EECE 370 Lecture, Spring 2008

A 1000-VA 230/115-¥ transformer has been tested to determine its equivalent circuit. The results of the tests are shown below. Open-circuit test Short-circuit test Voc = 230 V Vso = 19.1 V Toco =OA5A Isc =8.7 A Poo = 30 W Pso = 42.3 W All data given were taken from the primary side of the transformer. (a) Find the equivalent circuit of this transformer referred to the low-voltage side of the transformer. (b) Find the transformer’s voltage regulation at rated conditions and (1) 0.8 PF lagging, (2) 1.0 PF, (3) 0.8 PF leading. (c) Determine the transformer’s efficiency at rated conditions and 0.8 PF lagging. SOLUTION (a) OPEN CIRCUIT TEST: 045A Yox| = |G. — JBy| = = 0.001957 Heel = [Ge ~ JBy| = 9 @=cos™ Foo =cos™ 30W = 73.15° Vooloo (230V\0OA5A) Vax = Go — jBy = 0.001957Z — 73.15° mho = 0.000567 - j0.001873 mho Ro= 1 =17632 G, SHORT CIRCUIT TEST: 191¥ [Ziq] =|Req + Xzo| =F 722 @=cos Fee cos! 423 W = 75,3° Veoloc (19.1 V)(B7 A) Zug = Rag + (Xgq = 2.20275.3° 2 =0.558+ j2.128Q Reg = 0.5582 Xgq = j2.1282 To convert the equivalent circuit to the secondary side, divide each impedance by the square. of the turns ratio (a = 230/115 = 2). The resulting equivalent circuit is shown below: al, F I col Ras Kegs SS +0 l AM, SYN. O+ "Sk PS v. a ac o -o l R oO- (b) Regs = 3 + Rs »¢ Xe = = + Xz Reg, = 0.1402 Xag, = j0.532Q R,, = 4410 X yy, =I (b) To find the required voltage regulation, we will use the equivalent circuit of the transformer referred to the secondary side. The rated secondary current is 7, = 1000VA _g aga *T1SV We will now calculate the primary voltage referred to the secondary side and use the yoltage regulation equation for each power Factor. () 0.8 PF Lagging: V, =V,+Zpqly =HS20°V + (0.140 + j0.532.2)(8.72—36.87° A) V, -1188A14°V = NEES 100% =3.3% (2) 1.0 PR: vy, =Y, + Zugls =11520° V + (0.140 + j0.532Q)(8.7.20° A) V,. =116.322.28° V _ 1163-115 ~ 115 (3) 0.8 PF Leading: VR x 100% = 1.1% Vv, =¥, +Zagls =11520° V + (0.140 + j0.532 Q)(8.7236.87° A) V, =113.322.24° V _ 113.3415 115 (ce) At rated conditions and 0.8 PF lagging, the output power of this transformer is =V,1, cos @=(115 V)(8.7 A)(0.8)= 800 W VR x 100% =-1.5% P OUT The copper and core losses of this transformer are Poy = 1g? Rags =(8.7 AY (0.140 Q)=10.6 W r\2 (V, | (118.8 V)’ Po = =+—""/ _320W R, 441 Q Therefore the efficiency of this transformer at these conditions is Four x100% = 800 W = 94.9% Pour + Peay + Fore 800 W + 10.6 W + 320W R= 2-6 A.15 kVA 8000/230 V distribution transformer has an impedance referred to the primary of 80 + (300 ©. The companents of the excitation branch referred to the primary side are R,=350kKQ and Xy =70kO (a) If the primary voltage is 7967 V and the load impedance is Z7, = 3.2 1 71.5 Q, what is the secondary voltage of the transformer? What is the voltage regulation of the transformer? (b) If the load is disconnected and a capacitor of —/3.5 Q is connected in its place, what is the secondary voltage of the wansformer? What is its voltage regulation under these. conditions? SOLUTION (a) lhe easiest way to solve this problem is to refer all components to the primary side of the transformer. The turns ratio is a — 8000/230 — 34.78. Thus the load impedance referred to the primary side is £,, (34.78) (3.24 1.5 £2)— 3871+ f815.Q The referred secondary current is ’ Fad 1 I, = : D6TACTV = PTO Vy ag 798.2" A (80+ 7300 Q)+(38/14 JLS15 QQ) 4481.228.2° O and the referred secondary voltage. is V, =I, 2, =(1782 28.2? A\(38711 j1815.2)=76102 3.1°V The actual secondary voltage is thus ’ 3610/-3.1°V vy, aN -76102-3.PV siggy 3c “a 34.78 The voltage regulation is _ 7967-7610 7O1G VR «100% — 4./% (b) The easiest way to solve this problem is to refer all components to the primary side of the transformer. The turns ratio is again a = 31.78. Thus the load impedance referred to the primary side is Z, ~(34.78) (-)3.5 Q1--j4234.2 The referred secondary current is ‘ T9867 70° V T9657 SOY V 1, = oO = = 2U25Z88.8° A (80+ 7300 2)+(— 74234 Q) 39352-88.8° A and the referred secondary voltage is V, =1, Z, =(2.25288.8" A)(—j4234 Q) = 85732 -1.2° V The actual secondary vollage is ubus ? V, — 8573Z2-1.2°V V,=—~= = 246.52-1.2° V a 34.78 The voltage regulation is vr = 287-8523 100% = 7.07% 8573 2-8. A 200-MVA 15/200-EV single-phase power transformer has a per-unit resistance of 1.2 percent and a per- unit reactance of 5 percent (data taken from the transformer’s nameplate), The magnetizing impedance is J80 per unil. (a) Find the equivalent circuit referred to the low voltage side of this transformer. (&) Calculate the voltage regulation of this transformer for a full-load current at power factor of 0.8 lagging. ic) Assume. that the primary voltage. of this transformer is a constant 15 KV, and plot the secondary vcltage as a function of load current for currents from no-load to full-load. Repeat this process for power factors of 0.8 lagging, 1.0, md 0.8 leading. SoLurion (a) The base impedance of this transformer referred to the primary (low-voltage) side is 2 (skvy Zoae Moet MISEYY 5.0 Si. 200 MVA so Ry = (0.012\(1.125 Q)=0.01335 2 Xyo = (0.05) (1.125 2)= 0.0553 2 X yy = (100)(1.125 Q)= 112.5 The equivalent circuit is I 1, iP R x a Rap = 9.0135 Q Xyggp = J0,0563 Q R, =not specified Xy =112.52 (b) Tf the load on the secondary side of the transformer is 200 MVA at 0.8 PF lagging, and the referred secondary voltage is 15 kV, then the referred secondary current is 1 — Plow __200MVA ° "y, PF (15kV)(0.8) I, =16,6672—36.87° A = 16,667 A The voltage on the primary side of the transformer is yo V,=V, +1, Zeop V, =15,000/0° V+ (16, 667—3687° A)(0.0135+ j0.0563.O)=15,755/2.24° V Therefore the voltage regulation of the transformer is _ 15,755-15,000 15,000 (c) This problem is repetitive in nature, and is ideally suited for MATLAB. A program to calculate the secondary voltage of the transformer as a function of load is shown below: VR X100% = 5.03% M-file: prob2_8.m M-file to calculate and plot the secondary voltage of a transformer as a function of load for power factors of 0.2 lagging, 1.0, and 0.2 leading. These calculations are done using an equivalent circuit referred to the primary side. oe oe oe ee 88 oe % Define values for this transformer VP 15000; % Primary voltage (V) amps = 0:166.67:16667; % Current values (A) Req = 0.0125; % Equivalent R (ohms) Xeq = 0.0563; % Equivalent X (ohms) Calculate the current values for the three power factors. The first row of I contains the lagging currents, the second rew contains the unity currents, and the third row contains 22 20 6° oe %® the leading currents I(1,:) = amps .* ( 0.8 - j*0.6); % Lagging I(2,:) = amps .* ( 1.0 ); % Unity I(3,:) = amps .* ( 0.8 + j*0.6); % Leading % Calculate VS referred to the primary side % for each current and power factor. aVS = VP - (Req.*I + j.*Xeq.*I); % Refer the secondary voltages back to the %® secondary side using the turns ratio. VS = aVS * (200/15); 2 %® Plot the secondary voltage (in kV!) versus load plot (amps,abs(VS(1,:)/1000),'b-','LineWidth',2.0); hold on; plot (amps,abs(VS(2,:)/1000),'k--','LineWidth' ,2.0); plot (amps,abs(VS(3,:)/1000),'r-.','LineWidth' ,2.0); title ('\bfSecondary Voltage Versus Load'); xlabel ('\bfLoad (A)'); ylabel ('\bfSecondary Voltage (kV)'); legend('0.8 PF lagging','1.0 PF','0.8 PF leading’) ; grid on; hold off; The resulting plot of secondary voltage versus load is shown below: Secondary Voltage Versus Load 206 T T T T T T 204 202 ix) Ss 3 198 Secondary Voltage (kV) © a 194 192 190 ’ i l i i i ° 2000 4000 6000 8000 10000 12000 Load (A) i 14000 Ll 16000 18000 2-10. (a) A 13,800/480 V three-phase Y-A-connected transformer bank consists of three identical 100-KVA 79G7/480-V tramsfouuers. It is supplied with power directly rom a large coustant-vollage bus. Lo the short-cirenit test, the recorded valnes on the high-voltage side for one of these transformers are. Vig = 560 V I,, =12.6A Py. = 3300 W (a) Tf this hank delivers a rated load at 0.85 PF lagging anc rated voltage, what is the line-to-line voltage on the primary of the transtormer bank? (b) Whaat is the veltage regulation uncker these ccrdditions? (ec) Assume that the primary voltage of this transformer bank is a constant 13.8 kV, and plot the secondary voltage as a function of load current for currents from no-load to full-load. Repeat this process for power facters of 0.85 lagging, 1.0, and 0.85 leading. (@) Plot the voltage regulation of this transformer as a fimction of lead current for currents from no-load to full load. Repeat this process for power factors of 0.85 lagging, 1.0, and 0.85 leading. SoLutTion From the short-circuit information, it is possible to determine the per-phase impedance of the transformer bank referred to the high-voltage side. The primary side of this transformer is Y-ccnnccted, so the short-circuit phase voliage is r 56 Vyge = 8 = OY =3233V 3 3 the short-circuit phase current is Eyeo = tag =12.6A and the power per phase is Po. Py se >: 1100 W Thus the per phase impedance is 323.3 V9 Zee) =|Ray + JX ge | = ——— = 25.66 2 val =|Roe + Aa 12.6A { @=cost Pye = 1100 W =743° Veolon wee 3V)(12.6 A} Zan = Reg + aq = 25.66271.3° = 6.914 fU7Q Rey = 6.94 2 Nig = fATO If this Y-A transformer bank delivers rated kVA (300 kVA) at 0.85 power factor lagging while the secondary voltage is at rated value, then each transformer delivers 100 kVA at a voltage of 480 V and 0.85 PF lagging. Referred to the primary side of one of the transformers, the load on each transformer is equivalent to 100 kVA at 7967 V and 0.85 PF lagging. The equivalent current flowing in the secondary of one transformer referred to the primary side is + 100 kVA = =1255A ee 7967 V 9 712.5582 -31.79° A The voltage on the primary side of a single transformer is thus Vie =Vys tIys Z , 98 MEQP Vy» = 796720" V+ (12.552 —31.79° A)(6.94 + 724.7 Q)=820721.52° V ‘The line-to-line voltage on the primary of the transformer is Vine = V3 Vp = V3(8207 V)= 14.22 kV (2) The voltage regulation of the transformer is _ 8207-7967 7967 100% = 3.01% Note: It is much easier to solve problems of this sort in the per-unit stem, as we shall see in the next problem. (c) This sort of repetitive operation is best performed with MATLAB. A suitable MATLAB program is shown below: M-file: prob2_10c.m M-file to calculate and plot the secondary voltage of a three-phase Y-delta transformer bank as a function of load for power factors of 0.85 lagging, 1.0, and 0.85 leading. These calculations are done using an equivalent circuit referred to the primazy side. dP AP BP AP BP AP oe Define values for this transfcrmer VL — 13900; % Primary line voltage (V) VPP = VL / sqrti3); % Primary phase voltage (V) amps = 0:0.0126:12.6; % Phase current valuss (A) Reg = 6.94; % Equivalent Ek (ohms) Meq = 24.7; % Equivalent ¥ (ohms) Calculate che current values for the three power factors. The first row of I contains * the lagging currents, tne seccnd row contains %& the unity currents, and the third row contains % the leading currents. re = 0.85; im = sin(acos(re)); ( -* ( re - j*im); % Lagging ( * (1.0 ); % Unity ( .* ( re + j*im); % Leading % Calculate secondary phase voltaye referred % to the primary side for each current and % power factor. aVSP = VPP - (Req.¥I + 3. ¥*Xeq.*I); Refer the secondary phase voltages back to the secondary side using the turns ratio Because this is a delta-connected secondary, this is also the line voltage. VSP = aVSP * (480/7967); ae dP 30 8 % Plot the secondary voltage vereus lead plot (amps,abs(VSP(1,:)),'b-','LineWidth’,2.0); held on; plot (amps,abs(VSP(2,-)),'k--','LineWidth',2.0); plot (amps,abs(VSP(3,:)),'r-.','LineWidth',2.0); title ('\bfSecondary Voltage Versus Load’); xlabel ('\bfLoad (A)'); ylabel ('\bfSecondary Voltage (V)'); legend ('0.95 PF lagging','1.0 PP','0.95 PP leading'); grid on; held off; The resulting plot is shown below: Secondary Voltage Versus Load 490 T T T T T 485 ae 0.85 PF lagging 1.0PF 0.85 PF leading ‘Secondary Voltage (V) 470 465 ° (a) This sort of repetitive operation is best performed with MATLAB. A suitable MATLAB program is shown below: & M-file: prob2_10d.m of a three-phase Y-delta transformer bank as a dP dP dP dP de ae Define values for this transformer VL = 13800; VPP = VL / sqrt(3); amps = 0:0.0126:12.6; Req 6.94; geq = 24.7; de dP do Phase current values Equivalent R (ohms) Equivalent X (ohms) oe ote Calculate the current values for the three power factors. The first row of I contains the lagging currents, the second row contains the unity currents, and the third row contains the leading currents. dP dP dP dP ade re = 0.85; im = sin(acos(re)); I(1,:) = amps .* ( re - j*im); % Lagging I(2,:) = amps .* ( 1.0 ); % Unity I(3,:) = amps .* ( re + j*im); % Leading Calculate secondary phase voltage referred to the primary side for each current and power factor. 2 3 2 so 2 7 aVSP = VPP - (Req.*I + j.*Xeq.*I); % Calculate the voltage regulation. VR = (VPP - abs(aVSP)) ./ abs(aVSP) .* 100; % Plot the voltage regulation versus load plot(amps,VR(1,:),'b-','LineWidth',2.0); hold on; plot(amps,VR(2,:),'k--','LineWidth',2.0); plot(amps,VR(3,:),'r-.','LineWidth',2.0); title ('\bfVoltage Regulation Versus Load’); xlabel ('\bfLoad (A)'); ylabel ('\bfVoltage Regulation (%)'); M-file to calculate and plot the voltage regulation function of load for power factors of 0.85 lagging, 1.0, and 0.85 leading. These calculations are done using an equivalent circuit referred to the primary side. Primary line voltage (V) Primary phase voltage (V) (A) legend ('0.85 PF lagging','1.0 PF','0.85 PF leading') ; grid on; hold off; The resulting plot is shown below: Voltage Regulation Versus Load 35 T T T T T T 3h — 0.85 PF lagging 4 --- 1.0 PF — = 0.85 PF leading Voltage Regulation (%) Oo 2 4 6 8 10 12 14 2-11. A 100,000-KVA 230/115-KV A-A three-phase power transformer has a per-init. resistance of 0.02 pn and a pet-unit reaclanwe of 0.055 pu. The eacilation branch ekments are R, =110 pu and X,, = 20 pu. (a) Lf this transformer supplies a load of 8) MYA at U.85 PEF lagging, draw the phasor diagram of one phase of the transformer. () What is the voltage regulation of the transformer bank under these conditions? (fc) Sketch the equivalent circuit referred to the low-voltage side of one phase of this transformer. Calentate. all of the transfermer impedances referred to tha low-voltage side. SoLur:on (a) The transformer supplies a load of 80 MVA at 0.85 PF lagging. Therefore, the secondary line current of the transformer is S _ 80,000,000 VA Lig =e = 02 A *NBV,, V3(115,000 V) The base value of the secondary line current is I Sra — _ 100,000,000 VA ane = eee =“ SO2A sr F3V ese V3(11S,000V) 30 the per-unit secondary current is lig _402A om TOA I Leos (0.85) =0.82—-31.8°