A lift is a means of vertical transportation for both passengers and loads. This mode of transportation has been employed by mankind since the ancient time. During these periods, electric machine was not yet invented. So in order to operate it, animal, human power or by the water-driven mechanism was used.
With the growing needs of moving greater amount of materials and the increase of tall buildings, the lift technology has known a rapid evolution. In the 1880, with the advent of electricity, the first electric was used in order to power the lift (HowStuffWorks, 2010).Today, almost all tall buildings are equipped with elevators. They have eased the life of people and they now form part of the life of people especially those accessing tall buildings.
Elevators contribute to the energy consumption of the building by 3-5%(Shmitt+ Sohn, 2014). To further decrease this consumption, the regenerative drive was introduced in the lift system. When the cab moves up with light load or moves down with heavy load, the system produces more power than it consumes(Debbie Sniderman, 2012).This is because when lightly loaded, the mass of the counterweight is larger than mass of the cab. Therefore it is the counterweight which does the work. Power is in turn generated. For the second case, during the downward motion of a heavy load, the mass of the load is greater than that of the counterweight. It is therefore the weight of the car which does the work causing power to be generated.However, Thisenergy is wasted as heat in the braking resistor. The latter dissipates the regenerated power in order to keep the bus voltage from exceeding the rated limit of the drive(Powerohm, 2014). Instead of wasting it, this energy can be recovered using the regenerative braking method by either storing it in the battery or simply by returning it to the main supply. The use of this drive not only cuts the consumption of the building but it also reduces the power demands during the peak period andreducesthe main cable requirements. A regenerative drive can save the building’s elevator energy consumption by 20-40%(GREEN-ENERGY-BUILDING.COM, 2011). ‘
Working Principle of an elevator
Elevators consist of 2 main units namely the control system and the mechanical lifting device(Buck M, Lawson B, 2003).
There are two types of elevators. These are the hydraulic elevator and the traction elevator. However the traction elevator is the most common one. It is also referred as a roped elevator. In this typical design, the car is raised and lowered by traction steel ropes. One end of the ropes is attached at the top of the car and the other end is attached to the counterweight. It is looped around by a sheave which is a pulley. The sheave contains grooves around its circumference. So, when the sheave rotates, the ropes move(HowStuffWorks, 2010).
The counterweight plays an important role in this system. Its weight is almost same to 40% -45% weight of the filled capacity of the car(Debbie Sniderman, 2012). The counterweight provides additional accelerating force when the car is ascending. It also provides retarding force when the latter is descending. This takes less motor power. It also balances the energy and thus reduces the energy consumption.
The sheave is connected to the induction motor. So when the motor turns in one direction, the sheave raises the car and when it turns in the other direction, it lowers the car.
It is important to know that the elevator car and the counterweight ride in the guide rail along the elevator shaft’s sides. This keeps the car and the counterweight to sway back and forth.
Characteristics of the lift
Figure 1.1 shows the four quadrants operation of an ac induction motor. The motor is driving a hoist which consists of a cage loaded or unloaded for a lift application. The Figure shows the direction of the motor torque TM and that of the load torque TL.
The first quadrant refers to the forward motoring of a fully loaded car moving up. The second quadrant is the forward braking of a fully loaded car moving down. The third quadrant refers to the reverse motoring of the empty car moving down and the fourth motoring is that of the empty car moving up.
Figure 1.1
Lift Operation
The speed Characteristics of the lift with respect to time is shown in Figure 1.2(Saha S, Kosaka T, Matsui N, Sundarsingh V P, 1998). It is divided into different periods namely the acceleration period, the constant speed period and the deceleration period.
‘
General structure of the regenerative braking induction motor drive system of a lift
The lift system constitute of several stages. These include the induction motor, the inverter, the field oriented control stage, the space vector modulation and the mechanical system.
Induction motor
The induction motor is one the simplest type of ac motor. About 95% of all industrial applications require an electric drive which is an induction motor.
It consists of the stator and the rotor. The stator is the stationary part whereas the rotor is the rotating part of the motor. Current is induced in the rotor by the rotating magnetic field of the stator. This current in turn produces a rotor magnetic field. This magnetic field interacts with the stator magnetic field and thus produces a torque. It is this torque which causes the sheave to rotate.
If the applied voltage and the frequency is less than that of the generated Counter Electromotive Force voltage (CEMF) the motor acts as a generator.
Inverter
For an adjustable speed drive system especially one employing the induction motor, three phase inverter is an important element. It supplies supply voltages and currents of adjustable frequency and magnitude to the stator. Inverters convert dc power to ac power. They can be classified as voltage source inverters (VSI) or current source inverter (CSI). The VSI can be voltage or current controlled. For this project, the VSI is used and it is voltage controlled.
Induction Machine Control
There are two types of controllers which can be used for the induction drives. These are the volt per hertz (v/f) controller and the vector flux controller. In the v/f control, the voltage and the frequency magnitudes are kept in proportion. However, its performance is not so good since the rate of change of voltage and frequency should be low. So vector flux controller is a more successful controller as it controls both the amplitude and the phase of the AC excitation.
This concept is used to accomplish the decoupling control of both flux and torque.
Space Vector Pulse Width Modulation(SVPWM)
The SVPWM is the best of all PWM techniques which can be used for variable speed drive application. It not only produces less harmonic distortion in the output voltage and current but also provides a more efficient use of the DC supply voltage.
The space vector modulation is a special switching sequence based on the upper switches of the three phase matrix converter. It considers the sinusoidal voltage as a phasor vector rotating at a constant angular frequency of w. All the three modulating voltages are considered as one unit. The vector summation of these voltages is referred as the reference voltage.
The aim of the SVM is mainly to approximate this reference voltage from the switching topologies.
Mechanical system
In this project, a roped elevator will be considered. This elevator is comprised of an elevator car, a counterweight and a rope.A 2:1 gearless dynamic model is used. For this model, few assumptions are made. These include mass-less hoisting ropes. The dynamic compensating roping and the governor roping are ignored
Aims and Objectives
The aim of the project is to model, to analyse and to demonstrate the regenerative braking system of an induction motor of a lift. In addition, the energy output is determined and the speed control is analysed.
The objectives of the project are as follows:
The regenerative system should be modelled using the computer software Matlab Simulink.
The energy output and speed profile of the system should be well analysed.
Dissertation Outline
Chapter 2 consists of the modelling of the different stages of the regenerative braking system of the induction motor of the lift. It consists of the modelling of the inverter, the vector control, the mechanical system and the regenerative system. It also contains figures enhancing the understanding of the different stages.
Chapter 3 is about the analysis of the different part of the lift system. Parameters are set to the model and the circuit is implemented using simulating scheme. After proper simulation, the results are generated.
Chapter 4 includes all the simulation results. From these results, necessary discussions are made.
Chapter 5 consists of the conclusion on the work that has been done throughout the project. It also constitutes the goals and the objectives that have been accomplished. The recommendations for future work have also been mentioned in this chapter.
Chapter 6 contains all the references that have been used while doing this project.
CHAPTER 2:
MODELLING OF THE REGENERATIVE BRAKING SYSTEM OF THE INDUCTION MOTOR OF THE LIFT
Introduction
This chapter deals with the modelling of the regenerative braking system of an induction motor of a lift. With the help of Figures and equations, thedifferent stages involved in the modelling are explained. This chapter also allows a better understanding of the topic of the dissertation.
Induction Motor
The induction motor drive system consists of three major components of namely the induction machine, the power converter and the controller. Induction motor is the most widely used machine among all the types of ac machines. The squirrel cage induction motor is the most commonly one. [3]
Figure 2.1 is a representation of all the three phase voltages va , vb and vc respectively.
The equations used to describe asynchronous machines in this chapter are given below [3].
From this diagram, the equations 2.1, 2.2 and 2.3 can be derived.
v_as=V_m sin'(wt) (2.1)
v_bs=V_m sin'(wt-120) (2.2)
v_cs=V_m sin'(wt+120) (2.3)
The above equations 2.1, 2.2 and 2.3 can be represented in matrix form as shown below.
[‘(v_as@v_bs@v_cs )]=[‘(cos'(wt)&sin'(wt)@cos'(wt-120)&sin'(wt-120)@cos'(wt+120)&sin'(wt+120))][‘(v_qs^s@v_ds^s )]
(2.4)
Converting vds s and vqs sto the as-bs-cs component give the matrix:
[‘(v_qs^s@v_ds^s )]=[‘(cos'(wt)&cos'(wt-120)&cos'(wt+120)@sin'(wt)&sin'(wt-120)&sin'(wt+120))][‘(v_as@v_bs@v_cs )]
(2.5)
Figure 2.2 shows the dynamic de-qe equivalent circuit of the machine is shown.
From Figure 2.4 and Figure 2.5 the stator and the rotor equations can be written.
‘ v’_qs=i_qs R_s+(w_e ) _ds+d/dt _qs (2.6)
v_qr=i_qr R_r+(w_e-w_r ) _dr+d/dt _qr (2.7)
v_ds=i_ds R_s-(w_e ) _qs+d/dt _ds (2.8)
v_dr=i_dr R_r-(w_e-w_r ) _qr+d/dt _dr (2.9)
The flux linkage expression can be written in terms of current.
_qs=L_ls i_qs+L_m (i_qs+i_qr ) (2.10)
_qr=L_lr i_qr+L_m (i_qs+i_qr ) (2.11)
‘ ‘_ds=L_ls i_ds+L_m (i_ds+i_dr ) (2.12)
‘ ‘_dr=L_ls i_dr+L_m (i_ds+i_dr ) (2.13)
_dm=L_m (i_ds+i_dr ) (2.14)
‘ ‘_qm=L_m (i_qs+i_qr ) (2.15)
Note that:
L_s=L_ls+ L_m (2.16)
L_r=L_lr+ L_m (2.17)
Using the equation 4 to equation 13, the voltage expression in terms of currents can be obtained.
v_qs=(R_s+’SL’_s ) i_qs+w_e L_s i_ds+SL_m i_qr+w_e L_m i_dr (2.18)
v_qr=’SL’_m i_qs+(w_e-w_r ) L_m i_ds+(R_r+’SL’_r ) i_qr+(w_e-w_r ) L_r i_dr (2.19)
v_ds=-w_e L_s i_qs+(R_s+’SL’_s ) i_ds-w_e L_m i_qr+SL_m i_dr (2.20)
v_qr=’-(w_e-w_r )L’_m i_qs+’SL’_m i_ds+(w_e-w_r ) i_qr+(R_r+’SL’_r ) i_dr (2.21)
The equation 14 to equation 17 can be represented in matrix form as shown below.
[‘(v_qs@v_ds@'(v_qr@v_dr ))]=[‘(‘((R_s+’SL’_s )@w_e L_m )&'(w_e L_s@(w_e-w_r ) L_m )@-w_e L_s&(R_s+’SL’_s )@-(w_e-w_r ) L_m&SL_m )'(‘(SL_m@(R_r+’SL’_r ) )&'(w_e L_m@(w_e-w_r ) L_r )@’-w’_e L_m&SL_m@(w_e-w_r )&(R_r+’SL’_r ) )][‘(i_qs@i_ds@'(i_qr@i_dr ))] (2.22)
The equation 18 shows the relationship between the electromagnetic torque and the currents.
T_M=(3/2)(P/2)(_dm i_qs-_qm i_ds )
(2.23)
From these equations, the induction motor can be modelled.
Vector control
The vector control is also referred as the field oriented control. This method consists of controlling the stator currents which are represented by a vector.
This control enables the motor to operate smoothly over a full range of speed. It can as well generate full torque at a speed of zero and it enables a fast deceleration and a fast acceleration.
This technique is mostly used in the induction motor and AC synchronous applications.
Since the vector control is based on projections transforming three phase time dependent system into d-q coordinates time invariant system, the control structure handles the instantaneous quantities very well, thus making it accurate in all working operation. The modelling of the vector control involves numerous transformations namely Park transformation, Clark transformation and the inverse park transformation as shown in Figure 2.3.
Park Transformation
The park transformation is the most important transformation in the Field oriented control system. This projection transforms a two phase orthogonal system that is and into the d,q rotating frame. The input currents are is and is respectively and the output currents are isd and isq respectively.
The following equations are considered for the transformation(Texas Instruments, 1998).
i_sd=i_s cos+i_s sin (2.24)
‘ i’_sq=i_s sin+i_s cos (2.25)
These equations determined the flux and torque component of the current vector.
.
Clark Transformation
The space vector (a, b, c) can be transformed in another frame with only two orthogonal axes namely (, ). Axis a is assumed to be in the same direction as the axis . Figure 2.4 represents the Clark Transformation module. Currents, ia and ib are the input of the module and currents, is and is are the output projections.
The following equations show the transformation occurring in the module [13].
i_s=i_a (2.26)
i_s=1/’3 (‘2i’_a+i_b ) (2.27)
Inverse Park Transformation
This transformation involves the modification of the voltages in d,q rotating reference frame in a two phase orthogonal system (,).
The following equations show the voltage transformation [13].
‘ v’_Sref=v_Sdref cos-v_Sqref sin (2.23)
v_Sref=v_Sdref cos+v_Sqref sin (2.24)
From the above equations, the Simulink circuit can be modelled.
Current model
This module takes the currents: isd and isq and the rotor electrical speed as input. The angle is the output of the module.
The current model consists of the implementation of the following equations[13]:
i_ds=T_R ‘di’_mR/dt+i_mR (2.25)
‘ f’_s=1/w_b d/dt=n+i_qs/(T_R i_mR w_b ) (2.26)
Such that:
T_R=L_R/R_R
Discretizing both equations give:
‘ i’_(‘ds’_k )=T_R i_(‘mR’_k )+i_(‘mR’_(k+1) ) (2.27)
‘ f’_(s_k ) = 1/w_b _k=n_((k+1))+i_(‘qs’_((k+1)) )/(T_R i_(‘mR’_(k+1) ) w_b ) (2.28)
From the above equations, two Simulink models circuit can be implemented.
Since
‘ ‘_(‘cm’_(k+1) )=_(‘cm’_k )+w_b f_(s_k ) T (2.29)
From this equation, the block diagram to obtain _(‘cm’_(k+1) ) can be derived.
The angle is then fed to the Park, Clark and Inverse Park transformationmodule respectively.
Field weakening
Field weakening is a very important at high speed since it tends to be economical.
The aim of this principle is to increase the control speed range beyond that of the nominal speed.
Since
P=Tw
Therefore under the nominal load, the mechanical power increases linearly with the speed up to the nominal power of the respective nominal speed.It is important to note that in this operating range, the flux is equal to its nominal value.
The Figure 2.5 shows the Field weakening operation.
According to [12], an approximate maximum value of the torque can be calculated as follows:
T_max=’3Z’_p/’2w’^2 V^2/(L_ls+L_lr ) (2.30)
Since,
w=2f_s (2.31)
Therefore,
T_max=’3Z’_p/’2(2f_s)’^2 V^2/(L_ls+L_lr ) (2.32)
From the above equation, it can be deduced that Tmax is inversely proportional to the square of the angular speed.
In order to obtain the continuous field weakening, the following output polynomial [13] can be used. Note that this polynomial is in the 8.8 format.
i_(‘sd’_ref )=’-5n_ref’^3+”5n’_ref’^2-‘209n’_ref+300 (2.33)
PI Regulator
For an induction motor which is based on the vector control requires two constants as its parameters. These are the torque reference component, Isqref and the flux reference component, Isdref. A classic numerical Proportional and Integral (PI) regulator is a good torque and flux feedback regulator. It can reach the value of the references when the proportional term constant and the Integral term constant are properly set. Figure 2.6 shows the diagram for the PI regulator.
Space Vector Pulse Width Modulation(SVPWM)
Space Vector Pulse Width Modulation (SVPWM) is a special sequence of the three upper device of a three phase voltage source inverter [14]. It supplies the induction motor with the desired phase voltages. In this method, when the upper transistor is switched ON, that is, S1 or S3 or S5 is ‘1’ , its corresponding lower transistor is switched OFF, that is, S2 or S4 or S6 is ‘0’. Figure 2.7 shows the voltage source inverter.
From these on-off states, eight switching combinations are obtained which determine eight different phase voltage configurations. These switching combinations are shown in Figure 2.8.
There are therefore six non-zero space vectors and two zero space vector. This divides the vectors into six sectors.
SVPWM uses a vector as reference namely Vref.
The modelling of a SVPWM involves numerous steps. It involves [10]:
Finding magnitude and angle of Vref.
Determination of sector
Determination of the time duration
Determination of the switching time
Generation of the inverter output voltage
Finding magnitude of angle of Vref
The magnitude and angle of Vref are calculated using equations and respectively[2].
|V_ref |='(‘V_’^2+’V_’^2 ) (2.34)
=’tan’^(-1) (V_/V_ ) (2.35 )
The reference voltage can be written as follows:
V_+’jV’_=2/3 (V_a+cos(2/3) V_b+cos(2/3) V_c )+j 2/3 (sin(2/3) V_b-sin(2/3) V_c )
= 2/3 (V_a+1/2 V_b+1/2 V_c )+j 2/3 (‘3/2 V_b-‘3/2 V_c )
The equation can be written in matrix form.
[‘(V_@V_ )]=2/3 [‘(1&-1/2&-1/2@0&’3/2&-‘3/2)][‘(V_a@V_b@V_c )]
Determination of sector
Each vector has a different phase angle. Based on this angle, the sector in which the vector lies can be identified [11].
The Table 2.1 shows for which range of angle a vector is said to be in which particular sector.
Range of angle, Sector
0>’60 1
60>’120 2
120>’180 3
-180>’-120 4
-120>’-60 5
-60>’0 6
Determination of the time duration
taand tb are the period during which the vector V4 and V6 are applied respectively and t0 is the period during which the zero vector is applied. These time periods can be calculated as follows [9] :
t_a=T_ch 3|V_ref |sin(2/3 k-)/(2V_DC sin(/3) ) (2.36)
t_b=T_ch 3|V_ref |sin()/(2V_DC sin(-/3 (k-1) ) ) (2.37)
t_0=t_7=(T_ch-t_a-t_b)/2 (2.38)
The Table 2.2 shows the switching on state for each sector and its corresponding time period of Tch.
Sector State Time period of Tch
1 Sa = 0 ta + tb + t0
Sb =0 tb + t0
Sc=1 t0
2 Sa = 0 ta+t0
Sb =1 ta + tb + t0
Sc=1 t0
3 Sa = 0 t0
Sb =1 ta + tb + t0
Sc=0 tb + t0
4 Sa = 1 t0
Sb =1 ta + t0
Sc=0 ta + tb + t0
5 Sa = 1 tb + t0
Sb =0 t0
Sc=0 ta + tb + t0
6
Sa = 1 ta + tb + t0
Sb =0 t0
Sc=1 ta + t0
Determination of switching time
Each sector consists of 7 switching states for each cycle and it starts and ends with a zero vector.
For sector 3, the switching states are as follows: 000, 100, 110, 111, 110, 100, 000. This occurs during the time Tch. Note that the uneven numbers travel counter clockwise in each sector and the sector also travel clockwise [14].
The time Tchis then divided into time for 7 switching states. Equation ‘ shows the sum of the switching times.
‘ T’_Ch=T_0/4+T_1/2+T_2/2+T_0/2+T_2/2+T_1/2+T_0/4 (2.39)
From Figure 2.8, the switching states for the others sectors can be obtained.
For the switch to know that it should be switched ON for specific times, a timer is required. A ramp or a repeated sequence can also be used to provide this information to the switches.
Mechanical system
Figure 2.9 shows the mechanical system of the lift.
The entire system from the perspective of the motor is based on equation 2.34[9]:
T_M=J_M ‘dw’_m/dt+’B.w’_m+T_L (2..40)
The load torque equation is expressed in equation 2.35.
T_L=R_P.F_L+J_P (dw_m)/dt (2.41)
The elevator is assumed to move upwards. The load force, FL thus becomes :
F_L=1/2 [g.(M_c-M_cw )+M_c ‘du’_c/dt] (2.42)
Where the car speed,uc is defined in equation 2.37
u_c=1/2 R_P.w_m (2.43)
Replacing equation 2.37 in equation 2.36, the load force equation is obtained in terms of rotor speed as defined in equation 2.38.
F_L=1/2 [g.(M_c-M_cw )+M_c (d( 1/2 R_p.w_m ))/dt] (2.44)
To obtain the load torque, equation 2.38 is substituted in equation 2.3. It is defined in equation 2.39.
T_L=1/2 R_P [g.(M_c-M_cw ) ]+[‘1/4 ‘R_p’^2 M’_c+J_P ] ‘dw’_m/( dt) (2.45)
The electromagnetic torque, Tem is defined in terms of the rotor speed in equation 2.40.
T_M=1/2 R_P [g.(M_c-M_cw ) ]+[‘1/4 ‘R_p’^2 M’_c+J_P+J_M ] ‘dw’_m/dt +’B.w’_m (2.46)
From equation 2.40 and equation 2.39, the Simulink model of the mechanical system can be constructed.
When the car moves downward, the counterweight moves upward. The load force equation thus is defined in equation 2.41.
F_L=1/2 [g.(M_cw-M_c )+M_cw (d( 1/2 R_p.w_m ))/dt] (2.47)
`The load torque equation becomes:
T_L=1/2 R_P [g.(M_cw-M_c ) ]+[‘1/4 ‘R_p’^2 M’_cw+J_P ] ‘dw’_m/( dt) (2.48)
Therefore the electromagnetic torque is defined as in equation 2.44.
‘ T’_L=1/2 R_P [g.(M_c-M_cw ) ]+[‘1/4 ‘R_p’^2 M’_c+J_P ] ‘dw’_m/( dt) (2.49)
By modifying the Simulink model of the mechanical system of the car moving upward, the Simulink model of car moving downward can be obtained.
Regenerative capacity of the inverter
The drive will regenerate in two conditions.
When car is empty and is moving up and
When car moves down with gravity.
During these two conditions, the motor is braking. This implies that the motor torque is in opposite direction to that of the speed.
It is the inverter which will return the regenerative energy from the induction motor to the battery or the mains.
Figure 2.11 shows the VSI-PWM with regenerative capacity and power factor control[1].
When the motor drives, the current follows the path shown in Figure 2.12. The current’s path is shown for only one phase. In the line side converter, the current passes through the diode and in the machine side, the current passes through the thyristor.
When the motor generates, the current follows the path shown in Figure 2.13 to return to the main supply. Only one phase is considered in this Figure. In the machine side converter, the current passes through the diode and in the line side it passes through the thyristor to enter the main supply.
Energy calculation
The Energy regenerated or drawn during acceleration, constant speed and deceleration of the cage is calculated for each quadrant. The equations used are given below [12].
1st quadrant- fully loaded cage moving up
During acceleration
E_1mf=E_1p+E_1k
= (M_L-M_Cw ) ‘gS’_1+0.5(M_L-M_Cw ) ‘V_ms’^2 ( 2.50)
E_1if=E_1mf/(_am’?_i’?_mt ) (2.51)
At Constant speed
E_2mf=E_2p+E_2k
= (M_L-M_Cw ) ‘gS’_2 (2.52)
E_2if=E_2mf/(_fm’?_i’?_mt ) (2.53)
During deceleration
E_3mf=E_3p+E_3k
= (M_L-M_Cw ) ‘gS’_3-0.5(M_L-M_Cw )0.99’V_ms’^2 (2.54)
E_3if=E_2mf/(_fm’?_i’?_mt ) (2.55)
2nd Quadrant- Fully load car moving down
During acceleration
E_1gr= (M_L-M_Cw ) ‘gS’_1 (2.56)
‘ E’_1sr=E_1gr’?_ag’?_i’?_gt (2.57)
At Constant speed
‘ E’_1gr= (M_L-M_Cw ) ‘gS’_2 (2.58)
‘ E’_2sr=E_2gr’?_fg’?_i’?_gt (2.59)
During deceleration
Assumption: When the motor’s speed decreases by 10%, the mechanical brakes are applied.
E_3gr=E_3p+E_3k
= (M_L-M_Cw ) ‘gS’_3+0.5(M_L-M_Cw )0.99’V_ms’^2 (2.60)
E_3sr=E_3gr’?_ag’?_i’?_gt (2.61)
Third Quadrant-Empty car moving down
In this case, the mass of the counterweight Mc is greater than the mass of the empty car ME. This will cause the car to move up. For the car to move down, the induction motor will have to operate as a motor.
During acceleration
E_1mr= (M_Cw-M_E ) ‘gS’_1+0.5(M_Cw-M_E ) ‘V_ms’^2 (2.62)
E_1ir=E_1mr/(_am’?_i’?_mt ) (2.63)
At Constant speed
E_2mr= (M_Cw-M_E ) ‘gS’_2 (2.64)
E_2ir=E_2mr/(_fm’?_i’?_mt ) (2.65)
During deceleration
E_3mr= (M_Cw-M_E ) ‘gS’_3+0.5(M_Cw-M_E )0.99’V_ms’^2 (2.67)
E_3ir=E_2mr/(_fm’?_i’?_mt ) (2.68)
Fourth Quadrant-Empty car moving up
In this case, the ac machine operates as a generator. Note that the motor torque and the speed of the machine will be in opposite direction as compared to the second quadrant.
During acceleration
E_1gf= (M_C-M_E ) ‘gS’_1 (2.69)
E_1sf=E_1gf’?_ag’?_i’?_gt (2.70)
At Constant speed
E_1gf= (M_Cw-M_E ) ‘gS’_2 (2.71)
E_2sf=E_2gf’?_fg’?_i’?_gt (2.72)
During deceleration
E_3gf= (M_Cw-M_E ) ‘gS’_3+0.5(M_Cw-M_E )0.99’V_ms’^2 (2.73)
E_3sf=E_3gf’?_ag’?_i’?_gt (2.74)