NoteThis is a collection of my undergraduate assignments that I translated to English myself in the Basic Electrical Power Engineering course. This assignment has never been published anywhere and I, as the author and copyright holder, license this assignment customized CC-BY-SA where anyone can share, copy, republish, and sell on condition to state my name as the author and notify that the original and open version available here. General Electric Energy Supply Process![]() Electrical ConversionConversion in the field of electric power is the conversion from other energy to electrical energy. This conversion is usually used to generate electricity, producing this electricity is called a generator, in other words as a producer. Here are example sources of electrical energy: Hydroelectric Power Plant, Steam Power Plant, Nuclear Power Plant, Gas Power Plant, and Diesel Power Plant. Other generators include: Solar Power, Wind Power, Chemical Reactions, Geothermal Energy and others. Nuclear Energy etc → Reactor Converter etc → Alternate → Transformer Step Up; Reactor Converter etc → Generate Electric TransmissionElectric transmission is the process of sending electricity. The delivery goes through the power grid. Step-Up Transformer → High Voltage Network → Step-Down Transformer → Medium Voltage Network → Factory Electricity DistributionElectricity distribution is the process of sharing electricity to consumers. Medium Voltage Network → Step-Down Transformer → Low Voltage Network → Consumer ![]() ![]() Basic TheoriesOhm LawOhm law: V=IR, V = voltage (volt), I = current (ampere), R = resistance (ohm). Inverter, Rectifier, UPS![]() TransformatorTransformers are devices for changing the voltage. The step-down transformer lowers the voltage while the step-up transformer increases the voltage. ![]() ![]() Magnetic Field![]() dH = (IdI x aR)/(4πR2)(A/m), H = ∮(IdI x aR)/(4πR2) I = current (A), dl = length, aR = vector, R = vector length. Magnetic Field Flux B = dΦ/ds = μH, Φ = ∮Bds B = magnetic field flux density (weber/area), Φ=magnetic field flux (weber), ds = area (m2), H = magnetic field strength (A/m) Direct Current Generator Working PrincipleThe working principle of a direct current generator is based on Faraday's law: e = -N dΦ/dt Where, N: number of turns, Φ: magnetic flux, e: induced voltage, emf (electromotive force). ![]() ![]() The commutator functions as a switch. The commutator is in the form of a split ring attached to the end of the anchor. When the anchor rotates, the ring will rotate. When the coil has rotated half a turn, the brush will close the ring gap so that the voltage becomes zero. Because the ring continues to rotate, the gap will open again and create tension again. If the voltage period is the same as the ring rotation period, the voltage that arises is the full wave direct current voltage. Types of Direct Current Generators Viewed from its Field Winding![]() Self-Strengthening Generator![]() ![]() Compound![]() ![]() Some Exercises![]() ![]() Ish = 600V/75Ω = 8A, I = 300000W/600V = 500A, Ia = Ish+I = 8A+500A = 508A ∆Vtotal = ∆V+∆Va = Ia R+Ia Ra = Ia (R+Ra) = 508A(0.03Ω+0.02Ω) = 25.4V emfV = V+∆Vtotal = 600V+20.4V = 625.4V emfP = (emfV)(Ia) = (625.4V)(508A) = 317703.2W 2. Transformer 1ɸ is given a 50 Hz supply step down 2200 V → 250 V with an area of 36 cm2 and a flux density of 6 wb/m2, look for the primary and secondary windings. Known: f = 50 Hz E1 = 2200 V E2 = 250 V A = 0.0036 m2 Bm = 6 wb/m2 Asked: N1 and N2 Answer Φm = (Bm)(A)=(6 wb/m2)(0.0036 m2) = 0.0216 wb E = (4.44)(f)(N)(Φm) = (4.44)(f)(N)(Bm)(A) Primary Winding = E1/((4.44)(f)(Φm)) = 2200/((4.44)(50 Hz)(0.0216 wb)) = 458.79 Secondary Winding = E2/((4.44)(f)(Φm)) = 250/((4.44)(50 Hz)(0.0216 wb)) = 52.135 3. The power of the transformer 1ɸ is 25 KVI, the primary winding is 500, the secondary winding is 50, the primary winding is connected to a voltage of 3000 V with a frequency of 50 Hz. Find the load currents in the primary and secondary winding, secondary emf and maximum flux. (ignore the drop voltage) Known P = 25 VI N1 = 500 N2 = 50 E1 = 3000 V f = 50 Hz Asked I1, I2, E2 and Φm Answer E = (4.44)(f)(N)(Φm) = (4.44)(f)(N)(Bm)(A) Φm = (4.44)(f)(N1)/E1 = (4.44)(50 Hz)(500)/(3000 V) = 37 wb E2/E1 = (4.44)(f)(N2)(Φm)/(4.44)(f)(N1)(Φm) = N2/N1 E2=N2/N1 E1 = 500/50 3000 V = 30000V P=(V)(I) I1 = P/E1 = (25 W)/(3000 V) = 8.3mA I2 = P/E2 = (25 W)/(30000 V) = 0.83mA
0 Comments
Leave a Reply. |
Archives
August 2022
Categories
All
source code
old source code Get any amount of 0FP0EXP tokens to stop automatic JavaScript Mining or get 10 0FP0EXP tokens to remove this completely. get 30 0FP0EXP Token to remove this paypal donation. get 20 0FP0EXP Token to remove my personal ADS. Get 50 0FP0EXP Token to remove my NFTS advertisements! |