Download Energy Conservation (Ch 18)

Energy Conservation (Ch 18)
All types of energy are measured in Joules
Potential Energy
PE = mgh
m = mass (kg)
g = gravity accel (m/s2)
h = height (m)
Spring Energy
SE = 0.5kx2
k = Spring constant (N/m)
x = displacement or deformation (m)
Kinetic Energy (Linear)
KE = 0.5mv2
m = mass (kg)
v = velocity (m/s)
Kinetic Energy (Rotational)
KE = 0.5I2
I = Mass moment of Inertia (kgm2)
 = velocity (rad/s)
Work (Linear)
W=FS
F = Force (N)
S = displacement (m)
Work (Rotation)
W = T
T = Torque (Nm)
 = Angular displacement (rad)
Power = Work / time
(Watts) = (Joules)/(second)
P = F v (Linear)
F = Force (N)
v = velocity (m/s)
P = T (Rotary)
T = Torque (Nm)
 = angular velocity (rad/s)
Conservation of Energy.
(Energy at point 1) = (Energy at point 2)
PE1 + KE1 + SE1 +/- W = PE2 + KE2 + SE2
+W = Positive work = Force moves the object (motor). Force in same
direction as motion.
-W = Negative work = Brakes (the motor is working as a generator).
Friction is always negative. Force opposes motion.
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Rail Car Spring
Thursday, 10 May 2012
11:12 AM
PE1 + KE1 + SE1 = PE2 + KE2 + SE2
KE1 = SE2
0.5mv^2 = 0.5kx^2
KE = 0.5mv^2 = 0.5*9400*(12/3.6)^2 = 52222 J
In engineering we usually use kJ and kW.
Find k in SE = 0.5kx^2
But KE1 = SE2
k = SE/0.5x^2 = 52222 / (0.5*0.37^2) = 762922 N/m
This is the spring modulus (k) which is normally given in
N/mm, but you must calculate in N/m in formulas.
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Charpy Energy
Tuesday, 8 May 2012
6:44 PM
Job for CAD...
1341.4078 mm
PE = mgh
= 33*9.81*1.3414
= 434.2514 J
Velocity of hammer before impact?
v = (2gh)^0.5 = (2*9.81*1.3414)^0.5
= 5.1301 m/s
Velocity of hammer after impact?
PE after: PE = mgh
= 33*9.81*0.32253 = 104.41264 J
KE after = 104.41264
v = (2KE/m)^0.5
= (2*104.41264/33)^0.5
= 2.51556 m/s
Check: v = (2gh)^0.5 = (2*9.81*0.32253)^0.5
= 2.51556 m/s
Finding height by
CAD.
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Charpy apparatus
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Weight falls on a spring.
Tuesday, 17 May 2011
6:26 PM
Example 18.5. Mass 15kg, falls 350mm to spring.
Modulus 10N/mm. Find max compression.
1. At rest and height = 350mm.
2. At speed and touching spring.
3. At rest and spring compressed.
PE1 + KE1 + SE1 +/- W = PE2 + KE2 + SE2
PE1 = KE2
PE1 = KE2
mg(h) = 0.5mv2
15*9.81*0.35 = 0.5 *15*v2
v2 = 51.5025 /(0.5*15) = 6.867
v = 6.867 ^0.5 = 2.6205 m/s (Check (2*9.81*0.35)^0.5 = 2.6205m/s)
PE2 + KE2 + SE2 +/- W = PE3 + KE3 + SE3
PE2 + KE2 = SE3
mgx + 0.5mv2 = 0.5kx2
Quadratic equation...
0.5kx2 - mgx - 0.5mv2 = 0
0.5*10*1000x2 - 15*9.81x - 0.5*15*2.62052 = 0
5000 x2 - 147.15 x - 51.5027 = 0
a = 5000, b = -147.15, c = - 51.5027
Quadratic equation solution…
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Quadratic equation solution…
Read the book. Page 258
So answers are x = 117mm or x = -87mm (????)
This quadratic formula assumes the mass and spring are stuck
together. Imagine dropping the mass, then it latches onto the spring
for good.
-87mm is how high the stuck mass will pull the spring ABOVE the
initial start position of the spring.
If it doesn't latch, the mass will bounce back to the start (350mm).
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Spring and Mass
Tuesday, 31 March 2015
6:48 PM
Q10: (cont) Spring modulus 1.1N/mm and a mass
2.9kg hooked on and then released suddenly. (b) What
is the maximum stretch of the spring?
PE1 + KE1 + SE1 +/- W = PE2 + KE2 + SE2
PE1 = SE2
mgh = 0.5kx2
mgx = 0.5kx2
mg = 0.5kx
x= 2mg/k = 2*2.9*9.81/(1.1*1000) = 0.0517 m
= 0.0517 * 1000 = 51.7 mm
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Rollercoaster
Tuesday, 17 May 2011
5:33 PM
Example 18.2: Roller Coaster.
Mass = 500kg, starting height A=18m, lowest point B=0m, up to C=13m.
From A to B
PEA + KEA = PEB + KEB
From B to C
PEB + KEB = PEC + KEC
Straight from A to C
PEA + KEA = PEC + KEC
PEA = PEC + KEC
mgha = mghc + 0.5*m*vc2
500*9.81*18 = 500*9.81*13 + 0.5*500*v2
88290 = 63765 + 250*v2
250*v2 = 88290 - 63765 = 24525 J
v2 = 24525/250 = 98.1
v = 98.1 ^ 0.5 = 9.9045 m/s (9.9045*3.6 = 35.6562 km/h)
Check with h=5... v = (2*9.81*5)^0.5 = 9.9045 m/s
PE = KE is a common problem:
mgh = 0.5mv2
mgh = 0.5mv2
v = 2gh
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Rail-Car Buffer
Tuesday, 8 May 2012
8:16 PM
PE1 + KE1 + SE1 +/- W = PE2 + KE2 + SE2
PE1= SE2
mgh = 0.5*k*x^2
x = (2mgh/k)^0.5
= (2*9400*9.81*2.6/640000)^0.5 = 0.8656 m
Note: We could not do this using linear motion & F=ma
because the spring does not give a constant force (therefore not
constant accel)
e.g. velocity before impact..(2*9.81*2.6)^0.5 = 7.1423 m/s
Assume this stops in 0.8656m, by linear motion...
v2-vo2 = 2as, so a = (0-7.1423^2)/(2*0.8656) = -29.4665 m/s2
From F=ma, F = 9400*29.4665 = 276.98 kN from spring
How many mm? F=kx x=F/k = 276980/640 = 432.7813 mm
This shows the average is half of the maximum.
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Rolling Inertia
Tuesday, 17 May 2011
6:02 PM
Example 18.3. Cylinder 100mm,
Mass = 30kg, height difference 1.2m
Find linear/rotary speed at bottom.
PE1 + KE1 + SE1 +/- W = PE2 + KE2 + SE2
PE1 = KE2
mgh = 0.5mv2 + 0.5I2
Calculate mass moment of inertia I = 0.5mr2 (solid cylinder)
= 0.5*30*0.05^2
= 0.0375 kgm2
30*9.81*1.2 = 0.5*30*v2 + 0.5*0.0375*2
Oh no! We have 2 variables in 1 equation!!!!
No probs. Use v=r
30*9.81*1.2 = 0.5*30*v2 + 0.5*0.0375* (v/r)2
353.16 = 15v2 + 0.5*0.0375* (v/0.05)2
353.16 = 15v2 + 0.5*0.0375* (v2/0.052)
353.16 = v2(15 + 0.5*0.0375/0.052)
v2 = 353.16/ 22.5 = 15.696
v= 15.696 ^0.5 = 3.9618 m/s
=v/r 3.9618/0.05 = 79.236 rad/s
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Note: Compare to not spinning. v=(2gh)^0.5
= (2*9.81*1.2)^0.5 = 4.8522 m/s
(or with a zero mass moment of inertia I)
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Multiple Bodies - energy
Tuesday, 17 May 2011
7:03 PM
First solve this question
WITHOUT friction.
PE1 + KE1 + SE1 +/- W = PE2 + KE2 + SE2
PE1 = PE2 + KE2
mAgh1 + mBgh1 = mAgh2 + mBgh2 + 0.5mAv^2 + 0.5mBv^2
PE1 = (1.3+3.3)*9.81*1.6 = 72.2016 J
PE2 = 1.3*3.2*9.81 = 40.8096 J
KE2 = PE1-PE2 = 72.2016 - 40.8096 = 31.392 J
0.5*v2*(1.3+3.3) = 31.392
V2 = 31.392/(0.5*(1.3+3.3)) = 13.6487
V = 13.6487 ^0.5 = 3.6944 m/s
Compare to Freefall: v= (2*9.81*1.6)^0.5 = 5.6029 m/s
Include friction this time…
PE1 + KE1 + SE1 +/- W = PE2 + KE2 + SE2
PE1 - W = PE2 + KE2
PE1 = (1.3*9.81*1.6+3.3*9.81*1.6) = 72.2016 J
PE2 = (1.3*9.81*3.2+3.3*9.81*0) = 40.8096 J
W = T
Find angle: s=r  so = s/r = 1.6/0.04 = 40.0 rad
In linear W=FS or in rotary motion; W = T
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In linear W=FS or in rotary motion; W = T
W = 0.340*40 = 13.6 J
PE1 - W - PE2 = KE2
72.2016 - 13.6 - 40.8096 = 17.792
KE = 0.5*mv^2
v = (2KE/m)^0.5 = (2*17.792/(1.3+3.3) )^0.5 = 2.7813 m/s
Check: Freefall drop 1.6m v = (2*9.81*1.6)^0.5 = 5.6029
Our answer is slower than freefall, so looks good.
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Multiple Bodies - force method
Tuesday, 17 May 2011
7:03 PM
Compare to solving by the
FORCE METHOD…
\First solve this question
WITHOUT friction.
Set up FBD for each body:
Bodies are NOT in equilibrium!!!
Body A:
Fy = ma (Newton's 2nd law)
Ta - mg = ma
Ta = ma+mg
= 1.3a + 1.3*9.81
= 1.3a + 12.753
Body B:
Fy = ma (Newton's 2nd law)
Tb - mg = -ma (Accel DOWN)
Tb = mg - ma
= 3.3*9.81 - 3.3a
= 32.373 -3.3a
Because the pulley has no friction,
Ta=Tb
So
1.3a + 12.753 = 32.373 -3.3a
4.6a = 32.373 -12.753 = 19.62
a = 19.62/4.6 = 4.2652 m/s2
Linear motion question:
S = 1.6m
V0 = 0 m/s
V1 = ?
a = 4.2652 m/s2
t=
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t=
Linear motion (no time): V2 - V02 = 2as
V2 = 2*4.2652*1.6 = 13.6486
V = 13.6486^0.5 = 3.6944 m/s
This time the friction will alter the tension across the pulley
Find force due to friction on pulley;
Torque = force x radius
F = T/r = 0.34/0.04 = 8.5 N
So all the same until
Ta=Tb - 8.5
Etc..
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Coal Loader
Tuesday, 15 May 2012
4:47 PM
This is NOT an
inclined plane
problem. The
conveyor is
curved, and
even if it was a
straight line, we
don't know the
angle!
...but no worries, we can solve using conservation of energy!
Coal loaded at 560 t/hr = 560/60 = 9.3333 tonnes/min
Tonnes/sec = 560/3600 = 0.1556 or 155.6 kg/s
Belt carries 44kg per m
1 second = 155.6/44 = 3.5364 m
Velocity = 3.5364 m/s
Do as 1 hour's worth...
PE1 + KE1 +/- W = PE2 + KE2
The coal gains KE at the start and
has the same KE at the end, so...
KE1 = KE2 so...
PE1 + KE1 +/- W = PE2 + KE2
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KE1 = KE2 so...
PE1 + KE1 +/- W = PE2 + KE2
or
Work: W = PE2
= mg(h2-h1)
= 560 *1000* 9.81*(25-2.5) = 123606000 J (123.606 MJ)
1 second's worth...
Work: W = 123606000/3600 = 34335 J (34.335kJ)
Power: P = 34.335 kJ/s = 34.335 kW
3 phase motor: I = 34.335*1000/(3^0.5*415) = 47.767 Amps
For 1 second: (m=155.6m, v=3.5364 m/s)
KE = 0.5*m*v^2 = 0.5*155.6*3.5364^2 = 972.9765 J
Power = 972.9765 W
Total Power = gravity + KE + friction
= 34.335 + 0.973 + 8.1
= 43.408 kW
PE1 + KE1 +/- W = PE2 + KE2
We know:
Power, Diam, Belt Velocity
P, D, v... Find T
From W = T, so P = Tor T = P/
Find : From v=r, =v/r = 3.5364/0.43 = 8.2242 rad/s
So from T = P/43408 / 8.2242 = 5278.08 Nm
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Car on a hill
Tuesday, 15 May 2012
6:31 PM
Total Force = Fdrag + Frolling + Fgravity
Velocity = 95/3.6 = 26.3889
Fdrag = 0.5*1.2*26.3889^2*0.27*2.19 = 247.0596 N
Convert to power: P = Fv = 247.0596 * 26.3889 = 6.519 kW
Frolling = FN = 0.013*1700*9.81 = 216.801 N
Convert to power: P = Fv = 216.801 * 26.3889 = 5.72114 kW
Actual value: FN cos = 0.013*1700*9.81*cos(3.5)= 216.397 N
Fgravity = mgsin = 1700*9.81*sin(3.5) = 1018.106 N
Convert to power: P = Fv = 1018.106 * 26.3889 = 26.8667 kW
Total Force = 247.0596 + 216.801 + 1018.106 = 1481.9666 N
Convert to power: P=Fv = 1481.9666 * 26.3889 = 39107.47 W
(39.107 kW)
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