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Consider a three-phase generator rated
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Chapter 3 Solutions
MindTap Engineering, 1 term (6 months) Printed Access Card for Glover/Overbye/Sarma's Power System Analysis and Design, 6th
- An ideal transformer has no real or reactive power loss. (a) True (b) Falsearrow_forwardThe ratings of a three-phase three-winding transformer are Primary(1): Y connected 66kV,15MVA Secondary (2): Y connected, 13.2kV,10MVA Tertiary (3): A connected, 2.3kV,5MVA Neglecting winding resistances and exciting current, the per-unit leakage reactances are X12=0.08 on a 15-MVA,66-kV base X13=0.10 on a 15-MVA,66-kV base X23=0.09 on a 10-MVA,13.2-kV base (a) Determine the per-unit reactances X1,X2,X3 of the equivalent circuit on a 15-MVA,66-kV base at the primary terminals. (b) Purely resistive loads of 7.5 MW at 13.2 kV and 5 MW at 2.3kV are connected to the secondary and tertiary sides of the transformer, respectively. Draw the per- unit impedance diagram, showing the per-unit impedances on a 15-MVA,66-kV base at the primary terminals.arrow_forwardA single-phase, 50-kVA,2400/240-V,60-Hz distribution transformer has the following parameters: Resistance of the 2400-V winding: R1=0.75 Resistance of the 240-V winding: R2=0.0075 Leakage reactance of the 2400-V winding: X1=1.0 Leakage reactance of the 240-V winding: X2=0.01 Exciting admittance on the 240-V side =0.003j0.02S (a) Draw the equivalent circuit referred to the high-voltage side of the transformer. (b) Draw the equivalent circuit referred to the low-voltage side of the transformer. Show the numerical values of impedances on the equivalent circuits.arrow_forward
- A bank of three single-phase transformers, each rated 30MVA,38.1/3.81kV, are connected in Y- with a balanced load of three 1, Y-connected resistors. Choosing a base of 90MVA,66kV for the high-voltage side of the three-phase transformer. spify the base for the low-voltage side. Compute the per-unit resistance of the load on the base for the low-voltage side. Also, determine the load resistance in ohms referred to the high-voltage side and the per-unit value on the chosen base.arrow_forwardTwo 100-kW, single-phase transformers are connected in parallel both on the primary and secondary. One transformer has an ohmic drop of 0.5% at full-load and an inductive drop of 8% at full-load current. The other has an ohmic drop of 0.75% and an inductive drop of 2%. Show how they will share a load of 180 kW at 0.9 p.f.arrow_forward4. An Open-Wye/Open-Delta transformer, as shown in Figure, is formed by two identical single-phase transformers. The primary line-to-line voltage is 12.47 kV, and the secondary line-to-line voltage is 208 V. Takearrow_forwardTwo transformersA and B are connected in parallel to a load of (2+j1.5) ohm.Their impedances in Secondary terms ZA = (0.15+j0.8) ohm and Zb = (0.1+j30.6)ohm Their no-load termind voltages are EA=200<0 V and EB=210<0 Volt.Find the power and Power factor of each transformer? %3D %3Darrow_forward1 Two transformers A and B are connected in parallel to a load of (2+j1.5)2. Their impedances in secondary terms are ZA = (0.15 + j0.5)2 and ZB= (0.1 +j 0.6)2. Their no-load terminal voltage are EA 207 /0° volt and EB = 205/0° volt. Find the power output and power factor of each transformer.arrow_forwardNo-load operation and short circuit tests were performed on three-phase, 50 Hz 480/220 delta / Y connected transformer. In these testsThe line voltage, line current and three-phase total power have been measured from the high voltage windings and the test results are below.pictures. 100 A DC current when 1.08 V DC voltage from non-transferable software is applied to the low voltage windingshas attracted.Find the single phase equivalent circuit parameters of this transformer.idle test:Voltage: 480V Power:114w current: 0.71Ashort circiut test:V: 10v P: 78w c: 18.3aarrow_forwardThe primary of a certain transformer, with and open-circuit secondary, takes 1A at a power factor of 0.4 when connected to a 200V 50Hz supply. The turns ratio is Np/Ns = 2. A load taking 50A at a power factor of 0.8 lagging is now connected to the secondary. Sketch the phasor diagram and calculate the primary current. Neglect leakage flux and winding resistancesarrow_forwardTwo single-phase transformers whose phase electrical equivalent circuits are given below are connected in parallel and ZL= 0.8+j0.6 ohm is feeding a worth of load. Idle operating voltage of Transformer 1 is 220 V and Transformer 2 idle operating voltage is 240V. a.) Find the currents flowing from transformers to the load. b.) Find the active powers that transformers transfer to the load.arrow_forward3. At three winding transformer at no-load case, the current flows through the tertiary windings is large current load current secoundary current no-current 4.We can correct P.f (Power Factor) by using: * impedance Inductance capacitance non 5. There are two basic types of capacitor installations which used in power factor correction which are: used in power factor correction which are: series, shunt series, individual individual capacitors, and banks of fixed or automatically switched capacitors.arrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
- Power System Analysis and Design (MindTap Course ...Electrical EngineeringISBN:9781305632134Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. SarmaPublisher:Cengage Learning