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

[David144]

Homework Statement

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.

Homework Equations

Power = Torque*angular velocity
Angular velocity=angular acceleration*time
Power=torque*(angular acceleration*time)
Angular velocity =initial velocity*angular acceleration*time
Angular acceleration = ( final velocity - inital velocity )/time

The Attempt at a Solution

I have worked through the first parts of the question and am now stuck on the last part of the question

I have an input torque value of 39.84Nm Assuming that the torque value remains constant, the power is then calculated by multiplying by the variable angular velocity

If after 60 seconds the input shaft is rotating 60 revs min-1 to calculate the angular acceleration do I assume that after 30 seconds that the angular velocity would also be half?

I seem to going around in circles between the angular velocity as a function of time and angular acceleration

Any advice please?

 

[NFuller]

If the torque is constant, then what do you know about the angular acceleration?

 

[David144]

The angular acceleration varies.

Knowing that P=T*W

and
W= angular acceleration x time

I know that I need to use the equation for angular velocity as a function of time

Wf =Wo*a*t

After I have found the angular velocity then I should be able to then substitute back into the original equation.

Where I'm becoming stuck is to calculate angular acceleration. I want to use the equation

a=(W2-W1)/t

If you know from the answer from the earlier question that the input shaft is rotating at 60rpm after 60 seconds then using the above equation the angular acceleration will be fixed, which cannot be correct?

Thank you for the patience

 

[CWatters]

Torque = moment of inertia * angular acceleration

So if the torque is constant what happens to the angular acceleration ?

 

[David144]

Angular acceleration must also be constant

 

[David144]

If I'm to use the equation

Torque = moment of inertia * angular acceleration

I will need to calculate it for the output shaft and then divide by the gearbox ratio and allow for the efficiency to give me the input power?

 

[NFuller]

Angular acceleration must also be constant

That's right.

I will need to calculate it for the output shaft and then divide by the gearbox ratio and allow for the efficiency to give me the input power?

You should calculate the toque at the output and find the power required to accelerate the flywheel first, then you can use the efficiency of the gearbox to find the power required at the input.

 

[David144]

Ok thank's

I believe that the output torque is 7.33Nm and the input torque 39.84Nm

Therefore for 300rpm (31.41rad/s) at the output side
angular acceleration is calculated

a=(31.41-0)/60
a=0.5235

Angular velocity is a reversal of the equation

Wf=W0*a*t
31.41=0*0.5235*60

Power output=T*W
230W=7.33Nm*31.41rads

230/5=46W gearbox reduction
46/0.92 gearbox efficiency

input power = 50W

If this makes sense?

 

[NFuller]

I believe that the output torque is 7.33Nm

I cannot get this answer. I assumed the flywheel was a disk since you did not give its width, but this could be wrong. Can you show how you got this number?

Wf=W0*a*t
31.41=0*0.5235*60

I assume you meant $\omega_{f}=\omega_{0}+\alpha t$.

Power output=T*W
230W=7.33Nm*31.41rads

The question asks you to plot power as a function of time, it is not constant. You should plug in the equation for $\omega$ into $P=\tau\omega$ to get $P$ as a function of time.

 

[David144]

The input torque was calculated by first dividing the stated output of 300 revs/min by 5 to give the input 60 revs/min.

The moment of inertia I have calculated using
I=½M(R*R)
=0.5*50*(0.75*0.75)
I=14.0625Kgm

Output torque I've calculated as
τ=I(ω1/t)
τ=14(10π/60)
τ=7.33Nm

Input torque found by multiplying the output torque by 5 to account for the gearbox and then dividing again by 0.92 to allow for the efficiency
39.84Nmknowing that from previous comments that angular acceleration is constant, I have calculated this as 0.5235 and with the variable being time, I have then calculated the angular velocity using time values from 60, 55,50 all the way to 0 seconds.

ωf=ω0+αt

knowing the values of ωf from 60 seconds (31.41rads/s), 30 seconds (15.705rads/s) down to 0 I have then calculated the output power using

P=τ*ωf

knowing the power output value I can then calculate the input power. Looking at my previous post I have realized that each output power value for a given time should be multiplied by 5 to account for the gearbox and then divide by 0.92 to compensate for efficiency losses.

for example @ 60sec

P=7.33*31.41
P=230W

Power input = power input*5=1150W
1150W/0.92=1250W input power

For 40 sec
P=7.33*20.94= 153W

153W *5=765W
765W/0.92=831W input power

Plotting a graph this should show the increase from 0 sec to 60 sec of power from 0 to 1250W

Once again many thanks for the help in working through this.

 

[NFuller]

Output torque I've calculated as
τ=I(ω1/t)
τ=14(10π/60)
τ=7.33Nm

Was $I$ given to you or did you calculate it from somewhere else?

knowing the power output value I can then calculate the input power. Looking at my previous post I have realized that each output power value for a given time should be multiplied by 5 to account for the gearbox and then divide by 0.92 to compensate for efficiency losses.

The gearbox ratio only changes $\tau$ and $\omega$, but does not change the power.

 

[GSXR-750]

Input torque found by multiplying the output torque by 5 to account for the gearbox and then dividing again by 0.92 to allow for the efficiency 39.84Nm


Should this not be output torque divided by 5. The output shaft is running at 5 times the speed of the input shaft?

 

[CWatters]

Input torque found by multiplying the output torque by 5 to account for the gearbox and then dividing again by 0.92 to allow for the efficiency 39.84Nm

Should this not be output torque divided by 5. The output shaft is running at 5 times the speed of the input shaft?

No.
If we ignore losses in the gearbox..

Power In = Power Out
so
Input Torque * Input angular velocity = Output Torque * Output Angular Velocity

rearrange to give

Input Torque / Output Torque = Output Angular Velocity / Input angular velocity

Output Angular Velocity/Input angular velocity = 5
so
Input Torque/Output Torque = 5
or
Input Torque= 5 * Output Torque

 

[GSXR-750]

Thanks CWatters, I see this now.

I have calculated very similar to the above now within a few degrees of rounding accuracy.Is Davids method for calculating input power correct?

 

[CWatters]

Is Davids method for calculating input power correct?

I've just had another look and no it isn't.

The input torque was calculated by first dividing the stated output of 300 revs/min by 5 to give the input 60 revs/min.

The moment of inertia I have calculated using
I=½M(R*R)
=0.5*50*(0.75*0.75)
I=14.0625Kgm

Output torque I've calculated as
τ=I(ω1/t)
τ=14(10π/60)
τ=7.33Nm

I agree down to that point but the bit about multiplying the power by 5 is wrong.

Efficiency (%) = 100 * Power Out/Power In

So I would calculate the power required to turn the flywheel at the 300rpm point using...

Power = Torque * Angular velocity
= 7.33 * 10pi = 230.3W

which is the same as David got. But then..

Power In = Power Out * 100/Eff(%)
= 230.3 * 100/92 = 250.3W

Aside: I know this is a step up gearbox (1:5) but if you had a step down gearbox (5:1) then multiplying the output power by the gear ratio would reduce the input power required. That would mean that a 46W motor could deliver 230W to the load. Clearly that's wrong as a gearbox cannot make energy (the law of conservation of energy also applies to a gearbox!).

 

[GSXR-750]

I calculated input Power using the input torque of around 39.73Nm

Power = Torque * AngVel
= 39.73 * 10pi
= 1248W

That would give full speed i.e. 300rpm

Then for 30sec

Power = Torque * AngVel
= 39.73 * 5pi
= 624W

 

[CWatters]

I calculated input Power using the input torque of around 39.73Nm

Power = Torque * AngVel
= 39.73 * 10pi
= 1248W

That appears to be something like the Input Torque * Output Angular velocity? That's not correct. The input power is the Input Torque * Input Angular velocity.

 

[GSXR-750]

Rookie error, thanks.

 

[David144]

Cwatters many many thanks.

Using

Power In = Power Out * 100/Eff(%)

I have now drawn up a table which shows @ 60 seconds 250.26W, @40 seconds 166.85W, @10 seconds 41.75W etc.

The 5:1 gearbox and efficiency confused me.