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The Per Unit System (Examples) Example 1 A three-phase, wye-connected system is rated at 50MVA and 120kV. Express 40MVA of three-phase apparent power as a per unit value referred to: a) The three-phase system as MVA base, and b) The per-phase system MVA as base. Solution a) For the three-phase system MVA as base, S B 3φ = 50 MVA S pu 3φ =



V B (line ) = 120 kV



S actual 40 = = 0.8 pu S B 3φ 50



b) For the per phase base S B 3φ



50 M = 16.67 MVA 3 3 VB ( line ) 120 k V B ( phase ) = = = 69.28kV 3 3 1 40 S pu 1φ = × = 0.8 pu 3 16.67 S B 1φ =



=



Example 2 Find the per unit value for XT1, XT2 and XT if the base values are 11kV and 60MVA. 60 MVA 11kV X G=0.5 EG=1.5pu



T1



T2 X T=20Ω



G 60 MVA 12/132kV X T1=10%



30 MVA 132/33kV X T2=10%



Solution Step 1: Draw the section of the network.



1



60 MVA 11kV X G=0.5 EG=1.5pu



T1



T2 XT=20Ω



G



30 MVA 132/33kV X T2=10%



60 MVA 12/132kV XT1=10%



Step 2: Find the base value of the voltage for each section. S B = 60 MVA V B1 = 11kV VB 2 =



V2 132 × V B1 = × 11kV = 121kV V1 12



VB 3 =



V3 33 × VB 2 = × 121kV = 30.25kV V2 132



Step 3: Find the per unit values of each component Transformer T1 and T2 Since both transformer voltage base are the same as their rated values, their p.u reactance on a 60MVA are: 2



 V old   S new  ×  Bnew  ×  Bold   VB   S B 



Z



new pu



=Z



X



new T1



 12   60  = 0.1 ×   ×   = 0.12 pu  11   60 



old pu



2



:: Looking at the LV side of T1



OR 2



 132   60  X Tnew  ×   = 0.12 pu :: Looking at the HV side of T1 1 = 0.1 ×   121   60  2



 132   60  X Tnew  ×   = 0.24 pu :: Looking at the HV side of T2 2 = 0.1 ×   121   30  OR 2



X



new T1



 33   60  = 0.1 ×   ×   = 0.24 pu :: Looking at the LV side of T2  30.25   30 



NOTE: The per unit value for the transformer impedance is the same whether it is seen in the HV or LV side of the transformer- one of the advantages of per unit system.



2



Line Impedance Since the impedance given in actual value, we have to find the base value for the impedance.



(V B 2 ) 2



X T ( base ) = X T ( pu ) =



=



SB *



X T ( actual ) X T ( base )



(121k ) 2



=



60 M



= 244Ω



20 = 0.082 pu 244



Note that the line impedance has only the resistive value. Therefore the complex power conjugate value is the same since θ is equal to 0.



Example 3 The one-line diagram of three-phase power system is shown below. Select a common base of 100 MVA and 22 kV on the generator side. Draw an impedance diagram with all impedance including the load impedance marked in per-unit. 50 MVA 22/220 kV X = 10%



1



T1



40 MVA 220/11 kV X = 6.0%



T2



2 Line 1 220 kV



66.5 MVA 10.45 kV X = 18.5%



X = 48.4 Ω



G 90 MVA 22 kV X = 18%



4



M T3



T4



3 Line 2 110 kV



40 MVA 22/110 kV X = 6.4%



X = 64.43 Ω



Load 40 MVA 110/11 kV X = 8.0%



57 MVA 0.6 pf lag 10.45 kV



Solution Step 1: Draw the zone in the circuit diagram



3



50 MVA 22/220 kV X = 10%



40 MVA 220/11 kV X = 6.0%



T1



T2



2



1



4 Line 1 220 kV



66.5 MVA 10.45 kV X = 18.5%



X = 48.4 Ω



G



M



90 MVA 22 kV X = 18%



57 MVA 0.6 pf lag 10.45 kV



3 Line 2 110 kV T3



X = 64.43 Ω



40 MVA 22/110 kV X = 6.4%



Load T4



40 MVA 110/11 kV X = 8.0%



Step 2: Find the base voltage for each zone. Vb1 = 22kV Vb 2 =



V2 220 × Vb1 = × 22k = 220 kV V1 22



Vb 3 =



V3 110 × Vb1 = × 22k = 110kV V1 22



Vb 4 =



V4 11 × Vb 2 = × 220 k = 11kV V2 220



Step 3: Find the per unit value Generator and Transformer Since generator & transformer voltage base are the same as their rated values, their p.u reactance on a 100 MVA 2



 V Bold   S Bnew  Z = Z  new   old   VB   S B  new  new old  S B  old  Z pu = Z pu  SB   100  Z Gnew = 0.18  = 0.2 pu  90   100  Z Tnew  = 0.2 pu 1 = 0.10  50   100  Z Tnew  = 0.15 pu 2 = 0.06  40   100  Z Tnew  = 0.16 pu 3 = 0.064  40  Motor  100  Z Tnew  = 0.2 pu 4 = 0.08  40  new pu



old pu



4



S Bnew = 100 MVA



S Bold = 66.5MVA



V Bnew = 11kV



V Bold = 10.45kV



old Z motor = 0.18 2



Z



new motor



=Z



old motor



 V old  S new ×  Bnew  × Bold SB  VB  2



 10.45   100  = 0.18 ×   ×   11   66.5  = 0.25 pu



Line Impedance Line 1 V B 2 = 220 kV



Line 2



S B = 100 MVA Z B line1 =



(V )



2



B2



=



SB



Z line1( pu ) =



( 220k )



V B 3 = 110 kV



2



100 M



S B = 100 MVA



= 484Ω



Z B line 2 =



Z actual 48.4 = = 0.1 pu Z B line1 484



(V B 3 ) 2



Z line 2 ( pu ) =



SB



=



(110k ) 2 100 M



= 121Ω



Z actual 65.43 = = 0.54 pu Z B line 2 121



Load



S 3φ = 57 MVA VL = 10.45kV pf = 0.6 lagging



θ = cos −1 0.6 = 53.13° S 3φ load = 57 ∠53.13°MVA Z load ( act ) = Z load (base )



VL



(10.45k ) 2



2



S 3φ load *



=



57 M∠53.13°



2 2 ( ( VB 4 ) 11k ) = =



Z load ( pu ) =



SB



Z load ( act ) Z load ( base )



100 M



=



= 1.1495 + j1.5327 Ω



= 1.21Ω



1.1495 + j1.5327 Ω = 0.95 + j1.267 pu 1.21Ω



Step 4: Draw the per unit impedance diagram. 5



XG= j0.2p.u



XT1= j0.2p.u



ZL1= j0.10p.u



XT3= j0.16p.u



ZL1= j0.54p.u



XT2= j0.15p.u



XT4= j0.20p.u



Zmotor= j0.25p.u



M



G ZLoad= 0.95+j1.267



Example 4 Using base values of 30kVA and 240V in the generator side, draw the per unit circuit, and determine the per unit impedances and the per unit source voltage. Then calculate the load current both in per unit and in amperes. Transformer winding resistance and shunt admittance branches are neglected. T1 30 kVA VG=220∟0ºV



Xline=2Ω



G 30 kVA 240/480V X T1=0.1pu



T2



20 kVA 460/115V XT2=0.1pu



Zload=0.9+j0.2Ω



Solution Step 1: Draw the zone in the network. T2



T1 30 kVA VG=220∟0ºV



Xline=2Ω



G 30 kVA 240/480V X T1=0.1pu



20 kVA 460/115V XT2=0.1pu



Zload=0.9+j0.2Ω



Step 2: Draw the per unit circuit



6



IS(pu)



VS(pu)



XT1(pu)



Zline(pu)



XT2(pu)



Iload(pu)



Zload(pu)



G



Step 3: Find the base values SB for the entire network is 30kVA. Find the base voltages and impedances of each zone. 2



ZB =



VB SB



V B1 = 240V



240 2 = 1.92Ω 30k 480 2 = = 7.68Ω 30k



Z B1 =



VB 2 =



V2 480 × V B1 = × 240 = 480V V1 240



Z B2



VB 3 =



115 × 480 = 120V 460



Z B3 =



120 2 = 0.48Ω 30k



Find base current in zone 3 [!:to calculate the load current later] S 30k I B3 = B = = 250 A V B 3 120 Step 4: Find the per unit values



7



Vs ( pu ) =



Vs 220∠0° = = 0.9167 ∠0° pu V B1 240



old X Tnew 1 = X T 1 = 0.1 pu



X line ( pu ) =



X line 2 = = 0.2604 pu Z B2 7.68 2



X



new T2



=X



old T2



 V Bold   S Bnew  ×  new  ×  old   VB   S B  2



 460   30  = 0.1 ×   ×   = 0.1378 pu  480   20  or



:: calculation using base value



2



X



new T2



=X



old T2



 V old   S new  ×  Bnew  ×  Bold   VB   S B  2



Z load(pu)



 115   30  = 0.1 ×   ×   = 0.1378 pu :: calculation using voltage ratio  120   20  Zload 0.9 + j 0.2 = = = 1.875 + j 0.4167 pu ZB3 0.48



Step 5: Find the load current I load ( pu ) = I s ( pu ) = =



Vs ( pu ) Z total ( pu ) Vs ( pu )



j ( X T 1 + X line + X T 2 ) + Z load ( pu )



0.9167 ∠0° j (0.1 + 0.2604 + 0.1378) + (1.875 + j 0.4167 ) 0.9167 ∠0° = 2.086∠26.01° = 0.4395∠ − 26.01° pu = I load ( pu ) × I B 3 =



I load



= 0.4395∠ − 26.01° × 250 = 109.9∠ − 26.01° A



8