## Sunday, May 26, 2013

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### Motor Starting

The worked example here is a very simple power system with two voltage levels and supplied by a single generator. While unrealistic, it does manage to demonstrate the key concepts pertaining to motor starting calculations.

### Step 1: Construct System Model and Collect Equipment Parameters

 Simplified system model for motor starting example

The power system has two voltage levels, 11kV and 415V, and is fed via a single 4MVA generator (G1). The 11kV bus has a standing load of 950kVA (S1) and we want to model the effects of starting a 250kW motor (M1). There is a standing load of 600kVA at 415V (S2), supplied by a 1.6MVA transformer (TX1). The equipment and cable parameters are as follows:
Equipment Parameters
Generator G1
• $S_{g1} ,$ = 4,000 kVA
• $V_{g1} ,$ = 11,000 V
• $chi_{d}^{} ,$ = 0.33 pu
• $cos phi ,$ = 0.85 pu
Generator Cable C1
• Length = 50m
• Size = 500 mm2
(R = 0.0522 Ωkm, X = 0.0826 Ωkm)
• $S_{s1} ,$ = 950 kVA
• $V_{s1} ,$ = 11,000 V
• $cos phi ,$ = 0.84 pu
Motor M1
• $P_{m1} ,$ = 250 kW
• $V_{m1} ,$ = 11,000 V
• $I_{LRC} ,$ = 106.7 A
• $I_{LRC}/I_{FLC} ,$ = 6.5 pu
• $cos phi_{m} ,$ = 0.85 pu
• $cos phi_{s} ,$ = 0.30 pu
Motor Cable C2
• Length = 150m
• Size = 35 mm2
(R = 0.668 Ωkm, X = 0.115 Ωkm)
Transformer TX1
• $S_{tx1} ,$ = 1,600 kVA
• $V_{t1} ,$ = 11,000 V
• $V_{t2} ,$ = 415 V
• $u_{k} ,$ = 0.06 pu
• $P_{kt} ,$ = 12,700 W
• $t_{p} ,$ = 0%
Transformer Cable C3
• Length = 60m
• Size = 120 mm2
(R = 0.196 Ωkm, X = 0.096 Ωkm)
• $S_{s2} ,$ = 600 kVA
• $V_{s2} ,$ = 415 V
• $cos phi ,$ = 0.80 pu

### Step 2: Calculate Equipment Impedances

Using the patameters above and the equations outlined earlier in the methodology, the following impedances were calculated:
Equipment Resistance (Ω) Reactance (Ω)
Generator G1 0.65462 9.35457
Generator Cable C1 0.00261 0.00413
11kV Standing Load S1 106.98947 69.10837
Motor M1 16.77752 61.02812
Motor Cable C2 0.1002 0.01725
Transformer TX1 (Primary Side) 0.60027 4.49762
Transformer Cable C3 0.01176 0.00576
415V Standing Load S2 0.22963 0.17223

### Step 3: Referring Impedances

11kV will be used as the reference voltage. The only impedance that needs to be referred to this reference voltage is the 415V Standing Load (S2). Knowing that the transformer is set at principal tap, we can calculate the winding ratio and apply it to refer the 415V Standing Load impedance to the 11kV side:
$n = frac{415 left( 1 + 0%
ight)}{11,000} = 0.03773 ,$
The resistance and reactance of the standing load referred to the 11kV side is now, R = 161.33333 Ω and X = 121.00 Ω.

### Step 4: Construct the Equivalent Circuit

 Equivalent circuit for motor starting example
The equivalent circuit for the system is shown in the figure to the right. The "Near" Thevenin equivalent circuit is also shown, and we now calculate the equivalent load impedance $Z_{eq} ,$ in the steady-state condition (i.e. without the motor and motor cable impedances included):
$Z_{eq} = Z_{C1} + left[ Z_{S1} || left( Z_{C3} + Z_{TX1} + Z_{S2}
ight)
ight] ,$
$= 64.59747 + j 44.80458 ,$
Similarly the equivalent load impedance during motor starting (with the motor impedances included) can be calculated as as follows:
$Z_{eq,s} = Z_{C1} + left[ Z_{S1} || left( Z_{C3} + Z_{TX1} + Z_{S2}
ight) || Z_{C2} + Z_{M1}
ight] ,$
$= 20.371997 + j 31.22116 ,$

### Step 5: Calculate the Initial Source EMF

 "Near" Thevenin equivalent circuit for motor starting example

Assuming that there is nominal voltage at the 11kV bus in the steady-state condition, the initial generator emf can be calculated by voltage divider:
$E_{0} = V_{n} left( 1 + frac{Z_{G1}}{Z_{eq}}
ight) ,$
$= 11,821.25 + j 1,023.33 = 11,865 ,$ Vac

### Step 6: Calculate System Voltages During Motor Start

Now we can calculate the transient effects of motor starting on the system voltages. Firstly, the current supplied by the generator during motor start is calculated:
$I_{G1} = frac{E_{0}}{Z_{eq,s} + Z_{G1}} ,$
$= 138.8949 - j 219.36166 = 259.64A ,$
Next, the voltage at the 11kV bus can be found:
$V_{11kV} = E_{0} - I_{G1} ( Z_{G1} + Z_{C1} ) ,$
$= 9,677.024 - j 132.375 = 9,677.9 ,$ Vac (or 87.98% of nominal voltage)
The voltage at the motor terminals can then be found by voltage divider:
$V_{M1} = V_{11kV} frac{Z_{M1}}{Z_{C2} + Z_{M1}} ,$
$= 9,670.597 - j 118.231 = 9,671.3 ,$ Vac (or 87.92% of nominal voltage)
The voltage at the low voltage bus is:
$V_{415V} = V_{11kV} frac{Z_{S2}}{Z_{C3} + + Z_{TX1} + Z_{S2}} ,$
$= 9,521.825 - j 280.698 = 9,525.6 ,$ Vac, then referred to the LV side = 359.39Vac (or 86.60% of nominal voltage)
Any other voltages of interest on the system can be determined using the same methods as above.
Suppose that our maximum voltage drop at the motor terminals is 15%. From above, we have found that the voltage drop is 12.08% at the motor terminals. This is a slightly marginal result and it may be prudent to simulate the system in a software package to confirm the results.