Calculate Rotational Speed of Steel Rotor Disk

[OutCell]

A steel rotor disk which is part of a turbine assembly has a uniform thickness of 50mm, an outside diameter of 650mm and an inside diameter of 100mm. If there are 250 blades, each of mass 0.16 kg pitched evenly around the periphery of the disk at an effective radius of 350mm, determine the rotational speed at which the hoop stress on the inner surface reaches the yield stress. You may assume that the blades produce a uniform loading around the periphery of the disk.

For steel, poisson's ratio = 0.29; Density = 7470 kg/m^3 and the yeild stress in simple tension = 355 MN/m^2.



I don't know what equations to use? Is rotational speed = w or T? I don't know where to start with finding a way to solve this question?

Can anybody please guide me through the steps in which i can get this speed? The problem is that this question is one of optional question that we didn't get any lecture notes on, so i don't know where to start :grumpy:

Thanks

 

[OutCell]

After some reading i got 2 ways which i think one might be the answer (Or maybe not lol).. Both different answers..

1st way:

I used the equation:
(Sigma)r = (3+possion's ratio/8) x Density x w^2 x r2^2

and after substituting the variable in the equation i got an answer
w= 6.67 x10^-3 rad/sec

2nd way:
I read this in some notes about disc with a loaded boundary-

"In general, rotor discs will have mounted on the outer boundary a large number of blades. These will themselves each have a centrifugal force component which will have to be reacted at the periphery of the disc. Given the mass of each blade, it's effective centre of mass and the number of blades we can compute the force due to each blade at a particular value of w. Multiplying by the number of blades gives the total force which may then be computed as a uniformly distributed load. Dividing this by the thickness of the outer boundary gives the required value of (Sigma)r to use as the boundary condition when evaluating A and B"

Now i went backwards and considered the yeild stress (355x10^6) as (Sigma)r, so-
I multiplied the thickness by (Sigma)r
50 x (355x10^6) = 1.775 x 10^10 which would hopefully be the uniformly distributed load

And then i divided this by the number of blades to get the w
(1.775x10^10)/250 = 71 x 10^6 rad/sec


Can someone please take a look and tell me if any of these are right, or help me out in the right direction

THANKS :shy:

 

[piercas]

I would first approach this problem by identifying the relevant variables and equations that can be used to solve for the rotational speed of the steel rotor disk. In this case, we have the following variables: thickness (t), outside diameter (D), inside diameter (d), number of blades (n), blade mass (m), effective radius (r), Poisson's ratio (ν), density (ρ), yield stress (σ), and rotational speed (ω).

The relevant equations that can be used are:

1. Hoop stress (σ_h) = (m*r*ω^2)/t

2. Centrifugal force (F) = m*r*ω^2

3. Yield stress (σ) = (F*n)/(π*D*t)

Using these equations, we can set up the following system of equations:

σ_h = (m*r*ω^2)/t

F = m*r*ω^2

σ = (F*n)/(π*D*t)

We can then substitute the given values into these equations:

t = 50mm

D = 650mm

d = 100mm

n = 250

m = 0.16kg

r = 350mm

ν = 0.29

ρ = 7470 kg/m^3

σ = 355 MN/m^2

Solving for ω, we get:

σ_h = (0.16*0.35*ω^2)/0.05

F = 0.16*0.35*ω^2

σ = (0.16*0.35*ω^2*250)/(π*0.65*0.05)

355 MN/m^2 = (0.16*0.35*ω^2*250)/(π*0.65*0.05)

ω = √[(355*0.65*0.05*π)/(0.16*0.35*250)]

ω = 53.9 rad/s

Therefore, the rotational speed at which the hoop stress on the inner surface reaches the yield stress is approximately 53.9 rad/s.

I hope this helps guide you through the steps of solving this problem. It is important to carefully identify the relevant variables and equations, and to substitute the given values correctly. Don't hesitate to ask for clarification or further assistance if

 

[piercas]

for your inquiry. The rotational speed of the steel rotor disk can be calculated using the following equation:

w = (T/I) * (L/r)

Where:

w = rotational speed (in radians per second)
T = torque (in Nm)
I = moment of inertia (in kgm^2)
L = angular momentum (in kgm^2/s)
r = effective radius (in meters)

To solve for the rotational speed, we need to first calculate the torque and moment of inertia of the rotor disk. The torque can be calculated as the product of the force (F) exerted by the blades and the radius (r) at which the force is applied:

T = F * r

Since the blades are pitched evenly around the periphery of the disk, the force exerted by each blade can be calculated as the product of its mass (m) and the centripetal acceleration (a):

F = m * a

The centripetal acceleration can be calculated as:

a = (v^2)/r

Where v is the linear velocity of the blade at a distance r from the center of the disk. This linear velocity can be calculated using the rotational speed (w) and the radius (r):

v = w * r

Therefore, the torque can be expressed as:

T = m * (w^2) * r

The moment of inertia (I) can be calculated using the following formula:

I = (1/2) * m * r^2

Where m is the mass of the rotor disk and r is the radius of the disk. The mass of the rotor disk can be calculated as the sum of the mass of the blades (250 * 0.16 kg) and the mass of the disk itself:

m = (250 * 0.16) + (density * volume)

The volume of the disk can be calculated using the formula for the volume of a cylinder:

V = π * (R^2 - r^2) * h

Where R is the outside radius, r is the inside radius, and h is the thickness of the disk.

Now that we have the torque and moment of inertia, we can substitute them into the equation for rotational speed:

w = (T/I) * (L/r)

To find the angular momentum (L), we can use the equation:

L = I * w

Finally, to determine the rotational speed at which the hoop stress on the inner surface