Calculating Power Transferred in a 2-Pulley System

In summary, the first pulley has a diameter of 150mm and driven by an electric motor at 250 rev min -1. The second pulley has a diameter of 400mm and is driven by a v-belt, at a load of 200NM. The distance between the two pulleys is 600mm. The coefficient of friction is 0.4 between belt and pulley. The belt has an ultimate strength of 8kN.
  • #1
stackemup
26
0

Homework Statement


From the below information provided, I have to calculate the power which has been transferred to the second pully.

2 pulley system, first pulley has a diameter of 150mm and driven by an electric motor at 250 rev min -1. , second pulley has a diameter of 400mm and is driven by a v-belt, at a load of 200NM. The distance between the two pulleys is 600mm.
Included angle of the pulley groove is 40°. Coefficient of friction is 0.4 between belt and pulley. The belt has an ultimate strength of 8kN.

Homework Equations


Power = Torque x Angluar Velocity

The Attempt at a Solution


From one of the other threads I found the below -

"Driver speed = 250 revs min-1 = 25pi/3 rad s-1 = 26.18 rad s-1
Driven Speed (angular Velocity) = (25pi/3) * (0.075/0.2) = 25pi/8 rad s-1 = 9.817 rad s-1
Power dissipated at pulley 2 = (25pi/8) * 200 = 625*pi W
= 9.817 * 200 = 1963.4 W = 1.9634 kW"

I need to understand how the angular velocity has been calculated, as I am unsure where the figures have come from.
 
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  • #2
Anyone :)
 
  • #3
Are you asking for an explanation of this:
stackemup said:
Driven Speed (angular Velocity) = (25pi/3) * (0.075/0.2)
?
If the driving pulley is turning at rate ##\omega_1## and has radius r1, what is the rate of travel of the belt?
If the driven pulley has radius r2, what is its angular rate?
 
  • #4
I'm unsure on how to calculate the above.

I have found that to calculate the power, the net torque is multiplied by the angular velocity and radius. Is this correct ?

Motor speed = 250 rev min -1

Angular velocity = (250 x 2pi) / 60
= 26.18 rads -1

Is this correct so far?
 
  • #5
stackemup said:
I'm unsure on how to calculate the above.
If a pulley radius r turns through an angle ##\theta## (in radians), how far does the belt move?
If it does that in time t, what is the average angular velocity (##\omega##) and what is the average belt speed?
stackemup said:
Motor speed = 250 rev min -1

Angular velocity = (250 x 2pi) / 60
= 26.18 rads -1

Is this correct so far?
Yes.
 
  • #6
Still struggling to understand.

If I take the angular velocity calculated above, I need to multiply this by the torque and effective radius to achieve the power transmitted into the second pulley, correct?

I'm assuming that in the answer previously calculated in the other thread, uses the effective radius of 0.075 (rad 1) / 0.2 (rad 2?) = 0.375, but I need an understanding of why these are used as I cannot find anything through internet sources etc.

The final answer quoted in my original post was calculated by another member so this is irrelevant, I just need an understanding of the full working out process.
 
  • #7
Can anyone advise?
 
  • #8
stackemup said:
If I take the angular velocity calculated above, I need to multiply this by the torque and effective radius to achieve the power transmitted into the second pulley,
No. Check the dimensions:
angular velocity dimension = T-1
torque dimension = force * distance = ML2T-2
radius dimension = L
multiply together: ML3T-3
power dimension = energy/time = ML2T-3
So what do you think you might have wrong?
stackemup said:
uses the effective radius of 0.075 (rad 1) / 0.2 (rad 2?) = 0.375, but I need an understanding of why these are used
I was trying to explain that to you. Please try to answer my questions:
haruspex said:
If a pulley radius r turns through an angle θ (in radians), how far does the belt move?
If it does that in time t, what is the average angular velocity (ω) and what is the average belt speed?
 
  • #9
I have finally worked it out and now understand it. Can you please review my below answer to find the power transferred if the max tension is halved to 4kN.

F1 = 4000N
F1/F2 = e^(μθ/sin⁡α )
4000/F2 = e^((0.4 x 2.722)/sin⁡20 )
4000/F2 = e^3.1834
4000/F2 =24.1296
F2= 4000/24.1296
F2= 165.7715

T= (F1 - F2 ) r
T= (4000 -165.7715) x 0.075
T=287.5671 Nm

P= Tω
ω= (250 x 2π)/60 = 26.1799

∴ using a torque load of 287.5671 Nm the actual power transmitted:
P= 26.1799 x 287.5671
P= 7.5285 kW
 
Last edited:
  • #10
stackemup said:
F1 = 4000N
That's the max tension, but at this stage we don't know if that will be reached. Start with the actual load.
stackemup said:
e^(μθ/sin⁡α )
Please define all your variables, otherwise I have to guess what you are doing. What are these angles?
 

What is a 2-pulley system?

A 2-pulley system is a mechanical system that uses two pulleys and a belt or rope to transfer power and motion from one point to another. It is commonly used in machines and devices to change the direction of force and transmit power efficiently.

How is the power transferred calculated in a 2-pulley system?

The power transferred in a 2-pulley system can be calculated by multiplying the tension in the belt or rope by the velocity of the belt. This can be represented by the equation: Power = Tension x Velocity.

What factors affect the power transferred in a 2-pulley system?

The power transferred in a 2-pulley system is affected by several factors, including the tension in the belt or rope, the velocity of the belt, the size and weight of the pulleys, and the friction between the belt and the pulleys. Any changes in these factors can affect the efficiency of power transfer in the system.

What is the relationship between the number of pulleys and power transferred in a 2-pulley system?

In a 2-pulley system, the power transferred is directly proportional to the number of pulleys. This means that the more pulleys there are in the system, the more power can be transferred. However, adding more pulleys also increases the complexity and potential for mechanical losses in the system.

Are there any limitations to using a 2-pulley system for power transfer?

Yes, there are limitations to using a 2-pulley system for power transfer. As the number of pulleys increases, the efficiency of power transfer decreases due to friction and mechanical losses. Additionally, the system may become more complex and expensive to maintain, making it less practical for certain applications.

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