# How to calculate required hp for machine

3
one rubber coated roll weight 1000 kg. dia. 300 mm & face length of 6360 mm. bothe end shaft dia. 65 rotating in antifrictional bearings.

i want to rotate the roll at 100 rpm within 60 seconds.

so please help me out to find the HP required of geared motor of 1400 rpm input of motor & gearbox ratio is approx 25:1

its very urgent

2. ### huntoon

147

this is just lazy: "bothe end shaft dia. 65"

have respect for people trying to help you. tell us what you have done to solve the problem. what approaches have you made so far? did you at least look it up in wikipedia? was that not enough?

3. ### huntoon

147
"i want to rotate the roll at 100 rpm within 60 seconds."

you realize RPM stands for rotations per minute, right?

4. ### huntoon

147
1 hp = 745.7 watts

1 watt = 1 W = 1 J/s = 1 N·m/s = 1 kg·m2/s3

5. ### zezo010

2
but how can you calculate 1 hp ?

6. ### OldEngr63

734
We are going to need to know more about the makeup of the roll. Do you have a drawing with dimensions? We need to be able to calculate a mass moment of inertia for the roll.

7. ### Bandit127

216
Can we not approximate a solid cylinder of 300 mm dia and 6,360 mm in length? And 1,000 kg. I suspect he doesn't need the answer correct to 2 d.p. (unless it is for homework).

I feel we ought to do something for the OP who may have been frightened off of further posting by some of the 'forthright' replies.

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9. ### huntoon

147
Do you have a variable speed motor?

If you have 1400 RPM, and you reduce it by a factor of 25:1 (through the gearbox) the resulting RPM will be 56. You can't get 100 RPM with this setup unless you can vary the input rpm. To get 100 RPM, you need a 2500 RPM motor hooked up to the 25:1 gearbox. Or, you could use a 1000 RPM motor hooked up to a 10:1 gearbox.

Your motor's RPM is decreased by reduction ratio of the gearbox. Your motor's torque is increased by the same ratio.

3
thank u but sorry my mistake was there motor rom is 2500
i had attached a drawing of the roll earliar for referance & to calculate MOI

11. ### OldEngr63

734
Bandit127, go right ahead if you want to mislead the man. If yoou look at the drawing that he posted, treating the roll as a solid cylinder will produce a grossly erroneous MMOI value. That is why it is necessary to ask for information.

12. ### Bandit127

216
Yes, I take your point - a solid cylinder would have underestimated the power required. Perhaps even by a gross amount.

In fairness, my post was before we had the drawing though.

I did have a look at the maths but it was too far from my field for me to get a good hold of it, so I will leave the calculations to the experts.

Good to see you are still in the thread though harshadeng. While I am sure they weren't meant that way, some of the initial posts read as unnecessarily critical to me and could have put some people off.

13. ### Bob S

The moment of inertia for a solid cylinder is $I=\frac{1}{2}mr^{2}$, while the moment of inertia of a hollow cylinder is $I=mr^{2}$. The energy of a rotating cylinder is $E=\frac{1}{2}I\omega^2$. Dividing E by the acceleration time τ gives the power required (in watts). So the power required is about
$$E=\frac{1}{2\tau}m r^2 \omega^2$$
Using τ=60, ω=2π100/60, r=0.15, and m=1000, The power to required accelerate to 100 RPM in 60 seconds is about 21 watts (constant power solution).

If the acceleration rate is constant, the torque is $T=I\frac{d\omega}{dt}=m r^2\frac{d\omega}{dt}=3.9 \text{ Newton-meters}$ and the maximum power is $P=T\omega= 3.9\omega_{max}=41 \space watts$

14. ### OldEngr63

734
Would you like to comment on the realism of your results, Bob S?

15. ### Bob S

What means realism? Isn't this the moment of inertia of a solid cylinder?
$$I=\int_{V}\rho r^2 dV=2\pi\int_{0}^{r}\rho \ell r^3 dr=\frac{1}{2}Mr^2$$

16. ### OldEngr63

734
Realism means that you really believe that 41 watts will do this job. I don't. Look at your overall calculation and tell us that the whole scenario is realistic, if you believe it is.

17. ### Bob S

What do you think the total rotational energy of the cylinder is, in Joules?

18. ### OldEngr63

734
I have estimated the total MMOI as 16.5 kg-m^2, and on that basis, Bob_S value of 41 watts appears to be a reasonable estimate for the total power required to accelerate the roll (I calculated slightly less). The total energy of the spinning roll at 100 rpm is 903 J.

As to realism, now, these numbers are so small that I rather think that they are close to meaningless. The friction losses in the gear train, the bearings, the windage, etc. will consume more than what we have computed for the change in the energy state of the rotating mass. Thus I think that considerably more power will be required to overcome the system losses than to power the roll. Therefore, I think that these calculations are not realistic for evaluating the power required.

Last edited: Apr 10, 2012
19. ### Bob S

First, you have to do the math.
If the stored rotational energy after 60 seconds is
$$E=\frac{1}{2}I\omega^2=\frac{1}{2}mr^{2}\omega^2 = 1234 \text{ Joules}$$ then you have to add energy at the rate of 1234/60 joules/sec = 21 watts for 60 seconds.
This does not include motor inefficiency, friction losses, not using the right gear ratios, etc. Please check my math.

20. ### OldEngr63

734
@ Bob_S: I did do the math, in considerably more detail than it appears you have done. It appears that you are simply saying that
I = MR2 = 17.956 kg-m2
Note that the radius value you used is to the outer diameter of the rubber layer, well past the last of the steel structure. This gives an excessively large estimate for R.
By a more detailed reckoning, based on the drawing, I came up with the figure that I gave previously
I = 16.5 kg-m2
I think that you have somewhat over estimated the MMOI by putting it all at the rubber layer, well outside the steel, considering that part of it is internal structure.

N = 100 rpm → ω = 10.47 rad/s
ω2 = 109.66 1/s2

Because of the difference in the estimates for I, we differ substantially in the total energy content of the rotating roll. This is where our differences lie.

My comments about these values being minor compared to the losses in the system and therefore not useful for determining drive power requirements still stand.