Types and Calculations Elements of Power Types and Calculations Computer Integrated Manufacturing © 2013 Project Lead The Way, Inc.

Elements of Power CIM Elements of Automation Power What is Work and Power? Work is the energy transferred when a force displaces a mass Power is the rate at which work is done Typically mechanical, electrical, or fluid Work is the energy transferred when a force, F, is applied to an object moving through a distance, d. Power is the rate at which work is done. There are many examples of devices that use power in one form and convert it into another. This presentation will focus on mechanical, electrical, and fluid devices. Project Lead The Way, Inc. Copyright 2008

Mechanical Calculations Elements of Power CIM Elements of Automation Power Mechanical Calculations W = F ∥ ∙ d W = work, measured as ft-lb P = power measured as ft-lb/s t = time measured as s 1 hp = 550 ft-lb/s Work is calculated by multiplying force times displacement. It is expressed as ft-lb. Power is the rate with which work is done and is calculated as a ratio of work divided by time. It is expressed as ft-lb/s. If the amount of work remains constant then power and time are inversely proportional. In other words, if less is power used then more time is needed. Inversely, if more is power used then less time is needed. One horsepower is the amount of power required to displace 33,000 pounds one foot in one minute. To calculate the work in one second, you would divide by 60 since there are 60 seconds in one minute. Therefore, 33,000 / 60 = 550. So the power required to displace 33,000 pounds one foot in one minute is equal to the power required to displace 550 pounds one foot in one second. Project Lead The Way, Inc. Copyright 2008

Elements of Power CIM Elements of Automation Power Example If Jessica moves her 112 pound (lb) body up 8 feet (ft) of stairs in 2.3 seconds (s), how much work did she perform? How much power did she use? ( ) ft-lb 1 hp W = F ∥ ∙ d W = 112 lb (8 ft) W = 896 ft-lb Php = 390 s 550 ft-lb s Php = 0.71 hp The units ft-lb/sec is converted into horsepower (HP) using the conversion factor of 550 as presented previously. W P = t 896 ft-lb P = 2.3 s ft-lb s P = 390 Project Lead The Way, Inc. Copyright 2008

Elements of Power CIM Elements of Automation Power Your Turn An escalator moves 84 people per minute from one floor of a mall to the second floor. The mall’s floors are 24 ft apart, and the average passenger weighs 154 lb. How much power does the escalator need? Write the formula. Then identify and substitute your unknowns. Project Lead The Way, Inc. Copyright 2008

( ) Solution F = 154 lb/person  84 people F = 12,936 lb W = F ∥ ∙ d Elements of Power CIM Elements of Automation Power Solution F = 154 lb/person  84 people F = 12,936 lb W = F ∥ ∙ d W = 12,936 lb (24 ft) W = 310,464 ft-lb W P = t 310,464 ft-lb P = 60 s P = 5174 ft-lb s ( ) ft-lb 1 hp P (hp) = 5174 s 550 ft-lb s P (hp) = 9.4 hp Project Lead The Way, Inc. Copyright 2008

Torque Torque,  , is a twisting force on a shaft Elements of Power CIM Elements of Automation Power Torque Torque,  , is a twisting force on a shaft Torque measured as in.-lb or ft-lb  = dFsinθ A force applied at 90 degrees allows the formula to be simplified to  = dF Examples of torque can be seen in a wrench or an engine. The Greek letter  (Tau), which is usually pronounced so that it rhymes with now but can also be pronounced so that it rhymes with law, is used to denote torque. r is the length of the radius, defined as the distance from the pivot to the point at which the force is being applied. This is why r can be changed to d for distance, not diameter. F is the force applied. sin  is a trigonometric function, relating the sides of a right triangle, which would be formed if a vertical line were drawn from the lever to the ending line. For a 90 degree angle, sin  = 1, which is why that part of the equation is reduced with simplification. r  = 90 F Project Lead The Way, Inc. Copyright 2008

Elements of Power CIM Elements of Automation Power Finding Torque 116 pounds of force is applied at the end of a lever 2.4 feet long. The force is applied at an angle of 90° to the lever. What is the torque produced? 90 116 lb 2.4 ft pivot Project Lead The Way, Inc. Copyright 2008

Torque Calculation 116 lb 2.4 ft 90 pivot d= 3 ft 𝛕=dFsinθ F=100 lb Elements of Power CIM Elements of Automation Power Torque Calculation 90 116 lb 2.4 ft pivot d= 3 ft 𝛕=dFsinθ F=100 lb 𝛕=2.4 ft 116 lb sin90° 𝛕=278 ft−lb Project Lead The Way, Inc. Copyright 2008

Elements of Power CIM Elements of Automation Power Power Calculations Using the previous problem, how much power will be generated if the force is applied at a rate of 42 rpm? Convert rpm to rad/s ω=42 𝑟𝑒𝑣 𝑚𝑖𝑛 1 𝑚𝑖𝑛 60 𝑠 2𝜋 𝑟𝑎𝑑 1 𝑟𝑒𝑣 ω=4.40 𝑟𝑎𝑑 𝑠 P=1222 ft−lb sec 1 hp 550 ft−lb s Converting torque to horsepower is not as straightforward as ft-lb/sec. The angular velocity is measured in radians per seconds so must convert rpm into rad/sec to use the power formula. P = 𝛕 ω P=278 ft−lb 4.40 𝑟𝑎𝑑 𝑠 P= 2.2 hp P=1222 ft−lb 𝑠 Convert hp Project Lead The Way, Inc. Copyright 2008

Elements of Power CIM Elements of Automation Power Your Turn A shaft has a force of 32 pounds applied 5.2 feet from the pivot point. What is the resulting torque? How much power is generated if the torque is produced at 2,850 rpm? Project Lead The Way, Inc. Copyright 2008

Solution F=32 lb 𝛕=dFsinθ r=d = 5.2 ft 𝛕=5.2 ft 32 lb sin90° Elements of Power CIM Elements of Automation Power Solution F=32 lb 𝛕=dFsinθ r=d = 5.2 ft 𝛕=5.2 ft 32 lb sin90° 𝛕=166 ft−lb Rate of rotation= 2,850 rpm Convert rpm to rad/s ω=2,850 𝑟𝑒𝑣 𝑚𝑖𝑛 1 𝑚𝑖𝑛 60 𝑠 2𝜋 𝑟𝑎𝑑 1 𝑟𝑒𝑣 ω=298 𝑟𝑎𝑑 𝑠 P=49,637 ft−lb s 1 hp 550 ft−lb s P = 𝛕 ω P=166 ft−lb 298 𝑟𝑎𝑑 𝑠 P= 90 hp P=49,637 ft−lb 𝑠 Convert to horsepower Project Lead The Way, Inc. Copyright 2008

Electrical Work and Power Elements of Power CIM Elements of Automation Power Electrical Work and Power Commonly measured as watt (W) Also measured as joule or horsepower A joule is a measurement of electrical work in the International System of Units (SI). One watt is equal to one joule per second. Project Lead The Way, Inc. Copyright 2008

Electrical Calculations Elements of Power CIM Elements of Automation Power Electrical Calculations Electrical power is determined by the formula P = IV P = power Measured as watt (W) and horsepower (hp) Students should know that kW means for kilowatt, or 1000 W. Note that power can also be expressed as a function of electrical resistance measured in ohms. Apply the equation V = IR for two alternate equations: P = (I^2)R or P = (V^2)/R 1 hp = 745.7 W 1 W = 0.00134 hp V = voltage measured as volt (V) I = current measured as ampere (A) Project Lead The Way, Inc. Copyright 2008

Electrical Calculations Elements of Power CIM Elements of Automation Power Electrical Calculations A house is wired such that a typical circuit can allow 15 A. Which two appliances can be used simultaneously without blowing a circuit? P = IV Hair dryer (1500 W, 110 V) Microwave (1600 W, 120 V) Toaster (400 W, 110 V) Coffee maker (250 W, 120 V) Dishwasher (1300 W, 120 V) Flat iron (72 W, 110 V) 13.6 A 13.3 A 3.6 A 2.1 A 10.8 A 0.65 A The hairdryer and microwave must be used alone. All other combination of two appliances would work. Project Lead The Way, Inc. Copyright 2008

Elements of Power CIM Elements of Automation Power Your Turn Assume that a shop is wired so that a typical circuit is 20 amps. Which appliances can be used simultaneously? P = IV 1 hp = 745.7 W 7 A 1.7 A 16 A 3.3A 2.1 A 7.3 A Band saw (1 hp, 110 V motor) CNC laser (200 W, 120 V) Laser engraver (3500 W, 220 V) Light (each) (400 W, 120 V) Computer (250 W, 120 V) 3D printer (1600 W, 220 V) If more than one light is used, then the combinations would change. This activity gives you practice in applying electrical power calculations; however in practice heavy machinery often requires a dedicated circuit. Combination ABDE, ABEF, ADEF, BDEF, BC, CD, CE or any subset Project Lead The Way, Inc. Copyright 2008

Fluid Power Operated by compressed fluid Low load-carrying capacity Elements of Power CIM Elements of Automation Power Fluid Power Operated by compressed fluid Low load-carrying capacity Requires mechanical stops if not used with servomotors Two types: hydraulic and pneumatic Pneumatic systems are operated by air or other compressed gas. Advantages of pneumatic power: - light weight - compressed air is readily available Disadvantages of pneumatic power: - low efficiency - low stiffness - low accuracy Hydraulic systems are operated by the action of water or other fluids of low viscosity. Advantages of hydraulic power: - provides a large amount of power - moves high loads at reasonable speeds - moderate noise level - Usually more efficient Disadvantages of hydraulic power: - more expensive - less accurate - requires an energy storage system, such as pumps or accumulators - susceptible to leakage Project Lead The Way, Inc. Copyright 2008

Fluid Power Definitions The use of a fluid to transmit power from one location to another Hydraulics The use of a liquid flowing under pressure to transmit power from one location to another Pneumatics The use of a gas flowing under pressure to transmit power from one location to another

Fluid Calculations p= F A Elements of Power CIM Elements of Automation Power Fluid Calculations p = Pressure, measured in pounds per square inch (psi) F = Force, measured in pounds A = Area of the base of the fluid container p= F A F A piston has the geometric shape of a circle so how is area calculated? A The area of a circle is r2 If the piston diameter is 4 in. then what is the area? 12.6 in.2 Project Lead The Way, Inc. Copyright 2008

A = Area to which force is being applied Elements of Power CIM Elements of Automation Power F = Force being applied A = Area to which force is being applied The area is defined as a circle for a piston. Project Lead The Way, Inc. Copyright 2008

Elements of Power CIM Elements of Automation Power Your Turn 62 pounds of force is applied to a ¾ inch pneumatic cylinder. How much pressure is created in the cylinder? An accepted industry standard is that a 1 inch cylinder means the cylinder has a diameter of 1 inch. Project Lead The Way, Inc. Copyright 2008

Solution r = ½ D r = ½ (0.75 in.) r = 0.375 in. A = r2 Elements of Power CIM Elements of Automation Power Solution r = ½ D r = ½ (0.75 in.) r = 0.375 in. A = r2 A = (0.375 in.)2 A = 0.44 in.2 p= F A p= 62 lb 0.44 in.2 p=140 psi Project Lead The Way, Inc. Copyright 2008

Elements of Power Your Turn CIM Elements of Automation Power A hydraulic pressure of 175 psi is applied to a piston that has a 3.75 inch diameter and 8 inch stroke. What is the power produced by the piston that moves its full stroke in 2.25 s? Project Lead The Way, Inc. Copyright 2008

( ) Solution p= F A F=pA F=175 psi (11.0 in.2) F=1932 lb A = r2 Elements of Power Solution CIM Elements of Automation Power A = r2 A = (1.875 in.)2 A = 11.0 in.2 W r = ½ D r = ½ (3.75 in.) r = 1.875 in. P = t 1288 ft-lb P = p= F A 2.25 s P = 572 ft-lb F=pA s ( ) F=175 psi (11.0 in.2) ft-lb 1 hp P (hp) = 572 F=1932 lb s 550 ft-lb s P (hp) = 1.04 hp W = F ∥ ∙ d W = 1932 lb (8 in.) W = 15,455 in.-lb W = 1288 ft-lb Project Lead The Way, Inc. Copyright 2008

Image Resources Microsoft, Inc. (2012). Clip art. Retrieved from http://office. microsoft.com/en-us/clipart/default.aspx