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ELECTRICANS ARE

ELECTRICANS ARE. “LOADED”. WITH MATH!. This is one of my created lessons for electrical. We practiced calculating lighting loads by use literacy strategies to review concepts and the PDE Math Council T Charts to review and incorporate academic math.

elizabeth-garner
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ELECTRICANS ARE

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  1. ELECTRICANS ARE “LOADED” WITH MATH! This is one of my created lessons for electrical. We practiced calculating lighting loads by use literacy strategies to review concepts and the PDE Math Council T Charts to review and incorporate academic math. I started the lesson with a logic reading (next page) to show them how it is important to read the problem and draw a picture of the problem. They drew the picture of the crime scene to show how to solve a problem with pictures. I also had them underline the only part of the problem they needed. (Extra information needs to be eliminate.)

  2. Start The Area of a Rectangle is A = length x width The perimeter of a rectangle is Perimeter equals 2length + 2width circle perimeter and its formula is C= 2πr or C= πd. The r and d in this formula are the measure of the radius and diameter of the circle. The area of a circle measures the space inside the circle A = πr2 The area of a Triangle is A = ½ bh (because a triangle is ½ of a rectangle.) The perimeter of a triangle is found by adding To find the area of an irregular figure, separate the figure into shapes of which you can calculate the area. The sum of the areas of each smaller part of the irregular figure is the area of the irregular figure. The circumference of a circle means it is the

  3. the three sides together . P = a + b + c The Pythagorean Theorem states that c2 = a2 + b2 The hypotenuse is the diagonal side of the triangle. The reason electricians need to know all of be known. Since most houses are not perfect squares, or rectangles. With this said, all living space must be calculated. The General Lighting load equation (VA) = these formulas is that when an electrician has to calculate a single-dwelling lighting load for an entire residence, the total square footage of the dwelling must # square feet X 3 (VA) per square feet. This is information that seniors will use to take the

  4. NOCTI exam. Juniors will use this information to pass their PSSA exams. END! This is for the game to have students match other students content. Each student gets one cut out card. The card has a beginning of a sentence and each has an ending. They must listen to each other to get the correct order of the cards. I did this once with them and then I timed them. The PM tried to beat the AM score. After the second try, my AM asked if they could have 5 minutes to discuss the reading and come up with a quicker way to do this? (I was amazed they asked, but I said great!) So they planed and discussed the questions. One student in the PM, had trouble reading “Pythagorean”. I would not allow them to switch cards, but another student worked with him to help him pronounce it. I was very impressed with this game. The idea for it came from Lenape Tech. School.

  5. Example 1: A two-story house has a first floor measuring fifty feet wide by 35 feet deep. It has a second floor measuring 40 feet wide and thirty feet deep. What is the square footage of the house? Calculate the lighting load of the structure, then provide the minimum number of 15 amp lighting branch-circuits required for this house.

  6. Step 1: Find the total square footage First Floor A = lw Second Floor A = lw A = 50 x 35 A = 1750 sq ft. A = 40 x 30 A = 1200 sq ft. Total Square Footage = 2950 sq. ft.

  7. Step 2: Find the lighting load Step 3: Minimum # of 15 amp circuits 2950 sq ft x 3 VA per sq ft = 8850 VA 8850 VA / 120 volts = 73.75 amps 73.75 / 15 = 4.91666 (round up to 5) 5 15-amp branch

  8. Think – Pair – Share You will work with a partner. Each of you will do one of these problems. When you are finished, turn to your partner and explain your answer. 1. A ranch house has a width of 70 feet and a length of 25 feet. What is the total square footage? Calculate the General Lighting Load, and the minimum number of 15 amp branch circuits for this dwelling. 2. A ranch house has a width of 92 feet and a length of 49 feet. What is the total square footage? Calculate the General Lighting Load, and the minimum number of 15 amp branch circuits for this dwelling. Total Sq Footage = 4,508.00 Lighting Load 13,524 VA 13524 VA/120 volts = 112.7 amps 112.7 amps / 15 = 7.51 = 8 branch circuits Total Sq Footage = 3,784.00 Lighting Load 11,352 VA 11352 VA/120 volts = 94.6 amps 94.6 amps / 15 = 6.3 = 7 branch circuits

  9. Example #2: An electrician needs to wire the perimeter of a very unique addition. The drawing is below. What is the amount of wire he will need to purchase to wire the perimeter of this construction? In addition to the wire he needs, he has decided to purchase an extra 20 feet of wire to have on hand. c = 37 ft a b = 24 ft

  10. Step 1: We need to find the value of the variable a. a2 + b2 = c2 a2 + 242 = 372 a2 + 576 = 1369 a2 + 576 -576= 1369 -576 a2 = 793 a = √793 = 28.2 ft a2 + b2 = c2 Pythagorean Theorem Substitute what you know: c = 37 ft a b = 24 ft

  11. Step 2: We need to find the circumference of ½ of a circle. Circumference of a semicircle: Circle = 2πr or πd Semicircle = ½ (πd) The diameter = 28.2 ft Solve the equation: Semi C = ½ (πd) ½ (3.14)(28.2) Semi C = 44.3 feet c = 37 ft a = 28.2 ft b = 24 ft

  12. Step 3: Find the perimeter of the entire addition. Perimeter = Add all sides together. • Solve the equation: • 44.3 ft • ft • 24 ft The total perimeter is 105.3 ft. You must then add 20 ft of wire and purchase 125.3 ft. 105.3 ft c = 37 ft a = 28.2 ft 44.3 ft b = 24 ft

  13. An electrician needs to wire the perimeter of a very unique addition. The drawing is below. What is the amount of wire he will need to purchase to wire the perimeter of this construction? In addition to the wire he needs, he has decided to purchase an extra 20 feet of wire to have on hand. c = 60 ft a b = 32 ft

  14. You will work with a partner to complete the following problem. An electrician needs to wire the perimeter of a very unique addition. The drawing is below. What is the amount of wire he will need to purchase to wire the perimeter of this construction? In addition to the wire he needs, he has decided to purchase an extra 10 feet of wire to have on hand. Solve the equation: Semi C = ½ (πd) ½ (3.14)(50.8) Semi C = 79.8 feet a2 + b2 = c2 a2 + 322 = 602 a2 + 1024 = 3600 a2 = 2576 a = √2576 = 50.8 ft Solve the equation: 60 ft 32 ft 79.8 ft 171.80 ft + 10 ft = 181.80 ft of wire

  15. Give homework in the T Charts page #4. (Adobe file in the folder.) Tell students to pick 1 from each area and solve. I used the math council T-Charts for my examples and for homework Thanks, Ms. Plesnarski and Mr. Monkiewicz

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