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    Conveyor - EX

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    Quỳnh Nguyễn
    Consider the AGVs layout in figure. Vehicles travel counterclockwise around the loop to deliver loads from the load station to the unload station. Loading time = 0.70 min and unloading time = 0.65 min. The following performance parameters are given: Vehicle velocity: 50m/min. Availability: 0.95. Traffic factor: 0.90. Worker efficiency: 0.97. The total of 50 deliveries/hour must be completed by the AGVs. Determine: a) Ideal delivery cycle time b) Number of vehicles required to satisfy the deliver

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    Conveyor - EX

    Uploaded by

    Quỳnh Nguyễn
    977 views9 pages
    Consider the AGVs layout in figure. Vehicles travel counterclockwise around the loop to deliver loads from the load station to the unload station. Loading time = 0.70 min and unloading time = 0.65 min. The following performance parameters are given: Vehicle velocity: 50m/min. Availability: 0.95. Traffic factor: 0.90. Worker efficiency: 0.97. The total of 50 deliveries/hour must be completed by the AGVs. Determine: a) Ideal delivery cycle time b) Number of vehicles required to satisfy the deliver

    Original Title

    Conveyor - EX(1)

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    Consider the AGVs layout in figure. Vehicles travel counterclockwise around the loop to deliver loads from the load station to the unload station. Loading time = 0.70 min and unloading time = 0.65 min. The following performance parameters are given: Vehicle velocity: 50m/min. Availability: 0.95. Traffic factor: 0.90. Worker efficiency: 0.97. The total of 50 deliveries/hour must be completed by the AGVs. Determine: a) Ideal delivery cycle time b) Number of vehicles required to satisfy the deliver

    Copyright:

    © All Rights Reserved
    Download as PPTX, PDF, TXT or read online from Scribd
    Download as pptx, pdf, or txt
    977 views9 pages

    Conveyor - EX

    Uploaded by

    Quỳnh Nguyễn
    Consider the AGVs layout in figure. Vehicles travel counterclockwise around the loop to deliver loads from the load station to the unload station. Loading time = 0.70 min and unloading time = 0.65 min. The following performance parameters are given: Vehicle velocity: 50m/min. Availability: 0.95. Traffic factor: 0.90. Worker efficiency: 0.97. The total of 50 deliveries/hour must be completed by the AGVs. Determine: a) Ideal delivery cycle time b) Number of vehicles required to satisfy the deliver

    Copyright:

    © All Rights Reserved
    Download as PPTX, PDF, TXT or read online from Scribd
    Download as pptx, pdf, or txt
    of 9
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    The key takeaways are the three principles of conveyor design: speed rule, capacity constraint, and uniformity principle.

    The three principles of conveyor design according to Kwo are the speed rule, capacity constraint, and uniformity principle.

    The factors that determine the spacing between containers on a conveyor system are the conveyor speed and the loading/unloading time.

    Conveyor EXERCISES

    EXERCISES
    1) An overhead trolley conveyor is configured as a
    closed 1oop. The delivery loop has a length of 150
    m and the return loop is 100 m. All parts loaded at
    the load station are unloaded at the unload station.
    Each hook on the conveyor can hold one part and
    the hooks are separated by 3 m. Conveyor speed :
    0.3 m/sec. Determine
    a) Number of parts in the conveyor system under
    normal operations,
    b) parts flow rate; and
    c) Maximum loading and unloading times that are
    compatible with the operation of the conveyor system.
    •delivery loop is Ld = 150 m
    •return loop is Le = 100 m
    •np = 1
    •Sc = 3 m
    •Vc = 0.3 m/sec.
    a) Number of parts in the conveyor?
    nc = L/Sc = (Ld + Le)/Sc
    b) parts flow rate?
    •Rf = np * Vc / Sc
    c) Maximum loading and unloading times
    •Td = Ld /Vc (delivery time on the forward loop
    2) A 300-ft long roller conveyor operates at a velocity : 30
    ft/min and is used to move parts in containers between load
    and unload stations. Each container holds 20 parts. One
    worker at the load station is able to load parts into containers
    and place the containers onto the conveyor in 30 sec. It takes
    20 sec to unload at the unload station. Determine
    (a) center-to-center distance between containers,
    (b) number of containers on the conveyor at one time, and
    (c) hourly flow rate of parts.
    (d) By how much must conveyor speed be increased in
    order to increase flow rate to 3,600 parts/hr?
    •Ld = length of conveyor = 300 ft
    •Vc = 30 ft/min
    •np = 20
    •TL = loading time = 30 sec
    •TU = Unloading time = 20 sec
    (a) center-to-center distance
    •Sc = Vc * TL
    (b) number of containers on the conveyor at one time
    •nc = L / Sc
    (c) hourly flow rate of parts

    d) conveyor speed?
    •Vc = Rf * Sc/np
    •(Rf = 3,600parts/hr)
    3) A roller conveyor moves tote pans in one direction at 300
    ft/min between a load station and an unload station, a distance of
    450 ft. With one worker, the time to load parts into a tote pan at
    the load station is 5 sec per part. Each tote pan holds 15 parts. In
    addition, it takes 12 sec to load a tote pan of parts onto the
    conveyor. Determine
    (a) spacing between tote pan centers flowing in the conveyor
    system and
    (b) flow rate of parts on the conveyor system.
    (c) Consider the effect of the Unit Load Principle. Suppose the
    tote pans were smaller and could hold only one part instead of 15.
    Determine the flow rate of parts in this case if it takes 5 sec to load
    a tote pan onto the conveyor (instead of 12 sec for the larger tote
    pan), and it takes the same 5 sec to load the part into the tote pan.
    •Ld = length of conveyor = 450 ft
    •Vc = 300 ft/min
    •np = 15
    •TL = loading time = ( 5 * 15 + 12) sec
    (a) spacing between tote pan centers
    •Sc = Vc * TL
    (b) flow rate of parts
    •Rf = Vc * np/ Sc
    (c) Flow rate
    •np = 1
    •TL = loading time = 5 sec + 5 sec
    •Sc = Vc * TL
    • Rf = Vc / Sc
    4) A recirculating conveyor has a total length of
    700 ft and a speed of 90 ft/min. Spacing of part
    carriers 14ft. Each carrier holds one part.
    Automated machines load and unload the
    conveyor at the load and unload stations. Time to
    load a part is 0.10 min and unload time is the
    same. To satisfy production requirements, the
    loading and unloading rates are each 2.0 parts
    per min. Evaluate the conveyor system design
    with respect to the three principles developed by
    Kwo.
    •Ld = length of conveyor = 700 ft
    •Vc = 90 ft/min
    •np = 1
    •Sc = 14 ft
    •TL = loading time = TU = Unloading time = 0.1 min
    •Rf = 2 parts/min
    Evaluate the conveyor system design with the three principles
    •Speed Rule
    The lower limit
    The upper speed limit

    • Capacity Constraint

    •Uniformity Principle. parts (loads) should be uniformly distributed


    throughout the length of the conveyor
    •there will be no sections of the conveyor in which every carrier is full while
    other sections are virtually empty

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