Gearbox and Flywheel Homework: Calculating Torque & Power

[sponsoraw]

Homework Statement

A flywheel is attached on the output of the gearbox. 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 50kg, a diameter of 1.5m and is to accelerate from rest to 300 revs min-1 in 1min:

a) Calculate the torque required at input T1.

b) Calculate the magnitude and direction of the torque required to hold the gearbox stationary (holing 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

α=(ω₂-ω₁)/t
T=I*α
I=0.5*m*r²
η=(-T_o*ω_o)/(T_i*ω_i)

The Attempt at a Solution

a)
ω_i=-0.2ω_o
ω₂=300 revs min^-1=300*(2π/60)=10π rads^-1
ω_o=-10π rads-1

α=(ω₂-ω₁)/t=(10π-0)/60=π/6 rads^-2
I=0.5*m*r²=0.5*50*0.75²=14.062 kgm²
T_o=I*α=14.062*(π/6)=7.3628 Nm

ω_i=-0.2*-10π=2π rads^-1

η=(-T_o*ω_o)/(T_i*ω_i)
η(T_i*ω_i)=-T_o*ω_o
T_i*ω_i=(-T_o*ω_o)/η

T_i=(-T_o*ω_o)/(η*ω_i)=(-7.3628*-10π)/(0.92*2π)=36.814/0.92=40.0152 Nm

b)
T_i+T_o+T_h=0
T_h=-T_i-T_o=-40.0152-7.3628=-47.378 Nm

As the holding torque T_h has a negative sign the direction of it is clockwise which is opposite direction to the T_i and T_o.

The a) and b) seem to be not too difficult (if I got it right), however after spending several hours I can't workout c). Can someone put me in the right direction. I hope a) and b) are correct.

 

[SteamKing]

Homework Statement

A flywheel is attached on the output of the gearbox. 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 50kg, a diameter of 1.5m and is to accelerate from rest to 300 revs min-1 in 1min:

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

The a) and b) seem to be not too difficult (if I got it right), however after spending several hours I can't workout c). Can someone put me in the right direction. I hope a) and b) are correct.

1. What's the definition of power if you know the torque on a shaft and its angular velocity?

2. If you know the relationship between torque, mass moment of inertia, and angular acceleration, can't you calculate the change in angular velocity of the shaft versus time?

3. If you can calculate the change in angular velocity, can't you also calculate the angular velocity of the shaft at a particular time t = t₀ ?

4. Apply the results of 3.) above with the formula in 1.) and make a table of Power versus time.

5. Plot results in table of 4.) above.

 

[sponsoraw]

Thanks SteamKing for your reply. I will give it a go. Any comment about a) and b)?

 

[SteamKing]

Thanks SteamKing for your reply. I will give it a go. Any comment about a) and b)?

Not at this time.

 

[sponsoraw]

1. What's the definition of power if you know the torque on a shaft and its angular velocity?

2. If you know the relationship between torque, mass moment of inertia, and angular acceleration, can't you calculate the change in angular velocity of the shaft versus time?

3. If you can calculate the change in angular velocity, can't you also calculate the angular velocity of the shaft at a particular time t = t₀ ?

4. Apply the results of 3.) above with the formula in 1.) and make a table of Power versus time.

5. Plot results in table of 4.) above.

My attempt:
1. P=T*ω
2. T=I*α
so far easy
Change in angular velocity of the shaft versus time?
Is it α=(ω₂-ω₁)/t
4. T=I*[(ω₂-ω₁)/t]
as ω₁=0
T=I*(ω₂/t)
therefore P=(I*ω₂*ω_i)/t
as I=14.062 kgm^-2
ω2=10π rads^-1
ωi=2π rads^-1

P=(14.062*10π*2π)/t=2775.7275/t

Just need to now plot a graph of P for t between 0 and 60s.

Does it make any sense?

Any luck with a) and b) - sorry to be a pest.

 

[SteamKing]

My attempt:
1. P=T*ω
2. T=I*α
so far easy
Change in angular velocity of the shaft versus time?
Is it α=(ω₂-ω₁)/t
4. T=I*[(ω₂-ω₁)/t]
as ω₁=0
T=I*(ω₂/t)
therefore P=(I*ω₂*ω_i)/t
as I=14.062 kgm^-2
ω2=10π rads^-1
ωi=2π rads^-1

P=(14.062*10π*2π)/t=2775.7275/t

Just need to now plot a graph of P for t between 0 and 60s.

Does it make any sense?

No. There's some mistakes here.

1. P = T ω, not P = T ω₁ω₂ . I think you mistyped your equation here, multiplying the omegas instead of subtracting them.

2. Once you calculate the angular acceleration α of the flywheel, you want to calculate the ω of the shaft for a few intervals between t = 0 and t = 1 min. and the power applied to the shaft. ω = ω₀ + α*t . Set up a table of the values of time, the speed of the shaft, and the power. Plot Power versus time.

 

[sponsoraw]

SteamKing,

I think I'm to stupid for this, let's take it a step at a time.

You are correct, P=T*ω.
as T=I*α
and α=(ω₂-ω₁)/t
then T=[I*(ω₂-ω₁)]/t

as ω₁=0

so T=(I*ω₂)/t

substituting into power equation

P=(I*ω₂*ω_i)/t

 

[SteamKing]

OK, I see your derivation now.

Can you complete answering the question?

 

[sponsoraw]

As I=14.062 kgm²
ω₂=300 revmin^-1=10π rads^-1
ω_i=2π rads^-1

then, P=(14.062*10π*2π)/t=2775.7275/t

 

[sponsoraw]

Hi SteamKing, I had another think about it and my above solution is incorrect as I used the moment of inertia from the flywheel where I have to calculate the power input. I believe it should be:
P=T*ω

as α=(ω2-ω1)/t
ω1=0, so
α=ω2/t
ω2=α*t

Substituting this into power equation

P=T(α*t)
as T=40.0152 Nm and α=π/6 rads-2

P=40.0152*(π/6)*t=20.9519*t

The graph looks like this:

That is only if I got the question a) right.

 

[JohnnyS]

Hi sponsoraw,
I know this was a while ago but I am struggling with the same question. How did you get on? Do you have any helpful feedback?
Thanks

 

[gneill]

@JohnnyS,

@sponsoraw was last logged in in March 2016. It is unlikely (but not impossible) that you will get a response. I suggest that you start a new thread of your own if you wish to receive help with your work.

Since this is an old thread and the OP long absent, I am closing the thread.