Transcript document

Dynamics: POWER AND EFFICIENCY
Today’s Objectives:
1. Determine the power
generated by a machine,
engine, or motor.
2. Calculate the mechanical
efficiency of a machine.
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
(Section 14.4)
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
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
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 freebody 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.
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,
sB
SP
since the block moves with the pulley. From
SA
kinematics,
sP + 2 sA = l
 aP + 2 aA = 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 − (50kg)(9.81) = 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
CONSERVATIVE FORCES, POTENTIAL ENERGY
AND CONSERVATION OF ENERGY
Objectives:
1. Understand the concept of
conservative forces and
determine the potential
energy of such forces.
2. Apply the principle of
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
(Section 14.5)
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.
z
=
F
d
r
0
·
B
F

Thus, we say the work is conserved.
The work done by a conservative
force depends only on the positions
of the particle, and is independent of
its velocity or acceleration.
A
x
y
CONSERVATIVE FORCE (continued)
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 a linear 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.
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
Notice that the potential
function Ve always yields
positive energy.
CONSERVATION OF ENERGY
(Section 14.6)
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
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
CONCEPT QUIZ
Recall that the work of a spring is U1-2 = -½ k(s22 – s12) and
can be either positive or negative. The potential energy of a
spring is V = ½ ks2. Its value is __________
A) always negative.
C) always positive.
B) either positive or negative.
D) an imaginary number
Extra Credit: ATTENTION QUIZ
1. The power supplied by a machine will always be
_________ the power supplied to the machine.
A) less than
B) equal to
C) greater than
D) A or B
2. If the pendulum is released from the
horizontal position, the velocity of its
bob in the vertical position is _____
A) 3.8 m/s.
B) 6.9 m/s.
C) 14.7 m/s.
D) 21 m/s.