Motor Starting

STARTING OF LARGE MOTORS

WLSA -INDIA





WLSA-INDIA/GKT/2006

STARTING OF LARGE MOTORS > One of the challenge experienced by PP personnel is the question , whether a large capacity HT/LT motor can be started on the DG supply

> Situation generally under consideration include the impact of starting the motor i.e. application of Kick load • On a single source –DG/TG • one or more DG/TG in parallel with grid • More than one DG/TG in island parallel mode

> Factors that need to be considered while starting large motors are • Active kick load (KW) that the source is subjected to & hence the drop in speed/frequency • Reactive kick load( KVAr) that the source experiences & hence the voltage dip WLSA/INDIA/GKT/102006

STARTING OF LARGE MOTORS Data required Motor details & Type of starting 1. 2. 3. 4. 5.

Type of motor Capacity in KW or HP( to be converted to KW) Voltage rating Rated full load current Starting current ( Absol value or as a multiple of full load current) 6. Starting power factor 7. Start duration ( Time required for the motor to reach rated speed from rest) 8. Type of starting

WLSA/INDIA/GKT/102006

Sq Cage/Slip ring Kw VL Kv/Volt IFL IST Cos θs Ts DOL/ Star-Delta/ Auto transformer/ Liquid rotor rest/ Soft start, if any

STARTING OF LARGE MOTORS Data required Source details- Primover 1. Capacity 2. Base Load 3. Details of Kick Load capability • DG: App.40% of remaining capacity • STG: 10% of rated capacity

Kw/Mw Kw/Mw

Source details- Grid Capacity : ideally Infinity, but restricted by the intermediate power Transformer, in that case the details of the power transformer 1. Capacity KVA or MVA 2. Percentage Impedence Z%

WLSA/INDIA/GKT/102006

STARTING OF LARGE MOTORS Data required Source details- Alternator –DG or STG 1. Capacity (S) Kw/Mw 2. Voltage rating (V) Kv/Volts 3. Transient reactance X’d pu or %age 4. Sub transient reactance X”d pu or %age 5. Base load before starting motor Kw/Mw 6. Power factor of base load Cosθb 7. Number of sets N Note: If the N number of sets are all equal capacity sets with equal droop Settings then the load sharing will be equal If the sets are equal capacity with unequal droops, the load will be Shared in inverse proportion If the sets are of different capacity, with equal droops, then the Load will be shared in direct proportion to capacity WLSA/INDIA/GKT/102006

STARTING OF LARGE MOTORS Kick Load sharing 

If a kick load is applied on a system with one DGs/TGS in parallel with grid supplied through a power transformer with percentage impedance Z%, then the total kick load In KVA will be shared in inverse proportion as the percentage impedance of the transformer and sub-transient reactance X”d of the alternator



If a kick load is applied on a system with more than on DGs/TGS in parallel with grid supplied through a power transformer with percentage impedance Z%, then the total kick load In KVA will be shared in inverse proportion as the percentage impedance of the transformer and sub-transient reactances X”d of the alternator, as in the case of current division in a parallel circuits

WLSA/INDIA/GKT/102006

STARTING OF LARGE MOTORS Estimation of Active Kick load •

From the given data of motor, the active kick load (KWs)caused during motor start can be calculated using the relation KWs = √3 * VL * IST *Cos θs

Comments 

If the motor is started on grid, then the kick load has no impact on the supply frequency. But may cause voltage dip if the grid is weak or the grid transformer is of inadequate capacity



If the motor is started on single alternator then the active kick load due to motor starting should be less than 40% of the remaining capacity of the engine or 10% of the rated capacity of turbine. This is primarily to ensure that the transient speed drop is not to exceed about 5-10% of rated speed

WLSA/INDIA/GKT/102006

STARTING OF LARGE MOTORS Comments 

If the motor is started on alternators in parallel with grid, then, the part of kick load shared by each alternator should be within the primover capability as explained earlier



If the motor is started on a number of alternators in island parallel mode, then, the part of kick load shared by each alternator should be within the respective primover capability as explained earlier

WLSA/INDIA/GKT/102006

STARTING OF LARGE MOTORS Calculation of voltage drop with example Data avilable/ needed Generator (s) Rated capacity of alternator Sn-KVA - 8150 KVA Nominal Voltage Un-KV - 11 KV Subtransient reactacne X”d - 26.3% From test certificate Transient reactacne X`d - 38.9% From test certificate X”d weight factor X”d* - 0.2 X`d weight factor X`d* - 0.8 Equal reactance X eq - 0.2* X”d* + 0.8* X`d= 36.38 Number of generators N - 2 Total output Tot = N * Sn= 16300 KVA Base load details Generator load prior to starting So

= 4222 KVA

Generator PF prior to starting

pf

= 0.9

Generator reactive load

Qo

= So Sin (Cos inv 0.9) = 1840Kvar

WLSA/INDIA/GKT/102006

STARTING OF LARGE MOTORS Calculation of voltage drop with example Data avilable/ needed AC Motor Data Rated shaft output Pm-Kw - 400 Kw Motor Efficiency η - 95%= 0.95 Nominal power factor pfm - 0.8 Motor nominal input Snm- KVA - Pm/(η*pfm) = 526.32 KVA Nominal input current In - Snm/(√3*Un)= 28 Amps Motor start current factor I* - 6 In Motor starting KVA inrush Sk - Snm *6 = 3158 KVA Motor start power factor Spf - 0. 2 Motor starting Kvar inrush dQ = Sk Sin (Cos inv 0.2) = 3158 Sin 78.5 Deg = 3158 *0. 9799 = 3094 Kvar

WLSA/INDIA/GKT/102006

STARTING OF LARGE MOTORS Calculation of voltage drop with example Calculation of instantaneous voltage drop Inst.Voltage drop is given by dU

= = =

Xeq/100 *dQ/ (Xeq/100*(Qo+dQ) +Tot) * 100 36.38/100 * 3094/( 36.38/100 *( 1840+3094)+16300) *100 6.22%

Voltage drop in Kv

= 6.22% * 11000= 0.68 KV

Minimum Voltage level = 11000- 680 = 10.32 KV Recovery time to normal voltage will be almost equal to the motor Start duration

WLSA/INDIA/GKT/102006

STARTING OF LARGE MOTORS Calculation of voltage drop with example Comments 

If the instantaneous voltage drop is within 10% then it is safe to start the motor



If the instantaneous voltage drop is between 10 to 15%, there is a possibility of some of the existing loads may stop due to de energizing of their contactors

 If the instantaneous voltage drop is above 15% then it is risky to start the motor since, in addition to the stoppage of existing loads, the excitation current may increase to abnormal value, and the AVR may get damaged

WLSA/INDIA/GKT/102006

STARTING OF LARGE MOTORS Possible Solutions…..



Provision of Soft starter to reduce the starting inrush current



Connection of PF improving capacitor, which can reduce the starting Lagging Kvar load



Starting the motor while the sets are in parallel with grid & then go the island mode once the motor attains nominal load

WLSA/INDIA/GKT/102006

STARTING OF LARGE MOTORS

Questions ??

WLSA/INDIA/GKT/102006

STARTING OF LARGE MOTORS

THANK YOU

WLSA/INDIA/GKT/102006