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 system
30s – satisfactory system
• Quantity of service required
– The handling capacity within 5 minutes
• Estimation of arrival rate
Handling 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 seconds
H = highest call reversal floor
S = average no. of stops
tv= time to transit 2 adjacent floors at rated speed
in seconds
ts = time consumed when making a stop in
seconds
tp= passenger transfer time for entering or exiting
the lift car in seconds
P = 0.8xlift car capacity in person
Question: Lift Traffic Design
Design Input:
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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 floor
ii) Number of persons per arrival
iii) Nos. of lift trips per 5 minutes
iv) Car size required (person car)
v) Total travel distance per lift
vi) Transit time between 2 floors
vii) RTT (Round Trip Time)
viii) Nos. of car required
ix) 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