1. NAZARIN B. NORDIN nazarin@icam.edu.my

2. What you will learn: Simple machines Mechanical advantage (force ratio) Movement ratio (velocity ratio) Machine efficiency (in %) Gears: gear box ratio, gear efficiency Directions of rotations

3. Machines Is a device that can change magnitude or line of action, or both magnitude and line of action of a force. A simple machine usually amplifies an input force, call the effort, to give a larger output force, called the load. Examples; pulley systems, screw jacks, gear systems and lever systems.

4. Machines

5. A machine is a device that receives energy in some available form and converts it to a useable form. For example, a person may wish to lift a weight of 2 tonnes; this is not possible unaided but the use of a jack or a hoist permits the person to achieve their objective. A lever such as a pry bar is an example of a simple machine. A relatively small manual force can be converted into a large force in order to lift or move an object.

6. Levers Leverage and the use of levers occurs in the use of tools such as spanners, pry bars, pliers etc., and in many vehicle mechanisms such as clutch and brake pedals, throttle linkages and suspension units.

9. The mechanical advantage of a machine is the ratio of the load to the effort: Lifting machines such as jacks and cranes have a large mechanical advantage so that relatively small manual forces can be used to raise heavy weights. Mechanical advantage (force ratio) mechanical advantage, MA = load effort

10. Velocity ratio (movement ratio) The velocity ratio, VR, of a machine, or movement ratio, MR, is the ratio of the distance moved by the effort to the distance moved by the load: or velocity ratio, VR = distance moved by effort distance moved by load movement ratio, MR = distance moved by effort distance moved by load

11. Efficiency of a machine The efficiency of a machine = (energy output/energy input) in the same time and this also Efficiency of machine = MA ×100% VR

12. Example problem In the trolley jack example shown before an effort of 250 newtons is lifting a load of 2 tonnes. In lifting the load through a distance of 15 cm the operator performs 40 pumping strokes of the handle each of which is 50 cm long. Calculate the mechanical advantage, velocity ratio and efficiency of the jack. MA = 78.5 VR =133.3 The efficiency = 59%

13. Gears Leverage and gears The gear ratio in this case is 4:1gear ratio The gear ratio = revolutions of input gear revolutions of output gear.

14. Gearbox Gear trains Ratio of gear set = driven × driven driver driver = 35×28 = 1.225 25×32

15. Problem 1 A simple machine raises a load of 160 kg through a distance of 1.6 m. The effort applied to the machine is 200 N and moves through a distance of 16 m. Taking g as 9.8 m/s 2, determine i) the mechanical advantage (force ratio), ii) the velocity ratio (movement ratio) and iii) the efficiency of the machine.

16. Solution 1 Mechanical advantage (force ratio) = Load/effort = 160 kg/200 N = (160 x 9.8)N/200 N = 7.84 Velocity ratio (movement ratio) = distance moved by effort/distance moved by the load = 16 m/1.6 m = 10 Efficiency = (force ratio/movement ratio) x 100% = (7.84/10) x 100 = 78.4%

17. Problem 2 For the simple machine of Problem 1, determine: (a)the distance moved by the effort to move the load through a distance of 0.9 m, (b)the effort which would be required to raise a load of 200 kg, assuming the same efficiency, (c)the efficiency if, due to lubrication, the effort to raise the 160 kg load is reduced to 180N.

18. Solution 2 Distance moved by the effort = 10 x distance moved by the load = 10 x 0.9 mechanical advantage = 9 m Since the mechanical advantage is 7.84, effort = load/7.84 = (200 x 9.8) / 7.84 = 250 N

19. Solution 2 The new mechanical advantage (force ratio) load/effort = (160 x 9.8)/180 = 8.711 The new efficiency after lubrication = (8.711/10) x 100 = 87.11%

20. Problem 3 A driver gear on a shaft of a motor has 35 teeth and meshes with a follower having 98 teeth. If the speed of the motor is 1400 revolutions per minute, find the speed of rotation of the follower.

21. Solution 3 Speed of driverteeth on follower speed of followerteeth on driver 1400/speed of follower = 98/35 Speed of follower = (1400 x 35)/98 = 500 rev/min =

22. Problem 4 A compound gear train similar to that shown in last figure consists of a driver gear A, having 40 teeth, engaging with gear B, having 160 teeth. Attached to the same shaft as B, gear C has 48 teeth and meshes with gear D on the output shaft, having 96 teeth. Determine; (a) the movement ratio of this gear system (b) the efficiency ratio when the force ratio is 6

23. Solution 4 (a)The speed of D = speed of A x (T A /T B )x (T C /T D ) movement ratio =(N A /N D )=(T B /T A )x (T D /T C ) = (160/40) x (96/48) = 8 (b) The efficiency = (force ratio/movement ratio) x 100% = (6/8 ) x 100 = 75%

24. NAZARIN B. NORDIN nazarin@icam.edu.my

25. What you will learn: Torque and power Indicated power and brake power Volumetric efficiency Thermal efficiency Mechanical efficiency

26. Torque & power

27. Indicated power Indicated power (ip) is the power that is developed inside the engine cylinders. It is determined by measuring the pressures inside the cylinders while the engine is on test, on a dynamometer. Because the pressure varies greatly throughout one cycle of operation of the engine, the pressure that is used to calculate indicated power is the mean effective pressure.

28. Indicated mean effective pressure (Imep) Because the pressure varies greatly throughout one cycle of operation of the engine, the pressure that is used to calculate indicated power is the mean effective pressure.

29. Imep Example 6.1 An indicator diagram taken from a single- cylinder 4-stroke engine has an effective area of 600 mm 2. If the base length of the indicator diagram is 60 mm and the constant is 80 kPa/mm calculate the indicated mean effective pressure.

30. Indicated mean effective pressure (i.m.e.p) Solution Indicated mean effective pressure or (imep)= (effective area of indicator diagram/base length of the diagram)×constant imep = (600/60)×80 kPa/mm imep = 800 kPa = 8 bar

31. Calculation of indicated power Indicated power is calculated from the formula Indicated power (ip) = ( P × l × a × N)/1000 Unit as kW where P = mean effective pressure in N/m 2, l = length of engine stroke in m, a = cross sectional area of cylinder bore in m 2, N = number of working (power) strokes per second.

32. Calculation indicated power Example 6.2 In a test, a certain single-cylinder 4-stroke engine develops a mean effective pressure of 5 bar at a speed of 3000 rev/min. The length of the engine stroke is 0.12 m and the cross- sectional area of the cylinder bore is 0.008 m 2. Calculate the indicated power of the engine in kW.

33. Calculation of indicated power Solution The engine is single-cylinder 4-stroke, so there is one working stroke for every two revolutions. N, the number of working strokes per second N = (3000÷60)/2 = 25; P = 5 bar= 500 kPa=500,000 Pa = 500,000 N/m 2 ; l = 0.12 m; a = 0.008 m 2. Substituting these values in the formula gives ip = [(500 000×0.12×0.008×25)/1000] kW = 12 kW

34. Number of working (power) strokes As mentioned, each cylinder of a 4-stroke engine produces one power stroke for every two revolutions of the crankshaft. In a multi-cylinder engine, each cylinder will produce one power stroke in every two revolutions of the crankshaft. A formula to determine the number of power strokes per minute for a multi-cylinder, 4-stroke engine is: number of power strokes per minute = (number of cylinders/2)×rev/min

35. Indicated power Example 6.3 A 4-cylinder, 4-stroke engine develops an indicated mean effective pressure of 8 bar at 2800 rev/min. The cross-sectional area of the cylinder bore is 0.01 m 2 and the length of the stroke is 120 mm. Calculate the indicated power of the engine in kW. Solution

36. Brake power The engine power that actually reaches the output shaft or flywheel of an engine is known as the brake power. It is the power that is measured by a dynamometer and a dynamometer is also known as a brake, hence the term, brake power. The simplest form of dynamometer is shown in the Figure 14.1 on next slide:

37. Simple dynamometer

38. Work done per revolution of the flywheel = force × distance = (W−S)×2R joules (W−S×R also = T the torque that the engine is exerting. By substituting T for W−S×R the) Work done per revolution becomes = T×2π joules

39. Brake power Brake power = work done per second = 2π × Torque× number of revolutions per second =2π x T x n ie. 2π.T.N This is normally written as brake power (bp = 2πTN where N = number of revolutions per second.) When the torque is in newton metres (N m), the formula bp = 2πTN gives the power in watts. As engine power is given in thousands of watts, i.e. kilowatts, the formula for brake power normally appears as: b p = 2π.T.N/1000 kW

40. Brake power Example 6.4:

41. Volumetric efficiency

42. Example 6.5 A 4-cylinder 4-stroke petrol engine with a bore diameter of 100 mm and a stroke of 110mm has a volumetric efficiency of 74% at an engine speed of 4000 rev/min. Determine the actual volume of air at STP that flows into the engine in 1 minute. Solution

43. Thermal efficiency The thermal efficiency of an engine is a term that is used to express the effectiveness of an engine’s ability to convert heat energy into useful work. Thermal efficiency is the ratio of energy output of the engine to energy supplied to the engine in the fuel.

44. Thermal efficiency Example 6.6 During a 10-minute dynamometer test on a petrol engine, the engine develops a brake power of 45 kW and uses 3 kg of petrol. The petrol has a calorific value of 43 MJ/kg. Calculate the brake thermal efficiency.

45. Mechanical efficiency The mechanical efficiency of an engine is defined as brake power, b.p. indicated power, i.p. X 100 % An engine develops a brake power of 120 kW at a speed of 3000 revs/min. At this speed, the indicated power is 140 kW. Calculate the mechanical efficiency of the engine at this speed.