In the balanced three phase power measurement using two wattmeter method, each wattmeter reads 12 kW, when the pf is unity. What does each wattmeter reads when the pf falls to (i) 0.6 lag and (ii) 0.906 lag. The total three-phase power remains unchanged. (i) W1 at 0.6 lag (i) W2 at 0.6 lag (ii) W1 at 0.906 lag (ii) W2 at 0.906 lag

In the balanced three phase power measurement using two wattmeter method, each wattmeter reads 12 kW, when the pf is unity. What does each wattmeter reads when the pf falls to (i) 0.6 lag and (ii) 0.906 lag. The total three-phase power remains unchanged. (i) W1 at 0.6 lag (i) W2 at 0.6 lag (ii) W1 at 0.906 lag (ii) W2 at 0.906 lag

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter3: Power Transformers

Section: Chapter Questions

Problem 3.23P: Figure 3.32 shows the oneline diagram of a three-phase power system. By selecting a common base of...

See similar textbooks

Similar questions

do the discussion about Power measurement using Wattmeter, Ammeter and Voltmeter & Power measurement using Electronic Single phase measurement. Discuss the differences between both data. Why these problems occur and how to minimize it. Explain the role of power factor.
Q23/ In the balanced three phase power measurement using two wattmeter method, each wattmeter reads 18 kW , when the pf is unity. What does each wattmeter reads when the pf falls to (i) 0.6 lag and (ii) 0.422 lag. The total three-phase power remains unchanged.
A TWO-WATTMETER method is used to measure the total real power of the load in Problem , where wattmeter W1 is connected in line A and W2 is connected in line C. Line B being the common tapping point of the voltage coils of W1 and W2. Determine thefollowing:a. Reading of wattmeter W1b. Reading of wattmeter W2c. If a third wattmeter W3 is connected to line B and the two ends of the voltage coils areconnected to lines C and A, what is its reading?d. If the result in Part C is multiplied by 1.732 or square root of 3, what does this valuerepresents?
In the balanced three-phase power measurement using two wattmeter method, each wattmeter reads 10 kW, when the pf is unity. What does each wattmeter read when the pf falls to (i) 0.866 lag and (ii) 0.573 lag. The total three-phase power remains unchanged. (i) W1 at 0.866 lag (i) W2 at 0.866 lag (ii) W1 at 0.573 lag (ii) W2 at 0.573 lag
a. Must you see two wattmeters to measure the total three-phase power on a balancedthree-phase three-wire system? Explain.b. Can a wattmeter that has a current through its current coil and a potential across itsvoltage coil, indicate zero? Explain.
Q21 In an Industry, three phase power measurement is carried out, using two wattmeter method, each wattmeter reads 16 kW, when the pf is unity. What does each wattmeter reads when the pf falls to (i) 0.866 lag and (ii) 0.573 lag. The total three- phase power remains unchanged. (a) W1 at 0.866 lag (b) W2 at 0.866 lag (c) W1 at 0.573 lag (d) W2 at 0.573 lag
4. How will you calculate the average wattmeter reading in two wattmeter method for balanced load? a. Average wattmeter Reading = PT/3 b. Average wattmeter Reading = 2 X PW c. Average wattmeter Reading =(PW1-PW2)/(PW1+PW2) d. Average wattmeter Reading = PT/2
How will you calculate the average wattmeter reading in one wattmeter method for balanced load?
In the figure, the circuit is given as VRS =√3 . 60V and Zy = 6 + j8. Find the value of C to be connected in order to make the power drawn from the system in a phase 1, to make the load current 1, and to make cos(φ) 1.
With neat diagram, discuss the 3-phase power measurement using three wattmeter method and write down the drawbacks in this method. Note:please do not handwriting
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. 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.
Type your question here A power plant has to meet the following load demand: Load A: 100 MW from 8 AM - 6 PM Load B: 150 MW from 6 AM - 10 AM Load C: 50 MW from 6 AM - 10 AM Load D: 20 MW from 10 AM - 6 AM Plot the daily load curve and determine (a) diversity factor (b) load factor (c) daily energy produced.