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Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 6 Machine Equipment Power Requirements.

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Presentation on theme: "Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 6 Machine Equipment Power Requirements."— Presentation transcript:

1 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 6 Machine Equipment Power Requirements

2 POWER This is a critical question because: Speed affects cycle time Cycle time drives production Production determines cost “Why does a machine only travel at 10 mph when its top speed is 30 mph?”

3 Speed affects cycle time POWER Cycle time drives production Production determines cost

4 PAYLOAD The payload capacity of construction excavation and hauling equipment can be expressed either Volumetrically or gravimetrically.

5 ROLLING RESISTANCE Rolling resistance is a measure of the force (lb/ton) that must be overcome to rotate a wheel over the surface on which it makes contact.

6 ROLLING RESISTANCE Tire flexing Rolling resistance is caused by Internal friction Tire penetrating the surface

7 ROLLING RESISTANCE If tire penetration is known Rolling resistance (lb) = TP = tire penetration, inches (may be different for haul and return) GVW = gross vehicle weight, tons

8 ROLLING RESISTANCE If tire penetration is known Rolling Resistance (lb/ton) can be estimated from the information in Text Table 6.1

9 HAUL ROAD CONDITION water trucks, so there is a cost. If haul roads are well maintained rolling resistance is less and production improves. Good haul roads require graders and

10 GRADE RESISTANCE We seldom find a haul road which is level from point of load to point of dump.

11 GRADE RESISTANCE Grades are measured in % slope: the ratio between vertical rise (fall) and horizontal distance in which the rise/fall occurs. Rise Horizontal

12 GRADE RESISTANCE Grade example: 5 ft fall in 100 ft horizontal travel. 5 ft 100 ft

13 GRADE RESISTANCE Fig. 6.4 page 146

14 GRADE RESISTANCE You need to review the derivation of equation 6.8. What it tells us is that for small angles (% grade): GR = 20 lb/tn  % grade

15 GRADE RESISTANCE Example: A truck with a 23 tn GVW is moving up a 4% grade. What is the force required to overcome grade resistance? GR = 20 lb/tn  23 tn  4% grade GR = 1,840 lb

16 GRADE ASSISTANCE Gravity assists the machine when traveling down grade. That force is referred to as grade assistance.

17 GRADE ASSISTANCE Example: Our truck has dumped its load, the GVW is now 12 tn and on the return it is moving down the 4% grade. What is the force required to overcome grade resistance? GA = 20 lb/tn x 12 tn x -4% grade GA = -960 lb

18 POWER REQUIRED A machine must overcome the forces of rolling and grade resistance to propel itself. These can be expressed as: lb/ton % effective grade

19 TOTAL RESISTANCE Total Resistance = Rolling Resistance + Grade Resistance TR = RR + GR or TR = RR - GA

20 PRACTICAL EXERCISE A scraper is operating on an earth haul road which is poorly maintained. The grade from cut to fill is 2%. Calculate the total resistance in both pounds and equivalent grade.

21 PRACTICAL EXERCISE EVW of 96,880 lb Rated load of 75,000 lb This scraper has a:

22 PE Step 1 Calculate the operating weight in tons for the haul. Haul weight = Haul weight = 85.94 tn

23 PE Step 2 Calculate the operating weight in tons for return. Return weight = Return weight = 37.5 tn

24 PE Step 3 Calculate the rolling resistance. Earth haul road poorly maintained Table 6.1 Use an average value; 120 lb/tn Convert to equivalent grade (eq. 6.9)

25 PE Step 4 Calculate the grade resistance. Grade from cut to fill is 2%. Haul grade (GR) = 2% Return grade (GA) = -2% Haul Return

26 PE Step 4 Calculate the grade resistance. Haul grade (GR) = 2% Return grade (GA) = -2% Equation 6.8 Haul GR = 2%  40 lb/tn Return GA = -2%  -40 lb/tn

27 PE Step 5 Calculate Haul Total Resistance TR haul = 6% + 2%  8% TR haul = 120 lb/tn + 40 lb/tn = 160 lb/tn

28 PE Step 5 Calculate Return Total Resistance TR return = 6% - 2%  4% TR return = 120 lb/tn - 40 lb/tn = 80 lb/tn

29 PE Step 6 Calculate Haul Total Resistance in lb. TR haul = 8% TR haul = 160 lb/tn x 85.94 tn = 13,750 lb GVW

30 PE Step 6 Calculate Return Total Resistance lb. TR return = 4% TR return = 80 lb/tn x 37.5 tn = 3,000 lb EVW

31 POWER AVAILABLE Engine horsepower and operating gear are the primary factors that determine the power available at the drive wheels (drawbar) of a machine.

32 POWER AVAILABLE Horsepower involves a rate of doing work. One hp = 33,000 ft-lb per minute Therefore, must consider speed at which the machine travels when exerting a given amount of “pull.”

33 POWER AVAILABLE Performance charts are provided for machines enabling us to estimate machine speed. Text page 159-160 The charts relate rimpull (drawbar pull), GVW, speed and total resistance (%).

34 POWER AVAILABLE Page 230

35 POWER AVAILABLE Haul Empty (EVW) Loaded (GVW)

36 POWER AVAILABLE Haul

37 POWER AVAILABLE Haul

38 POWER AVAILABLE Speed  9 mph Haul

39 POWER AVAILABLE Speed  31 mph Return

40 POWER AVAILABLE What if the total resistance is negative? See Text page 163 and 231 Retarding Performance chart The effective grade numbers are negative numbers.

41 POWER USABLE The coefficient of traction is the ratio between the maximum amount of pull a machine exerts before slippage and the total weight on the drivers.

42 POWER USABLE Consider the scraper in the previous example. What is the weight on the drivers during the haul?

43 POWER USABLE Total weight is Table 8.1 p. 231 Weight distribution loaded: Drive axle 53% Also see Fig. 6.8

44 POWER USABLE Considering the rimpull necessary for the haul what is the minimum coefficient of traction allowable? Rimpull required 13,750 lb

45 POWER USABLE The haul road is a wet clay loam. Will coefficient of traction be sufficient?

46 POWER USABLE The haul road is a wet clay loam. Will coefficient of traction be sufficient? Table 6.4, page 156 Wet, clay loam - rubber tires Coefficient of traction 0.40-0.50 Should be ok, 0.40  0.15

47 ALTITUDE LIMITS POWER If equipment works at higher altitudes, where the air is less dense, the engine may produce at a reduced power output.

48 ALTITUDE LIMITS POWER Most machines with turbocharged engines will operate at altitudes above 2,500 before experiencing a loss of power. See Table 6.5 page 157


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