Help calculating power required to drive shaft

[Whatamiat]

Homework Statement

Calculate the motor size required to drive shaft A. (shown in schematic 1 diagram)

Drive shaft rotates at 75 Rpm.

Weight of plunger + attachments = 30kg's

Dimensions of plunger given in schematic 2.

Homework Equations

Work done = F x d (distance = radius of cam x 2 = .6m)
Power = work/time

The Attempt at a Solution

Neglecting the weight of the plunger and attachments to the drive shaft of the motor. I
Just looking at a direct relationship between maximum water displaced and power.

The maximum depth under water (Plunger profile) of 600mm x588 x 1219.2 displaces 215kg of water.

Force = 215 x 9.81 = 2109.15N

Work done = F x d = 2109.5 x.6 = 1265.5J

Power = Work/Time = 1265.5/(.4) = 3164 W or 3.164kW


Would I be right to recalculate this allowing for the weight of the plunger and attachments aiding the displacement of water? eg:
215-30=185kg

F = 185 x 9.81 ...etc?

Is my method here right? or am I way off on how this problem should be calculated?

 

[kerrick]

The problem seems tricky. How I would do it is:
-Note that you need to find the integral of force with respect to distance the wedge has been lifted as you draw the wedge out of the water
-Force is equal to weight of the block - weight of water displaced. You can determine the weight of the water displaced in terms of the distance lifted.
-Make an integral for the total work lifting the block out of water using these two facts and evaluate it. Add the amount of work it takes to get the block from just out of the water to the apex of its movement. Multiply by 75 since the motor must do this 75 times per second, and you'll have the total wattage the motor must output. This is not the only thing you must consider; I would also calculate the max torque and see if your motor can handle it.
If you double check your work using the method I outlined above you can be doubly confident in your answer.

 

[Whatamiat]

-Force is equal to weight of the block - weight of water displaced. You can determine the weight of the water displaced in terms of the distance lifted.

30 kg - 215kg = -185kg or 185 kg
Multiplying this x 9.81 to get force = 1815N

-Make an integral for the total work lifting the block out of water using these two facts and evaluate it. Add the amount of work it takes to get the block from just out of the water to the apex of its movement. Multiply by 75 since the motor must do this 75 times per second, and you'll have the total wattage the motor must output. This is not the only thing you must consider; I would also calculate the max torque and see if your motor can handle it.
If you double check your work using the method I outlined above you can be doubly confident in your answer.

I am unsure what you mean here.
What I did was make an integral for the Force with repsect to distance:
∫_0^xm▒( F)(x)dx

=1815 [x^2/2]_0^(.6)
=326.7 N


"Add the amount of work it takes to get the block from just out of the water to the apex of its movement"
Is this what I calulated in my original post?
= 326.7 + 1265.5 = 1592.2

"Multiply by 75 since the motor must do this 75 times per second, and you'll have the total wattage the motor must output."
The motor rotates at 75 rpm not second.
So this is 1.25 times a second.

= 1592.2x1.25 = 1990 W

What am I doing wrong here?

 

[stevoxbx]

not sure how to solve this but at 75rpm shouldn't drag forces be considered?

 

[Whatamiat]

I never even considered drag, but for this problem I would be happy to ignore its effect.
Anyone got any Ideas?