8.3.5power
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This document discusses power calculations in electrical circuits. It shows that total power (PT) in a circuit equals the sum of individual powers (P1 + P2 + P3...). It then calculates the power of two resistors in a circuit, which add up to 3kW total. It demonstrates that total power can be calculated from current and total resistance, or from total voltage and current. The document concludes by instructing the reader to complete review questions on power calculations.
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The cosine law describes the relationship between illuminance (E), luminous intensity (l), distance from the light source (d), and the angle of incidence (ฮธ). Specifically, it states that illuminance (E) equals the luminous intensity (l) divided by the distance from the light source squared (d2), multiplied by the cosine of the angle of incidence (Cos ฮธ).
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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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The document describes the circuits and loads for an electrical installation. It lists 19 circuits with various lighting, power outlet, motor, and appliance loads. It calculates the demand current in amps for each phase based on adding the full load or percentage of full load for each circuit based on rating and number of devices. The highest calculated demand is 153.25 amps on phase L2. With a 10% allowance for future additions, the total recommended maximum current is 168.575 amps.
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.
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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.
4.1.11 Light Demand 2 Table
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
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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.
4.1.7 Example Domestic 3
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.
4.1.5 Example Domestic 1
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.
4.1.1 Example Single
This document calculates the total demand current for an electrical installation consisting of:
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- 15 single sockets contributing 10A
- A 6kW oven contributing 0.5A
- A 2.4kW water heater contributing 0.33A
The total demand current calculated is 45.83A.
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The document discusses the time constant and final current value for an RL circuit. It states that:
1) The time constant for the circuit is 0.17 seconds based on the given inductance and resistance values.
2) It will take approximately 0.85 seconds (5 time constants) for the current to reach its final value.
3) Using Ohm's law, the approximate final current after 0.85 seconds is 2 amps.
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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14.5.8 Example 1 Part 5
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8.3.5power
- 1. PT = P1 + P 2 + P3 . . .P
- 2. P= I2 R or P = V I
- 3. R1 = (10)2x 10 = 1000 W = 1 kW
- 4. R2 = (10)2x 20 = 2000 W = 2kW
- 5. Pt = PR1 + PR2 = 1000 + 2000 = 3000 W = 3kW
- 6. Rt = 10 + 20 = 30 ฮฉ Pt = I2 Rt = (10) 2 x 30 = 3000 W
- 7. P t = Vt I = (100 + 200) x 10 = 3000W
- 8. Complete the Review Questions on page 70 of Electrical Trade Principles by Hampson & Hanssen.