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MSU - ECE 3614 - Study Guide - Midterm

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Department of Electrical and Computer Engineering Mississippi State University Fundamentals of Energy Systems (ECE 3614) Quiz 1 February 25, 2016 Name: NetID: Q 1 [10%]. What is the core loss in a magnetic circuit? Explain clearly. Answer: Core loss is the loss happening in the magnetic core material of a magnetic circuit wIf you want to learn more check out spanish ttu

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hen it is exposed to a changing field. It has two components: hysteresis and eddy current losses. Hysteresis losses are due to the hysteresis nature of B-H curve of ferro magnetic materials. The eddy currents are whirling electric currents generated inside the magnetic core due to the local voltages induced in the core. Q 2 [10%]. What is the reactive power? Why should industrial customers care about their reactive power? Answer: It is an oscillating (or pulsating) AC power that happens in AC loads and does not make up for any net amount of energy. It should be cared for because it causes extra losses in the utility lines and limits their capacity. Q 3 [10%]. Why is the AC power system prevalent as opposed to the DC? Why is it built in a three-phase structure not a single-phase? Answer: More efficient transmission using high voltage lines; more efficient construction of ma chines in AC than in DC. More efficient use of magnetic cores and transmission conductors; smooth ripple free power in balanced situation (lowering mechanical stress on the shaft of motors); stronger machines with higher power. Q 4 [5%]. State the Faraday’s law. Answer: A changing magnetic field when passing through a coil of wire induces a voltage in it according to eind(t) = ddtφ(t). Q 5 [5%]. What is a voltage regulator in the power system? Answer: Voltage regulators are tap-changing under load transformers with automatic control systems to change the taps in order to regulate the voltage in the power system.P 1 [20%]. A single phase transformer has the following ratings: 25 kVA, 7200/240 V, 60 Hz, series impedance: 0.01+j0.04 pu, excitation current: 0.05 pu. Find the following values: (a) the transformer’s rated copper loss in pu and in watts; (b) the transformer’s excitation current in Amperes; (c) the transformer’s voltage regulation when operating at full load and unity power factor; (d) the transformer’s voltage regulation when operating at full load and 0.8 power factor lagging. Answer: (a) The rated copper loss is 0.01 pu which is 0.01×25000=250 W. (b) The rated excitation current is 0.05 pu which is 0.05×25000 7200 =0.174 A. (c) −→V p =−→V s + (Req + jXeq)−→I = 1 + (0.01 + j0.04)1 = 1.01 + j0.04 = 1.0108∠2.27◦. VR=(Vp − 1) × 100 = 1.08%. (d) −→V p =−→V s + (Req + jXeq)−→I = 1 + (0.01 + j0.04)1e−j36.9◦= 1.0323∠1.44◦. VR=(Vp − 1) × 100 = 3.23%.P 2 [20%]. Consider a single-phase ac load with 0.866 PF lagging supplied at 120 V (rms), 50 Hz. Its current is measured at 30 A. 1) Calculate its real power, reactive power, and apparent power? Answer: P = V I cos φ = 120 × 30 × 0.866 = 3118 W. Q = V I sin φ = 120 × 30 × 0.5 = 1800 VAr. S = V I = 120 × 30 = 3600 VA 2) What is the impedance of the load? Answer: −→Z =−→V−→I=120 30∠(−30◦) = 4∠30◦ = 3.464 + j2 Ω 3) Draw a sketch of voltage and current waveforms. Specify values for the peaks, time-period, and the phase-shift between them, on your sketch. Answer: The current waveform lags the voltage 30 degrees of 50 Hz that is about 1.67 ms. Sketch is shown below. Notice that the voltage peak is 120 ×√2 = 170 V and peak of current is 42.4 A. 200 100 0 −100 −200 v(t) i(t) 1.67 ms 40 t (ms) 20 4) How much capacitance is needed to compensate the PF to unity? Answer: Q = CωV 2 = 1800 → C =1800 314×1202 = 398µFP 3 [20%]. A small industrial plant uses three-phase lines to power two of its loads. A single-line diagram of that is shown below. 480 V, 60 Hz ∼ Load 1 Load 2 100 kW, 0.8 PF lagging, ∆ 60 kW, unity PF, Y 1) How much is the total real power of the plant? Answer: P = 100 + 60 = 160 kW 2) How much is the total reactive power of the plant? Answer: Load 1: S1 =P1 0.8 = 125 kVA, Q1 =pS21 − P21 =√1252 − 1002 = 75 kVAR P F =100 Load 2: Q2 = 0 Total reactive power: Q = Q1 + Q2 = 75 kVAR. 3) How much is the total PF of the plant? Answer: Total apparent power: S =√1602 + 752 = 176.7 kVA and total power factor: P F =PS =160 176.7 = 0.90 lagging. 4) Calculate the current of each load and the total current. Answer: I1 = √S13V= √125000 3×480 = 150.35 A I2 = √S23V= √60000 3×480 = 72.17 A I = √S3V= √176700 3×480 = 212.5 ADepartment of Electrical and Computer Engineering Mississippi State University Fundamentals of Energy Systems (ECE 3614) Quiz 1 September 27, 2016 1. Complete this statement. [10%] Bulk electric energy is produced in .............. plants where a high-pressure, high-temperature steam is released to spin a turbine that is coupled to the shaft of a .............. . Power plants are often in remote areas to keep ................ off the urban centers and to lower the ................... costs. In order to transmit the power over long distance, the voltage is raised by a ................. . The bulk power is generated, transmitted and used in three-phase because (1) .................................................. (2) ................................................... (3) .......................................................... power, generator, pollution, fuel transportation, transformer, (1)-(3): 3-ph machines are more efficient and stronger; lower usage of conductors; smooth power on the shaft of a three-phase machine. 2. Why does the voltage levels in a power system should be tightly regulated? Why should the frequency be tightly regulated in an AC power system? [10%] Because electric machines are designed to operate at a rated voltage. This voltage is in the knee region of the magnetic curve in order to maximize the efficiency. If the voltage goes above that, it will push the operating point into the saturation region and will cause huge and distorted current. Frequency should be tightly regulated to ensure synchronized operation of all machines that are interconnected. 3. Specify and explain the two types of losses in a transformer. [10%] Copper loss: caused by the flow of current in wires; and Core loss: caused in the magnetic core. Core loss has two components: hysteresis loss (caused by hysteresis phenomenon of a magnetic material) and eddy current losses (caused by the swirling currents in the core). 4. A magnetic circuit comprises a core with relative permeability of 2000 and a winding of 200 turns. Its cross sectional area is 1 cm2 and its average length size is 20 cm. Calculate its inductance. [10%] L =µN2A ℓ =2000×4π×10−7×2002×1×10−4 20×10−2 = 0.050 H=50 mH 15. If a voltage supply of 12 V (rms), 60 Hz is applied to the inductor of problem 4, how much would be the current (rms)? Also, calculate the magnetic flux and magnetic flux density in the core (rms values). [15%] I =V Lω =12 Φ = V 0.050×377 = 0.64 A Nω =12 200×377 = 0.00016 Wb, B =ΦA =0.00016 0.0001 = 1.6 T 6. A single-phase load is specified as 1 kW, 120 V, 60 Hz, 0.8 PF lagging. Calculate its current. If we desire to boost its PF to unity, how much capacitance is needed in parallel to it? Calculate the current after compensation. [15%] V cos ϕ =1000 I =P 120×0.8 = 10.42 A Q = V I sin ϕ = 120 × 10.42 × .6 = 750 VAR Q = CωV 2 ⇒ C =750 377×1202 = 138 µF I =P V cos ϕ =1000 120 = 8.33 A 7. A three-phase load is characterized by 5 kW, 208 V, 60 Hz, 0.8 PF lagging, Y-connected. Calculate its current. Repeat if the load is ∆-connected. [5%] I = √P 3V cos ϕ= √5000 3×208×0.8= 17.35 A. Will be the same for ∆-connected! 28. A single-phase transformer has the following nameplate data: 10 kVA, 7200/240 V, 60 Hz, series impedance: 0.01+j0.05 pu. Calculate its copper loss in watts and its V.R. when operating at full load 0.8 PF lagging. [10%] Pcu = 0.01 × 10000 = 100 W Vp = Vs + (R + jX)−→I = 1 + (0.01 + j0.05) × 1 × (0.8 − j0.6) = 1.039∠·, V.R.= 3.9% 9. If in the transformer of problem 8, we measure the efficiency at 97%, find the values of its core loss (in watts) and core loss resistor Rc in pu. [10%] η =Pout Pin+Ploss=1×1×0.8 0.8+Ploss= 0.97 ⇒ Ploss = 0.0247 pu, Pcore = Ploss −Pcu = 0.0247−0.01 = 0.0147 pu=147 watts Rc ≈1 0.0147=68 pu 10. An autotransformer is used to convert 120 V to 100 V. Find its power advantage ratio (that is the ratio of the total power to the winding power). [5%] 100 = 1 + Ns 120 Nc ⇒ Ns SW= 1 + Nc Nc= 0.2; S Ns= 1 + 5 = 6 3Department of Electrical and Computer Engineering Mississippi State University Fundamentals of Energy Systems (ECE 3614) Quiz 1 September 22, 2015 Name: NetID: Q 1 [10%]. What are the two components of core losses in a magnetic core? Explain each one and describe how we can deal with them. Answer: Core losses include hysteresis and eddy current losses. Hysteresis losses are due to the hysteresis nature of B-H curve of ferromagnetic materials. The eddy currents are generated inside magnetic core due to the local voltages induced in the core (according to Faraday’s law). Q 2 [5%]. Why did AC power system become popular and not DC? Answer: Because AC voltage can be increased using transformer and transmitted over long distances without excessive transmission losses. Q 3 [5%]. Why is the AC power system built in three-phase structure not single-phase? Answer: More efficient use of magnetic cores and transmission conductors, smooth ripple free power in balanced situation (lowering mechanical stress on the shaft of motors), stronger machines with higher power.P 1 [20%]. Consider a single-phase ac load supplied at 240 V (rms), 50 Hz. Its current and power are measured at 100 A and 19200 W. 1) Calculate reactive power, apparent power and PF of this load? Answer: S = V I = 240 × 100 = 24000 VA PF=PS =19200 24000 = 0.8 lagging Q =√S2 − P2 =√240002 − 192002 = 14440 VAR 2) What is the impedance of the load? Answer: −→Z =→−V→−I=240 100∠(−36.9◦ ) = 2.4∠36.9◦ = 1.92 + j1.44 Ω 3) Draw a sketch of voltage and current waveforms. Specify values for the peaks, time-period, and the phase-shift between them, on your sketch. Answer: The current waveform lags the voltage 36.9 degrees of 50 Hz that is about 2.05 ms. Sketch is shown below. Notice that the voltage peak is 240 ×√2 = 340 V and peak of current is 141 A. 300 200 100 0 −100 −200 −300 v(t) i(t) 2.05 ms40 t (ms) 20 P 2 [20%]. The linear dc machine shown below has a magnetic flux density of 0.5 T, a resistance of 0.25 Ω, a bar length of 1 m, and a battery voltage of 100 V. E + i Sw R × × × + e B × × × frictionless rails × × × × × - - × × × × × × bar of conducting metal (mass m and length ℓ) 1. How much is the force on the bar at start when the switch gets closed and in what direction? What is the initial current flow? 2. What is the no-load steady state speed of the bar? 3. While running at no-load, if the bar is loaded with a force of 25 N opposite to the direction of motion, what is the new steady state speed? Is it a motor or a generator in this case? Why? Answer: 1. At the start: i =ER=100 0.25= 400 A, F = iℓB = 400 × 1 × 0.5 = 200 N The force is to the right. 2. At the no-load steady state: i = 0 ⇒ e = E = 100 V, and e = vℓB ⇒ v =eℓB =100 1 × 0.5= 200 m/s 3. In this case, the motor generates a force of F=25 N to balance the load. There fore, F = iℓB = 25 ⇒ i =25 1 × 0.5= 50 A e = E − Ri = 100 − 0.25 × 50 = 87.5 V e = vℓB ⇒ v =eℓB =87.5 1 × 0.5= 175 m/s It is motor because its speed is below the no-load speed and it draws current from the battery.P 3 [20%]. The magnetic circuit shown below has an iron core with the relative permeability of 5000. The core dimensions are A = 16 cm2, ℓg = 2 mm, ℓc = 80 cm. The coil has 500 turns and draws a current of 4 A from the source. Neglect φ i N ℓg air gap magnetic leakage and fringing effect. Calculate 1. the total magnetic flux, 2. the flux linkage of the coil, and 3. the coil inductance. Answer: 1. The equivalent circuit of this structure is shown below, where F = N i = 500 × 4 = 2000 At, Rc =ℓc 5000×4π×10−7×0.0016 = 7.96 × 104 At/Wb and Rg =ℓg µrµoA =0.8 4π×10−7×0.0016 = 9.95 × 105 At/Wb. µoA =0.002 + equivalent circuit: F - Therefore, φRc Rg F = Rφ ⇒ φ =FR=2000 10.75 × 105 = 0.0019 Wb. 2. λ = Nφ = 500 × 0.0019 = 0.93 Wb 3. L =λi=0.93 4= 0.233 HP 4 [20%]. A small industrial plant uses three-phase lines to power two of its loads. A single-line diagram of that is shown below. 480 V, 60 Hz ∼ Load 1 Load 2 100 kW, 0.8 PF lagging, ∆ 60 kW, unity PF, Y 1) How much is the total real power of the plant? Answer: P = 100 + 60 = 160 kW 2) How much is the total reactive power of the plant? Answer: Load 1: S1 =P1 0.8 = 125 kVA, Q1 =pS21 − P21 =√1252 − 1002 = 75 kVAR P F =100 Load 2: Q2 = 0 Total reactive power: Q = Q1 + Q2 = 75 kVAR. 3) How much is the total PF of the plant? Answer: Total apparent power: S =√1602 + 752 = 176.7 kVA and total power factor: P F =PS =160 176.7 = 0.90 lagging. 4) Calculate the current of each load and the total current. Answer: I1 = √S13V= √125000 3×480 = 150.35 A I2 = √S23V= √60000 3×480 = 72.17 A I = √S3V= √176700 3×480 = 212.5 A