(Power Factor Correction) In an industrial plant with a large number of induction motors, during certain loading conditions, the inductor motors have a large reactive component and the power factor is low. Capacitor banks and synchronous motors are two commonly used methods for power factor correction. a) Capacitor based power factor correction: A capacitor bank is connected in parallel with the induction motor in problem 2. Calculate the capacitance value needed to bring the total power factor to unity. (1 point) b) Synchronous motors can also be used for power factor correction. A synchronous motor with 5 kW power input is connected in parallel with the induction motor, what power factor should the synchronous motor operate at to bring the total power factor to unity? (1 point) c) Both capacitors and synchronous motors can achieve the goal of power factor correction. Discuss the implications of a low power factor. Compare these two methods and discuss the advantages and disadvantages of each method with considerations of social, environmental, economic factors. (2 points) d) For a MW level large industrial plant located in rural area, its low power factor has been causing high utility bills. This specific plant has enough budget to install a power factor correction system and they are looking into a solution with long service life and flexible power factor control, which solution should they choose, capacitor banks or synchronous motor? (1 point)
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The formula to calculate the capacitance is: C = Q / (V^2 * 2 * π * f) Where: C = capacitance in farads Q = reactive power in VAR (volt-ampere reactive) V = operating voltage in volts f = frequency in hertz (usually 50 or 60 Hz) Once we have the values for Q, Show more…
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Madhur L.
1. A balanced load of 6 + j8 ohms is connected in Wye. Determine the value of the capacitor per phase to be connected in parallel with the load to improve the power factor to 0.8 lagging. 2. A 440V, 50Hz induction motor takes a line current of 45A at a power factor of 0.8 lagging. Three delta-connected capacitors are installed to improve the power factor to 0.9 lagging. Calculate the KVA of the capacitor bank and the capacitance of each capacitor. 3. A factory takes the following balanced loads from a 440V, 3-phase, 50Hz supply. Load 1: 20KW lighting load Load 2: a continuous motor load of 30KVA at 0.5pf lagging Load 3: an intermittent welding load of 30KVA at 0.5pf lagging Determine the CKVAR rating of the capacitor bank required to improve the power factor of Load 1 and Load 2 together to 0.95. If the capacitor bank is connected in star, determine the capacitor required in each phase. What is the new overall power factor after the correction has been applied when the welding machine is switched on? 4. Three unequal single-phase loads are connected across a balanced 3-phase, 230V supply. Load 1 is connected between terminals a and b, takes a current of 100A at 0.8 pf lagging. Load 2 is connected between terminals b and c and takes a current of 150A, 0.707 leading. The third load between c and a takes a current of 120A, 0.77 lagging. Assuming CBA sequence and Vab as reference. a. Find the three line currents. b. If the load connected between a and b is disconnected, determine the line current Ic.
Sri K.
You manage a factory that uses many electric motors. The motors create a large inductive load to the electric power line as well as a resistive load. The electric company builds an extra-heavy distribution line to supply you with two components of current: one that is 90° out of phase with the voltage and another that is in phase with the voltage. The electric company charges you an extra fee for "reactive volt-amps" in addition to the amount you pay for the energy you use. You can avoid the extra fee and the need for two components of current by installing a capacitor between the power line and your factory. But, you need to convince the owners of the factory to spend the funds to purchase and install this capacitor. You decide to make a presentation to the owners, using a simple RL circuit as a demonstration device. In your demonstration circuit, you represent the power company with a 120 V (rms), 60.0 Hz source. This source is in series with a series combination of a 21.0 mH inductor and a 20.0 Ω resistor. This combination represents the inductive and resistive loads for your factory. (a) To impress the owners, you calculate for them the power factor for the circuit and show that it is not equal to 1. power factor = 0.911 (b) You then determine the capacitance (in µF) of a capacitor that will bring the power factor to 1. (c) Demonstrate to the owners the percentage of increased power delivered to the factory. (P_new - P_old) / P_old x 100% =
Adi S.
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