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Lift Design Example The following notes give an example for design of lifts

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**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)

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(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

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**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)

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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

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Results: i.e. Pop=1200/12=100 persons per floor •, i.e x0.8x100x10flr=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 = persons / 5min

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