My Undergraduate Electrical Power Engineering Assignment Collection

in #technology9 months ago

Note

This is a collection of my undergraduate assignments that I translated to English myself in the Basic Electrical Power Engineering course. Bahasa Indonesia dibawah / Indonesian language below. 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

Process of supplying electrical energy in general: Conversion (generation) → Transmission → Distribution.

Electrical Conversion

Conversion 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 Transmission

Electric 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 Distribution

Electricity distribution is the process of sharing electricity to consumers.

Medium Voltage Network → Step-Down Transformer → Low Voltage Network → Consumer

DC (Direct Current) electric graph, V(t) = 2 volt, V(0) = 0 volt

AC (Alternating Current) electric graph, V(t) = sin(wt)

Basic Theories

Ohm Law

Ohm law: V=IR, V = voltage (volt), I = current (ampere), R = resistance (ohm).

Inverter, Rectifier, UPS

AC (alternating current) electricity to DC, using a rectifier. DC (direct current) electricity to AC, using an inverter. If the electricity suddenly goes out, it can damage the computer, hardware and software (data). To anticipate this, a UPS was created. The UPS supplies AC power to the computer as well as stores electricity in DC form. If the power suddenly goes out, the computer will use the UPS stored power, at least giving it time to store data and shut down the computer safely.

Transformator

Transformers are devices for changing the voltage. The step-down transformer lowers the voltage while the step-up transformer increases the voltage.

Example DC step-down transformer, DC step-up transformer when the circuit is reversed. Vin = V1+V2 = I1 R1+I3 R2. I1 = I2+I3. Vout = V2 = Vin-V1.

AC Transformator. V1/V2 = N1/N2 = I2/I1.

Magnetic Field

If there is a current flowing I towards dl (long) then the magnetic field strength H.

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 Principle

The 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 process of forming an emf on the side of the generator coil.

The resulting voltage is an alternating current. Furthermore, the current will be rectified by the commutator.

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

Separated gain DC generator. Vf = If Rf, Ea = Vt + Ia Ra

Self-Strengthening Generator

Series. Vt = Ia Ra, Ea = Ia (Ra + Rf) + Vt + < Vsi

Shunt. Vt = If Rf, Ea = Ia Ra + Vt + < Vsi

Compound

Long. Ia = If1 = IL + If2, Ea = Vt + Ia(Ra + Rf1) + < Vsi

Short. Ia = If1 + If2 = IL + If2, Ea = Vt + ILRf1 + IaRa + < Vsi

Some Exercises

1. A long compound DC generator provides 300 kW of power at a terminal voltage of 600V. The parallel field resistance is 75 the armature resistance, the brush resistance is 0.03, the field winding commutation resistance is 0.011Ω the series field resistance is 0.012Ω, the divertor resistance is 0.036. When the generator is fully loaded calculate the voltage and power generated by the armature winding!

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

N1 and N2

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

I1, I2, E2 and Φm

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

Koleksi Tugas Dasar Teknik Tenaga Listrik Sarjana Saya

Catatan

Ini merupakan kumpulan tugas-tugas S1 saya di mata kuliah Dasar Teknik Tenaga Listrik. Tugas ini tidak pernah dipublikasi dimanapun dan saya sebagai penulis dan pemegang hak cipta melisensi tugas ini customized CC-BY-SA dimana siapa saja boleh membagi, menyalin, mempublikasi ulang, dan menjualnya dengan syarat mencatumkan nama saya sebagai penulis dan memberitahu bahwa versi asli dan terbuka tersedia disini.

Proses Pengadaan Energi Listrik Secara Umum

Proses pengadaan energi listrik secara umum: Konversi(pembangkit) → Transmisi → Distribusi.

Konversi Listrik

Konversi dalam bidang tenaga listrik adalah konversi dari energi lain ke energi listrik. Konversi ini biasanya dipakai dalam menghasilkan listrik, penghasil listrik ini disebut pembangkit. Dengan kata lain sebagai produsen. Sebagai sumber energi listrik yang sudah resmi antara lain: PLTA (Pembangkit Listrik Tenaga Air), PLTU (Pembangkit Listrik Tenaga Uap), PLTN (Pembangkit Listrik Tenaga Nuklir), PLTG (Pembangkit Listrik Tenaga Gas), PLTD (Pembangkit Listrik Tenaga Diesel) dan lain-lain. Pembangkit lain antara lain: Tenaga Surya, Tenaga Angin, Reaksi Kimia, Panas Bumi dan lain-lain.

Energi Nuklir dll → Converter Reaktor dll → Alternate → Transformator Step Up; Converter Reaktor dll → Generate

Transmisi Listrik

Transmisi listrik adalah proses pengiriman listrik. Pengiriman melewati jaringan listrik.

Transformator Step-Up → Jaringan Tegangan Tinggi → Transformator Step-Down → Jaringan Tegangan Menengah → Pabrik

Distribusi Listrik

Jaringan Tegangan Menengah → Transformator Step-Down → Jaringan Tegangan Rendah → Konsumen

Grafik listrik DC (Direct Current), V(t) = 2 volt, V(0) = 0 volt

Grafik listrik AC (Alternating Current), V(t) = sin(wt)

Teori-Teori Dasar

Hukum Ohm

Hukum ohm: V=IR, V = tegangan (volt), I = arus (ampere), R = hambatan (ohm).

Inverter, Rectifier, UPS

Listrik AC (alternating current) menjadi DC, menggunakan rectifier. Listrik DC (direct current) menjadi AC, menggunakan inverter. Jika listrik tiba-tiba mati, maka dapat merusak komputer, hardware maupun software (data). Untuk mengantisipasi hal ini, dibikin UPS. UPS mengalirkan listrik AC ke komputer sekaligus menyimpan listrik dalam bentuk DC. Jika listrik tiba-tiba mati, maka komputer akan menggunakan listrik simpanan UPS, setidaknya memberi waktu untuk menyimpan data dan mematikan komputer dengan aman.

Trafo (Transformator)

Transformator (trafo) adalah alat untuk mengubah tegangan. Trafo step-down menurunkan tegangan sedangan trafo step-up meningkatkan tegangan.

Contoh Trafo step-down DC, trafo step-up DC bila rangkaian dibalik. Vin = V1+V2 = I1 R1+I3 R2. I1 = I2+I3. Vout = V2 = Vin-V1.

Trafo AC. V1/V2 = N1/N2 = I2/I1.

Magnetic Field

Jika ada arus yang mengalir I ke arah dl (panjang) maka timbul kuat medan magnet H.

dH = (IdI x aR)/(4πR2)(A/m), H = ∮(IdI x aR)/(4πR2)

I = arus (A), dl = panjang, aR = vektor, R = panjang vektor.

Flux Medan Magnet

B = dΦ/ds = μH, Φ = ∮Bds

B = kerapatan flux medan magnet (weber/area), Φ=flux medan magnet (weber), ds = luar area (m2), H = kuat medan magnet (A/m)

Prinsip Kerja Generator Arus Searah

Prinsip kerja suatu generator arus searah berdasarkan hukum Faraday: e = -N dΦ/dt

Dimana, N: jumlah lilitan, Φ: fluksi magnet, e: Tegangan imbas, ggl(gaya gerak listrik).

Proses terbentuknya ggl pada sisi kumparan generator.

Tegangan yang dihasilkan adalah arus bolak-balik. Selanjutnya arus tersebut akan disearahkan oleh komutator.

Komutator berfungsi sebagai saklar. Komutator berupa cicin belah yang dipasang pada ujung jangkar. Bila jangkar berputar maka cincin akan berputar. Bila kumparan telah berputar setengah putaran, sikat akan menutup celah cincin sehingga tegangan menjadi nol. Karena cincin berputar terus, maka celah akan terbuka lagi dan timbul tegangan lagi. Bila perioda tegangan sama dengan perioda perputaran cincin, tegangan yang timbul adalah tegangan arus searah gelombang penuh.

Jenis-Jenis Generator Arus Searah Dilihat Dari Belitan Medannya

Generator DC penguatan terpisah. Vf = If Rf, Ea = Vt + Ia Ra

Generator Penguatan Sendiri

Seri. Vt = Ia Ra, Ea = Ia (Ra + Rf) + Vt + < Vsi

Shunt. Vt = If Rf, Ea = Ia Ra + Vt + < Vsi

Compound

Panjang. Ia = If1 = IL + If2, Ea = Vt + Ia(Ra + Rf1) + < Vsi

Pendek. Ia = If1 + If2 = IL + If2, Ea = Vt + ILRf1 + IaRa + < Vsi

Soal-Soal

1. Sebuah generator DC kompon panjang memberikan daya 300 kW pada tegangan terminal 600V. Resistansi medan parallel adalah 75 resistansi armature, resistansi brush adalah 0.03, resistansi komutasi belitan medan adalah 0.011 resistansi medan seri adalah 0.012, resistansi divertor adalah 0.036 Bila generator dibebani penuh hitunglah tegangan dan daya yang dihasilkan oleh beletan jangkar (armature)!

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. Trafo 1ɸ diberikan supply 50 Hz step down 2200 V→250 V dengan luas 36 cm2 dan kerapatn fluks 6 wb/m2, carilah belitan primer dan sekunder.

Diketahui:

f = 50 Hz

E1 = 2200 V

E2 = 250 V

A = 0.0036 m2

Bm = 6 wb/m2

Ditanya:

N1 dan N2

Jawab

Φm = (Bm)(A)=(6 wb/m2)(0.0036 m2) = 0.0216 wb

E = (4.44)(f)(N)(Φm) = (4.44)(f)(N)(Bm)(A)

belitan primer = E1/((4.44)(f)(Φm)) = 2200/((4.44)(50 Hz)(0.0216 wb)) = 458.79

belitan sekunder = E2/((4.44)(f)(Φm)) = 250/((4.44)(50 Hz)(0.0216 wb)) = 52.135

3. Daya trafo 1ɸ adalah 25 KVI, belitan primer adalah 500, belitan sekunder adalah 50, belitan primer dihubungkan dengan tegangan 3000 V dengan frekuensi 50 Hz. Carilah arus beban pada belitan primer dan sekunder, emf sekunder dan fluks maksimal. (abaikan drop voltage)

Diketahui

P = 25 VI

N1 = 500

N2 = 50

E1 = 3000 V

f = 50 Hz

Ditanya

I1, I2, E2 dan Φm

Jawab

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