Concept explainers
Figure 2.26 shows three loads connected in parallel across a
Load 1: Inductive load,
Load 2: Capacitive load,
Load 3: Resistive load,
(a) Determine the total kW, kvar, kva, and supply power factor.
(b) In order to improve the power factor to 0.8 lagging. a capacitor of negligible resistance is connected in parallel with the above loads. Find the kvar rating of that capacitor and the capacitance in
Comment on the magnitude of the supply current after adding the capacitor.
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Chapter 2 Solutions
Power System Analysis & Design
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- A SUBSTATION RATED 1250 KVA IS OPERATING AT FULL LOAD AT 0.8 PF LAGGING AND TO CARRY AN ADDITIONAL LOAD OF 170 KW AT 0.85 PF LAGGING. WITH THE MENTIONED POWER FACTOR, THE SUBSTATION IS EXPERIENCING AN INEFFICIENT ELECTRICAL SYSTEM: HIGH POWER LOSSES, HIGH VOLTAGE DROPS IN THE ELECTRICAL EQUIPMENT, AND HIGH ELECTRICITY BILLS. TO REDUCE POWER LOSSES AND TO REDUCE THE ELECTRICITY BILL, POWER FACTOR CORRECTION IS TO BE OF HIGH IMPORTANCE AND CONSIDERATION. ANSWER: CITE A STEP BY STEP SUMMARY OF HOW TO DETERMINE THE POWER FACTOR CORRECTION REQUIRED IN THE GIVEN PROBLEM TO ARRIVE AT THE ALTERNATIVE THAT SHALL BE SYSTEMATICALLY CARRIED OUT TO GUARANTEE A SUCCESSFUL DECISION IMPLEMENTATION,arrow_forwardA substation rated 1250 kVA is operating at full-load at 0.8 pf lagging and to carry an additional load of 170 kW at 0.85 pf lagging. With the mentioned power factor, the substation is experiencing inefficient electrical system: high power losses, high voltage drops in the electrical equipment, and high electricity bill. To reduce power losses and to reduce the electricity bill, power factor correction is to be of high importance and consideration. With this condition, the substation is considering of two alternatives namely: Alternative A: To increase the capacity of the substation Alternative B: To add shunt power capacitor In this regard, the owner requested to prepare a summary report consisting of the following details: 1. A rightful power phasor diagram.arrow_forwardA SUBSTATION RATED 1250 KVA IS OPERATING AT FULL LOAD AT 0.8 PF LAGGING AND TO CARRY AN ADDITIONAL LOAD OF 170 KW AT 0.85 PF LAGGING. WITH THE MENTIONED POWER FACTOR, THE SUBSTATION IS EXPERIENCING AN INEFFICIENT ELECTRICAL SYSTEM: HIGH POWER LOSSES, HIGH VOLTAGE DROPS IN THE ELECTRICAL EQUIPMENT, AND HIGH ELECTRICITY BILLS. TO REDUCE POWER LOSSES AND TO REDUCE THE ELECTRICITY BILL, POWER FACTOR CORRECTION IS TO BE OF HIGH IMPORTANCE AND CONSIDERATION. Make a total power triangle (phasor diagram) for Alternate A additional substation capacity in KVA and Alternate B shunt capacitor in KVAarrow_forward
- A substation rated 1250 kVA is operating at full-load at 0.8 pf lagging and to carry an additional load of 170 kW at 0.85 pf lagging. With the mentioned power factor, the substation is experiencing inefficient electrical system: high power losses, high voltage drops in the electrical equipment, and high electricity bill. To reduce power losses and to reduce the electricity bill, power factor correction is to be of high importance and consideration. Find: power phasor diagram. computation of the following: additional substation capacity in KVA shunt capacitor KVA. Which alternative is more economical if cost/kVA of additional substation capacity is $400.00 and the cost/kVA of shunt capacitor is $180?arrow_forwardBorduria Generation owns three generating units that have the following cost functions: Unit A: 15 + 1.4 PA + 0.04 P2 A $/h Unit B: 25 + 1.6 PB + 0.05 P2 B $/h Unit C: 20 + 1.8 PC + 0.02 P2 C $/h How should these units be dispatched if Borduria Generation must supply a load of 350 MW at minimum cost?arrow_forwardDon't use other platforms answer and also don't use chatgpt. A 3 phase 6000 kVA, 6600V, Y connected 2 pole 60 Hz alternator has the following characteristics: a. Field current= 125A, Open Circuit Voltage line value=8000V, Short circuit current=800A At full load the resistance drop is 3%. Find voltage regulation at power factor of 0.8 lagging.arrow_forward
- A 20 KVA single phase, which can be taken as of unity power factor, has a full load efficiency of 95.3%, the copper loss then being twice the iron loss Calculate its all-day efficiency at following daily cycle • quarter load for 10 hours. . half load for 8 hours. full load for 6 hoursarrow_forwardGiven the secondary value of Iactual = 5 A compute for the per unit value of a current assume a based values of S = 1000VA, V = 2400/240 (assume single phase). round answer to two decimal places.arrow_forwardA three-phase system has balanced conditions so that the per-phase circuit representation can be used as shown in Figure 1 Select the turns ratio of the step-up and step-down transformers that the system operates with an efficiency greater than 99 percent. Moreover, find the complex power (received or given) of all components in the circuit and the V1 and V2 voltages. The load voltage is specified as 4 kV rms, and the load impedance is 4/3 ohm. ..arrow_forward
- Power System Analysis and Design (MindTap Course ...Electrical EngineeringISBN:9781305632134Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. SarmaPublisher:Cengage Learning