The incremental cost (in Rupees/MWh) of operating two generating units are functions of theirrespective power P₁ and P2 in MW, and are given bydC₁/dP₁ = 0.2P₁ + 50dC2/dP2 = 0.24P2 + 40Where20MW ≤ P₁ ≤ 150MW20MW ≤ P₂ ≤ 150MWSFoe a certain load demand, P₁ and P2 have been chosen such that dC₁/dP₁ = 76 Rs./MWh anddC₁/dP₁ = 68.8 Rs./MWh. IF the generations are rescheduled to minimize the total cost, then P2 is

Answered: Image /qna-images/answer/5ae7b086-b343-4b4e-8bfc-9414785ee974.jpg

The incremental cost (in Rupees/MWh) of operating two generating units are functions of their respective power P₁ and P2 in MW, and are given by dC₁/dP₁ = 0.2P₁ + 50 dC2/dP2 = 0.24P2 + 40 Where 20MW ≤ P1 ≤ 150MW 20MW ≤ P2 ≤ 150MW Foe a certain load demand, P₁ and P2 have been chosen such that dC₁/dP1 = 76 Rs./MWh and dC₁/dP1 = 68.8 Rs./MWh. IF the generations are rescheduled to minimize the total cost, then P2 is

The incremental cost (in Rupees/MWh) of operating two generating units are functions of their respective power P₁ and P2 in MW, and are given by dC₁/dP₁ = 0.2P₁ + 50 dC2/dP2 = 0.24P2 + 40 Where 20MW ≤ P1 ≤ 150MW 20MW ≤ P2 ≤ 150MW Foe a certain load demand, P₁ and P2 have been chosen such that dC₁/dP1 = 76 Rs./MWh and dC₁/dP1 = 68.8 Rs./MWh. IF the generations are rescheduled to minimize the total cost, then P2 is

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

See similar textbooks

Similar questions

Vbase_zonel Sbase_zonel . * For the power system shown in Figure Q1b, calculate the per unit impedances of all components with given zone 1 voltage base 22kV and power base 100 MVA (for the system). = 22 kV = 100 MVA Generator: . YG₁ Zone 1 100 MVA 22 kV (0.1+ j0.5) pu 1 . E 1 T₁ YY Transformer 1: 100 MVA 25/130 kV (0.1+j0.5) pu Zone 2 L = (10+j100) S2 Figure Q1b. T₂ Zone 3 YA Transformer 2: 50 MVA 110/33 kV (0.1+j0.5) pu M₂Y Load (motor): ● 50 MVA . 33 kV (1+j0.5) pu
I need the answer as soon as possible
a question by considering the annual load time curve for an electric station as a straight line from 20mw to 3mw, and to cover these loads, then operate three generating units, the first and second, each with a capacity of 10mw, while the third has a capacity of 5mw, and illustrate the load time curve by drawing, then find: 1- the total capacity of the station 2- the maximum demand 3- capacitance coefficient 4- pregnancy coefficient 5 utilizatio
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,
A generating station has a maximum demand of 30,000 kW. Calculate the cost per unit generated from the following data: Capital cost = P 130M; Annual load factor = 50%; Annual cost of fuel and oil = P7.5M ; Taxes, wages and salaries etc. = P7M and Interest and depreciation = 12%
Q)The question is that how kW is converted to kVA in complex form and how kV square is 2.4 for T1,T2,T3 can you plz explain it by seeing the solution
A 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. ..
Three identical Y-connected loads, each having a resistance of 20 22 in parallel with an inductor, are connected to a three phase supply of 400 V and 50 Hz frequency. The power absorb by the load is 8 kW and power factor at the load is 0.9919. (i) Determine the current that flows through the inductor. Please state the assumptions used to solve the question. Calculate load inductance in Henry (H).
Please answer completely with accurate diagrams.
Question is attached
can you please show me the sol'ncan you please show me the sol'ncan you please show me the sol'ncan you please show me the sol'ncan you please show me the sol'n