Download ert146 lect on power and effeciency

POWER AND EFFICIENCY
APPLICATIONS
Engines and motors are often
rated in terms of their power
output. The power output of the
motor lifting this elevator is
related to the vertical force F
acting on the elevator, causing it
to move upwards.
Given a desired lift velocity for the
elevator (with a known maximum
load), how can we determine the
power requirement of the motor?
APPLICATIONS (continued)
The speed at which a truck
can climb a hill depends in
part on the power output of
the engine and the angle of
inclination of the hill.
For a given angle, how can we determine the speed of this
truck, knowing the power transmitted by the engine to the
wheels? Can we find the speed, if we know the power?
If we know the engine power output and speed of the truck, can
we determine the maximum angle of climb of this truck ?
POWER AND EFFICIENCY
Power is defined as the amount of work performed per unit
of time.
If a machine or engine performs a certain amount of work,
dU, within a given time interval, dt, the power generated can
be calculated as
P = dU/dt
Since the work can be expressed as dU = F • dr, the power
can be written (from yesterday’s lecture)
P = dU/dt = (F • dr)/dt = F • (dr/dt) = F • v
Thus, power is a scalar defined as the product of the force
and velocity components acting in the same direction.
POWER
Using scalar notation, power can be written
P = F • v = F v cos q Work = F * s cosq
where q is the angle between the force and velocity vectors.
So if the velocity of a body acted on by a force F is known,
the power can be determined by calculating the dot product
or by multiplying force and velocity components.
The unit of power in the SI system is the Watt (W) where
1 W = 1 J/s = 1 (N · m)/s .
In the FPS system, power is usually expressed in units of
horsepower (hp) where
1 hp = 550 (ft · lb)/s = 746 W .
EFFICIENCY
The mechanical efficiency of a machine is the ratio of the
useful power produced (output power) to the power supplied
to the machine (input power) or
e = (power output) / (power input)
If energy input and removal occur at the same time, efficiency
may also be expressed in terms of the ratio of output energy
to input energy or
e = (energy output) / (energy input)
Machines will always have frictional forces. Since frictional
forces dissipate energy, additional power will be required to
overcome these forces. Consequently, the efficiency of a
machine is always less than 1.
PROCEDURE FOR ANALYSIS
• Find the resultant external force acting on the body causing
its motion. It may be necessary to draw a free-body diagram.
• Determine the velocity of the point on the body at which the
force is applied. Energy methods or the equation of motion
and appropriate kinematic relations, may be necessary.
• Multiply the force magnitude by the component of velocity
acting in the direction of F to determine the power supplied
to the body (P = F v cos q ).
• In some cases, power may be found by calculating the work
done per unit of time (P = dU/dt).
• If the mechanical efficiency of a machine is known, either
the power input or output can be determined.
Force in the direction of displacement
EXAMPLE
Given: A 50 kg block (A) is hoisted by the pulley
system and motor M. The motor has an
efficiency of 0.8. At this instant, point P
on the cable has a velocity of 12 m/s
which is increasing at a rate of 6 m/s2.
Neglect the mass of the pulleys and
cable.
Find: The power supplied to the motor at this
instant.
Plan:
1) Relate the cable and block velocities by defining position
coordinates. Draw a FBD of the block.
2) Use the equation of motion to determine the cable tension.
3) Calculate the power supplied by the motor and then to the
motor.
EXAMPLE (continued)
Solution:
1) Define position coordinates to relate velocities.
Datum
sm
Here sP is defined to a point on the cable. Also
sA is defined only to the lower pulley, since the
sB
SP
block moves with the pulley. From kinematics,
SA
sP + 2 s A = l
 aP + 2 a A = 0
 aA = − aP / 2 = −3 m/s2 (↑)
Draw the FBD and kinetic diagram of the block:
2T
mA aA
=
A
WA
A
EXAMPLE
(continued)
2) The tension of the cable can be obtained by applying the
equation of motion to the block.
+↑ Fy = mA aA
2T − 490.5 = 50 (3)  T = 320.3 N
3) The power supplied by the motor is the product of the force
applied to the cable and the velocity of the cable.
Po = F • v = (320.3)(12) = 3844 W
The power supplied to the motor is determined using the
motor’s efficiency and the basic efficiency equation.
Pi = Po/e = 3844/0.8 = 4804 W = 4.8 kW
TUTORIAL ON POWER AND EFFICIENCY
(Don’t submit)
F14-7 (pg 196); F14-12 (pg 196); 14-61 pg 199)
OWN READING ON CENTROID AND CENTRE OF
MASS
CONSERVATIVE FORCES, POTENTIAL
ENERGY
AND CONSERVATION OF ENERGY
APPLICATIONS
The weight of the sacks resting on
this platform causes potential energy
to be stored in the supporting springs.
As each sack is removed, the platform
will rise slightly since some of the
potential energy within the springs
will be transformed into an increase
in gravitational potential energy of the
remaining sacks.
If the sacks weigh 100 lb and the equivalent spring constant
is k = 500 lb/ft, what is the energy stored in the springs?
APPLICATIONS (continued)
The boy pulls the water balloon launcher back, stretching each
of the four elastic cords.
If we know the unstretched length and stiffness of each cord,
can we estimate the maximum height and the maximum range
of the water balloon when it is released from the current
position ?
APPLICATIONS (continued)
The roller coaster is released from rest at the top of the hill. As
the coaster moves down the hill, potential energy is
transformed into kinetic energy.
What is the velocity of the coaster when it is at B and C?
Also, how can we determine the minimum height of the hill
so that the car travels around both inside loops without
leaving the track?
CONSERVATIVE FORCE
A force F is said to be conservative if the work done is independent of the
path followed by the force acting on a particle as it moves from A to B. This
also means that the work done by the force F in a closed path (i.e., from A to
B and then back to A) is zero.
Vertical displacement – along Z direction
 F · dr = 0
z
F
B
Thus, we say the work is conserved.
A
x
y
CONSERVATIVE FORCE
The work done by a conservative force depends only
on the positions of the particle, and is independent of
its velocity or acceleration; eg are the weight
(depends only on vertical displacement of weight or
spring (elongation or compression).
Friction force is non-conservative, depends on path –
the longer displacement, the greater the work. Work
is dissipated from the body in the form of heat
CONSERVATIVE FORCE
A more rigorous definition of a conservative force makes
use of a potential function (V) and partial differential
calculus, as explained in the text. However, even without
the use of the these mathematical relationships, much can be
understood and accomplished.
The “conservative” potential energy of a particle/system is
typically written using the potential function V. There are two
major components to V commonly encountered in mechanical
systems, the potential energy from gravity and the potential
energy from springs or other elastic elements.
Vtotal = Vgravity + Vsprings
POTENTIAL ENERGY
Potential energy is a measure of the amount of work a
conservative force will do when a body changes position.
In general, for any conservative force system, we can define
the potential function (V) as a function of position. The work
done by conservative forces as the particle moves equals the
change in the value of the potential function (e.g., the sum of
Vgravity and Vsprings).
It is important to become familiar with the two types of
potential energy and how to calculate their magnitudes.
POTENTIAL ENERGY DUE TO GRAVITY
The potential function (formula) for a gravitational force, e.g.,
weight (W = mg), is the force multiplied by its elevation from a
datum. The datum can be defined at any convenient location.
Vg = ± W y
Vg is positive if y is above the
datum and negative if y is
below the datum. Remember,
YOU get to set the datum.
ELASTIC POTENTIAL ENERGY
Recall that the force of an elastic spring is F = ks. It is
important to realize that the potential energy of a spring, while
it looks similar, is a different formula.
Elongated position
Ve (where ‘e’ denotes an
elastic spring) has the distance
“s” raised to a power (the
result of an integration) or
1 2
=
Ve
ks
2
Compressed position
Notice that the potential
function Ve always yields
positive energy.
CONSERVATION OF ENERGY
When a particle is acted upon by a system of conservative forces, the
work done by these forces is conserved and the sum of kinetic energy
and potential energy remains constant. In other words, as the particle
moves, kinetic energy is converted to potential energy and vice versa.
This principle is called the principle of conservation of MECHANICAL
energy or SIMPLY principle of conservation of energy and is expressed
as
T1 + V1 = T2 + V2 = Constant
T1 stands for the kinetic energy at state 1 and V1 is the
potential energy function for state 1. T2 and V2
represent these energy states at state 2. Recall, the
kinetic energy is defined as T = ½ mv2.
EXAMPLE
Given: The 2 kg collar is moving down
with the velocity of 4 m/s at A.
The spring constant is 30 N/m. The
unstretched length of the spring is
1 m.
Find:
The velocity of the collar when
s = 1 m.
Plan:
Apply the conservation of energy equation between A and
C. Set the gravitational potential energy datum at point A
or point C (in this example, choose point A—why?).
Solution:
EXAMPLE
(continued)
Note that the potential energy at C has two parts.
VC = (VC)e + (VC)g
VC = 0.5 (30) (√5 – 1)2 – 2 (9.81) 1
The kinetic energy at C is
TC = 0.5 (2) v2
Similarly, the potential and kinetic energies at A will be
VA = 0.5 (30) (2 – 1)2, TA = 0.5 (2) 42
The energy conservation equation becomes TA + VA = TC + VC .
[ 0.5(30) (√5 – 1)2 – 2(9.81)1 ] + 0.5 (2) v2
= [0.5 (30) (2 – 1)2 ]+ 0.5 (2) 42
 v = 5.26 m/s
PROBLEM SOLVING
Given: The 800 kg roller
coaster starts from
A with a speed of
3 m/s.
Find: The minimum height, h, of the hill so that the car
travels around inside loop at B without leaving the track. Also
find the normal reaction on the car when the car is at C for this
height of A.
Plan: Note that only kinetic energy and potential energy due
to gravity are involved. Determine the velocity at B using the
equation of equilibrium and then apply the conservation of
energy equation to find minimum height h .
PROBLEM SOLVING (continued)
Solution:
1) Placing the datum at A:
TA + VA = TB + VB
 0.5 (800) 32 + 0
= 0.5 (800) (vB)2 − 800(9.81) (h − 20)
(1)
2) Find the required velocity of the coaster at B so it doesn’t
leave the track.
Equation of motion applied at B:
NB  0
2
v
 Fn = man = m r
(vB)2
=
800 (9.81) = 800
10
man
mg
 vB = 9.905 m/s
PROBLEM SOLVING (continued)
Now using the energy conservation, eq. (1), the minimum h
can be determined.
0.5 (800) 32 + 0 = 0.5 (800) (9.905)2 − 800(9.81) (h − 20)
 h= 24.5 m
3) To find the normal reaction at C, we need vc.
TA + VA = TC + VC
 0.5 (800) 32 + 0 = 0.5 (800) (vC)2 − 800(9.81) (24.5 − 14)
 VC = 14.66 m/s
Equation of motion applied at B:
v2
14.662
 Fn = m r  NC+800 (9.81) = 800
7
 NC = 16.8 kN
NC
=
mg
man