How much horsepower required for motor to rotate cylinder?

  • Thread starter niket548
  • Start date
  • #1
10
0

Main Question or Discussion Point

I done this calculation But something wrong.


cylinder weight: 17000 lb weight , Diameter = 36"= 3 ft, length = 5 ft. , 3000 rpm , time=90 sec.


Aceee = Change in angular velocity / time

Angular velocity = 2xPi (3.14) 3000/60 = 314rad / sec

acee = ( 314-0) / 90


= 3.48 Rad / sec2.


Torque = force x radius


= 17000 x 1.5 = 25500 lb-ft

Power = torque x omega

= 25500 x 314

= 8007000 lb-ft / sec

1 hp = 550 lb-ft / sec

Hp = 8007000 / 550 = 14558 HP

Actuaalyy i need to rotate that thing in 90 second so where can i find that? 14558 hp is too much i think . i am doing something wrong somewhere please help me on that
 

Answers and Replies

  • #2
OmCheeto
Gold Member
2,121
2,543
I done this calculation But something wrong.


cylinder weight: 17000 lb weight , Diameter = 36"= 3 ft, length = 5 ft. , 3000 rpm , time=90 sec.


Aceee = Change in angular velocity / time

Angular velocity = 2xPi (3.14) 3000/60 = 314rad / sec

acee = ( 314-0) / 90


= 3.48 Rad / sec2.


Torque = force x radius


= 17000 x 1.5 = 25500 lb-ft

Power = torque x omega

= 25500 x 314

= 8007000 lb-ft / sec

1 hp = 550 lb-ft / sec

Hp = 8007000 / 550 = 14558 HP

Actuaalyy i need to rotate that thing in 90 second so where can i find that? 14558 hp is too much i think . i am doing something wrong somewhere please help me on that
I would imagine it is your "Power = torque x omega" that has thrown things off.
Omega changes from zero to 314 rad/sec over the 90 seconds.
"Power = torque x omega" may only be good for steady state conditions.
 
  • #3
tiny-tim
Science Advisor
Homework Helper
25,832
251
welcome to pf!

hi niket548! welcome to pf! :smile:

is the question asking for the power needed to accelerate a cylinder of weight 17000 lb from 0 to 3000 rpm in 90 seconds, exerting a force on the edge of the cylinder, 1.5 ft from the axle?

if so, your angular acceleration, α (3.48), looks correct, but you don't seem to have used that result :confused:

your equation …
Torque = force x radius


= 17000 x 1.5 = 25500 lb-ft
… is wrong … 17000 is not the force


you can only find the force by first finding the torque using the rotational version of Newton's second law … τ = Iα :wink:

try again :smile:
 
  • #4
10
0
Hi Tiny tim and omcheeto ,

Thanks for reply. so i am confusing where can i use that angular acce .

If I use in Torque = Moment of inertia x Alpha (angular accee)

= 1/2(Radiusx radius 0x Weight X 3.48 Rad/sec^2
= 1/2x1.5x1.5x17000x3.48
= 66555 lb-ft^2 / sec^2

Hp = Torque X Rpm / 5252 = 66555 x3000 / 5252 = 38016 hp

It seems wrong also .

what is eq for finding hp?

Where can i use that angular acce to finding horsepower?

Please reply
and thanks again.
 
  • #5
tiny-tim
Science Advisor
Homework Helper
25,832
251
hi niket548! :smile:
Hp = Torque X Rpm / 5252 = 66555 x3000 / 5252 = 38016 hp
(what's 5252? is that something to do with horses? :confused:)

you need to convert rpm to radians per second! :biggrin:
 
  • #6
10
0
Hi Tiny tim ,


Hp = Torque x RPM / 5252

= 66555x 2x 3.14x 3000 / 5252 = 3979 .10 hp

is that correce ?
becoz i find torqe = i x alpha
= lb-ft^2 X Rad /Sec^2

Torque unit is lb.ft

Now i am cofusing about units ?

Thank you very much for yr prompt response.
 
  • #7
10
0
If torque is in lbf·ft and rotational speed in revolutions per minute, the equation gives power in ft·lbf/min. The horsepower form of the equation is then derived by applying the conversion factor 33000 ft·lbf/min per horsepower:


hp = torque x rpm / 5252
 
  • #8
tiny-tim
Science Advisor
Homework Helper
25,832
251
ah, so that's what the 5252 was! :biggrin:

(where i live, the horses have no power! :smile:)

in that case, i can't see what's wrong with your solution
 
  • #9
OmCheeto
Gold Member
2,121
2,543
ah, so that's what the 5252 was! :biggrin:

(where i live, the horses have no power! :smile:)

in that case, i can't see what's wrong with your solution
Wait! What?

What's the solution? 3979 .10 hp?

Gads. I'm glad I bowed out then. I was off by a factor of 6.63. :frown:
 
  • #10
10
0
Hey omcheeto can u help me to find right solution please ?
 
  • #11
10
0
i think i am trouble with unit when i found torque = I x a
= lb-ft^2 x rad / sec^2


Torque unit lb-ft , lb-inch , n-m
 
  • #12
OmCheeto
Gold Member
2,121
2,543
Hey omcheeto can u help me to find right solution please ?
I thought Tiny-Tim implied that you had the correct solution?

Is the answer not in the back of the book?
 
  • #13
10
0
the power needed to accelerate a cylinder of weight 17000 lb from 0 to 3000 rpm in 90 seconds, exerting a force on the edge of the cylinder, 1.5 ft from the axle?

can u help me from starting?
 
  • #14
OmCheeto
Gold Member
2,121
2,543
the power needed to accelerate a cylinder of weight 17000 lb from 0 to 3000 rpm in 90 seconds, exerting a force on the edge of the cylinder, 1.5 ft from the axle?

can u help me from starting?
Well, rather than jerk you around in some quantum physics, multi-pathed, "there are an infinite number of ways of solving this" kind of thing, I bowed out when Tiny jumped in, as he is much smarter than I, and it was obvious to me that the paths of our solutions to your problem were very different.

Anyways....

So how do you know that your answer is still wrong?
 
  • #15
diazona
Homework Helper
2,175
6
OK, let's recap what you've done so far:
the power needed to accelerate a cylinder of weight 17000 lb from 0 to 3000 rpm in 90 seconds, exerting a force on the edge of the cylinder, 1.5 ft from the axle?
You found the angular acceleration required,
Aceee = Change in angular velocity / time
Angular velocity = 2xPi (3.14) 3000/60 = 314rad / sec
acee = ( 314-0) / 90
= 3.48 Rad / sec2.
[tex]3000\text{ rpm}\times\frac{2\pi\text{ rad}}{\text{rev}}\times\frac{\text{min}}{60\text{ s}} = 314\frac{\text{rad}}{\text{s}}[/tex]
[tex]\alpha = 314\frac{\text{rad}}{\text{s}} / 90\text{ s} = 3.5\frac{\text{rad}}{\text{s}^2}[/tex]
(here we're using lb as a unit of mass)

Then you found the torque required,
If I use in Torque = Moment of inertia x Alpha (angular accee)
= 1/2(Radiusx radius 0x Weight X 3.48 Rad/sec^2
= 1/2x1.5x1.5x17000x3.48
= 66555 lb-ft^2 / sec^2
[tex]\tau = \frac{1}{2}MR^2\alpha = \frac{1}{2}(17000\text{ lb})(1.5\text{ ft})^2\left(3.49\frac{\text{rad}}{\text{s}^2}\right) = 67000\frac{\text{lb ft}^2}{\text{s}^2}[/tex]
(note that the radian is a "dummy unit" - it's really just equal to 1, so you can remove it)

Then you attempted to find the power by multiplying torque by angular speed, but you forgot about what OmCheeto pointed out:
I would imagine it is your "Power = torque x omega" that has thrown things off.
Omega changes from zero to 314 rad/sec over the 90 seconds.
"Power = torque x omega" may only be good for steady state conditions.
What you need to do is multiply the torque by the angular speed at each individual time and add up all those contributions to the torque - in other words, do an integral. In this particular case, though, since the torque is constant, you can just find the average angular speed and multiply that by the torque. But don't use the final angular speed!

becoz i find torqe = i x alpha
= lb-ft^2 X Rad /Sec^2
Torque unit is lb.ft
Now i am cofusing about units ?
Remember that we are using lb as a unit of mass! Torque has to have a unit of
[tex]\frac{(\text{mass})(\text{distance})^2}{(\text{time})^2}[/tex]
Notice that lb*ft doesn't work, since that's just mass*distance. You need lb*ft^2/s^2.

In some other places you may see people write torque in units of lb*ft (foot-pounds), but what they really mean is lbf*ft, foot-pounds-of-force. The lbf is a unit of force,
[tex]\text{lbf} = 32.2\frac{\text{lb}\cdot\text{ft}}{\text{s}^2}[/tex]

In order to calculate the power required in units of horsepower, you will need to multiply the torque,
[tex]67000\frac{\text{lb ft}^2}{\text{s}^2}[/tex]
by the correct angular speed. There are two ways to do this. Either you could leave the angular speed in revolutions per minute, then you will get an answer of
[tex]\text{(number)}\frac{\text{lb ft}^2}{\text{s}^2}\times\frac{\text{rev}}{\text{min}}[/tex]
You can then use the conversion factor
[tex]\text{hp} = 5252\text{lbf}\cdot\text{ft}\times\frac{\text{rev}}{\text{min}}[/tex]
You will also need to use the definition of the foot-pound-of-force (four equations up) as a conversion factor.

The alternative way is to convert the angular speed into radians per second. If you do that, you will get an answer of
[tex]\text{(number)}\frac{\text{lb ft}^2}{\text{s}^3}[/tex]
(remember that the radian is a "dummy unit"). From here, you can use the conversion factor
[tex]\text{hp} = 17696\frac{\text{lb ft}^2}{\text{s}^3}[/tex]
which you could find from Google, among other sources.
 
  • #16
10
0
cylinder weight: 17000 lb weight , Diameter = 36"= 3 ft, length = 5 ft. , 3000 rpm , time=90 sec.
the power needed to accelerate a cylinder of weight 17000 lb from 0 to 3000 rpm in 90 seconds, exerting a force on the edge of the cylinder, 1.5 ft from the axle?

See I done this calculation.

Aceee = Change in angular velocity / time

Angular velocity = 2xPi (3.14) 3000/60 = 314rad / sec

acee = ( 314-0) / 90


= 3.48 Rad / sec2.

Torque = Moment of inertia X Angular acceleration

MOment of inertia I = 1/2 m r^2

= 1 /2 (17000/32.16) 1.5x1.5(((( W= Massx g))))

= 594.68 lb.ft^2

Torque = 594.68 X 3.48 = 2069 lbft^2 /Sec^2.


Hp = torque X Rpm / 5250

= 2069 X 3000 / 5250 = 1182 hp . But i have confusion in torque unit.
Can you guys help me on that ? is that right solution ya not ?
 
Last edited:
  • #17
10
0
Thank yo very much diazona.


"What you need to do is multiply the torque by the angular speed at each individual time and add up all those contributions to the torque - in other words, do an integral. In this particular case, though, since the torque is constant, you can just find the average angular speed and multiply that by the torque. But don't use the final angular speed!"


Average angular speed = (314-0)/2 = 157 rad/sec.

So I have to Multiply torque by avg angular speed.

Power = Torque X 157

= 67000 x 157 = 10519000 lb-ft^2 / sec^3

1 hp = 17696 lb-ft2 / sec3 so 10519000/17696 = 594 HP right Diazona.

I have one more confusion In moment of inertia I = 1/2 mass x r^2 , in thr i have to take mass = 17000 lb or 17000/gravity ?
 
  • #18
10
0
Thank you very much diazona .

I am solving this thing for 5 days i can't even sleep properly . thanks a lot.

Thanks evrybody else also.
 
  • #19
diazona
Homework Helper
2,175
6
I have one more confusion In moment of inertia I = 1/2 mass x r^2 , in thr i have to take mass = 17000 lb or 17000/gravity ?
I've been using the pound as a unit of mass. So the mass is 17000 lb.
 

Related Threads on How much horsepower required for motor to rotate cylinder?

Replies
4
Views
1K
Replies
10
Views
11K
Replies
2
Views
1K
Replies
5
Views
789
  • Last Post
Replies
1
Views
1K
Replies
5
Views
494
Replies
5
Views
18K
Replies
3
Views
727
Replies
16
Views
14K
Replies
10
Views
3K
Top