2010, 12th International Conference on Optimization of Electrical and Electronic Equipment, OPTIM 2010
Determining Start Time for Three-Phase Cage Induction Motors that Drive Belt Transport Conveyers Gabriel Nicolae Popa, Member, IEEE, Iosif Popa, Corina Maria Dinis, Angela Iagar Department of Electrotechnical Engineering and Industrial Informatics Politehnica University of Timisoara Revolutiei str., no. 5, Hunedoara, Romania
[email protected] Abstract- The paper establishes the analytical relationship of the
start time for a work machine with constant torque resistance driven by a three-phase cage induction motor powered directly from the grid. The medium torque of the induction motor was obtained by integrating the simplified Kloss formula from the natural mechanical characteristic of the driving motor. It is used the medium torque of the motor, the reduce resistance torque at the shaft, the reduced moment of inertia and the speed of the motor for horizontal and inclined belt transport conveyers, without and with deviation drums.
I.
time exceeding their allowable stall time are: impedance or distance relays, thermal relays, and solid-state protective relays (with microcontroller that monitor the current, voltage, and temperature in the motor) [13]. To determine the starting time for belt transport conveyers, it is necessary to compute the mechanical tensions from the belt of transport conveyor, in the case of conveyers without and with deviation drums, after which the reduced resistant moments the motors’ drive shafts can be determined. II. STARTING TIME OF INDUCTION MOTORS. GENERAL ASPECTS
INTRODUCTION
Belt transport conveyers have constant resistance torque during work time. To power belt transport conveyers with low and medium capacity three-phase cage induction motors, can be used [1],[2],[3],[4],[5]. If a downhill high power conveyor system equipped with a regenerative a.c. induction motor with a wound rotor, it is possible to generate an important amount of electrical energy in the network [6],[7],[8]. A well known method to minimize the effects of starting currents for high power induction motors is to use soft starting devices with silicon-controlled rectifiers [9]. Knowing the starting currents and working torques of induction motors are required in any applications. The high currents in the starting time cause the heating of the motor windings. Protection of the induction motors is crucial and it depends on the type of motor and the type of load. During start time the protective devices must be configured correctly so that the starting current is not interpreted as an overload. If the load has high inertia (i.e. the belt transport conveyers), the protective devices must be able to recognize the normal operation or a fault of the induction motor (protection against overloading and overheating of the induction motor windings) [10],[11],[12]. The starting time of induction motors depends on: the torque developed by the motor, the load torque connected at the motor shaft, the moment of inertia for the motor-load groups and the magnitude of the voltage applied to the motor during starting. The protective devices characteristics must be close to the motor’s thermal curve (time depending on current). The protection devices of the motor having a starting
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The method used to determine the start time of a threephase induction motors which drive different types of work machines, can is based on the fundamental equation of torque for electrical drives with a constant inertia moment. If M (N⋅m) is the motor torque, Mrr (N⋅m) is the reduced resistance torque of the drive shaft, Jr (kg⋅m2) is the equivalent inertia moment reduced at motor shaft, and dΩ/dt (s-2) is the acceleration at the motor shaft, than: M = M rr + J r ⋅
dΩ dt
.
(1)
In fig.1: 1 is motor torque, 2 is reduced resistance torque of the belt conveyer to the motor shaft, and 3 is dynamic torque.
447
Fig. 1. The start time explanation for a work machine with constant resistance torque driven by an induction motor.
If Mk (N⋅m) is the motor’s critical torque, s the slip, and sk is the slip at the critical torque, the motor torque is given by Kloss's simplified formula [1],[2]: 2⋅Mk M = , (2) s sk + sk s λ is the motor overload coefficient obtained from the catalog, Pn (W) is the motor nominal power, n0 (rpm) and Ω0 (s-1) are the synchronous speed, respectively the angular speed, nn (rpm) and Ωn (s-1) are the nominal speed, respectively the angular speed, n (rpm) and Ω (s-1) are the instant speed, respectively the angular speed and, sn the slip proper at the nominal torque Mn. These values are calculated with the following formulas: P 30 ⋅ Pn , (3) M k = λ ⋅ M n;M k = λ ⋅ n ;M k = λ ⋅ Ωn π ⋅ nn s k = s n ⋅ ⎛⎜ λ + λ 2 − 1 ⎞⎟ , ⎝ ⎠ n0 − n n ⎛ sk = ⋅ ⎜ λ + λ 2 − 1 ⎞⎟ , ⎠ ⎝ n0
(4)
Ω0 − Ωn ⎛ ⋅ ⎜ λ + λ 2 − 1 ⎞⎟ , ⎠ ⎝ Ω0 Ω0 − Ω , n0 − n ;s = s= Ω0 n0 Ω − Ωn . n − nn sn = 0 ; sn = 0 Ω0 n0 sk =
(5) (6)
If Mr (N⋅m), Mrn (N⋅m) and Mr0 (N⋅m) are the resistant torque, the instant nominal torque and free of work moment, ΩML (s-1) and ΩnML (s-1) are the current and nominal angular speed at the main shaft of work machine (through which it is driven), x is a variable whose values impose the work machine type that the motor drives (x= -1, 0, 1, 2), ηR is the mechanical transmission efficiency (reduction device, transmission belt), i is the transmission ratio, and PnML (W) is the nominal power of the work machine. The mechanical characteristic of the work machine is expressed through this general equation [2],[4]: ⎛Ω ⎞ M r = M r 0 + (M rn − M r 0 ) ⋅ ⎜⎜ ML ⎟⎟ ⎝ Ω nML ⎠
x
.
(7)
The reduced resistant torque for the motor shaft is: M rr =
x ⎡ ⎛Ω ⎞ ⎤ ⋅ ⎢ M r 0 + (M rn − M r 0 ) ⋅ ⎜⎜ ML ⎟⎟ ⎥ , ηR ⋅ i ⎢ ⎝ Ω nML ⎠ ⎥⎦ ⎣
1
(8)
where: PnML , Pn Ωn . i= Ω nML
ηR =
(9)
The reduced inertia torque at the motor shaft is determined based on kinematics system configuration and its mass and size of the elements in rotation and translation movement. If Md is the dynamic torque, (1) can be written like: (12) M = M rr + M d , (13)
From (13): Jr (14) ⋅ dΩ , Md and after that, the start time of motor ts can be calculated with: dt =
ts
Ωn
= Jr ⋅
dΩ ;t = Jr ⋅ Md p
∫ 0
Ωn
∫ 0
dΩ M − M rr .
(15)
III. DETERMINING THE STARTING TIME The next step is to consider that a three-phase cage induction motor is driving a work machine where for drive a constant drive torque is necessary. For determining the start time is using (15). Usually, the driven motor does not run with nominal charger, and in this case, the integration may be done between the limits of speed: 0 (s-1) and ΩF (s-1). With (15) the start time can be calculated, if the value of dynamic torque is known. The motor torque is calculated with Kloss's simplified formula (2), (4) and the values of Mk, sk, sn and s with expressions (5) and (6), where: π ⋅ nn (s-1), (16) Ω = n
30
Ω0 =
π ⋅ n0 ( S -1),
(17)
Ω=
π ⋅ n (s-1).
(18)
30
30
The reduced resistant torque of motor shaft is calculated with the relation (11). The sizes variation M, Mrr, and Md in start time, which depends on the motor slip is given in fig.1. The angular speed of the motor driven rotor and work machine in stabilized regime with value ΩF, is usually different then the nominal speed Ωn, but close to it. To solve this integral (15) is calculated the medium dynamic torque Mdm by taking the difference between electromagnetic medium torque of motor Mm and reduced resistant medium torque Mrrm: (19) M dm = M m − M rrm (N⋅m). With relations (15) and (19) ts is obtained:
(10)
The reduced resistant torque at the motor shaft (for x=0) is calculated using (8): M (11) M rrx = 0 = rn . ηR ⋅ i
dΩ . dt
M d = Jr ⋅
ts =
Jr ⋅ ΩF (s). M m − M rrm
(20)
To calculate the electromagnetic medium moment, s is changed (2) with Ω keeping in the first plan equation (5). The mechanical characteristic equation of three-phase induction motor becomes:
448
M =
2 ⋅ M k ⋅ s k ⋅ Ω 0 ⋅ (Ω 0 − Ω ) (N⋅m). Ω 2 − 2 ⋅ Ω ⋅ Ω 0 + Ω 02 ⋅ 1 + s k2
(
)
The electromagnetic medium torque is calculated with: 1 ΩF 1 ΩF 2⋅ M ⋅ s ⋅ Ω ⋅ (Ω − Ω) . dΩ Mm = ⋅ ∫ MdΩ;Mm = ⋅ ∫ 2 k k 0 20 ΩF 0 ΩF 0 Ω − 2⋅ Ω0 ⋅ Ω + Ω0 ⋅ 1+ sk2 After solving integral (22): ⎡ ⎛ Ω ⎞⎤ Ω F2 ⋅ ⎜⎜ 1 − 2 ⋅ 0 ⎟⎟ ⎥ ⎢ ΩF ⎠⎥ . Ω ⎝ M m = − M k ⋅ s k ⋅ 0 ⋅ ln ⎢1 + ⎢ ⎥ ΩF Ω 02 ⋅ 1 + s k2
( )
(
⎢ ⎣⎢
)
(21)
(22)
(23)
⎥ ⎦⎥
The medium resistant torque is calculated with (11) because Mrrm=Mrr for work machines that have a constant resistant moment. With (11), (23), (19) and (20): J ⋅ΩF , (24) ts = r A where:
⎧ ⎫ ⎡ Ω0 ⎞ ⎤ 2 ⎛ ⎟⎟ ⎥ ⎪ ⎪ ⎢ Ω F ⋅ ⎜⎜1 − 2 ⋅ Ω F ⎠ ⎥ M r ⎪ . (25) ⎛ Ω0 ⎞ ⎢ ⎪ ⎝ ⎟⎟ ⋅ ln 1 + + A = −⎨M k ⋅ sk ⋅ ⎜⎜ ⎬ Ω02 ⋅ 1 + sk2 ⎥ ηR ⋅ i ⎪ ⎝ ΩF ⎠ ⎢ ⎪ ⎢ ⎥ ⎪ ⎪ ⎢⎣ ⎥⎦ ⎩ ⎭
(
Fig. 2. The belt inclination transport conveyer without deviation drum.
)
IV. THE DETERMINATION OF REDUCED RESISTANT MOMENTS AT MOTOR SHAFT
In the particular case, when ΩF = Ωn, the relations (24) and (25) become: J ⋅Ωn , (26) ts = r A
⎧ ⎫ ⎡ Ω0 ⎞⎤ 2 ⎛ ⎟⎟ ⎥ ⎪ ⎪ ⎢ Ω n ⋅ ⎜⎜1 − 2 ⋅ Ω n ⎠ ⎥ M rn ⎪ . (27) ⎛ Ω0 ⎞ ⎢ ⎪ ⎝ ⎟⎟ ⋅ ln 1 + + A = −⎨M k ⋅ sk ⋅ ⎜⎜ ⎬ Ω 02 ⋅ 1 + sk2 ⎥ ηR ⋅ i ⎪ ⎝ Ωn ⎠ ⎢ ⎪ ⎢ ⎥ ⎪ ⎪ ⎢⎣ ⎥⎦ ⎩ ⎭ If expressions (24), (25), (26), and (27) are substituted for angular speeds Ω0, Ωn, and ΩF and the speeds n0, nn and nF are obtained in the case of the work machine with any speed (nF ≠ nn): π ⋅ nF ⋅ J r , (28) ts =
(
)
30 ⋅ A
⎧ ⎫ ⎡ n0 ⎞ ⎤ 2 ⎛ ⎪ ⎪ ⎢ nF ⋅ ⎜⎜1 − 2 ⋅ ⎟⎟ ⎥ nF ⎠ ⎥ M r ⎪ . ⎛ n0 ⎞ ⎢ ⎪ ⎝ + A = −⎨M k ⋅ sk ⋅ ⎜⎜ ⎟⎟ ⋅ ln 1 + ⎬ n02 ⋅ 1 + sk2 ⎥ η R ⋅ i ⎪ ⎝ nF ⎠ ⎢ ⎪ ⎥ ⎢ ⎪ ⎪ ⎦⎥ ⎣⎢ ⎩ ⎭ If nF = nn, the start time of the driven motor is: π ⋅ nn ⋅ J r ts = , 30 ⋅ A where:
(
⎧ ⎪ ⎪ A = −⎨M k ⋅ sk ⎪ ⎪ ⎩
)
⎫ ⎡ n0 ⎞ ⎤ 2 ⎛ ⎪ . ⎢ nn ⋅ ⎜⎜1 − 2 ⋅ ⎟⎟ ⎥ nn ⎠ ⎥ M rn ⎪ ⎛n ⎞ ⎝ + ⋅ ⎜⎜ 0 ⎟⎟ ⋅ ln ⎢1 + ⎬ n02 ⋅ 1 + s k2 ⎥ η R ⋅ i ⎪ ⎝ nn ⎠ ⎢ ⎢ ⎥ ⎪ ⎣⎢ ⎦⎥ ⎭
(
(29)
)
The tension Sx at the point where the conveyer belt is detach from the motor driven drum is unknown (fig.2). qrp (N/m) and qrg (N/m) are uniformly distributed forces for the mobile parts of the superior train rolls, respectively for the inferior, qb (N/m)and qî (N/m) are uniformly distributed forces of belt transport and for material load, w it is the specific resistance to movement of the band (w=0.03,…,0.05 for the pipe rolls) and kî it is a coefficient which shows the contribution of return and deviation drums, at modify the band tensions. The tensions in the belt, for an inclined transport conveyer with β>0, without deviation drum (fig.1), in points 1, …, 4 are: S1 = Sx (N) , (32) S 2 = S x + qb + q rg ⋅ L ⋅ w ⋅ cos β − qb ⋅ L ⋅ sin β (N), (33)
(
)
S3 = kî ⋅ S 2 (N) ,
S4 = S3 + (qb + qî + qrp ) ⋅ L⋅ w⋅ cosβ + (qb + qî ) ⋅ L⋅ sinβ (N). (35) If Grp (N) and Grg(N) are the forces in the moving parts of the superior train rolls, respectively of the inferior train rolls, the uniformly distribute forces qrp and qrg can be calculated with: Grp , (36) qrp =
(30)
(31)
(34)
qrg =
l1 Grg
.
(37)
l2
The coefficient kî takes values between 1.05…1.07 for the wrap up angles of 1800, and 1.03,…,1.05 for the wrap up angles of 900, and 1.02…1.03 the wrap up angles smaller than 900 [14]. kf is the safety coefficent so that the motor drum does not slip (kf=1.2,…,1.3), e is the base of naturals logarithms, µ is the
449
(kî2=1.07,…,1.09) [1],[2]. The tensions in the belt, in points 1,…, 8, for β>0 are: S1 = Sx (N), (45) S2 ≈ S1 (N), (46) (47) S 3 = k î 1 ⋅ S 2 (N), (48) S4 = S3 + qb + qrg ⋅ L2 ⋅ w ⋅ cos β − qb ⋅ L2 ⋅ sin β (N),
(
)
(
)
S5 = kî1 ⋅ S4 (N),
(49)
S6 ≈ S5 (N), S7 = kî 2 ⋅ S6 (N),
(50) (51)
S8 = S7 + qb + qî + qrp ⋅ L1 ⋅ w ⋅ cos β + (qb + qî )⋅ L1 ⋅ sin β (N). (52)
Fig. 3. The belt inclination transport conveyer with deviation drum.
friction coefficient between the conveyer belt and the driven drum (µ=0.25,…,0.35) and α(rad) is the wrap up angle of the belt on the motor drum [1],[2]. Euler’s equation for the condition no slip of the band on the motor drum is: (38) k f ⋅ S 4 = S 1 ⋅ e µ ⋅α (N) . With relations (32),…, (35) and (37): k f ⋅ L⋅ w⋅ cosβ ⋅ qb ⋅ (1+ kî ) + kî ⋅ qrg + qî + qrp + sinβ ⋅ [qb ⋅ (1−kî ) + qî ] S1 = eµ⋅α −k f ⋅ ki
[
{
]
}
(N),
(39) S S4 = kî ⋅ S1 + c kf
(40)
(N),
i is the transmission ratio of reduction devices, Ωm(s-1) and ΩT(s-1) are the angular speeds of the motor and the driven drum: Ω . (41) i= m ΩT The reduced resistant torques for the motor shaft can be obtained with: (S − S ) D (42) M rr = 4 1 ⋅ T (Nm) , ηR ⋅ i 2 where the tensions in band S1 and S2 are taken from expressions (39) and (40). For horizontal transport conveyers, without deviation drums when β=0 and from expressions (39) and (40) results:
S1
k
=
f
[
⋅ L ⋅ w ⋅ q b ⋅ (1 + k e
µ ⋅α
î
)+
− k
k î ⋅q f
⋅k
rg
+ qî + q
rp
[
With these expressions (45),…, (53) can be computed with: S1 =
î
]}
k f ⋅S A
e µ ⋅α − k î21 ⋅ k î 2 ⋅ k f S8 = S1 ⋅ kî21 ⋅ kî 2 + S A (N),
(54)
(N),
(55)
where SA has this formula: SA = w⋅ cosβ ⋅ qb ⋅ (L1 + kî1 ⋅ kî2 ⋅ L2 ) + qrg ⋅ kî1 ⋅ kî2 ⋅ L2 + L1 ⋅ qî + qrp +
(
(
))
(56) + sinβ ⋅ (L1 ⋅ (qb + qî ) − kî1 ⋅ kî2 ⋅ qb ⋅ L2 ). For a horizontal transport conveyer (β=0) with belt deviation drums, the tensions S1 and S8 from the band are: k f ⋅ SB (N) (57) S1 = e µ ⋅α − kî21 ⋅ kî 2 ⋅ k f
S8 = S1 ⋅ kî21 ⋅ kî 2 + S B (N) where SB can be calculated with:
SB
[
(58)
(
)]
= w ⋅ qb ⋅( L1 +kî 1 ⋅kî 2 ⋅L2 )+qrg ⋅kî 1⋅kî 2 ⋅L2 + L1⋅ qî +qrp .
(59)
For these two particular situations (β>0 and β=0) the reduced resistant torques at the motor shaft can be calculate with: (S − S ) D (60) M rr = 8 1 ⋅ T (Nm). ηR ⋅ i 2 V. START TIME SIMULATIONS OF THREE-PHASE CAGE INDUCTION MOTORS THAT DRIVE BELT TRANSPORT CONVEYERS
]
(N), (43) (N). (44) S4 = L ⋅ kî ⋅ S1 + w ⋅ qb ⋅ (1 + kî ) + kî ⋅ qrg + qî + qrp
{
The belt loading coefficients that are passing over the deviation drums have the some values, because the wrap up angles of the belt on these drums are equal. The no slip condition of belt on the motor drum is: (53) k f ⋅ S 8 = S1 ⋅ e µ ⋅α .
To compute the start time (used for motor protection relays) of different types of belt transport conveyers, a computer program written in TURBO Pascal language was designed.
The reduced resistance torques at the motor shaft, can be determined with (41) where S1 and S4 are given by expression (43) and (44). The reduced resistance torques at the motor drive shaft, can be determined for the belt transport conveyer with deviation drum (fig.3). The coefficient kî1 is the band loading coefficient because the band is passing over deviation drums (kî1=1.02…1.03), and kî2 it is the band loading coefficient over the return drum
450
Fig. 4. The start time ts depending on motor nominal power P without and with deviation drum, for the same length of belt transport conveyer (80 m) and the same load (31 t/h) – horizontal conveyer.
Fig. 7. The start time ts depending on belt transport conveyer load Q without and with deviation drum, for the same motor power (7.5 kW, 960 rpm) and the same length (80 m) – horizontal conveyer.
Increasing of the motor power will decrease the start time of belt transport conveyers (fig.4). Increasing inclination degree, the length and the load will increase the start time of belt transport conveyers (fig.5,6,7). VI. CONCLUSIONS
Fig. 5. The start time ts depending on belt transport conveyer inclination β without and with deviation drum, for the same motor power (11 kW, 960 rpm), length of belt transport conveyer (80m) and load (31 t/h).
An original analytical method to determine the start time (used for system protection) of three-phase cage induction motors is presented. This method is based on the calculation of the dynamic medium moment difference between the active medium moment of the driving motor and the reduced resistant moment (constant) of the work machine at the drive shaft. In comparison with other methods (i.e. graphic-analytical) the procedure used to determine the start time is quicker and the margin of error in the calculations is acceptable for industrial standards. Equation (20) is used to compute the start time where the electromagnetic medium torque is computed with (22). The reduced resistant medium torque is calculated using (42), the tensions are (43) and (44) when analyzing an inclined or horizontal belt transport conveyer without deviation drum, and using (60), the tensions are (57) and (58) for a inclined or horizontal belt transport conveyer with deviation drum. APPENDIX Input data - computing the start time for horizontal and inclination belt transport conveyers For simulation from fig.4,5,6,7
Fig. 6. The start time ts depending on belt transport conveyer length L without and with deviation drum, for the same motor power (7.5 kW, 960 rpm) and the same load (31t/h) – horizontal conveyer.
451
TABLE I ELECTRICAL MOTORS CHARACTERISTICS - S1 TYPE OF LOAD 3 4 5.5 7.5 11 15 Nominal power of the motor Pm (kW) 1000 1000 1000 1000 1000 1000 Synchronous speed of motor n0 (rot/min) 955 960 960 960 960 960 Nominal speed of motor nn (rot/min) Overload 2 2 2 2 2 2 coefficient λ (-)
The number of shafts reduction: 2 Reduction efficiency ηR = 0.875 Speed for shaft 1 n1 (rot/min): 186.0 Inertia moment for shaft 1 J1 (kg⋅m2): 0.01 Speed for shaft 2 n2 (rot/min): 58.125 Inertia moment for shaft 2 J2 (kg⋅m2): 1.37 Drum diameter DT (m): 0.4 Full roll diameter Drp (m): 0.102 Full roll force Grp (N): 115 Emptiness roll diameter Drg (m): 0.102 Emptiness roll force Grg (N): 75 Distance between full rolls l1 (m): 1.2 Distance between emptiness rolls l2 (m): 1.2 Uniform distribute force of belt conveyer qb (N/m)=54 Thickness of the belt conveyor δ (m) : 0.01 Equivalent inertia moment of rotor J (kg⋅m2) = 0.084 Material height hi (m):0.15 Equivalent inertia moment of drum JT (kg⋅m2) = 1.36 kf=1.15 w=0.03 µ=0.35 α ( 0)=180 for conveyers without deviation drum α ( 0)=210 for conveyers with deviation drum kî=1.03 for conveyers without deviation drum Deviation drum diameter DTD (m): 0.4 Inertia moment for deviation drum diameter JTD (kg⋅m2): 1.36 kî1(-)=1.02 for conveyers with deviation drum kî2(-)=1.07 for conveyers with deviation drum
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