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Tutorial 4 (Solution)

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3. A large `weigh tank' is to be used in the calibration of a flow metre. Measurements of

weights as a function of time are to be made. Water enters the tank vertically from the

flow metering system at a speed of 20 ft/sec through a 1.5 in. diameter pipe. If the weight

of the empty tank is 50 lbf, determine the scale reading at t = 10 sec. What is the weightof the water and tank at t = 10 sec?

3. Solution: Since we are asked to find the scale reading, so we will select our control volume

that will wrap the weight tank with the interface between the tank and the scale. Apply

Momentum equation, which has the following form

Consider in the y direction,

(Steady flow)

The reason that this is a steady flow problem is because once the water jet is shot into the control

volume, upon hitting the bottom of the tank, the water will be splashed on the xz plane. As we

are considering in the y direction and thus there will not be any momentum change with time.

Furthermore, the cross sectional area of the tank is much larger than the incoming jet, so even

though if there is velocity moving up in the y direction, it will be very small. Another reason can

be considered is that as the control volume is not having any movement, so there will not be any

momentum change within the Control Volume. In fact, when the control volume is not having

any movement, we can always say the this is steady flow. This term is to look at the momentum

change within the CV, as there is no movement with the CV, so there is no movement inside the

CV. Because, if there is any momentum change in the CV, it will cause its movement.

Substituting all these back into the Momentum equation, we have

This is the force acted by the SR on to the CV, the force acted by CV to SR will be the negative


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of it and will also give the scale reading.

The mass flow rate

It is given, Wt= 50 lbf, v = 20 ft/sec, d = 1.5/12 ft, t = 10 sec, = 62.4 lbm/ft3.

As for the true weight, the momentum flux does not contribute, so it will be

is a force from SR to CV, the force by CV on SR (scale) will be opposite to it. Due to

this force acting on the scale, the scale will give a reading. So it seems that the answer should

be negative of the force. But the true weight is definitely negative, if we go along with it, then

the reading will be negative as well. So the reading is kind of neutral.

This problem also assumes that the water jet is open to atmosphere.


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7. The jet engine shown in Fig.T.4.7 has produced a thrust force of 52000 lbfon the test

stand. A typical installation is shown, together with some test data. Fuel enters the top

of the engine vertically at a rate equal to 2% of the mass flow rate of the inlet air. For the

given conditions, compute the air flow rate through the engine.7. Solution: We can take our control volume to be wrapping the whole engine. Let section 1 be

where air comes in, section 2 be where exhausted gas leaves, section f be where fuel comes in.

From Steady Continuity equation, we have

It is given that , so we have

Now we may apply Momentum equation, which has the following form,

Consider in the x direction,

(No body force in x direction)

(Steady flow)

The surface force consists of the shear stress and the pressure force. That is,

Where Rxis the force acted by the SR to CV, and thrust is a force acted by CV to SR, so it is

negative of Rx, P1gA1is the gage pressure force at section 1, P2gA2is the gage pressure force at

section 2. (The reason for using gage pressure is due to the selection of our CV. For this

question, we are asked to find the mass flow rate, so it is possible to use a CV wrapping the fluid

inside the engine or a CV wrapping the engine. But thrust force is a force given by the engine

on the test stand, in order to make use of this piece of info, we need to select CV that wraps the

engine and cutting the stand. With this CV, we will see the atmospheric pressure force acting on

the CV and therefore we have to consider it.)

Substituting all these back into the Momentum equation, we have


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9. A small lawn sprinkler is shown in Fig.T.4.9. The sprinkler operates at a gage pressure

of 140 kPa. The total flow rate of water through the sprinkler is 4 litres per minute. Each

jet discharges at 17 m/sec relative to the sprinkler arm in a direction inclined 30oabove

the horizontal. The sprinkler rotates about a vertical axis. Friction in the bearing causesa torque of 0.18 NCm opposing rotation. Evaluate the torque required to hold the

sprinkler stationary.

9. Solution: The CV is selected to wrap the whole sprinkler. For this problem, we have to use

Moment of momentum equation, which has the following form,

Consider in the z direction, we have

The moment due to body force is 0, because the body force is symmetric on

both side.

Since the incoming stream is flowing along the z axis, therefore the cross product will be zero.

We can concentrate on the outgoing stream. We have to first determine each individual vector.

vrel is in negative direction, (remember the order of r z r z,

), also vrelis relative to the moving sprinkler

arm where the sprinkler arm is moving with a velocity in the direction of R.

As we are only interested in the z direction,


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Where Tzis the torque acted by SR on CV, (- Tf)zis the torque caused by friction, which opposes

rotation.

Substituting every term back into the Moment of momentum equation, we have

After rearrangement,

Now we are to hold the sprinkler stationary, then that means it is not to rotate, thenis actually

zero. From this, we can work out the torque required to hold the sprinkler stationary to be


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10. Water at 20oC flows at 30 gal/min through the 0.75 in diameter double pipe bend of

Fig.T.4.10. The pressures are P1= 30 psi and P2= 24 psi. Compute the torque T at point

B necessary to keep the pipe from rotating.

10. Solution: For this problem, we are asked to find the moment to hold the pipe in place, wehave to select our control volume to be wrapping the piping system, then use the Moment of

momentum equation, which has the following form,

Consider in the z direction, we have

The moment due to body force is 0, because the gravitation flow field

acts in z direction and hence the cross product will not be in z direction.

where , , ,

For vectors being in the same direction, the cross product will be simply zero. So although the

position vector in x direction is not defined, we dont have to worry about it.

So

Now we may consider the moment which is contributed by the pressure force. Pressure force at

the exit end is P2gA2, the direction is .

Substituting every terms back into the moment of momentum equation, we have


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= 8.4727 + 12.3258 = 20.7985 lbf-ft

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