physics 231 introductory physics i lecture 20. heat engine refrigerator, heat pump last lecture q...
TRANSCRIPT
Heat Engine Refrigerator,
Heat Pump
Last Lecture
Qhot
engine
Qcold
W
Qhot
fridge
Qcold
W
€
e =W
Qhot< eCarnot =1−
TcoldThot
€
W =Qhot −Qcold€
COP =Qc,hW
< COPCarnot =Tc,h
Thot −Tcold
Entropy• Measure of Disorder of the system
(randomness, ignorance)
• Entropy: S = kBlog(N) N = # of possible arrangements for fixed E and Q
0
100
200
300
400
500
600
700
800
900
1000
(0,12) (1,11) (2,10) (3,9) (4,8) (5,7) (6,6) (7,5) (8,4) (9,3) (10,2) (11,1) (12,0)
Number of ways for 12 molecules to arrange themselves in two halves of container.
S is greater if molecules spread evenly in both halves.
On a macroscopic level, one finds that adding heat raises entropy:
Temperature in Kelvin!
ΔS =Q /T
2nd Law of Thermodynamics(version 2)
The Total Entropy of the Universe can never decrease.
(but entropy of system can increase or decrease)
Why does Q flow from hot to cold?
• Consider two systems, one with TA and one with TB
• Allow Q > 0 to flow from TA to TB
• Entropy changes by:
ΔS = Q/TB - Q/TA
• This can only occur if ΔS > 0, requiring TA > TB.
• System will achieve more randomness by exchanging heat until TB = TA
Carnot Engine
Carnot cycle is most efficient possible, because the total entropy change is zero. It is a “reversible process”.
For real engines:
€
ΔS = ΔSenvironment =QcoldTcold
−QhotThot
> 0 ⇒
€
e =W
Qhot=1−
QcoldQhot
<1−TcoldThot
= eCarnot
• When x is positive ,F is negative ;
• When at equilibrium (x=0), F = 0 ;
• When x is negative ,F is positive ;
Hooke’s Law Reviewed
F =−kx
QuickTime™ and aAnimation decompressor
are needed to see this picture.
Amplitude: A
Period: T
Frequency: f = 1/T
Angular frequency:
A
T
Period and Frequency
€
x = Acosωt
€
T = 2π⇓
€
T =2π
ω, f =
ω
2π
Phases
Often a phase is included to shift the timing of the peak:
Phase of 90-degrees changes cosine to sine
cos t−π2
⎛⎝⎜
⎞⎠⎟=sin t( )
€
x = Acos ωt −φ( )
= Acos ω(t − t0)( ) for peak at
€
t = t0
a
x
v
• Velocity is 90 “out of phase” with x: When x is at max,v is at min ....
• Acceleration is 180° “out of phase” with x a = F/m = - (k/m) x
Velocity and Acceleration vs.
time
T
T
T
v and a vs. t
Find vmax with E conservation
Find amax using F=ma
x =Acostv=−vmaxsinta=−amax cost
1
2kA2 =
12
mvmax2
vmax =Akm
−kx = ma−kAcosωt = −mamax cosωt
amax = Ak
m
Connection to Circular Motion
QuickTime™ and aAnimation decompressor
are needed to see this picture.
circular motion with constant angular velocity
Simple Harmonic Motion
Projection on axis
What is ?
=k
mQuickTime™ and aAnimation decompressor
are needed to see this picture.
Circular motion
Angular speed: Radius: A
=> Speed: v=A
Simple Harmonic Motion
Cons. of E:
€
vmax = Ak
m
€
=k
m
Example13.1
An block-spring system oscillates with an amplitude of 3.5 cm. If the spring constant is 250 N/m and the block has a mass of 0.50 kg, determine
(a) the mechanical energy of the system
(b) the maximum speed of the block
(c) the maximum acceleration.
a) 0.153 J
b) 0.783 m/s
c) 17.5 m/s2
Example 13.2
A 36-kg block is attached to a spring of constant k=600 N/m. The block is pulled 3.5 cm away from its equilibrium positions and released from rest at t=0. At t=0.75 seconds,
a) what is the position of the block?
b) what is the velocity of the block?
a) -3.489 cm
b) -1.138 cm/s
Example 13.3
A 36-kg block is attached to a spring of constant k=600 N/m. The block is pulled 3.5 cm away from its equilibrium position and is pushed so that is has an initial velocity of 5.0 cm/s at t=0.
a) What is the position of the block at t=0.75 seconds?
a) -3.39 cm
Example 13.4a
An object undergoing simple harmonic motion follows the expression,
x(t)=4 + 2cos[π (t−3)]
The amplitude of the motion is:a) 1 cmb) 2 cmc) 3 cmd) 4 cme) -4 cm
Where x will be in cm if t is in seconds
Example 13.4b
An object undergoing simple harmonic motion follows the expression,
x(t)=4 + 2cos[π (t−3)]
The period of the motion is:a) 1/3 sb) 1/2 sc) 1 sd) 2 se) 2/π s
Here, x will be in cm if t is in seconds
Example 13.4c
An object undergoing simple harmonic motion follows the expression,
x(t)=4 + 2cos[π (t−3)]
The frequency of the motion is:a) 1/3 Hzb) 1/2 Hzc) 1 Hzd) 2 Hze) π Hz
Here, x will be in cm if t is in seconds
Example 13.4d
An object undergoing simple harmonic motion follows the expression,
x(t)=4 + 2cos[π (t−3)]
The angular frequency of the motion is:a) 1/3 rad/sb) 1/2 rad/sc) 1 rad/sd) 2 rad/se) π rad/s
Here, x will be in cm if t is in seconds
Example 13.4e
An object undergoing simple harmonic motion follows the expression,
x(t)=4 + 2cos[π (t−3)]
The object will pass through the equilibrium positionat the times, t = _____ seconds
a) …, -2, -1, 0, 1, 2 … b) …, -1.5, -0.5, 0.5, 1.5, 2.5, …c) …, -1.5, -1, -0.5, 0, 0.5, 1.0, 1.5, …d) …, -4, -2, 0, 2, 4, …e) …, -2.5, -0.5, 1.5, 3.5,
Here, x will be in cm if t is in seconds