Lift Design Example

• The following notes give an example for design of lifts

Traffic Design - Nos. and Size of Lifts

• Estimation of population

• Quality of service required– The up-peak interval of an office tower:

20s or less – excellent system30s – satisfactory system

• Quantity of service required– The handling capacity within 5 minutes

• Estimation of arrival rateHandling capacity = (5minx60x0.8xLift Car Capacity in nos. of persons)/(up peak interval x population above terminal floor of zone)

(cont.)

• Round Trip Time (RTT) by up peak model

RTT = 2Htv + (S +1)ts+ 2Ptp

Where,

RTT = round trip time in secondsH = highest call reversal floorS = average no. of stopstv= time to transit 2 adjacent floors at rated speed in secondsts = time consumed when making a stop in secondstp= passenger transfer time for entering or exiting the lift car in secondsP = 0.8xlift car capacity in person

Question: Lift Traffic Design

Design Input:

• An office building of 10 floors above the main terminal is to be built, each floor of 1200 sqm of net space.

• The inter-floor distance is 3.3m.• The required up peak interval is 30s (i.e. satisfactory grade)• Estimation of population: assume 12sqm per person per floor,• Assume 12.5% population peak arrival rate• Assumed lift rated speed = 1.6m/s

Given that:

• ts=7.7s (from code)• tp=1.2s (assumed)• H=9.5, S=6.7 (from code)

Design Output:

Determine the following information:

i) Persons per floorii) Number of persons per arrivaliii) Nos. of lift trips per 5 minutesiv) Car size required (person car)v) Total travel distance per liftvi) Transit time between 2 floorsvii) RTT (Round Trip Time)viii) Nos. of car requiredix) Handling Capacity (persons per 5 mins)x) What is the difference between quantity of service

and quality of service

Results:

i.e. Pop=1200/12=100 persons per floor•, i.e. 0.125x0.8x100x10flr=100 persons• Qty of service: Nos of trip in 5 minutes = 5minx60/30s=10• Nos. of person per trip = 100/10 = 10• The required car size is 10/0.8 = 13-person car• Total travel distance = 3.3x10=33m• Assumed rated speed = 1.6m/s• tv=3.3/1.6=2.1s• ts=7.7s (from code)• tp=1.2s (assumed)• P=13x0.8=10.4 persons• H=9.5, S=6.7 (from code)

• RTT=(2x9.5x2.1)+(6.7+1)7.7+(2x10.4x1.2)=124.2s• Since up peak interval required is 30s, i.e. 4 cars are required• The up peak interval is 124.2/4 = 31.1s• The up peak handling capacity is (300/124.2)x10.4x4 = 100.5 persons / 5min