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**MECHANICS OF FLUIDS**

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**Problem statement**A jet of water having a velocity of 10 m/s and a diameter of 50 mm is di?rected against a trough

in a horizontal plane, as shown in Fig. The trough can have an angle shown as (theta) with

line A-A which is normal to the jet. The water splits up along the trough such that the volume

flow rates Q1 and Q2 are related as QI = Q2 (cos (theta))^2. We will neglect friction and gravity

effects on the speed of the water at all times. Plot the algebraic sum of the force components at the

hinge B for (theta) going from 0 in steps of 2 degrees. Also, what is the torque at B?

The trough is held in equilibrium by B at each setting of (theta).

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**Strategy**In Fig. , we choose a control volume that is at the interface between the water and the trough surface

while cutting the pin at B and the incoming and the exiting jets. We will use the linear momentum

and angular momenlum equa?tions in simple forms at each stationary angular setting of the trough.

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**Figures**.

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**Code**here theta=x

pi/180 -> c % it is for convertion. .019635/(1+1/((cos(x*c))^2)) -> q1 % This is the volume flow to the top at all times as a % function of theta. This was solved using % conservation of mass. q1/((cos(x*c))^2) - > q2 %We know how ql and q2 are related so we can solve %for q2 once we have ql. Now that we know the flow %going to either end at all times as a function of %theta all that remains is finding the forces and %moments involved with these flows. f=fx+fy= -196.3495- (10000*q1*sin(x*c)+(10000*q1*sin(x*c) /(cos(x*c))^2))+ 10000*q1*cos(x*c)-(10000*q1/cos(x*c))

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**Debriefing**The plot of total force vs. angle is shown in Fig.

Torque at the base of the trough will not change throughout the rotation of the trough.

Since the flows toward the end of the trough have no moment arm,

the only flow that will cause a torque is the constant incoming flow.

The force from this flow will not change throughout,

and the moment arm will not change during rotation either.

It will always remain length/2, which in this case is 4.5 m.

The constant torque value of 883.573 was determined in the following way:

4.5*10*(1000*10*(pi/4)*0.05A2)=883.573

where 4.5 equals the moment arm when the trough is 9 m long,

10 is the velocity of the incoming water, 1000 is the density of the water,

and the remaining terms give the area of the incoming water stream.

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**plot**now we plot the Total force on the trough vs. Theta

be sure the angle is in Radian.

Variable range :

theta=x = 2..60

y=-210..-150

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*Eactivity files*:

here is the Eactivity .

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we could change the subtitle before . why it is not possible now.

I wanted to change

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