Hello, professor,solved in a detailedway, the samesolution, steps, but itis a detailed solutionfor mathematics.EXAMPLE 6-3The Force to Hold a Reversing Elbow in PlaceThe deflector elbow in Example 6-2 is replaced by a reversing elbow suchthat the fluid makes a 180° U-turn before it is discharged, as shown inFig. 6-21. The elevation difference between the centers of the inlet and theexit sections is still 0.3 m. Determine the anchoring force needed to holdthe elbow in place.SOLUTION The inlet and the outlet velocities and the pressure at the inletof the elbow remain the same, but the vertical component of the anchoringforce at the connection of the elbow to the pipe is zero in this case (FR = 0)since there is no other force or momentum flux in the vertical direction (weare neglecting the weight of the elbow and the water). The horizontal com-ponent of the anchoring force is determined from the momentum equationwritten in the x-direction. Noting that the outlet velocity is negative since itis in the negative x-direction, we haveatmFRFR + P pgA, = Byn(-V,) – B,inV, = -Bim(V, + V,)1, gageSolving for FRx and substituting the known values,gagoFRE = -Bim(V2 + V) – P1, page A|%3D1N-(1.03)(14 kg/s)[(20 + 1.24) m/s]- (202,200 N/m²)(0.0113 m²)1 kg-m/s² )= -306 – 2285 = - 2591 N

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Description: Hello, professor,solved in a detailedway, the samesolution, steps, but itis a detailed solutionfor mathematics.EXAMPLE 6-3The Force to Hold a Reversing Elbow in PlaceThe deflector elbow in Example 6-2 is replaced by a reversing elbow suchthat the fl

Given Pressure, P = 202200 N/m2 Area, A = 0.0113 m2 Mass flow rete, m =…

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The deflector elbow in Example 6-2 is replaced by a reversing elbow such that the fluid makes a 180° U-turn before it is discharged, as shown in Fig. 6-21. The elevation difference between the centers of the inlet and the exit sections is still 0.3 m. Determine the anchoring force needed to hold the elbow in place.

The deflector elbow in Example 6-2 is replaced by a reversing elbow such that the fluid makes a 180° U-turn before it is discharged, as shown in Fig. 6-21. The elevation difference between the centers of the inlet and the exit sections is still 0.3 m. Determine the anchoring force needed to hold the elbow in place.

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Hello, professor,
solved in a detailed
way, the same
solution, steps, but it
is a detailed solution
for mathematics.
EXAMPLE 6-3
The Force to Hold a Reversing Elbow in Place
The deflector elbow in Example 6-2 is replaced by a reversing elbow such
that the fluid makes a 180° U-turn before it is discharged, as shown in
Fig. 6-21. The elevation difference between the centers of the inlet and the
exit sections is still 0.3 m. Determine the anchoring force needed to hold
the elbow in place.
SOLUTION The inlet and the outlet velocities and the pressure at the inlet
of the elbow remain the same, but the vertical component of the anchoring
force at the connection of the elbow to the pipe is zero in this case (FR = 0)
since there is no other force or momentum flux in the vertical direction (we
are neglecting the weight of the elbow and the water). The horizontal com-
ponent of the anchoring force is determined from the momentum equation
written in the x-direction. Noting that the outlet velocity is negative since it
is in the negative x-direction, we have
atm
FR
FR + P pgA, = Byn(-V,) – B,inV, = -Bim(V, + V,)
1, gage
Solving for FRx and substituting the known values,
gago
FRE = -Bim(V2 + V) – P1, page A|
%3D
1N
-(1.03)(14 kg/s)[(20 + 1.24) m/s]
- (202,200 N/m²)(0.0113 m²)
1 kg-m/s² )
= -306 – 2285 = - 2591 N

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