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# Question: the images mught not appear but it is just of...

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The images mught not appear, but It is just of what is written, Dont think you need them to solve the question...

The operator in Figure 1 operates a conveyor belt that holds granule material ready to be despatched.

Figure 1 – Manually controlled conveyor

The concepts of operation can be described in 3 steps:

1. Start the conveyor to reach preferred speed
2. Release granule material from hopper when conveyor reaches speed
3. Stop granule material from hopper when material reaches end of conveyor (full conveyor)

The state of the conveyor will finish as shown in Figure 2.

Figure 2 – Conveyor loaded with granule material stopped

$F=\frac{2Tl}{D}$                                                    $u=w1\frac{D}{2}$

Note that in Figure 2, the parameters are:

• D = diameter of conveyor roller
• u = linear velocity of conveyor in m/s
• ${\omega }_{l}$ = rotational speed of conveyor roller in rad/s
• ${\tau }_{l}$ = rotational torque delivered by the motor through the gearbox in Nm
• F = force delivered by the motor through the gearbox pushing the conveyor forward

The motor is a DC voltage control motor with the following characteristics:

• Stalled torque at 24V input voltage = 3.4286 Nm
• No load RPM at 24V input voltage = 515.66 RPM

The voltage control box is operated by a handle.

Figure 3 – Voltage control box and control handle

The right hand side diagram in Figure 3 illustrates the operator’s handle pulling the handle to change the angle $\alpha$.  The left hand side diagram in Figure 3 illustrates the potentiometer circuit that can be used to produce the variable output voltage v:

$v={K}_{p}\alpha$

where:

Kp = Voltage regulator constant

The user specified the following performance parameters:

• The length of granule dump on the conveyor L = 1 m.
• Operating cycle time should be as fast as possible but the conveyor must move smoothly, i.e. no vibration.

Assuming a smooth conveyor (frictionless), which set of system design would you recommend.

Formulae applicable to this question:

Motor characteristics is given by:

${K}_{m}{\tau }_{m}+{K}_{o}{\omega }_{m}=v$

where:

• Km = free running constant
• Ko = stall constant

The speedometer indicates speed in a scale of 0 to 100 by the following characteristics:

$s={K}_{s}u$

where:

• Ks = speedometer constant
• u = linear velocity of conveyor in m/s
• s = speedometer reading (integers between 0 to 100)

Equation of motion for the conveyor before material dumping is started:

${M}_{c}\frac{du}{dt}=F$

where:

• Mc = mass of empty conveyor
• u = linear velocity of conveyor in m/s
• F = force delivered by the motor through the gearbox pushing the conveyor forward

Equation of motion for the conveyor after granule material dumping has started:

$\left({M}_{c}+mt\right)\frac{du}{dt}=F$

where:

Mc = mass of empty conveyor

m = rate of granule material dumping on the conveyor (kg/s)

t = time in seconds since dumping starts

u = linear velocity of conveyor in m/s

F = force delivered by the motor through the gearbox pushing the conveyor forward

$\alpha$ = voltage control angle = 20 degrees

n = gear ratio = 1

D = diameter of conveyor roller = 0.32 m

$\alpha$ = voltage control angle = 60 degrees

n = gear ratio = 1

D = diameter of conveyor roller = 0.52 m

$\alpha$ = voltage control angle = 60 degrees

n = gear ratio = 3

D = diameter of conveyor roller = 0.52 m

$\alpha$ = voltage control angle = 20 degrees

n = gear ratio = 3

D = diameter of conveyor roller = 0.32 m

$\alpha$ = voltage control angle = 60 degrees

n = gear ratio = 2

D = diameter of conveyor roller = 0.52 m

$\alpha$ = voltage control angle = 20 degrees

n = gear ratio = 2

D = diameter of conveyor roller = 0.52 m