2. 1 Steps in Design of a Hoisting System ©Dr. B. C. Paul 1999 major revision 2012 With Credit to Dr. H. Sevim for Original Book

3. 2 Steps in Design of Hoisting System w Determine the performance requirements Usually means production can also involve figuring acceptable stopping distances - number of levels to be served under what conditions w Select hoist type to meet constraints

4. 3 Once Upon a Mine w Dark Black coal mine will produce 3 million tons of coal from a single level. The hoisting distance from loading pocket to dump bin is 1000 ft. The mine operates 250 days per year 3 production shifts per day with 7 hours of operation each production shift. The peak production will be about 5000 tons per shift. The average production is 4000 tons per shift.

5. 4 Your Mission Jim, Should you decide to accept it w Design a hoisting system for the Dark Black Coal Mine. w First Step is to establish the performance requirement. w The fundamental Capacity Equation is Q = P / T Q is requirement in Tons Per Hour P is Production per shift T is the average shift production time

6. 5 Which Production Number Do We Use? w Actual production is a distribution - not an average number w If we design on average then all the numbers above the mean will go past capacity - we’ll loose our high values and not meet production w If we design on a peak that is seldom achieved we’ll pay big bucks

7. 6 Decision Criteria w If peak approaches 2 times average - need to consider cost of work stoppage vs. take work stoppages w If peak is somewhat close to average then design for the peak w In example the peak is 125% of the average, which is not considered a significant deviation. Design for the Peak!

8. 7 What if Life had not been so Kind? w Calculate the cost of a production stoppage May be cost of lost production May include penalties on contracts May include idle labor cost w Calculate the amortized cost of the next increment of production capacity w Check multiple points and go for minimum total cost.

9. 8 Pick production capacity w Apply formula 714 tph = 5000 t/shift / 7 hours prod time w Other design decisions need to be made Is this a Keope or a Drum Hoist? One Level so Keope won’t be too tricky

10. Input My Hoist Distance and Production Rate Target 9 My Spreadsheet is Opti-Hoist. Next Lets look at some fixed Cycle time elements

11. Elements w How Long will it take for the ore pocket doors to open and drop a load into the skip? w Once the Skip is in position how long will it take to deposit the load into the dump point chute? About 8 seconds to load 10 seconds to dump is reasonable Your load and unload times may depend on other design elements of the system. 10

12. Creep Time w Hoists and Elevators slow to near stop as they line up with a set level Pull away fromthe loading point Or line up with dump point Will Usually take a bit more time to line up with the unload point 2 seconds to pull away from load and 4 to line up with dump is reasonable. 11 A fowl Beast

13. Now We Need to Pick a Peak Speed and a Rate of Skip Acceleration 12 The maximum speed we lift at is safety related. For men there are regulations. For materials there are guidelines (shown in pink)

14. Considerations w This is a Keope Hoist The rope is just sitting over a friction wheel If I “peel out” and the rope starts to slip I have a major hoisting accident w High speed and high acceleration increase production But they also cause a big increase in motor size and energy bill. 13

15. I’m going to go for modest speed and particularly modest acceleration 14

16. Geometry Considerations 15 Two Skips Skip and Counter-weight

17. Stopping Considerations 16 If my controls miss Stopping the skip at The dump point – how Far do I have for an Emergency stop before The over-wiend turns Into a disaster? Dump Chute Loading chute Shaft Bottom

18. Getting Our First Estimate 17 Using a Nordberg approximation of the cycle time the program estimates the size Of skip that will be needed to achieve the production target using the distance, Speed and acceleration conditions specified. (This part of the spreadsheet is independent of whether the hoist was a Keope or A drum hoist).

19. Our Next Task is to Get Our Exact Cycle Time 18 The only missing Piece of information Was what is the Creep speed (2 ft/sec Is reasonable). (I want’s you money for my fake Global warming initiative)

20. Next We Must Balance Skip Size, Weight, Rope Diameters and Wheel Sizes 19

21. First We Need to Pick A Skip Size 20 We have already been given a first guess skip size. We try about that size And then check the actual hourly production achievable in the red box (we Now have figured the exact cycle time too). As can be seen a 16 ton skip will get me my 720 tons/hour.

22. Next I’ll Go for Skip Weight 21 In general a skip heavy enough to handle the banging of ore loading and Unloading will weight about 75% of the load weight. Opti-Hoist estimates this for us but leaves us a yellow blank to choose the Weight. A higher weight usually means we are just adding weights to our Skip.

23. Next We Go For Rope Sizing 22 Keope hoists usually have multiple ropes in even numbers, 2, 4, 6, and 8 being Common. Where that 12 came from is unsure. We need to consider both hoisting ropes and tail ropes. Sometimes tail ropes Are used and worn-out hoist ropes which can cause tail ropes to be the same As hoist ropes.

24. My First Guess is 4 ropes (This is a relatively modest depth) 23 The red box estimates that I will need a 1.064 inch rope to reach needed Safety factors for this depth.

25. From Here I Need to Go And Enter My Rope Properties 24 I enter my rope size, it’s weight and it’s strength. (I have the advantage of Having an estimate of what size rope to try). The spreadsheet then compares the achieved factor of safety to what is required

26. Where Do I Get These Rope Properties 25 The spreadsheet has a Rope properties table Right below for me to Look things up on. I’m looking for 1.064 Flattened strand I go for 1.125 Weight 2.28 lbs/ft Strength 57.9 tons

27. Enter My Rope and Check My Factor of Safety 26 A 7 factor of safety clearly meets a 6.5

28. I Wonder If I Could Get Away with a 1 inch rope 27 Eeee – one inch rope is a nope.

29. What if I Use High Strength Steal? 28 Oh so close but still nope to the rope

30. With Hoist Rope Selected I can now pick the Wheel for My Hoist Frame 29 I need a 7.5 ft wheel (see pick recommendation) to avoid bending my rope to Sharp. (You can see why I wanted a smaller rope – it would have allowed me To use a smaller lower inertia wheel).

31. Now I Need to Deal with Tail Rope w Simple case is to get used rope w Also easiest to have just one tail rope so have less swinging and tangling. w Number of tail ropes is commonly less than number of hoist ropes. 30

32. I’m going to try straight across Using my worn-out hoist ropes for tail ropes. 31

33. Now I’ll Check Conditions at My Keope Wheel 32 T1 is the weight of the heavy loaded side. T2 is the weight of the lighter empty Side. Remember – only friction stops the rope from slipping. The ratio of T1 to T2 Must therefor not be more than 1.5 Of course 1.95 is greater than 1.5 so life is sucking right now.

34. Another Parameter is Tread Pressure 33 Keope wheels are normally lined with a leather like frictional material. Since we Don’t want the ropes to cut the material to pieces we need to limit the load to About 300 psi or less. Well at least one thing worked.

35. So What Do I Do Now w Either T1 is to big or T2 is too little w I could look at rope weight but the rope weight shifts back and forth from T1 to T2 depending on where we are The T1 and T2 ratios are picked by the spreadsheet to be worst case w I could make my skip lighter But a light duty skip could get beat to pieces And a lighter skip would also reduce T2 34

36. Idea w If my skips were heavier then the skips would account for a higher percentage of the weight. w Since skips are the for T1 and T2 making it a larger proportion will even the ratio w (I could make a similar argument for picking heavier ropes but ropes are expensive and big ropes for larger higher inertia wheels). 35

37. Putting 8 tons of Dead Weight on My skips made it more even 36 Of course I’m still not there yet.

38. Ok – Adding 15.5 dead weight tons to the skip did it!!! 37 Oh boy did it do it – take a look at that tread pressure.

39. So What Can I Do With My Tread Pressure w I can’t change my T1 and T2 w But I can spread the load over a greater area That unfortunately would mean getting a bigger Keope wheel 38

40. There – A 10 ft Wheel Spreads the Load 39 I know – the inertia situation sucks.

41. Come to Think About It – The Factor of Safety Sucks Too 40 How profound – I put more weight on the rope without strengthening the rope And I get into safety factor trouble. (Where the Money Goes)

42. Take A Little Weight Off the Skip and Put A Little More On the Rope 41

43. Time to Pick Out Our Motor 42

44. 43 Looking at Hoist Duty Cycle w Hoist doesn’t always run at a single speed w Initial acceleration - time it starts to move - but its by the loading pocket so we don’t “floor it” w Creep 1 - carefully creeps past the loading area to avoid tearing something up w Main acceleration - after clear of loading area - hit it up to full speed

45. 44 Hoist Duty Cycle

46. 45 Duty Cycle Continued w Run at Full Speed - until you get close enough to the top that you’d better slow down or you'll put the skip up someplace interesting w Main Deceleration - slow down to creep speed before you take out the dump bin w Creep 2 - move slowly into dump position w Final Deceleration - stop to dump

47. 46 What Size Motor? w Required horsepower for motor varies greatly through hoisting cycle w Adjustments are made by calculating the Root Mean Squared (RMS) Horsepower requirements w This requires taking horsepower duties at multiple points through the hoisting cycle

48. The Spreadsheet Does Your Calculations 47 The Motor Sizing is A function of something Called EEW – what is that.?

49. 48 The Mystery of the EEW Term w EEW stands for equivalent effective weight w Hoist contains motors, gears, and large wheels that contribute inertia to the system during acceleration w Could go through long hand and calculate inertia of everything (if you’re sadistic enough) w Alternative is to use manufactures tables that reduce inertia to an equivalent load on a rope.

50. 49 Nordberg Equivalent Effective Weight Chart

51. Reading the Chart 50 We know it is A Keope hoist So I will use The Keope line. I know I have A 10ft wheel So I start at 10 and read up To the Keope Line

52. I Then Read Over to the Equivalent Effective Weight 51 I’ll be conservative In my reading and Call it 36,000 lbs

53. Enter the Number and Get Motor Sizing 52 Of course understanding What all these HP1, HPA And TSL stuff is would Add a lot of understanding

54. Keope Hoist53 The Horsepower Demand of a Keope Hoist Over Time Looks Like This

55. 54 Understanding Keope Duty Cycles Q - Why is there a steady flat line for Horsepower required in a Keope Hoist Duty Cycle A - Horsepower is an energy output per unit time. It takes energy to lift the skip load up the shaft as it travels at full steady speed.

56. 55 Keope Duty Cycles Q - Why is there a sloped line leading upward from when the hoist starts A - When the hoist is operating at less than full speed the load is transported a lesser distance per unit time and thus the energy output per unit time is less. The line has a linear slope because the acceleration rate is a constant.

57. 56 The Keope Duty Cycle Q - Why is there a peak that drops sharply to the flat line for Horsepower to run the Keope A - You must add additional force to accelerate the load. At the end of the acceleration period the additional force is no longer needed.

58. 57 Keope Duty Cycles Q - Why is there a big drop in horsepower at the end of the full speed run for the hoist A - When the hoist decelerates, the momentum of the load provides part of the energy to keep the load moving up the shaft.

59. 58 Keope Cycles Q - Why does the line slope down at the end of the Hoist Cycle A - The load is slowing down and accumulating less potential energy per unit of time.

60. 59 Keope Cycles Q- Why the funny dashed lines that show more power being used at the start and less being recovered at the end. A - Frictional losses

61. Keope Hoist60 Approach to Attacking the RMS Horsepower Requirements w We will calculate components of Horsepower requirements HP 1 will be the horsepower to accelerate the load HP 3 will be the horsepower to move the load at full speed up the shaft HP 2 will be the horsepower recovered from momentum when the load is decelerated HP 6 will be the horsepower still required to lift the load after deceleration starts HP 4 and HP 5 will cover frictional losses

62. You Can See Those Horsepowers Calculated. 61

63. Keope Hoist62 Horsepower #1 (For Keope Hoists) w HP 1 = TSL * V 2 / (550 * g * T a ) Where TSL is the Total Suspended Load V is the Velocity g is the acceleration of gravity Ta is the acceleration time to get the hoist to full speed and includes time to accelerate to creep speed (initial acceleration t 1 ) and then to accelerate to full speed (t 3 ) T a = t 1 + t 3

64. Keope Hoist63 Total Suspended Load w TSL = EEW + 2000 * SL + 2*(2000*SW) + Rope Weight (both sides) SL and SW are the skip weight and load in tons R is the rope weight Because of tail rope there is a full length of rope on both skip sides

65. Keope Hoist64 Horsepower #3 - Power to Lift a Loaded Skip at Full Speed w HP 3 = V * 2000 * SL / 550 Note that the skip weight term is missing I have a skip going down to balance a skip coming up w

66. Keope Hoist65 Horsepower #2 - Negative Horsepower from Momentum During Deceleration w HP 2 = - TSL * V 2 / (550 * g * T r ) Where T r is time during deceleration

67. Keope Hoist66 Horsepower #4 - Losses in Gears and Drives w Derived empirically rather than by physics fundamentals w HP 4 = 0.111 * (2000 * SL * V / 550)

68. Keope Hoist67 Approach to RMS Horsepower Continued w We will add the different fundamental components of horsepower to get the horsepower needs at various cardinal points during the lift w We will label these cardinal points A through E Example D is the peak power required at the height of the acceleration phase w We will put the horsepower values at the cardinal points into the RMS horsepower equation and use that to size the motor.

69. Keope Hoist68 Calculate Horsepower at 3 Cardinal Points w Point A - Peak of the Acceleration Phase w HP A = HP 1 + HP 3 + HP 4 w Point B - During Full Speed Run w HP B = HP 3 + HP 4 w Point C - At Initiation of Deceleration w HP C = HP 2 + HP 3 + HP 4

70. Yip – There are the values 69

71. Keope Hoist70 One More Monster Power Sink w These Big Motors have an Armature to Sink a Battleship! - takes a lot of inertia to spin the thing up or down w HP 5 (to spin it up) = 0.75 * HPA * 1.2 / T a w HP 6 (to spin down) = -0.75 * HPA * 1.2 / T r

72. Those Calculations are Done Too 71

73. Keope Hoist72 Correcting Cardinal Points for Armature Acceleration w Point D is the revision of A peak of acceleration w HP D = HP A + HP 5 w Point E is revision of C initiation of Deceleration w HP E = HP C + HP 6

74. And the Calculations Are There 73

75. Keope Hoist74 RMS Equations Depend on Motor Type w For AC Motor w HPrms = [ ( HP D 2 * T a + HP B 2 * T fs + HP E 2 * T r )/ ( 0.5 * T a + T sf + 0.5 * T r + 0.25 * t r ) ] 0.5 Where T a is acceleration time T sf is full speed time T r is deceleration time t r is the rest time

76. 75 RMS Horsepower for DC Motors w Numerator is the same as AC w Denominator is changed to w ( 0.75 * T a + T sf + 0.75 * T r + 0.5 *t r ) w RMS HP DC = [ Numerator/ Denominator] 0.5

77. Looks Like I Need About 1200 HP 76

78. Time to Pick the Motors 77 I need to choose the number and size of motor and the inertia of the rotor

79. Consider Picks 78 I’d rather go AC with frequency control. I’d like to do 2 600 hp motors but With a 10 ft diameter Keope wheel I’ll still need gear reduction so I would Only get about 94% transmission – My pick a 1250 two pole AC

80. Plug-In My Motor Parameters and Gear Reduction Efficiency 79 Ok – That seems to work (Note that there are limits to how much you can turn down the speed of A motor with variable frequency drives)

81. Our Last Task is to Size the Brake 80

82. We Want Two Things From Our Brake w Hold the maximum possible ubalanced load with a 1.5 factor of safety w If a full speed load passes the ore dump speed it must perform an emergency stop before the skip crashes into the top of the headframe. 81

83. 82 Next Step is to Design the Braking System w During Clutching Operations the Brake must hold the load so it doesn't go to the shaft bottom w Design practice is to rate the brake and clutch to hold the maximum load plus 50% w BR = CR = (D/2) * ( 2000*SL + 2000*SW + H * W r * n) * (1.5) {Units are ft-lbs} D is drum or wheel diameter BR and CR are Brake and Clutch Ratings in ft*lbs torque Note this is the load on one side of a drum hoist if the other is clutched

84. The Spreadsheet Runs the Calculation 83 I need to set my brake rating to at least this size.

85. I Fill In the Numbers 84 Brake rating comes from the recommendation Mass of the rotor came from the motor specification list The gear ratio and speed were worked to get a workable Ratio and a motor speed that was within the turn-down Limit for the motor.

86. 85 Another Factor is Brake Performance During an Emergency Stop w Design is done on worst case scenario maximum unbalanced load traveling at full speed discovered with a minimum tolerance distance before you ride into the head frame or crash into the bottom w Must either design for a tolerance distance your brake can stop in or size up the brake for the tolerance you have.

87. 86 Design Concept w Assume the Brake must fight against the maximum unbalanced load Subtract unbalanced load from brake capacity This leaves the net force available for the emergency stop w Use Newtons Second Law Know the net force available Know total mass in motion Solve for the deceleration rate w Calculate the time and distance to stop

88. 87 Formula w Look at Maximum differential load W = (T 1 - T 2 ) * 2000 T 1 = Max load = SL + SW + (H * W r * n/2000) T 2 = Min Balancing Load = SW

89. The Spreadsheet Applies the Formula to Get the Differential Weight. 88

90. 89 The Mass that must be Stopped w To use Newton's Law to get Force Requirements the load must be in mass Means we must use slugs - some how a system made by Kings using slugs sounds wrong w M = [ ( EEW * R 2 + WR 2 m * GR 2 ) / R 2 + T 1 *2000 + T 2 * 2000 ] / 32.2 Where R is the Drum or Wheel Radius D/2 WR 2 m is the inertia of the motor rotor in ft 2 GR is the gear ratio of the motor to the drive Note that the R 2 terms are needed to convert rotating inertia to equivalent mass

91. 90 Solving for the Deceleration Rate w DR = ( B - W ) / M DR = Deceleration Rate M = Mass to be stopped W = Net unbalanced load (T 1 - T 2 ) * 2000 B = Brake Rating in lbs linear force Get B = BR / R –BR is the Brake Rating in Ft*lbs w Applying the Deceleration Rate Time to Stop = T = V / DR Braking Distance = S = (V/2) * T w Check Braking Distance Against Available or make sure you have the distance

92. Keope Hoist91 The Gear Ratio Problem w GR = VM / VD VM is rpm of motor at rated travel speed VD is rpm of Drum w VD = V / (pi * D ) { remember I need rpm}

93. Solving for the Mass to be Stopped Keope Hoist92

94. Check Out Our Stopping Distance 93 Yup – We Appear to Be OK