Two Pulley System (Driver/Driven) power transmission.

[JimmyTheBlue]

Homework Statement

A pulley 150 mm diameter is driven directly by an electric motor at 250 revs min-1. A V-belt is used to transmit power from this pulley to a second pulley 400 mm diameter against a load of 200 Nm.

The distance between the centre of the pulleys is 600 mm, the included angle of the pulley groove = 40° (2A), the coefficient of friction (μ) between belt and pulley is 0.4 and the ultimate strength of the belt is 8 kN.

Calculate the actual power transmitted to the second pulley.

Homework Equations

P = Tω
F1(Tight Tension) / F2(Slack Tension) = e^((μ*θ)/Sin(A))
T = (F1-F2)r

The Attempt at a Solution

General info/conversions:
Pulley 1 = 150mm = 0.075m radius
Pulley 2 = 400mm = 0.2m radius
Radial difference = 0.125m
Driver speed = 250 revs min-1 = 8.3333pi rad s-1
Centre difference = 0.6m

1st Step: Worked out angle of lap on the driver/driven pulleys. Confident is right.
Sin(Small angle) = 0.125/0.6
→ Small angle (SA) = 12.0247°
→ Angle of lap on P1 = 180-2xSA = 155.951° = 2.722 radians
→ Angle of lap on P2 = 180+2xSA = 204.049° = 3.561 radians

2nd Step: Have enough info to attack my 2nd given relevant equation.
Assumption: F1 = 8000N.
8000 / F2 = e^((0.4*2.722)/sin20)
→ 8000 / F2 = 24.13
→ F2 = 331.543N

3rd Step: Calculate P1 torque. 3rd given relevant equation.
T = (8000-331.543)*0.075 = 575.134Nm
Assumption: I should take away the 200Nm opposing load.
T = 575.34 - 200 = 375.134Nm

4th Step: Use 1st given relevant equation.
P = 375.134 * 8.3333pi = 9821W

That's about it. But I'm not left confident as to whether my approach was correct as on a later question I seem to run into issues and as this was the only previously worked section I'm forced to look back over it. Am I correct to remove the resisting load? Or have I just gone wrong completely.

Help me Physicsforums, you're my only hope. I can't remembering ever having owned a droid...

 

[sjones]

Hi Jimmy
I don't understand why you have equated the ultimate strength in the belt to be T1. Will T1 not be a result of tension due to the load on the second pulley ?
I am stuck on this question also, since I can`t decide whether T1 is the sum of the initial tension plus the tension due to the load or simply the tension due to the torque of 200Nm on the second pulley i.e 200/rad =200/.2m=100N.
I agree with the angles of lap for both pulleys.

Would anyone like to comment ?

 

[sjones]

Sorry that should be 1000N

 

[JimmyTheBlue]

Setting T1 to the maximum tensile strength seemed to be the books method of finding the maximum power input, but as you say that method pretty much calculates it without taking into account the load.

We can find the speed of the second pulley assuming no slippage, and thus the power dissipated due to the torque load...maybe that information can be thrown in somewhere? Either way I used the load wrong in that first attempt.

 

[JimmyTheBlue]

The system isn't under acceleration, so we can say the input and output powers are equal? If that's the case then...
Driver speed = 250 revs min-1 = 25pi/3 rad s-1
Driven Speed = (25pi/3) * (0.075/0.2) = 25pi/8 rad s-1
Power dissipated at pulley 2 = (25pi/8) * 200 = 625pi W (This is the answer?)
Torque at pulley 1 = 625pi / (25pi/3) = 75Nm
T = (F1-F2)r
75/0.075 = 1000 = F1-F2
From F1(Tight Tension) / F2(Slack Tension) = e^((μ*θ)/Sin(A)) we get
F1 = 24.13F2
1000 = 23.13F2
F2 = 43.23N
F1 = 1043.23N

Does that make any more sense? Calculating the tensions seems irrelevant though since we can reach the power rating almost instantly...

 

[mooneymagic]

Homework Statement

A pulley 150 mm diameter is driven directly by an electric motor at 250 revs min-1. A V-belt is used to transmit power from this pulley to a second pulley 400 mm diameter against a load of 200 Nm.

The distance between the centre of the pulleys is 600 mm, the included angle of the pulley groove = 40° (2A), the coefficient of friction (μ) between belt and pulley is 0.4 and the ultimate strength of the belt is 8 kN.

Calculate the actual power transmitted to the second pulley.

Homework Equations

P = Tω
F1(Tight Tension) / F2(Slack Tension) = e^((μ*θ)/Sin(A))
T = (F1-F2)r

The Attempt at a Solution

General info/conversions:
Pulley 1 = 150mm = 0.075m radius
Pulley 2 = 400mm = 0.2m radius
Radial difference = 0.125m
Driver speed = 250 revs min-1 = 8.3333pi rad s-1
Centre difference = 0.6m

1st Step: Worked out angle of lap on the driver/driven pulleys. Confident is right.
Sin(Small angle) = 0.125/0.6
→ Small angle (SA) = 12.0247°
→ Angle of lap on P1 = 180-2xSA = 155.951° = 2.722 radians
→ Angle of lap on P2 = 180+2xSA = 204.049° = 3.561 radians

2nd Step: Have enough info to attack my 2nd given relevant equation.
Assumption: F1 = 8000N.
8000 / F2 = e^((0.4*2.722)/sin20)
→ 8000 / F2 = 24.13
→ F2 = 331.543N

3rd Step: Calculate P1 torque. 3rd given relevant equation.
T = (8000-331.543)*0.075 = 575.134Nm
Assumption: I should take away the 200Nm opposing load.
T = 575.34 - 200 = 375.134Nm

4th Step: Use 1st given relevant equation.
P = 375.134 * 8.3333pi = 9821W

That's about it. But I'm not left confident as to whether my approach was correct as on a later question I seem to run into issues and as this was the only previously worked section I'm forced to look back over it. Am I correct to remove the resisting load? Or have I just gone wrong completely.

Help me Physicsforums, you're my only hope. I can't remembering ever having owned a droid...

I am currently trying to solve this question and so far i have the following :

Sin Angle = 12.02 °
Angle of Lap on Smaller Pulley = 155.96 ° = 2.72 Radians
Angle of Lap on Large Pulley = 204.04 ° = 3.5611 Radians

V = 1.963 m/s

Does anyone have any ideas where to go next with it?

 

[Big Jock]

Did anyone get further along with this question??... Would I be right in saying the power transmitted was 523.6W?? Hopefully someone replies...

 

[Big Jock]

Can anyone help shed some light on this please??...

 

[RuBear]

I'm struggling with this too Big Jock. It should be a relatively simple question!
I got 1047W (double yours, maybe a coincidence?)
I calculated the torque required to turn pulley 2 then used that to work out the power transmitted from pulley 1. This make any sense?

 

[Big Jock]

Rubear I eventually got an answer of 1.963kW for the power transmitted to the 2nd pulley if that helps?!...

 

[RuBear]

What are the relevant equations for that Big Jock?

 

[Big Jock]

Power = Tw is what I used

 

[RuBear]

If P=Tw then the power needed to turn pulley 2 is 200x26.18 = 5236W
Is this the power transmitted to pulley 2 then?

Having a load on the second pulley has thrown me, there are no examples of it in the books I'm using.

 

[RuBear]

D'oh! Wrong rotational speed. I see how you get your answer now Big Jock.
Did you take the 200NM load into consideration when calculating the maximum power in Q(b)?

 

[Big Jock]

for question b I got 7.529kW for my answer...

 

[RuBear]

Thanks Big Jock, same as me then :)

 

[oxon88]

Can anyone advise how part B is calculated?

 

[mib12]

Part a is much simpler than you think.

P=t*w(driven)

w(driven)= v/r(driven)

v=w(driver)*r(driver)

 

[SteamKing]

Part a is much simpler than you think.

P=t*w(driven)

w(driven)= v/r(driven)

v=w(driver)*r(driver)

Please don't post to old threads.

This one last saw any activity more than a year ago and was started more than two years ago, being semi-hijacked a couple times in between.

 

[mib12]

Why not? It may be useful for somebody being confused by the above.

 

[SteamKing]

Why not? It may be useful for somebody being confused by the above.

Because in the HW forums, people have finished the assignments and moved on. You don't spend two years working on a homework assignment.

Also, PF Rules ask that members not reply to old threads (which is called necro-posting).