not thin, we can calculate the force of the air by using the trick as F_air = −p_A(A₁ + A₁W)j + p_A(A₂ + A₂W)k, where we have included the wall areas in the terms. Upon gathering the various terms, our momentum balance for a thin-walled elbow becomes (−ρV^2_1 A₁j−ρV^2_2 A₂k) = (p_{1_{gage}} A₁j + p_{2_{gage}} A₂k) + F_E . Solving for the external force, we have  F_E = −(p_{1_{gage}} +ρV^2_1 )A₁j − (p_{2_{gage}} +ρV^2_2 )A₂k. The exit of the elbow is at atmospheric pressure so p_{2_{gage}} = 0, and we have

F_E = -(p_{1_{gage}} + ρV^2_1) − ρV^2_1 A₁k

This is the answer. It predicts that the weld must exert a force on the elbow to the left and down. To see if this is sensible, we can imagine that the exit of the elbow is plugged, so that the elbow is subjected to hydrostatic pressure on the inside. This would tend to force the elbow to the right, which requires an external force to the left. With the elbow open the pressure distribution inside will still force the elbow to the right, but also force the elbow up. Think of the elbow exit acting as a jet to produce thrust upward.