Energy
In Part 2, I discussed power. A light bulb rated at 100 watts
and 120 volts will use 100 watts of power when operating at
120 volts. If I wanted to operate ten 100-watt light bulbs on
a gasoline powered generator, the rated power output of the
generator would have to be at least 1000 watts. The rated
horsepower of the gasoline engine that runs the generator is
related to the rated power output of the generator. Energy is power times time. Electrical energy is
usually measured in watt hours abbreviated Wh, kilo-watt hours
abbreviated kWh, and mega-watt hours abbreviated MWh. If we
operate ten 100-watt light bulbs for 24 hours, we would have
used 1000 watts X 24 hours = 24,000 Wh or 24 kWh of energy.
The amount of gasoline consumed by the generator is related to
the amount of energy delivered by the generator.
The watt-hour
meter
The mechanical watt-hour meter on the side of your house is a
very accurate instrument for measuring energy. The heart of a
typical mechanical watt-hour meter is a specially designed
electric motor. The motor’s speed is directly proportional
to the product of the voltage and the current. If you would
like to know how much power is flowing through the watt-hour
meter at any point in time, just time how long it takes for
the disc to rotate one revolution. The disc is located in the
center of the meter rotating on a vertical shaft. If the disc
is rotating rapidly, it is sometimes easier to time how long
it takes the disc to rotate ten times. There is a black mark
at one point on the edge of the disc to help you count the
number of revolutions. After you have determined how long it
takes for the disc to rotate one revolution in seconds,
calculate how many revolutions it would make in an hour by
dividing 3600, the number of seconds in an hour, by the number
of seconds for one revolution. Next, look at the nameplate of
the meter. On the nameplate you will find Kh followed by a number.
Multiply this number by the number of
revolutions per hour. That will give you the power in watts
being served through the watt-hour meter. Let’s try some
examples. If one revolution of the disc takes 50 seconds, to
calculate how many revolutions in an hour, divide 3600 by 50
and you will get 72. The Kh for my watt-hour meter
at my house is 12. We multiply 72 times 12 = 864 watts. My
heater blower, some lights and my computer is on. If the disc
was rotating fast and you counted ten revolutions in 20
seconds, the time for each revolution is 20 divided by 10, or
2 seconds. The number of revolutions in an hour is 3600
divided by 2 = 1800. If the Kh is 12, the power is
1800 x 12 = 21,600 watts or 21.6 kW. A load of this magnitude
in a normal size house would not happen every day unless the
house has electric heat.
Photo
1. Watt-hour meter
Source resistance
In Part 4 of this series, we assumed the source voltage at the
outside receptacle was 120 volts and we calculated what the
voltage would be at a shed some 100 feet away from the house.
We were attempting to heat the shed with a 1500 W electric
heater on the end of a 100 foot extension cord made with 18
AWG copper wire. The assumption that the source voltage was
constant at 120 was not realistic. In reality, the voltage at
the receptacle will not stay constant because there is a lot
of resistance between the source transformer and the
receptacle. There is also some resistance within the source
transformer. So how much resistance are we talking about? Let’s
assume the wire between the receptacle and the load center is
fifty feet of 14 AWG copper. The resistance of 14 AWG copper
wire is 2.52 W/
1000 feet. For fifty feet, the resistance is 0.126 W in each conductor. Since there are two conductors, we multiply
this resistance by two to get 0.252 W.
Let’s assume the service-entrance cable is 25 feet of 2/0
AWG aluminum. The resistance of 2/0 AWG aluminum is 0.128 W/1000
feet. For 25 feet, the resistance is 0.0032 W in each conductor. Since there are two conductors, we multiply
this resistance by two to get 0.0064 W.
Let’s assume the aerial service to the house is eighty feet
of 1/0 AWG aluminum triplex with a 2 AWG aluminum neutral. The
resistance of the 1/0 AWG aluminum phase conductors is 0.161 W/1000
feet. For eighty feet, the resistance is 0.1288 W.
The resistance of the 2 AWG aluminum neutral is 0.256 W/1000
feet. For eighty feet, the resistance is 0.02048 W.
Since there are two conductors, one phase and one neutral, we
add these resistances together to get 0.03336 W.
Note that figure
1, the circuit diagram,
shows only three resistances between the source transformer
and the heater, service, house wiring and extension cord. The
service resistance is the sum of the aerial service resistance
and the service-entrance cable resistance. As we did in Part
4, we apply Kirchhoff’s voltage law to the circuit. For this
circuit, the voltage at the heater is equal to the source
voltage minus the voltage drop in the service, minus the
voltage drop in the service-entrance cable, minus the voltage
drop in the house wire, minus the voltage drop in the
extension cord. This time we assume the source voltage at the
transformer is 120 volts. The voltage at the heater is:
V = 120 – ((0.03336 W x I) + (0.0064 W x I) + (0.252 W x I) + (1.278 x I))
Since the conductor resistances are each
multiplied by the current I, we can consolidate the
equation to:
V = 120 – ((0.03336 + 0.0064 + 0.252 +
1.278) x I)
V = 120 – (1.56976 x I)
You will recall from Part 4, we found from
testing that for this electric heater, the relationship
between the heater current I and the heater voltage V was:
I = (0.0976 x V) + 0.45
By combining these two equations, we can
solve for the circuit current I and the heater voltage V.
V = 120 – (1.56976 x ((0.0976 x V) +
0.45))
V = 120 – (0.15320 x V) – 0.7063
V = 119.2937 – (0.15320 x V)
Adding (0.15320 x V) to both sides of the
equation we get:
V + (0.15320 x V) = 119.2937
1.15320 x V = 119.2937
If we divide both sides of the equation by
1.15320 we get:
V = 119.2937 / 1.15320
V= 103.44 volts
The circuit current I = (0.0976 x V)
+ 0.45 = (0.0976 x 103.44) + 0.45 = 10.54 amps. Note that the
calculation in Part 4 determined the heater voltage to be
106.18 volts. A more correct voltage is 103.44 volts.
Source voltage
variation
In this calculation, we assumed the voltage at the transformer
was 120 volts. Is that reasonable? For residential
120/240-volt service, most utilities are required by law to
serve the customer at the service point within 5 percent of
normal voltage. This means the phase-to-neutral voltage at the
service point must be between 114 and 126 volts.
The phase-to-phase voltage must be between
228 and 252 volts. In the above calculation, the voltage at
the service point is 120 volts minus the voltage drop in the
service or 120 – (0.03336 W x 10.54 A) = 119.64 volts. If the voltage at the service point
were only 114 volts, the voltage at the heater would be less
than one hundred volts and the heater would be producing even
less heat. Utilities try to regulate their primary line
voltages to maintain the customer’s service voltage within
the range dictated by law.
Please send me your comments on this
series. If you have questions about basic electricity or
general questions about the NESC, please e-mail me at dave.young@conectiv.com .
National Electrical Safety Code and NESC
are registered trademarks of the Institute of Electrical and
Electronics Engineers. National Electrical Code and NEC
are registered trademarks of the National Fire Protection
Association.
Dave is a consulting engineer with Conectiv Power
Delivery of Wilmington, Delaware, where he has been working with and
teaching all aspects of the NESC for over 33 years. He is a member of
the NESC Interpretations Subcommittee and represents the Edison Electric
Institute on the NESC Overhead Line Clearances Subcommittee 4. Dave is
also an inspector member of the IAEI. |
