sub unit 6.1 ”power in mechanical systems” introduction using @ a 10 minute video –video shoes...
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Sub unit 6.1”Power in Mechanical Systems”
• Introduction using @ a 10 Minute video– Video shoes real life scenarios, equations,
terms, and labels for both SI and English units.– After video we have the following power point
lecture that discusses terms and their interpretations, formulas and their labels, and we finish with sample problems to identify the students abilities to use the new information.
• Power: Rate of doing work
• Power rating: maximum rate at which each machine can do work
• www.teachertube.com/view_video.php?viewkey=b9e96fefd21d68ab729c
Subunit 1Power in Mechanical Systems
Linear Mechanical Systems
Work = Force applied to
object
x Distance object moves along the direction of the force
acting on it
W = f x dPower =Work done by applied forceTime to do the work
P = Wt
P = f x dt
P = f x dt
P = f x V
Power = force x speed
Rotational Mechanical Systems
Work =
Torque applied
x Angle moved throughW = T x θ
P = Wt
P =T x θt
Since ω = θ/t
P = T x ω
Power = torque x angular speed
Units for Mechanical Power
English SI
ft•lbssec
horsepower
Joulesec
N•msecwatt
1 horsepower = 746 watts
1 horsepower = 550 ft•lbssec
• Efficiency: ratio of work out to work in or ratio of power out (Pout) to power in (Pin)
Efficiency
Power In = Work Intime
Power Out = Work Outtime
% efficiency = Power OutPower In
X 100
Or…
% efficiency = Work OutWork In
X 100
• Given: A steel casting weighs 200 pounds. It is
raised 3 ft in 4 sec at a constant speed.
• Find : horsepower of cylinder used to lift the steel casting.
• 1st …
• 2nd …
Sample Problem
P = 200 lb x 3 ft4 sec
= 150 ft•lb/sec
P = f x V
P = f x d
tV = d/t = 3 ft
4 sec = .75 ft/sec
P = (200 lbs) (.75 ft/sec) = 150 ft•lb/sec
• Given: A crate weighs 980N and is lifted 2 meters
in 2 sec. The winch pulls the cable a dist.
Of 8 meters with a force of 272N in the
same 2 sec.
• Find :– A.) Input power (in watts) supplied to block and tackle by the winch
– B.) Output power (in watts) of block and tackle used in lifting crate
– C.) Efficiency of block and tackle
Sample Problem
• A.
• B.
• C.
Sample Problem Cont.
Pin = Fin x Din
Time= (272 N) (8m)
2 sec= =1088 N•m
sec
Pout = Fout x Dout
Time= (980 N) (2m)
2sec= 980 N•m
sec=
Efficiency = Pout
Pin X 100 = 980 watts
1088 watts= .90 x 100 =
980 watts
90%
1088 watts
• Given: An electric motor has a shaft torque of
.73 lb•ft when rotating at 1800 rpm.• Find : horsepower of the shaft
Sample Problem
P = T x ω
ω =
1800( )revmin ( )1 min
60 sec ( 6.28 rad.rev )= 188.4 rad/sec
P = .73 lb•ft x 188.4 rad/sec = 137.5ft•lbsec
P = 137.5ft•lbsec x
1 hp .550 ft•lb
sec
= .25 hp
• Given: a piston in a compressor has a
flywheel with a rate of 1100 rpm
(115rad/sec) Torque is 65 N•m
• Find : hp of flywheel
Sample Problem
P = T x ω
65 N•m x 115 rad/sec = 7475 watts
( )( )7475 watts 1hp746 watts
= 10.0 hp
ουσ
Lab Demos