# Input Power against time graph - Flywheel rest to 300rpm

1. Feb 28, 2013

### Moranovich

Hello,

1. The problem statement, all variables and given/known data

A gearbox and flywheel are as shown in FIGURE 4. The output shaft
rotates in the opposite direction to the input shaft at 5 times its speed.
The gearbox has an efficiency of 92%.
If the flywheel is solid, has a mass of 50 kg, a diameter of 1.5 m and is to
accelerate from rest to 300 revs min–1 in 1 minute:

(a) Calculate the torque required at input T1.

(b) Calculate the magnitude and direction of the torque required to hold
the gearbox stationary (holding torque Th). Show the direction of the
holding torque applied to the shaft with the aid of a sketch.

(c) Plot a graph of the input power against time when taking the
flywheel from rest to 300 revs min–1.

2. Relevant equations
I am having trouble with (c) not sure how to tackle this part have all the answers for (a) and (b)
3. The attempt at a solution

Regards.

Last edited: Feb 28, 2013
2. Feb 28, 2013

### SteamKing

Staff Emeritus
What's the relationship between input torque, RPM, and power?

3. Feb 28, 2013

### Moranovich

P=T. angular velocity

I was thinking today I have worked out the input torque and input power required when the output shaft is at 300rpm so do I work out the input power from 60rpm down to 0 I.e at rest. But how can I do this if I only know the rpm eg 50,40,30etc to zero and dont know power or torque?
Where does time come in is it to do with the revolutions per minute?

Last edited: Feb 28, 2013
4. Feb 28, 2013

### SteamKing

Staff Emeritus
If your input torque is constant, then the power input is obviously going to vary as the RPM versus time. Knowing what your angular acceleration is as a function of time, you should be able to calculate angular velocity as a function of time as well as the power.

5. Mar 1, 2013

### Moranovich

Ok so I worked out my constant acceleration from this kinematic equation
final ang speed=initial ang speed + acceleration . time using 60sec as per question for time. Then using the constant acceleration figure I changed the time in the equation to 5sec.10sec etc to work out all the angular speeds.

Now do I multiple these angular speeds against my Torque constant to get the Input power values?

Also seems to be that the equation to find the time is t=angular speed/constant acceleration

Thanks again.

6. Mar 1, 2013

### SteamKing

Staff Emeritus
You were doing fine, then you suddenly veered off the rails.

t is not found by dividing ang. vel. by ang. acc. (where did you learn that?)

t is the independent variable. The ang. acc.of the flywheel is the moment of inertia of the flywheel divided by the torque. For a constant torque, the ang. acc. is constant.

The ang. vel. as a function of time is the initial velocity (or 0 since the flywheel starts at rest) plus the ang. acc. times t. Remeber, you are being asked to make a graph of input power v. time while the flywheel accelerates to 300 RPM.

Therefore, for various t values, you need to calculate the current ang.vel. and the resulting power so that you can make this plot.

7. Mar 1, 2013

### Moranovich

I had worked out the ang. vel for each 5 sec interval and then was messing around to check it and input ang. vel for 300rpm divided by acceleration to get the time taken. So 10pi/0.167pi = 59.98sec ......

So is how I worked out each ang. vel correct?

i.e final ang vel=initial and vel + acceleration . time
eg 10pi=0 + 0.167pi . 60
or for 20sec ang vel= 0 + 0.167pi . 20

Then just multiple these ang. vel figures against my Input Torque figure?

When I do as you suggested -
The ang. acc.of the flywheel is the moment of inertia of the flywheel divided by the torque. For a constant torque, the ang. acc. is constant.
This gives me the figure I had worked out for the Output shaft Torque not the Input shaft Torque value!

Think I am tying myself in knots ....

8. Mar 1, 2013

### Moranovich

When I do as you suggested -
The ang. acc.of the flywheel is the moment of inertia of the flywheel divided by the torque. For a constant torque, the ang. acc. is constant.
This gives me the figure I had worked out for the Output shaft Torque not the Input shaft Torque value!

Disregard above I was getting confused.

So dividing my moment of inertia by Output shaft Torque = 1.91rads^-2

The ang. vel. as a function of time is the initial velocity (or 0 since the flywheel starts at rest) plus the ang. acc. times t. Remeber, you are being asked to make a graph of input power v. time while the flywheel accelerates to 300 RPM.

Therefore, for various t values, you need to calculate the current ang.vel. and the resulting power so that you can make this plot.

So initial vel 0 + 1.91 . 60s =114.6

9. Mar 1, 2013

### Moranovich

Hello Steamking,

Thanks very much for your help the penny has finally dropped!
I think I wouldn't have got there if you hadn't of pointed me in the right direction.

Thanks again much appreciated.

10. Mar 19, 2013

### JimmyTheBlue

Doing the same question now alas, for part B did you calculate the gearbox torque magnitude via the assumption that it dissipates 8% of the initial power?

Or just via the assumption that T(Driver) = T(Acceleration) + T(Frictional) which semi amounts to the same thing.

Last edited: Mar 19, 2013
11. Apr 26, 2015

### Mollycoddle

How did you work out B mate?