Vertical Transportation Design and Traffic Calculations – Part Eleven


Today we will explain the following:
  1. How to size the lift motor KW/HP,
  2. How to calculate the Lift Energy Consumption.


First: How To Size The Lift Motor KW/HP




There are many methods for sizing the lift motor KW/HP like:

  1. Professional Formulas Method,
  2. Baldor Formulas Method,
  3. Equivalent Weight Method,
  4. Buildings Energy Code (BEC)’S Tables Method,
  5. Baldor Tables Method,
  6. Curves Method.



1- Professional Formulas Method

This method has two checks to be made:
  1. Power Check,
  2. Acceleration Check.


1.1 Power Check

  • Power check is used to check that the motor supplies sufficient power to move the out of balance mass (The fully loaded passenger mass - counterweight mass) at the rated speed.
  • The worst case is taken as that of lifting a fully loaded car in the up direction at the rated speed (assuming that the counterweight ratio is 50% or less). This calculated in the following stages:


  1. The fully loaded passenger mass is reduced by the amount of counterweight. This represents the out of balance mass (m).
  2. This is multiplied by the acceleration of gravity, to determine the force in Newtons needed to move this out of balance mass against gravity in the up direction (F (Newtons) = g (m/s2) x m (Kg)) & ( g = 9.81 m/s2)
  3. Multiplying this force by the rated speed gives the rate of flow of energy, or in other words the power in watts. This represents the net output power of the system. (Net output P (watts) = F (Newtons) x s (m/s))
  4. This calculated net output power has to flow through the system, and thus a larger value of power has to be supplied to account for all losses in the shaft and gearbox. Thus, the net output power is divided by the shaft efficiency and the gearbox efficiency to provide the required motor power output rating. (Required motor power (watts) = Net output P (watts) / efficiency of shaft and gear box)
  5. The efficiency of the shaft and the gearbox (forward) can be taken as a combined figure of 85%, if no exact data is available. The forward efficiency of the gearbox is taken in this case because the motor is driving.


The formula for sizing the motor for a lift is as follows:



Where:
  • M Required motor power in KW
  • P is the rated passenger number in the car;
  • 75 stands for 75 kg/passengers;
  • 9.81 is the acceleration due to gravity;
  • s is the rated top speed;
  • CF is the counterweight factor (a factor less than 1);
  • η is the total efficiency of the installation (taken around 85%).


Note:
  • For a hydraulic lift, the same formula can be used by replacing CF by -1.


Example#1:

In a lift system which has an MG set supplying its DC hoist, calculate the size of the AC prime mover for a 49 passenger lift, running at 1.6 m/s, if the efficiency of the installation (including the MG set, the DC hoist motor and the shaft efficiency) is 70%, and the counterweight ratio is 40%.

Solution:

Applying the formula above gives:
M = (P x 75 x 9.81 x s x (1-CF)) / η = 49 x 75 x 9.81 x 1.6 x (1-0.4) / 0.7 = 49.44 KW
So, 50 kW (or 55 kW) motor can be selected.

Example#2:

An 8 passenger hydraulic lift, runs at a speed of 1 m/s, and has an overall efficiency of 80%. If the mass of the car is equal to the rated load in the car, then calculate the required minimum size of the motor for the pump unit.

Solution:

As the mass of the car and associated equipment is equal to the rated load, then CF can be taken as -Thus, applying the formula gives a motor size of:
M = (P x 75 x 9.81 x s x (1-CF)) / η = 8 x 75 x 9.81 x 1 x (1-(-1)) / 0.8= 14.7 KW
Thus a motor sized at 16 kW could be used.

Example#3:

A two speed lift has a rated speed of 1.2 m/s, and a car load of 13 passengers. If the overall system efficiency is 75%, and the counterweight ratio of 50% is used, calculate the size of the motor.

Solution:

Applying the above formula, gives:
M = (P x 75 x 9.81 x s x (1-CF)) / η = 13 x 75 x 9.81 x 1.2 x (1-0.5) / 0.75 = 7.65 KW
Thus, an 8.4 kW motor can be selected.

Example#4:

A VVAC driven lift has a rated speed of 1.0 m/s, and a car load of 78 passengers. If the overall system efficiency is 68%, and the counterweight ratio of 45%, calculate the size of the motor.

Solution:

Applying the above formula, gives:
M = (P x 75 x 9.81 x s x (1-CF)) / η = 78 x 75 x 9.81 x 1 x (1-0.45) / 0.68 = 46.4 KW
Thus, a 50 kW motor can be selected.


1.2 Acceleration Check

Acceleration Check is used to ensure that the torque rating of the motor is sufficient to accelerate the motor in the necessary time.

The maximum value of linear acceleration supplied by the motor can be calculated from the following formula:



Where:
  • a is the maximum value of linear acceleration supplied by the motor
  • Trated is the rated torque of the motor
  • P is the rated number of passengers in the car;
  • 75 is the mass per passengers in kg;
  • CF is the counterweight ratio.
  • n is the motor speed in rev/min;
  • v is the linear lift speed in m/s.
  • η is the efficiency of the installation.
  • Jmot is the moment of inertia of the motor;
  • Jgear is the moment of inertia of the gearbox;
  • Jcoupling is the moment of inertia of the coupling and the brake drum;
  • Jflywheel is the moment of inertia of the flywheel or handwheel.
  • Q is the rated load of the car;
  • C is the mass of the car;
  • C/W is the mass of the counterweight;


The values of accepted acceleration is 0.8-1.0 m/s2 And If the acceleration is less than 0.6 m/s2, then the motor is not adequate, and a larger size motor with a higher torque needs to be selected, or the masses have to be reduced.

Notes:

1- If the value of acceleration is more than 1.0 m/s2, this is still acceptable if the drive is a variable speed drive because they drive the exact required voltage to achieve the required acceleration.  Variable speed drives are like:
  • ACVV: Variable voltage AC drive.
  • VVVF: Variable voltage variable frequency drive
  • DC SCR: Solid state DC drive, employing thyristors


2- If the drive is a two speed drive, then a flywheel might be needed to reduce the value of acceleration.


Example#5:

A lift system is designed to run at 1.75 m/s, with a car capacity of 28 passengers. The car mass is 1000 kg, and the counterweight ratio is 50%. Select a motor which will run at a speed of 920 rpm from the table below.

Motor
Torque (Nm)
Power (kW)
Motor inertia (kgm2)
A
138
13.5
0.55
B
167
16
0.62
C
200
19.5
0.77
D
248
24
1.1
E
286
27.5
1.3
F
325
31
1.52


Solution:

First: Applying the power check, gives a required power of:
M = (P x 75 x 9.81 x s x (1-CF)) / η = 28 x 75 x 9.81 x 1.75 x (1-0.5) / 0.75 = 24 KW
The nearest motor is motor D.

Second: Applying acceleration Check:


Thus, the motor will also be capable of accelerating the system at the required acceleration.

Example#6:

Taking the installation in Example#4 again, find out the maximum value of linear acceleration it is capable of assuming the following additional parameters:
Trated= 729 Nm
Tmax = 2.34 Trated
Jmot = 2.1 kg m2
Jcoupling = 0.25 kg m2
n = 610 rpm
C = 8700 kg
C/W = 11365 kg

Solution:

Applying acceleration Check for calculating the linear acceleration gives:


Thus, the motor will also be capable of accelerating the system at the required acceleration.




2- Baldor Formula Method

For a given Elevator, the required motor HP may be calculated using the following formula:


Where:
  • LBS = Car capacity in pounds
  • FPM = Car speed in feet per minute
  • OCW = Over counter weight in % of car capacity (usually ranges from 40 to 50%)
  • EFF = Elevator mechanical efficiency (decimal)


Also, for a given Elevator, the required motor KW may be calculated using the following formula:


Where:
  • kg = Car capacity in Kilograms
  • m/s = Car speed in meters per second
  • OCW = Over counter weight in % of car capacity
  • EFF = Elevator mechanical efficiency (decimal) – see below Table-1.

  
Elevator Car Type
Elevator Efficiency
Slow Moving Cars
45%
Higher Speed Cars
70%
Geared Machines
55% - 70%
Gearless Machines
90%
Table-1: Elevator Mechanical Efficiency



3- Equivalent Weight Method

The following steps can be used to size the motor HP by using Equivalent Weight Method:

First: calculate the total weight W in KGs as follows:
  • The car weight in KGs is given or estimated.
  • Total Passengers weight in KGs = 75 x total numbers of Passengers
  • Total weight W in KGs = car weight in KGs + total Passengers weight in KGs


Second: calculate Equivalent weight in KGs = 0.5 Total weight W in KGs

Third: calculate Motor HP = Equivalent weight in KGs X 1.5/75


Example#7:

Calculate the size of the AC prime mover for 80 passenger lift while The car mass is 1000 kg By using Equivalent Weight Method.

Solution:

First: calculate the total weight W in KGs as follows:
The car weight in KGs = 1000 kg
Total Passengers weight in KGs = 75 x total numbers of Passengers = 75 x 80 = 6000 kg
Total weight W in KGs = car weight in KGs + total Passengers weight in KGs = 1000 + 6000 = 7000 kg

Second: calculate Equivalent weight in KGs
Equivalent weight in KGs = 0.5 Total weight W in KGs = 0.5 x 1600 = 800 kg

Third: calculate Motor HP
Motor HP = Equivalent weight in KGs X 1.5/75 = 800 x 1.5 / 75 = 16 HP



4- Buildings Energy Code (BEC)’S Tables Method

Buildings Energy Code (BEC)’S Tables are applied only for the following lift and escalator installations:

  • Passenger lift, bed passenger lift, freight lift, vehicle lift, escalator and passenger conveyor;
  • Fireman’s lift that operates under normal condition (i.e. Fireman’s Switch is off);
  • Lift and escalator installation attached to the façade of the building and owned by the building owner.


Buildings Energy Code (BEC)’S Tables are not applied for the following lift and escalator installations:

  • Mechanized vehicle parking system;
  • Service lift;
  • Stair lift;
  • Industrial truck loaded freight lift;
  • Lift in a performance stage;
  • Powered lifting platform;
  • Lift that is not operated on a traction drive by suspension ropes or not operated by a hydraulic piston. 


4.1 Buildings Energy Code (BEC)’S Tables for Traction Drive Lift

The running active electrical power of the motor drive of a traction drive lift carrying a rated load at its rated speed in an upward direction should not exceed the corresponding maximum allowable value given in Table -2 and Table-3.


Table -2


Table -3

The requirement in Tables-2&3 should not be applicable to:

  • A lift with rated speed not less than 9 m/s serving a zone of over 50-storey or over 175m between top/bottom-most landing and principal/ground landing, and designated as fireman’s lift or sky lobby shuttle serving two principal stops;
  • A lift with rated load at or above 5000 kg at rated speed of 3 m/s or above.


4.2 Buildings Energy Code (BEC)’S Tables for Hydraulic Lift

The running active electrical power of the hydraulic oil pump motor of a hydraulic lift carrying a rated load at its rated speed in an upward direction should not exceed the corresponding maximum allowable value given in Table-4.

Maximum Electrical Power (kW) of Hydraulic Lift at Rated Load
Rated Load L (kg)
Power (kW)
L < 1000 kg
25.3
1000 kg ≤ L ≤ 2000 kg
47.9
2000 kg ≤ L ≤ 3000 kg
67.7
3000 kg ≤ L ≤ 4000 kg
87.6
4000 kg ≤ L ≤ 5000 kg
109.3
L ≥ 5000 kg
0.022L
Table-4

4.3 Buildings Energy Code (BEC)’S Tables for Escalator

The running active electrical power of the steps driving motor of an escalator with nominal width W and rise R when operating under no-load condition at rated speed Vr should not exceed the corresponding maximum allowable value given in Table-5.


Table-5

4.4 Buildings Energy Code (BEC)’S Tables for Passenger Conveyor

The running active electrical power of the steps driving motor of a passenger conveyor with length L and nominal width W at an inclination up to 60 from horizontal when operating under no-load condition at rated speed Vr should not exceed the corresponding maximum allowable value given in Table-6.

Table-6



5- Baldor Tables Method

Baldor Tables-7&8 can be used to determine the elevator motor HP/KW to use for your application by making the following steps:

  1. Find the “Car Speed” column in the first row of the table.
  2. Follow that column down to find the “Car Capacity” row.
  3. Follow that row to the left and read the recommended HP/KW size of the motor.


Elevator
Motor
HP
Elevator Car Speed In Feet Per Minute
100
150
200
250
300
350
400
500
700
Assumed Overall Mechanical Efficiency - Gearless Efficiency - 90%
55%
58%
60%
62%
63%
64%
64%
67%
70%
Maximum Elevator Capacity – In Pounds LB
7.5
2300
1600
1250
1030
870
750
660
550
410
10
3000
2150
1660
1370
1150
1000
880
740
550
15
4500
3200
2500
2060
1730
1500
1310
1100
820
20
6050
4300
3300
2750
2300
2000
1750
1470
1090
25
7500
5400
4150
3400
2880
2500
2190
1840
1370
30
9100
6400
4950
4000
3470
3000
2620
2210
1640
40
12100
8600
6650
5450
4620
4000
3500
2950
2180
50
15125
10700
8300
6840
5760
5000
4370
3670
2730
60
18150
12870
9900
8200
6940
6000
5250
4430
3280
75
22685
16090
12375
10300
8650
7500
6560
5520
4100
Table-7: Motor HP Sizes for Geared Elevators


Elevator
Motor
KW
Elevator Car Speed In Meters Per Second
0.5
0.75
1.0
1.25
1.5
1.75
2.0
2.5
3.5
Assumed Overall Mechanical Efficiency - Gearless Efficiency - 90%
55%
58%
60%
62%
63%
64%
64%
67%
70%
Maximum Elevator Capacity – In Kilograms
5.6
1045
727
568
468
395
341
300
250
186
7.5
1364
977
755
623
523
455
400
336
250
11.2
2045
1455
1136
936
786
682
595
500
373
14.9
2750
1955
1500
1250
1045
909
795
668
495
18.6
3409
2455
1886
1545
1309
1136
995
836
623
22.4
4136
2909
2250
1818
1577
1364
1191
1005
745
29.8
5500
3909
3023
2477
2100
1818
1591
1341
991
37.3
6875
4864
3773
2945
2618
2273
1986
1668
1241
44.8
8250
5850
4500
3727
3155
2727
2386
2014
1491
56
10311
7314
5625
4682
3932
3409
2982
2509
1864
Table-8: Motor KW Sizes for Geared Elevators

Note:
  • The above Baldor Tables are for typical car capacity with 40% over counter weighting.




6- Curves Method

From Curve-1, you can find the motor power in KW as follows:


Curve-1

  • Step#1: Find the value of lift speed on the lift speed axis,
  • Step#2: Draw a perpendicular line to the “lift Speed” axis at the point determined in step#1 to intersect the curve represent the total weight of the lift,
  • Step#3: Draw a horizontal line from the point of intersection assigned in step#2 to intersect with the motor power axis,
  • Step#4: Read the intersection point to give the motor power in KW for gear-less or geared machines,
  • Step#5: Apply the demand factor from Table-9 as per the required number of lifts to the reading fromstep#4 to calculate the required actual KW of the motor.


Nos. of lifts
Demand factor
2
0.85
3
0.77
4
0.72
5
0.67
6
0.63
7
0.59
10
0.52
15
0.44
20
0.4
25 and over
0.35
Table-9: Demand Factor for Group of Lifts

Example#8:

Calculate the motor KW for each lift from group of 4 numbers Gear-less lifts with speed 3 m/s each noting that the total weight of each lift is 1500 kg.

Solution:


Step#1: Find the value of lift speed on the lift speed axis = 3 m/s
Step#2: Draw a perpendicular line to the “lift Speed” axis at the point determined in step#1 to intersect the curve represent the total weight of the lift,
Step#3: Draw a horizontal line from the point of intersection assigned in step#2 to intersect with the motor power axis,
Step#4: Read the intersection point to give the motor power in KW for gear-less or geared machines = 30 KW
Step#5: Apply the demand factor as per the required number of lifts to the reading fromstep#4 to calculate the required actual KW of the motor.
Demand factor for 4 nos. lifts = 0.72 x 30 KW = 21.6 KW for each lift


Second: How to calculate the Lift Energy Consumption



We have two methods for calculating the energy consumption of lift drives as follows:
  1. Schroeder Calculation Method,
  2. ISO standard calculation method.



1- Schroeder Calculation Method

Schroeder carried out a large number of measurements, and developed a general formula for calculating the daily energy consumption of a lift installation as follows:

The daily energy consumed Ed = R x ST x TP / 3600

Where:
  • Ed is the daily energy consumed in (kWh/day),
  • R is motor rating in (KW),
  • ST is daily number of starts (day-1),
  • TP is trip time factor.


Note:
  • The calculation depends on the accuracy of the trip time factor.


Schroeder defined the trip time factor according to the type of lift drive as in below Table-10:

Drive
Floors above ground
Typical trip time, TP (s)
Range
Mean
Hydraulic (without
counterweight)
˂6
5-7
6
Geared:



— AC 2 speed
6
9-12
10.5
— ACVV {high mass)
12
7-10
8.5
— ACVV (low mass)
12
5-8
6.5
Gearless:



— motor-generator
18
4-6
5
— thyristor
18
3-5
4
Table-10

The Ed value found is then used to calculate the yearly energy consumed per unit floor area (kWh/m2), as follows:

W= (Ed x number of days per year x 0.85) / (Population x floor area per person)

Where W is the yearly energy consumed per unit floor area (kWh/m2).

And

The Annual energy cost = nos. of working days per year x Ed x cost per KWH

Example#9:

A building has six intensive duty lifts (with gearless thyristor drive), each running at 4 m/s with 23 passengers. The motor size is 45 kW. The space allocation for the occupants is 20 m2/person and the building population is 2000 persons. Calculate:
  • The daily energy consumed,
  • The yearly energy consumed per unit floor area,
  • Annual energy cost, assuming an energy cost of £0.065 per kWh.


Solution:

From Table-10, TP is 4 s (the mean value).

The number of starts per day can be estimated by assuming two peaks of two hours each, during which the maximum value of 240 starts per hour is achieved, and an eight hour period of off-peak activity, during which there are 40 starts per hour. (This assumes a 12-hour day, 7:00-19:00). This gives the total number of starts per day as follows:

ST = (2 x 240) + (2 x 240) + (8 x 40) = 1280 starts per day

Then the total daily consumption per lift is:
Ed = R x ST x TP / 3600 = 45 x 1280 x 4 / 3600 = 64 kWh/day per lift

Hence, for all six lifts in the group:
Ed = 6 x 64 = 384 kWh/day

Hence, for 260 working days per year, total annual energy consumption per m2 of floor area:

W= (Ed x number of days per year x 0.85) / (Population x floor area per person) = 384 x 260 x 0.85 / 2000 x 20 = 2.12 kWh/m2

Annual energy cost = nos. of working days per year x Ed x cost per KWH = 260 x 384 x 0.065 = £6490




2- ISO Standard Calculation Method

Energy used (E) by a lift per year (kWh) can be calculated by ISO standard from the following formula:


Where:
  • Tp = trips/applicable day
  • D = number of applicable days
  • W = number of applicable weeks
  • tf(0.5xN) = the time, in seconds, for the lift to travel half the possible travel distance measured from doors closed to doors opening
  • R = motor rating (kW)
  • St = standby energy (kWh)


With the following Assumptions:
  • Average car occupancy is low.
  • The lift runs at rated speed over whole trip.
  • The average power load is the motor rating.
  • Distance of average trip is 0.5×N. Where: N is the total travel distance (m) of the lift
  • No allowance for regeneration (Reduce first term by 20% for regenerative systems).
  • No allowance made for traffic controller actions.


And the following Guidance values can be used:


Lift
Duty
Rating
(starts/
hour)
Trips/
day
Examples (days/week)
Low
60
<100
residential care (7), goods (5), library (6), entertainment
centers (7), stadia (intermittent)
Medium
120
300
office car parks (5), general car parks (7), residential (7),
university (5), hotels (7), low rise hospitals (7), shopping
centers (7)
High
180
750
office (5), airports (7), high rise hospitals (7)
Intensive
240
1000
HQ office (5)

Note:
  • The assumptions and guidance values are empirical and very simplistic. They must be used with care as in some circumstances some may not be valid.


In the next article, we will explain the Lift Traffic design calculations by using Software Programs. Please, keep following.
The previous and related articles are listed in below table:
Subject Of Previous Article
Article
Applicable Standards and Codes Used In This Course,
The Need for Lifts,
The Efficient Elevator Design Solution
Parts of Elevator System Design Process
Overview of Elevator Design and Supply Chain Process.
Vertical Transportation Design and Traffic Calculations – Part One
The Concept of Traffic Planning,
The (4) Methods of Traffic Design Calculations,
Principles of Interior Building Circulation:
A- Efficiency of Interior Circulation
Vertical Transportation Design and Traffic Calculations – Part Two
B- Human Factors
Vertical Transportation Design and Traffic Calculations–Part Three
C- Circulation and Handling Capacity Factors:
Corridor handling capacity,
Portal handling capacity,
Stairway handling capacity,
Escalator handling capacity,
Vertical Transportation Design and Traffic Calculations – Part Four
Passenger Conveyors (Moving Walkways and Ramps) handling capacity,
Lifts Handling Capacity.
D- Location And Arrangement Of Transportation Facilities
Vertical Transportation Design and Traffic Calculations – Part Five
Traffic design calculations:
1- Calculation of the Number of Round Trips for a Single Car,
2- Estimation of Population,
3- Calculation of the Average Number of Passengers per Trip (P),
4- Calculation of the Uppeak Handling Capacity (UPPHC),
5- Calculation of the Waiting Interval (Passenger Waiting Time),
6- Calculation of The percentage population served (%POP),
Vertical Transportation Design and Traffic Calculations – Part Six
7- Estimation of Arrival Rate,
8- Calculation of the Round Trip Time RTT,
Vertical Transportation Design and Traffic Calculations – Part Seven
9- Calculation of the quality of service (Grade of Service)
Vertical Transportation Design and Traffic Calculations – Part Eight
Methods for Lift Traffic Design Calculations:
First Method:  The Conventional Design Method
Vertical Transportation Design and Traffic Calculations – Part Nine
Second Method: The Iterative Balance Method
Vertical Transportation Design and Traffic Calculations – Part Ten 






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