ASP -Prep-Electrical_Calulations_tp.pptx

ASP -Prep-Electrical_Calulations_tp.pptx

1. Ohm's Law   Deals with the relationship between voltage and current in an ideal conductor. This relationship states that: The potential difference (voltage) across an ideal conductor is proportional to the current through it. The constant of proportionality is called the "resistance", R.

2. ASP Prep-Electrical Calculations Ohms Law Formula Wheel * Span ASP Exam Study Workbook volume 1 2015 3 rd edition

3. ASP Prep-Electrical Calculations Based on Ohm’s Law: V = I x R, where: V= voltage, I= current (amps), R= resistance (ohms) Power P (watts) = voltage ( V) × current ( I ) Example Questions*: Simple: If the voltage in a DC circuit is 120 volts and the resistance is 9 ohms, what is the current? V =IR 120=Ix9, I =120/9, I = 13.3 amps

4. ASP Prep-Electrical Calculations More complex question 1 : Q. Compute the total circuit resistance in this series circuit. A : solution R series =R₁ + R₂+ … Rn, answer: 232 ohm

5. ASP Prep-Electrical Calculations More complex question 2 : Q. Compute the total power consumed for this parallel circuit. P =VI Note: P=VI (voltage x current, so you need to find V, I& R before you can find P - solve for resistance 1/R parallel = 1/R₁ + 1/R₂+ … 1/Rn V =IR P =VI 1/R parallel = 1/10 + 1/10 + 1/10 = 3/10 1/R parallel = 0.3 (the source book SPAN, says to press the X-¹ key to get) R parallel = 3.33 ohms

6. ASP Prep-Electrical Calculations More complex question 2 : Q. Compute the total power consumed for this parallel circuit. Note: P=VI (voltage x current, so you need to find V, I & R before you can find P - solve for (current) V =IR, I =V/R I = 13.8/3.33 I = 4.14 amps Summary : R (ohms) 3.33 I (amps) 4.14 P=VI, P =13.8 x 4.14 P = 57.13 watts * Span ASP Exam Study Workbook volume 1 2015 3 rd edition

7. ASP Prep-Electrical Calculations Summary Ohms Law use for Electrical Calculations: V =IR Where : V = voltage , I = current(amps ), R = resistance(ohms ) & P = power(watts ) V  =  I x R  (Voltage = Current multiplied by Resistance ) R  =  V / I  (Resistance = Voltage divided by Current )  I  =  V / R  (Current = Voltage Divided by Resistance ) P = V / I (Power = Voltage Divided by Current) * Span ASP Exam Study Workbook volume 1 2015 3 rd edition

8. ASP Prep-Electrical Calculations Arc Flash * Span ASP Exam Study Workbook volume 1 2015 3 rd edition What is an Arc Flash ? Simply put, an arc flash is a phenomenon where a flashover of electric current leaves its intended path and travels through the air from one conductor to another, or to ground. The results are often violent and when a human is in close proximity to the arc flash, serious injury and even death can occur. Source : OSHA https:// www.osha.gov/dte/grant_materials/fy07/sh-16615-07/arc_flash_handout.pdf Recommended Resources: National Fire Protection Association's 70E; Institute of Electrical & Electronic Engineers (IEEE) standard P1584 Guide for performing Arc-Flash hazard calculations and more.

9. ASP Prep-Electrical Calculations Arc Flash

10. ASP Prep-Electrical Calculations Arc Flash Source: Electrical Diagnostic Surveys , 8401 Claude Thomas Rd., Suite 27, Franklin , OH 45005 [email protected]

11. ASP Prep-Electrical Calculations Arc Flash Hazard Category Table Source: SPAN ASP Exam Study Workbook Volume 1, 2015 Edition

12. ASP Prep-Electrical Calculations Arc Flash Source: http://electrical-engineering-portal.com/what-is-incident-energy-caused-by-arc-flash Incident Energy & Damage Level Incident Energy ( cal /cm2) Degree burn 1.2 2nd degree burn to bare skin 4 Ignite a cotton shirt 8 3rd degree burn to bare skin

13. ASP Prep-Electrical Calculations Arc Flash Calculating Incident Energy The IEEE standard 1584-2002 is the recognized standard regarding the calculation of incident energy output from an alternating-current three-phase arcing fault.

14. ASP Prep-Electrical Calculations Arc Flash Calculating Incident Energy The following formula for the IEEE 1584-2002 is used: log [ En ] = k1 + k2 + 1.081 log [ Ia ] + 0.0011 G and E = 4.184 Cf × En (t ÷ 0.2) × (610x ÷ Dx ) Where E = arc flash energy, En is the normalized arc flash energy, Ia is the arcing current, Cf is the calculation factor, t is the duration of the arc, D is the distance from the arc to the person, X is the distance X-factor, k1 and k2 are constants, and G is the conductor air gap. To facilitate the calculation of this data, several software companies have made arc hazard programs

15. ASP Prep-Electrical Calculations Arc Flash Questions: (source: SPAN ASP Exam Study Workbook, vol 1, 2015, 3 rd edition) Once you know the heat exposure level , you can choose the protective clothing to best protect your employees. Which of the following other the greatest level of arc flash protection ? Hazard Category 1 Hazard Category 2 Hazard Category 3 Hazard Category 4 Answer: D is the correct answer

16. ASP Prep-Electrical Calculations Arc Flash Questions: (source: SPAN ASP Exam Study Workbook, vol 1, 2015, 3 rd edition) 2. Equipment should only be maintained in the de-energized condition when incident energy exceeds: 1.2 cal /cm² 20 cal /cm² 40 cal /cm² 50 cal /cm² Answer: C is the correct answer

17. ASP Prep-Electrical Calculations If the equipment is maintained under de-energized condition, there is no arc flash hazard. NFPA 70E [17] states that energized electrical conductors and circuit parts that operate at less than 50 V to ground should not be required to be de-energized.

18. ASP Prep-Electrical Calculations Arc Flash Questions: (source: SPAN ASP Exam Study Workbook, vol 1, 2015, 3 rd edition) 3. The incident energy that will cause a curable second degree burn for bare skin was selected in solving the equation for the arc flash boundary. Which of the following represent the threshold incident energy level for a second degree burn for systems 50 volts or greater? a. Energy of 1.2 cal /cm² (5.0 J/ cal /cm²) b. Energy of 0.6 cal /cm² (2.5 J/ cal /cm²) c. Energy of 4.8 cal /cm² (20 J/ cal /cm²) d. Energy of 2.4 cal /cm² (10 J/ cal /cm²) Answer: A is the correct answer

19. ASP Prep-Electrical Calculations NFPA 70E [ Standard for electrical safety in the workplace, 2012 pg. 10] states that “ a second degree burn is possible by an exposure of unprotected skin to an electrical arch flash above the incident energy level of 1.2 cal /cm² (5.0 J/ cal /cm²)” and assumes Energy of 1.2 cal /cm² as the threshold incident energy level for second degree burn for systems 50 Volts and or greater.

20. ASP Prep-Electrical Calculations Circuit Circuit: (electrical) an electrical device that provides a path for electrical current to flow Electron move from a negative (pole) to a positive (pole) this is called the Current . Current refers to the movement of charges. When we limit the number of electrons crossing over the " circuit, " by letting only a certain number through at a time. And we can make electricity do something for us while they are on their way. For example, we can "make" the electrons "heat" a filament in a bulb, causing it to glow and give off light . When we limit the number of electrons that can cross over our circuit, we say we are giving it " resistance ". Before electrons can move far, however, they can collide with one of the atoms along the way. This slows them down or even reverses their direction. As a result, they lose energy to the atoms . This energy appears as heat, and the scattering is a resistance to the current.

21. ASP Prep-Electrical Calculations Circuit Properties of electricity: Must complete a circuit Seeks easiest and all paths to ground, that path could go through you.