Basic Electrical - Part 2

PIE Formula Circle

The PIE formula circle demonstrates the relationship between power, current, and voltage, and is expressed in the formula P = I x E. This formula can be transposed to I = P/E or E = P/I. In order to use these formulas, two of the values must be known.

Place your thumb on the unknown value and the two remaining variables will “show” you the correct formula.

Power Loss Example

Question: What’s the power loss in watts for two conductors that carry 12A and have a voltage drop of 3.60V?

What’s the question? What’s “P?”

What do you know? I = 12A, E = 3.60V.

The formula is P = I x E.

The answer is P = 12A x 3.60V.

The answer is 43.20W.

Current Example

Question: What’s the current flow in amperes through a 7.50 kW heat strip rated 230V when connected to a 230V power supply? Figure 1–23

What’s the question? What’s “I?”

What do you know? P = 7,500W, E = 230V.

The formula is I = P/E.

The answer is I = 7,500W/230V.

The answer is 32.60A.

Formula Wheel

The formula wheel is a combination of the Ohm’s Law and the PIE formula wheels. The formulas in the formula wheel can be used for direct-current circuits or alternating-current circuits with unity power factor.

Unity power factor is explained in Unit 3. For the purpose of this unit, we’ll assume a power factor of 1.0 for all alternating-current circuits.

Using the Formula Wheel

The formula wheel is divided into four sections with three formulas in each section. When working the formula wheel, the key to calculating the correct answer is to follow these steps:

Step 1: Know what the question is asking for: I, E, R, or P.

Step 2: Determine the knowns: I, E, R, or P.

Step 3: Determine which section of the formula wheel applies: I, E, R, or P and select the formula from that section based on what you know.

Step 4: Work out the calculation.

Example

Question: The total resistance of two 12 AWG conductors, 75 ft long is 0.30 ohms, and the current through the circuit is 16A. What’s the power loss of the conductors?

Step 1: What’s the question? What’s the power loss of the conductors “P?”

Step 2: What do you know about the conductors?

I = 16A, R = 0.30 ohms

Step 3: What’s the formula? P = I 2  x R

Step 4: Calculate the answer: P = 16A² x 0.30 ohms = 76.80W.

The answer is 76.80W.

Power Losses of Conductors

Power in a circuit can be either “useful” or “wasted.” Most of the power used by loads such as fluorescent lighting, motors, or stove elements is consumed in useful work. However, the heating of con-ductors, transformers, and motor windings is wasted work. Wasted work is still energy used; therefore it must be paid for, so we call wasted work “power loss.”

Example

Question: What’s the conductor power loss in watts for a 10 AWG conductor that has a voltage drop of 3 percent in a 240V circuit, and carries a current flow of 24A?

Step 1: What’s the problem asking you to find? What’s wasted “P?”

Step 2: What do you know about the conductors?

I = 24A

E = 240V x 3%

E = 240V x 0.03

E = 7.20V

Step 3: The formula is P = I x E.

Step 4: Calculate the answer: P = 24A x 7.20V = 172.80W.

The answer is 172.80W.

Cost of Power

Since electric bills are based on power consumed in watts, we should understand how to determine the cost of power.

Example

Question: What does it cost per year (at 8.60 cents per kWh) for the power loss of two 10 AWG circuit conductors that have a total resistance of 0.30 ohms with a current flow of 24A?

Step 1: Determine the amount of power consumed:

P = I 2 x R

P = 24A2 x 0.30 ohms

P = 172.80W

Step 2: Convert the answer in Step 1 to kW:

P = 172.80W/1,000W

P = 0.1728 kW

Step 3: Determine the cost per hour:

(Rs.25 per kWh) x 0.17280 kW =

4.32 Rs per hr

Step 4: Determine the dollars per day:

4.32 Rs per hr x (24 hrs per day) =

103.68 Rs per day

Step 5: Determine the dollars per year:

0.3567 dollars per day x (365 days per year) = Rs.37843 per year

Power Changes with the Square of the Voltage

The voltage applied to a resistor dramatically affects the power con-sumed by that resistor. Power is determined by the square of the volt-age. This means that if the voltage is doubled, the power will increase four times. If the voltage is decreased 50 percent, the power will decrease to 25 percent of its original value.

Power Example at 230V

Question: What’s the power consumed by a 9.60 kW heat strip rated 230V connected to a 230V circuit?

Step 1: What’s the problem asking you to find?

Power consumed by the resistance.

Step 2: What do you know about the heat strip?

You were given “P = 9.60 kW” in the statement of the problem.

Power Example at 208V

Question: What’s the power consumed by a 9.60 kW heat strip rated 230V connected to a 208V circuit? Figure 1–31

Step 1: What’s the problem asking you to find?

The power consumed by the resistance.

Step 2: What do you know about the heat strip?

E = 208V, R = E 2 /P

R = 230V x 230V/9,600W

R = 5.51 ohms

Step 3: The formula to determine power is: P = E 2 /R .

Step 4: The answer is:

P = 208V2/5.51 ohms

P = 7,851W or 7.85 kW

It’s important to realize that the resistance of the heater unit doesn’t change—it’s a property of the mate-rial through which the current flows and isn’t dependent on the voltage applied.

Thus, for a small change in voltage, there’s a considerable change in the power consumption by this heater.

Author’s Comment: The current flow for this heat strip is I = P/E.

P = 7,851W

E = 208V

I = 7,851W/208V

I = 38A

Power Example at 240V

Question: What’s the power consumed by a 9.60 kW heat strip rated 230V connected to a 240V circuit?

Step 1: What’s the problem asking you to find?

The power consumed by the resistance.

Step 2: What do you know about the resistance?

R = 5.51 ohms*

*The resistance of the heat strip is determined by the formula R = E 2 /P .

E = Nameplate voltage rating of the resistance, 230V

P = Nameplate power rating of the resistance, 9,600W R = E 2 /P

R = 230V 2 /9,600W R = 5.51 ohms

Step 3: The formula to determine power is: P = E 2 /R

Step 4: The answer is:

P = 240V x 240V/5.51 ohms

P = 10,454W

P = 10.45 kW

 The current flow for this heat strip is I = P/E.

P = 10,454W

E = 240V

I = 10,454W/240V

I = 44A

As you can see, when the voltage changes, the power changes by the square of the change in the voltage, but the current changes in direct proportion to the change in the voltage.