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Example Momentum equation applicationchange in flow direction

A horizontal water exits a nozzle with a uniform speed of V1=3m/s, strikes a vane, and is turned

through an angle . Determine the anchoring force needed to hold the vane stationary if the

gravity and viscous effects are negligible.

Solution

We select a control volume that includes the vane and a portion of the water and apply the

momentum equation to this control volume. Section 1 (the entrance) and 2 (the exit) are

considered. The x and z components of momentum equation become

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xFVAuVAu 111222

zFVAwVAw 111222

Where kwiuV

and

x

F and

yF are the net x and z components of force acting on

the contents of the control volume.

The waterenters and leaves the control volume as a free jet at atmospheric pressure. Hence,

there is atmospheric pressure surrounding the entire control volume, and the net pressure force

on the control volume surface is zero;

If we neglect the weight of the water and vane, the only forces applied to the control volume

contents are the horizonal and vertical components of the anchoring forces,Ax

F andAZ

F .

With negligible gravity and viscous effects, and since p1=p2, the speed of the fluid remains

constant so that V1=V2=3m/s (Bernoulli equation). Hence

At Section 1,11

Vu , 01 w

At Section 2, cos12

Vu , sin12

Vw .

By using these equations, momentum equations can be written as

AxFVAVVAV

111121cos

AzFVAVAV

111210sin

Above two equations can be simplified by using mass conservation, i.e.1122

VAVA or

12AA since

12VV , thus

N

VAVAVAFAx

)cos1(54)cos1)(3)(006.0)(1000(

)cos1(cos 2112

11

2

11

N

cVAFAz

)(sin54))(sin3)(006.0)(1000(

sin2

11

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Note The values of FAxand FAz as a function of are shown in the figure below. Note that if =0

(i.e. the vane does not turn the water), the anchoring force is zero. The invicid fluid merely slides

along the vane without putting any force on it. If =90deg, and then FAx=-54 N and FAz=54 N. It is

necessary to push on the vane (and, hence, for the vane to push on the water) to the left (FAxis

negative) and up in order to change the direction of flow of the water from horizontal to vertical.

This momentum change requires a force. If =180deg, the water jet is turned back on itself. This

requires no vertical force (FAz=0), but the horizontal force (FAx=-108 N) is two times that required

if =90deg. This horizontal fluid momentum change requires a horizontal force only.

Note that the anchoring force can be written in terms of the mass flowrate,11VAm , as

)cos1(1 mVFAx

sin1

mVFAz

In this example, exerting a force on a fluid flow resulted in a change in its direction only.