3.2.3 Demonstrate
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The document compares the inductive reactance (XL) of two inductors, 0.5H and 10mH, at frequencies of 50Hz and 1kHz. It shows that XL is higher at higher frequencies and for the larger inductor. Specifically, at 1kHz the 0.5H inductor has an XL of 3.14kΩ while the 10mH inductor has an XL of 62.8Ω. It also explains that inductive reactance is directly proportional to frequency and inductance.
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7.4.6 slip
This document describes how to calculate the rotor frequency of a two-pole, 50 Hz induction motor given the rotor speed of 2850 rpm. It shows that the slip speed is 3000 rpm, the slip percentage is 5%, and using the formula fs x %slip / 100, the rotor frequency is calculated to be 2.5 Hz.
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4.1.13 Light Demand 3
This short document does not provide any substantive information to summarize in 3 sentences or less. It only notes that an explanation is not contained in a workbook, but provides no other context or details.
4.1.12 Light Demand 2 Table Part2
The document outlines the demand calculations for 19 different circuit load groups across 3 phases. It lists the load description, current allowance calculation method, and resulting demand current for each phase. The total demand current per phase is calculated at the bottom, with values of 143.8 amps for phase 1, 153.25 amps for phase 2, and 145.05 amps for phase 3.
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The document contains a table that calculates the current demand per phase for various circuit load groups in an electrical system. It lists 19 load groups categorized by letters A through D, describing each load. It shows the current allowance calculation method and resulting demand in amps for each phase. The total demand current calculated per phase is 143.8 amps for L1, 153.25 amps for L2, and 145.05 amps for L3.
4.1.10 Light Demand 2
This 3 sentence document provides instructions to refer to a specific table on a specific page of a particular standard for additional explanatory information not contained in the current workbook. The instructions direct the reader to Table C2 on page 359 of AS/NZA 3000:2007 for an explanation that is not included in the current document.
4.1.9 Light Demand 1
The document describes the functions of 19 electrical circuits in a building. It lists the types of equipment connected to each circuit such as fluorescent lighting, outlets, motors, and appliances. It also indicates which of the 3 electrical phases (L1, L2, L3) each circuit is connected to.
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The document calculates the electrical load of communal services in an apartment building. It shows that 24 lighting points will draw 240 watts and 6 10A sockets will draw up to 12A, for a total demand of 18A per phase. The total demand current per phase for communal services is 18A.
4.1.6 Example Domestic 2
The document calculates the electrical load and demand for 6 living units per phase. It lists the types of loads in each unit, the quantity and allowance per unit, and uses this to calculate the total demand current for phases L1, L2 and L3, which is 154.4 amps for each phase. The key loads included are lighting, power outlets, cooking ranges, air conditioners and hot water systems.
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This document discusses the number of living units per phase of a project. It calculates that for 18 total living units divided into 3 phases, there would be 6 living units per phase.
4.1.4 Step 2
This document summarizes the electrical load calculations for 11 circuits in a home. It lists the load type and description for each circuit, the current allowance per unit, and calculates the demand in amps for circuits 1-10. The largest demands are 17.7 amps for an air conditioner and 15 amps for an off-peak hot water system. The total calculated demand current for each phase is 40.7, 39.5, and 46.5 amps respectively.
4.1.3 Example Solution Three
The document calculates the electrical load and demand current for 11 circuits in a home. It groups the loads into categories like lighting, outlets, appliances, and assigns each a description, allowance, and demand calculation. The total demand current per phase is summarized at the bottom, with the highest draw of 46.5 amps on phase L3.
4.1.2 Example Three
The document outlines the functions and ratings of 12 electrical circuits. Circuit 1 provides power for 13 indoor lighting points. Circuit 10 powers an air conditioner rated at 23.6 amps per phase and can draw power from circuits L1, L2, and L3. Circuits 11a and 11b each power a controlled load hot water unit rated at 3.6 kW.
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This document calculates the total demand current for an electrical installation consisting of:
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2) It will take approximately 0.85 seconds (5 time constants) for the current to reach its final value.
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14.5.4 Example 1 Part 1
This document calculates the apparent power, power factor, and phase angle for a circuit. It finds that the apparent power is 2.308 kVA by multiplying the current of 9.615 by the voltage of 240. This apparent power is larger than the actual power of 1.5 kW, indicating a poor power factor of 0.65 or a 49.46 degree phase angle between the current and voltage.
14.1.2 TRA Power
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3.2.3 Demonstrate
- 3. XL = 2πfL
- 4. We will use one inductor with a known inductance of 0.5 Henries and another smaller one of 10mH. Letʼs compare their Inductance at frequencies of 50Hz and then at 1kHz.
- 5. For the 0.5H Inductor: At 50 Hz: XL = 2 π f L = 2 x 3.1416 x 50 x 0.5 = 157 Ω
- 6. For the 0.5H Inductor: At 1kHz: XL = 2 π f L = 2 x 3.1416 x 103 x 0.5 = 3.14kΩ
- 7. For the 10mH Inductor: At 50 Hz: XL = 2 π f L = 2 x 3.1416 x 50 x (10x 103) = 3.14 Ω
- 8. For the 10mH Inductor: At 1kHz: XL = 2 π f L = 2 x 3.1416 x 103 x (10x 103) = 62.8Ω
- 9. Yo u w i l l n o t i c e t h a t a t t h e h i g h e r frequency(1kHz), the inductance is much higher. Compare the results for the 0.5H inductor against the results for the smaller 10mH one.
- 10. Frequency 0.5H Inductor 10mH Inductor 50Hz 157 Ω 3.14 Ω 1 kHz 3.14 kΩ 62.8 Ω
- 11. Proportional to or inversely proportional to one another
- 12. Inductive reactance is proportional to the frequency of the alternating current and the inductance value.
- 13. XL f= 2∏ L
- 14. Frequency is inversely proportional to inductance.
- 15. Example: What would be the supply frequency if a coil has an inductive reactance of 20 ohms, and its inductance is 0.06H?
- 16. f = XL / 2∏ L = 20 / 2 x 3.1416 x 0.06 = 53.05 Hz