2. Force, Work and Power 330:148 (g) Objectives • Understand the difference between force, work, and power. Machine Design • Recognize and be able to convert between mass and weight. • Convert between English and metric units for force, work, and power. Nageswara Rao Posinasetti • Understand the basic principles of fluid mechanics as they apply to hydraulic and air cylinders or similar products. • Look up and/or calculate moments of inertia and section modulus for different shape parts. • Apply the principles of work, force, and power to moving machines. August 15, 2007 1 August 15, 2007 2 Weight, Force and Mass Work and Power Weight, W = m g Work = Force × Distance ft-lb or N m W = weight, lb or N Work done, W = F d M = Mass, lb or kg Power is rate of doing work W g = acceleration due to gravity, Power, P = 2 2 t 32.2 ft/s , 9.81 m/s t = time Force, F = m a a = acceleration August 15, 2007 3 August 15, 2007 4 Power and Speed Power and Speed 1 hp = 550 ft-lb/s P = T ω = 6600 in-lb/s P = Power, ft-lb/s or W T = Torque, ft-lb or N m = 33,000 ft-lb/min ω = Rotational speed in radians/second = 396,000 in-lb/min P T = Power in SI Units is Watts ω 1 hp = 746 W = 0.746 kW ω = 2 π n 60 n = revolutions per minute August 15, 2007 5 August 15, 2007 6 1 Torque Torque is a twisting moment Calculate the amount of torque in a shaft Rotates a part in relation to other transmitting 750 W of power while rotating T = F d at 183 rad/s. F = force applied d = distance P = T ω Torque = 750 / 183 = 4.09836 N-m August 15, 2007 7 August 15, 2007 8 Example Problem 2-1: Torque Example Problem 2-1: Torque • An automobile engine creates a torque of 300 ft-lb at a rotational speed of 3,500 rpm. • In fourth gear, the transmission has a 1 to 1 ratio. • An automobile engine creates a torque of 300 ft-lb • The differential has a ratio of 3.6 to 1, which creates a torque in the axle of 1,080 ft-lb. at a rotational speed of 3,500 rpm. • What is the force being exerted on the road by the automobile tire if the diameter of the tire is • In fourth gear, the transmission has a 1 to 1 ratio. 30 inches? • The differential has a ratio of 3.6 to 1, which creates T = Fd a torque in the axle of 1,080 ft-lb. T F = • The tire exerts an • What is the force being exerted on the road by the d 1080 ft − lb 864-pound force on automobile tire if the diameter of the tire is 30 F = 30in ( ft) the road surface. inches? 2 (12 in) F = 864 lb August 15, 2007 9 August 15, 2007 10 Work W × H Power and Rotational Speed Power = = Time t W × 2 π × R P = T ω = 1 min 2 π n Since, ω = 60 Power = 2 π T n 2 π T n 2 π T n Horse power = We get, P = 396,000 60 T n Horse power = 63,000 T = torque (in-lb) n = rotational speed, rpm August 15, 2007 11 August 15, 2007 12 2 Example Problem 2-2: Work and Power Example Problem 2-2: Work and Power • If the automobile in Example • If the automobile in Example Problem 2-1 Problem 2-1 was going up a long, was going up a long, steep hill at this speed, steep hill at this speed, how much work was done in the period of a 30in ft min how much work was done in the period of a 3500 rpm second? π rev 12in 60 sec second? What is the power? d = • As work is force times distance, 3.6 •Force on the road, F = 864 lb calculate the distance traveled in one 1 second. d = 127 ft W = Fd (2-3) W = 864 lb 127 ft W = 109,728 ft − lb of work August 15, 2007 13 August 15, 2007 14 Example Problem 2-2: Work and Power (cont’d) Example Problem 2-2: Work and Power (cont’d) Fd W P = = (2-4) t t • Compare it to motor output: 109,728 ft-lb ft-lb P • The power to do P = = 109,728 T = 63,000 1 sec sec n this would then be: where: n = rpm or if: T = in-lb ft-lb 550 = 1 hp sec Tn (2-6) P = 63,000 ft-lb 109,728 sec hp = = 200 hp 300 ft-lb 12 in 3500 rpm ft-lb P = ft 63,000 550 sec hp P = 200 hp • Power output equaled the power at the tire/road surface. August 15, 2007 15 August 15, 2007 16 Pressure, Force and Area Example Problem 2-3: Force Pressure Relationship Pneumatic and Hydraulic Cylinders • For the automobile engine from the prior example problems, calculate the pressure required in the cylinder under the following assumptions: Force = Pressure × Area 1. The cylinder is 2 inches in diameter. 2. The crankshaft has an effective radius of 4 inches. Work done = Force × Distance moved 3. Maximum power is achieved when the piston is perpendicular to the crankshaft. Distance moved = stroke length 4. At this point, all the power comes from this piston alone. • If the T is 300 ft-lb and the effective length of the crankshaft is 4 inches, the force exerted by the piston would be: T = Fd (2-5) T 300 ft-lb F = = = 900 lb d ft 4 in 12 in August 15, 2007 17 August 15, 2007 18 3 Moments of Inertia and Section Example Problem 2-3: Force Pressure Relationship (cont’d) Modulus F = PA Properties of beams • Analyze the cylinder: F P = Moment of Inertia A Moment of Inertia 900 lb Section Modulus = P = π (2 in)2 Distance from the neutral axis 4 P = 286 lb/in2 = 286 psi See Appendix 3 (p 469) for different sections • What may be correct in our assumptions? • Why is it likely that this value would be much higher in a typical engine? August 15, 2007 19 August 15, 2007 20 Problem 2-1 Problem 2-1 (cont’d.) a. Force in cable at rest: = 1000 lb 1. The elevator system shown has a combined weight b. Work done, W = Fd for the car and occupants of 1,000 pounds. Ignore W = 1000 lb x 150 ft = 150,000 ft-lb forces to accelerate the car and occupants. a. What is the force in the cable? c. Power = W = Fd b. To raise the car 150 feet, how much work was done? t t c. If it took 10 seconds to travel this distance, what power P = 150,000 ft-lb = 15,000 ft-lb/s was required in ft-lb/sec, hp, and kilowatts? 10s d. If the effective diameter of the cable drum is 18 P = 15,000 ft-lb s hp inches, determine the torque in the input shaft. s 550 ft-lb e. What is the rotational speed of this shaft? f. Calculate the power to turn this shaft, and compare this P = 27.3 hp power to that determined above in part c. P = 27.3 hp .746 kw = 20.3 kw hp August 15, 2007 21 August 15, 2007 22 4.