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.

Subunit 1 Power in Mechanical Systems Power: Rate of doing work Power rating: maximum rate at which each machine can do work www.teachertube.com/view_video.php?viewkey=b9e96fefd21d68ab729c

Linear Mechanical Systems Work = Force applied to object x Distance object moves along the direction of the force acting on it Power = Work done by applied force Time to do the work W = f x d P = W t P = f x d t P = f x d t P = f x V Power = force x speed

Rotational Mechanical Systems Work = Torque applied x Angle moved through W = T x θ P = W t P = T x θ t Since ω = θ/t P = T x ω Power = torque x angular speed

Units for Mechanical Power 1 horsepower = 746 watts 1 horsepower = 550 ft•lbs sec English SI ft•lbs sec Joule sec N•m sec watt horsepower

Efficiency Efficiency: ratio of work out to work in or ratio of power out (Pout) to power in (Pin) Power In = Work In time Power Out = Work Out time % efficiency = Work Out Work In X 100 Or… % efficiency = Power Out Power In X 100

Sample Problem 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 … P = f x d t P = 200 lb x 3 ft 4 sec = 150 ft•lb/sec V = d/t = 3 ft 4 sec = P = f x V .75 ft/sec 150 ft•lb/sec P = (200 lbs) (.75 ft/sec) =

Sample Problem 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 Cont. A. B. C. Pin = Fin x Din Time = = 1088 N•m sec 1088 watts Pout = Fout x Dout Time = (980 N) (2m) 2sec = 980 N•m sec = 980 watts Efficiency = Pout Pin X 100 = 980 watts 1088 watts = 90% .90 x 100 =

Sample Problem Given: An electric motor has a shaft torque of .73 lb•ft when rotating at 1800 rpm. Find : horsepower of the shaft P = T x ω ω = 1800 ( ) rev min 1 min 60 sec 6.28 rad. = 188.4 rad/sec 137.5 ft•lb sec P = .73 lb•ft x 188.4 rad/sec = P = 137.5 ft•lb sec x 1 hp . 550 ft•lb = .25 hp

( )( ) Sample Problem ουσ 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 P = T x ω 65 N•m x 115 rad/sec = 7475 watts ( )( ) 7475 watts 1hp 746 watts = 10.0 hp ουσ

Lab Demos