ACADEMIC YEAR 2013-2014
PURPOSE
This booklet contains the basic lecture notes that will be delivered between January 20th and February 24th 2014.
These notes are not comprehensive and will need to be supplemented by your own notes taken during the lectures.
You are additionally advised to read around the topics to further enhance your learning opportunity.
On completion of this section you will be assessed on your learning by means of written assignment which will give you an opportunity to provide evidence for the following learning outcomes and assessment criteria:
LO4 Be able to determine the dynamicparameters of rotating systems. / 4.2 determine the balancing masses required
to obtaindynamic equilibrium in a rotating
system
4.3 determine the energy storage requirements
of a flywheel
A following section will be issued at a later date and will cover the assessment criteria 4.1 and 4.4
The issued assignment will cover ALL four assessment criteria.
FLYWHEELS
A flywheel is an energy storage device used in rotating machinery. Its purpose is to smooth out the variations in speed of rotation that would otherwise occur if it was not there.
Let us consider a simple analogy; you are riding a bicycle along a flat horizontal road when you come to a series of inclines. The first incline is down and so you will begin to accelerate. If we assume that there is no further input from you then you will accelerate on every down incline and decelerate on each up incline. If the road surface is at the same altitude when you leave the series of inclines as when you entered it then your speed will be the same entering and leaving.
This analysis makes some assumptions
1) there is no friction between tyres and road.
2) there is no air resistance.
3) there is no friction in the wheel bearings.
4) the peaks of inclines between entering and leaving are at a lower altitude.
If the variation between peaks and troughs is large then your speed variation will be great. If however the inclines are only slight undulations, then you will maintain a fairly consistent speed.
So it is with flywheels: to maintain a reasonably constant speed of rotation the energy fluctuations must be small in comparison with the energy contained within the flywheel.
We will start by reminding ourselves of some of the basic definitions in linear and angular motion.
MOMENTUM
The property of a moving object that requires a force to change it.
You will recall that Newton determined that the rate of change of momentum of a body was directly proportional to the applied force, and in the same direction.
Linear momentum is the product of the mass of the body, m, and its linear speed, v.
Momentum = mv.
Angular momentum precludes us from using m and v as there is no translation of the body. We must use the rotational equivalent of m which is the Moment of inertia, I, and the angular speed, w.
Angular momentum = Iw
Comparison: Linear Angular
m kg I kg.m2
v m/s w rad/s
mv kg.m/s Iw kg.m2/s
KINETIC ENERGY
Energy possessed by a body as a result of its speed.
Linear kinetic energy mv2
Angular kinetic energy Iw2
Comparison Linear Angular
m kg I kg.m2
v m/s w rad/s
KE kg.m2/s2 KE kg.m2/s2
as kg.m/s2 =N the units of KE are Nm or joules in both cases.
CALCULATION OF MOMENT OF INERTIA
For a solid circular disc of radius r metres the Moment of Inertia is given by
I = mr2/2
So if a flywheel has a radius r, a thickness t and material density ρ
then it will have a mass given by
m=πr2tρ
giving a Moment of Inertia I=πr4tρ
Example
Determine the Moment of Inertia of a steel disc 50cm diameter and 2cm thick given that the density of steel is 7800kg/m3
Also calculate the angular momentum and angular KE when it is rotating at 2400rev/min.
Solution
r=25cm or 0.25m
t=2cm or 0.02m
ρ=7800kg/m3
Inserting the values gives I= π(0.25)4(0.02).7800
I=0.9572kg.m2
The speed 2400rev/min converted to rad/s is 2400.(2.π/60)
2400rev/min = 251.3274rad/s
Angular momentum = Iw = (0.9572)x(251.3274)
= 240.57kg.m2/s
Kinetic Energy = Iw2 = (0.9572)(251.3274)2
= 30231j
TURNING MOMENT DIAGRAMS
When attempting to determine the details of a flywheel for a particular use it is essential to have information regarding the cyclic variations in the torque from the driving member and also the resisting torque of the driven member.
The diagram which shows the torque variations against angular rotation is known as the TURNING MOMENT DIAGRAM.
Two turning moment diagrams are shown below.
MEAN TORQUE
The mean torque acting can be found from the turning moment diagram.
Plot the diagram to some suitable scale with Torque on the vertical axis and rotation on the horizontal axis over one cycle of operation.
Determine the areas above () and below () the horizontal axis.
Determine the net area of the diagram and divide this by the baseline length.
Example
If we look a little more carefully at the diagram of the single cylinder four stroke engine we can make the following additions.
The Mean Torque is found by is found by determining the area of the diagram over each stroke, i.e. 1800 or π rad.
The power stroke puts energy into the system, whereas the other three require energy from the system. The net energy requirement per cycle is area A minus area B and the mean torque will be found by dividing this net energy by the baseline length, in this case 4π.
In this engine the flywheel will speed up to the point where the torque curve crosses the mean torque curve and thereafter it will slow down.
In practical applications we do not want the flywheel to have too large a speed variation as this implies a large variation in engine running speed.
COEFFICIENT OF FLUCTUATION OF SPEED, KS
This term indicates the variation of the speed of the flywheel from a specified value. Suppose we wish the engine to vary less than15 rev/min either way when running at 3000 rev/min.
KS = maximum speed – minimum speed
mean speed
so in this case KS = 3015 -2985 = 30 =0.01
3000 3000
COEFFICIENT OF FLUCTUATION OF ENERGY, KE
This relates to the way in which the energy demands over the cycle vary, which is again related to the speed variation.
The definition is that KE = the greatest fluctuation of energy/cycle
work done/cycle
KE = I(w1)2 - I(w2)2
W
Let us look at an example.
Example
The turning moment diagram for an engine was drawn to a base of crank angle. The mean torque line was added and the areas above and below the line measured as +5.5; -1.5; +1.7 and -5.7 cm2
Vertically 1cm represents 1000Nm, and horizontally 1cm represents 150
Determine the dimensions of a steel flywheel of radius 0.7071m which will keep the speed between 295 and 305 rev/min.
Steel has a density of 7800kg/m3
Solution
As 1cm represents 1000Nm vertically and 150 horizontally we must firstly convert the degrees to radians.
150 = 15x2π/360 rads =0.2618 rad
therefore 1cm2 of turning moment diagram represents 261.8j of energy.
Multiplying the given areas above and below the mean torque line (which must sum to zero) by this figure gives the following situation.
At A surplus energy = 0
At B surplus energy = 1440j
At C surplus energy = 1440j – 393j =1047j
At D surplus energy = 1047j + 445j = 1492j
At E surplus energy = 1492j – 1492j = 0
So the speed will be a maximum at D and a minimum at A and E
(which are the same point).
The maximum fluctuation of energy in the cycle is 1492j
We know that the maximum speed is to be 305 rev/min and the minimum speed is to be 295 rev/min. Convert these to rad/s.
305 rev/min = 305x2π/60 rad/s = 31.94 rad/s,
295 rev/min = 292x2π/60 rad/s = 30.89 rad/s
The maximum fluctuation of energy is I(31.942-30.892)
Substituting values, 32.986I = 1492
I = 45.23kgm2
As the radius is specified as 0.7071m, we have that
πr4tρ = 45.23
2
From which t = 90.46
π7800(0.25)
t = 14.77mm
and the mass will be 180.92kg.
BALANCING
We are all familiar with rotating machinery of some kind e.g. alternators, turbines, petrol engines etc. What is important is that when rotation occurs the rotating parts are balanced, or excessive vibration leading to eventual failure will occur.
We will examine the conditions surrounding rotational components and evolve the conditions for balance to occur.
SINGLE PLANE BALANCING
Consider four masses m1, m2, m3 and m4 at radii of r1, r2, r3 and r4 from the axis of rotation. In this case all the masses are in the same plane.
See the diagram below.
For static balance we need that the moments of all the masses about the axis sums to zero.
Considering m1, the vertical component of that mass is m1gr1cosa1
If the angle is taken as 0 due east, and the convention is anticlockwise, then we can see that for static equilibrium
∑mgrcosa = 0
As g is common to all it follows that a vector diagram of mr must close for equilibrium.
Example
The following masses are placed at the specified radii from the axis of rotation. The angles specified are measured from due east anticlockwise.
m / r / a0.3 / 0.4 / 25
0.8 / 0.3 / 103
0.5 / 0.5 / 233
1.2 / 318
Determine the radius required for static balance by both a graphical and an analytical method.
Solution
1. Analytical
We know that ∑mrcosa must be equal to 0
For each line calculate mrcosa
Line 1 0.10876
Line 2 -0.05399
Line 3 -0.15045
Line 4 ?
By adding the values we have, we can determine the missing value.
Total = -0.09568
For equilibrium Line 4 = 0.09568
As we know mcosa = 0.89177, then r =0.10729.
2. Drawing
The mr diagram is drawn to scale at the specified angles.
As the diagram must close the length mr can be found.
Interrogation shows r to be 0.1066
For dynamic balance we are considering the effects of rotation. If the masses are rotating about the axis with an angular speed of w, then each mass will experience a centripetal force of mrw2 acting inwards. The axis experiences an equal and opposite centrifugal force acting outwards.
For equilibrium the vector sum of these forces must be zero i.e.
∑mrw2cosa = 0
As w is common to each mass, then the condition that
∑mrcosa = 0 must apply.
This is as before, so if the system is statically balanced it will also be dynamically balanced, FOR SINGLE PLANE CONDITIONS.
EXERCISE
m / r / a2 / 0.6 / 56
4 / 0.3 / 155
1 / 0.5 / 222
3
For this system determine the radius and angle to maintain equilibrium by drawing.
Using the angle that you have found by drawing confirm the radius analytically.
BALANCING IN DIFFERENT PLANES
The problem here is that we need to balance the rotating masses so that equilibrium exists both for the masses and the couples as these masses are in different planes.
The method is best explained by means of an example. Consider the line diagram below.
M1, M2, M3 and M4 are all at different angular positions as seen from an end view.
The requirement is that the vector diagram of the couples caused by this loading closes.
In effect this means that the SMrw2Lcosϕ = 0
We need to select a reference plane, any suitable one will do.
If the plane containing M1 is selected then the couple due to M1 will be zero. The couple due to M2 will be M2r2(L2-L1)ω2, and the couple due to M3 will be M3r3(L3-L1)ω2 and so on. These must add vectorially. Any out of balance couple will cause a cyclical force on the bearings which must be eliminated.
NOTE: static balance of a non-coplanar system does not ensure dynamic balance.
EXAMPLE
Consider the system shown below.
We will determine the dynamic reaction on each bearing at 1200rev/min.
We will ignore the effect of speed to begin with as we can bring that in at the end.
M, kg / R, mm / MR, kgm / L, m / MRL, kgm240 / 20 / 0.8 / 0.1 / 0.08
60 / 10 / 0.6 / 0.35 / 0.21
We can now draw the MRL diagram to scale and find the out of balance MRL.
MRL diagram.
The out of balance couple is 0.1836 kgm2 at an angle of 352.1730.
To get the actual value we must multiply by ω2, which is (2π1200/60)2. The actual value is 2890Nm.
The reaction on the right hand bearing will be the couple/distance from the reference plane i.e. 2890/0.5 = 5780N.
To determine the load on the left hand bearing we now draw the MR vector
The value of 0.3672 is obtained by dividing the out of balance couple by the length, in this case 0.1836/0.5m
The out of balance MR of 0.57166 is converted into a dynamic reaction of 9001N by multiplying by ω2.
Static Loads on bearings
Treating the system as a simply supported beam with two point loads gives RR and RL of 490.5N
Maximum and minimum loads on each bearing are the dynamic load plus or minus the static load.
RR=6270.5 to 5289.5 and RL= 9491.5 to 8510.5.
EXERCISE
Determine the dynamic loads on the bearings for the case below.
Assume a speed of 1800 rev/min/.
BALANCING CONCLUDED
We have examined the effect upon the bearings of having an out of balance system. The dynamic loading coupled with the static loading can cause severe damage to the bearings.