AN IMPROVED DESIGN PROCEDURE 1 The Iterative Balance Method

In the previous two examples the handling capacity was larger than that required and in Example 6.5 the required interval was not achieved. What should the designer do? Clearly in Example 6.5 there was too much handling capacity and a poorer interval than desired. In effect there were not enough

intending passengers to use the capability of the installed system.

< previous page

page_142

next page >

Page 142

Should the designer modify the component parts of the RTT expression to achieve a balance? The first term can only be altered by changing the rated speed and the effect would be small. The second term can be altered by changing the single floor flight time or door timings. Thus a lower specification door gear could produce a matching handling capacity. The third term can be altered by changing the rated car capacity, but this will alter S and H and can be counterproductive. Experienced designers will use intuitive procedures incorporating combinations of the above to establish a suitable design to cater for the desired handling capacity.

Using the conventional design method above, a designer would propose initial values for the dynamic parameters and estimate the rated car capacity of the lifts based on experience. It is then assumed that the average number of passengers (P) carried per trip is 80% of the rated car capacity and values for the expected number of stops (S) and the average highest reversal floor (H) are evaluated. Hence the round trip time (RTT), interval (INT) and handling capacity (HC) are calculated.

At this stage, the designer compares the calculated value of HC with the number of passengers to be moved in the peak five minutes. If HC is greater than or equal to this number of passengers, then the designer is satisfied that the system will cope with the traffic. The configuration will be trimmed if the handling capacity is too large, and should it be smaller than the required value, then the designer must repeat the evaluation for more and (or) bigger and (or) faster cars. However, the values calculated for RTT, INT and HC are exact only if there is a perfect match between the arrival rate and handling capacity.

The procedure suggested by Tregenza (1972) as Equation (5.17) is significant as he presents the idea of matching the lift handling capacity to the desired handling capacity exactly. This is achieved by not rigidly fixing P as a percentage of rated car capacity. P is allowed to take the most appropriate value. From Tregenza, P is equal to λINT.

A new design procedure, the Iterative Balance Method (Barney and Dos Santos, 1975), can now be proposed, which can be used with either the conventional or Poisson formulae. For simplicity, it is presented as a series of steps (Table 6.7).

To obtain values for H and S for the number of passengers to be carried in the car, Table 5.1 could be used by interpolating between the 80% values shown in parenthesis. These values follow a nonlinear sequence and make it difficult to estimate intermediate values. To assist, Table 6.8 is provided which gives values for integer values of P from five to 20 persons.

The Iterative Balance Method is a classical two point boundary problem, where the start and end results are known—in this case the balancing interval. To arrive at an answer, a two point boundary problem has to iterate, ie: change the start value to converge on the end value. To do this a suitable algorithm has to be chosen. Pick the wrong algorithm and the start and end values diverge rapidly, leading to infinite values, ie: no solution. The algorithm chosen in step (7), which is simply “twice the new value minus the old value”, does converge.

In step (8) it is important to remember that, where an average car load is much greater than 80%, poor passenger service might result, ie: long waiting times and queues. The designer must select a suitable car size to meet desired economic and operating conditions. It is important not to make any

simplifications either for the sake of arithmetic ease or to meet a preconceived idea of a suitable lift installation at any stage of the calculation until step (8) is reached. Always calculate primary and secondary values as precisely as possible.

The procedure outlined in Table 6.7 allows the RTT to be calculated for any values of P, H and S. Only when the initial (estimated) interval matches the final (calculated) interval are any decisions made.

Page 143

Table 6.7 New design procedure StepProcedure

1 Decide on λ rate of passenger arrivals over 5 minutes 2 Obtain or decide upon lift system data:

N Number of floors tv the interfloor time ts the operating time

tp the passenger transfer time 3 Estimate an appropriate interval 4 Obtain:

P average car load H average reversal floor S expected number of stops 5 Calculate RTT

6 Select L, the number of lifts to produce an interval close to that estimated in step (3)

7 Compare the estimated interval in step (3) with the calculated interval in step (6) and if significantly different, estimate another value for the interval and then iterate from step (4). A possible new trial could be:

New INT=INT (step (6))+[INT (step (6))−INT (step (3))]

8 Select a suitable standard car capacity, which allows approximately 80% average car load 6.10.2 Example 6.6

In Example 6.5 there was too much handling capacity and a poorer interval than desired. Using the Iterative Balance Method indicated in Table 6.7, the following is obtained:

(1) (2) (3) Let

(4) P=λINT=0.4×25=10 persons

H=9.5 (calculated from Equations (5.12) and (5.5) S=6.5 or from interpolation in Table 6.8)

(5)

(6) Let L=5, then

< previous page

page_144

next page >

Page 144

Table 6.8a Values of H and S for different values of P from 5 to 12 persons

Floors 5 6 7 8 9 10 11 12 N H S H S H S H S H S H S H S H S 5 4.6 3.4 4.7 3.7 4.8 4.0 4.8 4.2 4.9 4.3 4.9 4.5 4.9 4.6 4.9 4.7 6 5.4 3.6 5.6 4.0 5.7 4.3 5.7 4.6 5.8 4.8 5.8 5.0 5.9 5.2 5.9 5.3 7 6.3 3.8 6.4 4.2 6.5 4.6 6.6 5.0 6.7 5.3 6.7 5.5 6.8 5.7 6.8 5.9 8 7.1 3.9 7.3 4.4 7.4 4.9 7.5 5.3 7.6 5.6 7.7 5.9 7.7 6.2 7.8 6.4 9 8.0 4.0 8.2 4.6 8.3 5.1 8.4 5.5 8.5 5.9 8.6 6.2 8.7 6.5 8.7 6.8 10 8.8 4.1 9.0 4.7 9.2 5.2 9.3 5.7 9.4 6.1 9.5 6.5 9.6 6.9 9.6 7.2 11 9.6 4.2 9.9 4.8 10.1 5.4 10.2 5.9 10.3 6.3 10.4 6.8 10.5 7.1 10.6 7.5 12 10.5 4.2 10.7 4.9 11.0 5.5 11.1 6.0 11.2 6.5 11.3 7.0 11.4 7.4 11.5 7.8 13 11.3 4.3 11.6 5.0 11.8 5.6 12.0 6.1 12.1 6.7 12.3 7.2 12.3 7.6 12.4 8.0 14 12.1 4.3 12.5 5.0 12.7 5.7 12.9 6.3 13.0 6.8 13.2 7.3 13.3 7.8 13.4 8.2 15 13.0 4.4 13.3 5.1 13.6 5.7 13.8 6.4 14.0 6.9 14.1 7.5 14.2 8.0 14.3 8.4 16 13.8 4.4 14.2 5.1 14.5 5.8 14.7 6.5 14.9 7.0 15.0 7.6 15.1 8.1 15.2 8.6 17 14.6 4.4 15.0 5.2 15.3 5.9 15.6 6.5 15.8 7.1 15.9 7.7 16.0 8.3 16.1 8.8 18 15.5 4.5 15.9 5.2 16.2 5.9 16.5 6.6 16.7 7.2 16.8 7.8 16.9 8.4 17.1 8.9 19 16.3 4.5 16.8 5.3 17.1 6.0 17.4 6.7 17.6 7.3 17.7 7.9 17.9 8.5 18.0 9.1 20 17.1 4.5 17.6 5.3 18.0 6.0 18.2 6.7 18.5 7.4 18.6 8.0 18.8 8.6 18.9 9.2 21 18.0 4.5 18.5 5.3 18.8 6.1 19.1 6.8 19.4 7.5 19.6 8.1 19.7 8.7 19.8 9.3 22 18.8 4.6 19.3 5.4 19.7 6.1 20.0 6.8 20.3 7.5 20.5 8.2 20.6 8.8 20.8 9.4 23 19.6 4.6 20.2 5.4 20.6 6.2 20.9 6.9 21.2 7.6 21.4 8.3 21.5 8.9 21.7 9.5 24 20.5 4.6 21.1 5.4 21.5 6.2 21.8 6.9 22.1 7.6 22.3 8.3 22.5 9.0 22.6 9.6

N is the number of floors above the main terminal.

Page 145

Table 6.8b Values of H and S for different values of P from 13 to 20 persons

Floors 13 14 15 16 17 18 19 20 N H S H S H S H S H S H S H S H S 5 4.9 4.7 5.0 4.8 5.0 4.8 5.0 4.9 5.0 4.9 5.0 4.9 5.0 4.9 5.0 4.9 6 5.9 5.4 5.9 5.5 5.9 5.6 5.9 5.7 6.0 5.7 6.0 5.8 6.0 5.8 6.0 5.8 7 6.9 6.1 6.9 6.2 6.9 6.3 6.9 6.4 6.9 6.5 6.9 6.6 6.9 6.6 7.0 6.7 8 7.8 6.6 7.8 6.8 7.9 6.9 7.9 7.1 7.9 7.2 7.9 7.3 7.9 7.4 7.9 7.4 9 8.7 7.1 8.8 7.3 8.8 7.5 8.8 7.6 8.8 7.8 8.9 7.9 8.9 8.0 8.9 8.1 10 9.7 7.5 9.7 7.7 9.8 7.9 9.8 8.1 9.8 8.3 9.8 8.5 9.8 8.6 9.9 8.8 11 10.6 7.8 10.7 8.1 10.7 8.4 10.7 8.6 10.8 8.8 10.8 9.0 10.8 9.2 10.8 9.4 12 11.6 8.1 11.6 8.5 11.6 8.7 11.7 9.0 11.7 9.3 11.7 9.5 11.8 9.7 11.8 9.9 13 12.5 8.4 12.5 8.8 12.6 9.1 12.6 9.4 12.7 9.7 12.7 9.9 12.7 10.2 12.8 10.4 14 13.4 8.7 13.5 9.0 13.5 9.4 13.6 9.7 13.6 10.0 13.7 10.3 13.7 10.6 13.7 10.8 15 14.4 8.9 14.4 9.3 14.5 9.7 14.5 10.0 14.6 10.4 14.6 10.7 14.6 11.0 14.7 11.2 16 15.3 9.1 15.4 9.5 15.4 9.9 15.5 10.3 15.5 10.7 15.6 11.0 15.6 11.3 15.6 11.6 17 16.2 9.3 16.3 9.7 16.4 10.2 16.4 10.6 16.5 10.9 16.5 11.3 16.6 11.6 16.6 11.9 18 17.2 9.4 17.2 9.9 17.3 10.4 17.4 10.8 17.4 11.2 17.5 11.6 17.5 11.9 17.6 12.3 19 18.1 9.6 18.2 10.1 18.2 10.6 18.3 11.0 18.4 11.4 18.4 11.8 18.5 12.2 18.5 12.6 20 19.0 9.7 19.1 10.2 19.2 10.7 19.3 11.2 19.3 11.6 19.4 12.1 19.4 12.5 19.5 12.8 21 19.9 9.9 20.0 10.4 20.1 10.9 20.2 11.4 20.3 11.8 20.3 12.3 20.4 12.7 20.4 13.1 22 20.9 10.0 21.0 10.5 21.1 11.1 21.1 11.5 21.2 12.0 21.3 12.5 21.3 12.9 21.4 13.3 23 21.8 10.1 21.9 10.7 22.0 11.2 22.1 11.7 22.2 12.2 22.2 12.7 22.3 13.1 22.3 13.5 24 22.7 10.2 22.9 10.8 22.9 11.3 23.0 11.9 23.1 12.4 23.2 12.8 23.2 13.3 23.3 13.8

N is the number of floors above the main terminal.

< previous page

page_146

next page >

Page 146

(7) The value calculated of 24.3 s does not closely match the start value of 25 s. Try a new value viz: (4′)

(5′) (6′) Let

(7′) This is a close enough match (error <0.1 s)

(8) Standard rated car capacities are 10, 13 and 16. The 10 passenger car size would be too small and 16 passenger car size too large. Try 13 person cars.

Examining the result above, it can be seen that the arrival rate and the lift handling capacity exactly balance. This was achieved by the number of passengers in the car changing until the resulting system handling capacity was equal to the demand represented by the passenger arrival rate. The interval is less than the specified 25 s, but not a lot smaller. This is an improvement.

6.10.3 Example 6.7

As the Iterative Balance Method is important, a further example is presented. The arrival rate is 110 calls/5-minutes and an interval of 30 s is required. Relevant data are:

(1) (2)

(3) Let INT=30 s

(4) P=λINT=0.37×30=11 persons

H=15.1 (calculated from Equations (5.12) and (5.5) S=8.1 or from interpolation in Table 6.8)

(5)

(6) Let L=4, then INT=124.5/4=31.1 s

Page 147

(7) The value calculated of 31.1 s does not match the start value of 30 s. Try a new value viz: (4′)

(5′) (6′) Let

(7′) This is a close enough match (error <0.1 s)

(8) Standard rated car capacities are 13, 16 and 21. The 13 passenger car size would be too small and 21 passenger car size too large. Try 16 person cars.

This is a satisfactory system. Remember that the most suitable lift configuration chosen would be based on economic grounds, the cost of the system and performance.

6.11 EPILOGUE

In this chapter a number of definitions have been proposed. It is important that each design be

prepared against a set of mutually understood definition of terms. Unfortunately most designers and lift companies have their own set. A set of over 50 terms, widely used in the USA, are those produced by the NEII (2000). Another set of nearly 2000 terms and cross references can be found in Barney et al. (2001).

< previous page

page_148

next page >

Page 148

This page intentionally left blank.

Page 149