### 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: Professional Formulas Method, Baldor Formulas Method, Equivalent Weight Method, Buildings Energy Code (BEC)’S Tables Method, Baldor Tables Method, 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

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: Schroeder Calculation Method, 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. The Concept of Traffic Planning, The (4) Methods of Traffic Design Calculations, Principles of Interior Building Circulation: A- Efficiency of Interior Circulation B- Human Factors C- Circulation and Handling Capacity Factors: Corridor handling capacity, Portal handling capacity, Stairway handling capacity, Escalator handling capacity, Passenger Conveyors (Moving Walkways and Ramps) handling capacity, Lifts Handling Capacity. D- Location And Arrangement Of Transportation Facilities 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), 7- Estimation of Arrival Rate, 8- Calculation of the Round Trip Time RTT, 9- Calculation of the quality of service (Grade of Service) Methods for Lift Traffic Design Calculations: First Method:  The Conventional Design Method Second Method: The Iterative Balance Method