Question

A 94-ft3/s water jet is moving in the positive x-direction at 18 ft/s. The stream hits a stationary splitter, such that half of the flow is diverted upward at 45° and the other half is directed downward, and both streams have a final average speed of 18 ft/s. Disregarding gravitational effects, determine the x- and z-components of the force required to hold the splitter in place against the water force. Take the density of water as 62.4 lbm/ft3.

Answer 1

FRx = 960.37 lbf (←)

FRz = 0 lbf

Step-by-step explanation:

Given:

Q = 94 ft³/s

vx = 18 ft/s

ρ = 62.4 lbm/ft³

∅ = 45°

Assumptions:

1. The flow is steady and incompressible.

2 . The water jet is exposed to the atmosphere, and thus the pressure of the water jet before and after the split is the atmospheric pressure which is disregarded since it acts on all surfaces.

3. The gravitational effects are disregarded.

4. The flow is nearly uniform at all cross sections, and thus the effect of the momentum-flux correction factor is negligible, β ≅ 1.

Properties: We take the density of water to be ρ = 62.4 lbm/ft³

Analysis: The mass flow rate of water jet is

M = ρ*Q = (62.4 lbm/ft³ )(94 ft³/s) = 5865.6 lbm/s

We take the splitting section of water jet, including the splitter as the control volume, and designate the entrance by 1 and the outlet of either arm by 2 (both arms have the same velocity and mass flow rate M). We also designate the horizontal coordinate by x with the direction of flow as being the positive direction and the vertical coordinate by z.

The momentum equation for steady flow is

∑ F = ∑ (β*M*v) out - ∑ (β*M*v) in

We let the x- and y- components of the anchoring force of the splitter be FRx and FRz, and assume them to be in the positive directions. Noting that

v₂ = v₁ = v and M₂ = (1/2) M, the momentum equations along the x and z axes become

FRx = 2*(1/2) M*v₂*Cos ∅ - M*v₁ = M*v*(Cos ∅ - 1)

FRz = (1/2) M*(v₂*Sin ∅) + (1/2) M*(-v₂*Sin ∅) = 0

Substituting the given values,

FRx = (5865.6 lbm/s)*(18 ft/s)*(Cos (45°) - 1)(1 lbf / 32.2 lbm*ft/s²)

⇒ FRx = - 960.37 lbf

FRz = 0 lbf

The negative value for FRx indicates the assumed direction is wrong, and should be reversed. Therefore, a force of 960.37 lbf must be applied to the splitter in the opposite direction to flow to hold it in place. No holding force is necessary in the vertical direction. This can also be concluded from the symmetry.

In reality, the gravitational effects will cause the upper stream to slow down and the lower stream to speed up after the split. But for short distances, these effects are negligible.