Transcript Power and Energy

Lesson 3: Power, and Energy
(Chapter 4)
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Learning Objectives
• Describe the relationship between battery capacity,
current drain and battery’s useful life.
• Calculate the total cost given a rate of energy
consumption.
• Calculate power supplied/dissipated in a circuit.
• Calculate the power efficiency of a circuit.
• Demonstrate how to measure current, voltage, and
resistance in a circuit.
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Batteries
• Batteries are the most common direct current (DC)
source.
• Capacity of a battery is the product of the drain
current and the length of time the battery can provide
that rate of current drain (i.e. 110 amp-hrs).
• Battery capacity is measure in ampere-hours.
capacity (Ah)  life (hours)*current drain [ampere-hours, A-hrs]
capacity (Ah)
life (hours) 
[hours]
current drain (A)
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Battery
Example Problem
• How long can a 1.5 V flashlight battery provide a
current of 250 mA to light the bulb if the amper-hour
rating is 16 Ah?
life (hours) 
capacity (Ah)
[hours]
current drain (A)
16 (Ah)
life (hours) 
250 (mA)
life (hours) 
16 (Ah)
250 *103 (A)
life = 64h
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Power
• In general, the term power is applied to provide an
indication of how much work (energy conversion) can
be accomplished in a specified amount of time; that
is, power is a rate of doing work.
• Since energy is measured in joules (J) and time in
seconds (s), power is measured in joules/second (J/s).
• The electrical unit of measurement for power is the
watt (W), defined by:
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Power
• Power is defined as the rate of doing work or as the
rate of energy transfer.
W
P
t
[watts, W]
• The SI unit of power is the watt (W) or joules per
second.
• The English unit of power is horsepower (hp).
1 hp  746 watts
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Power in
Electrical Systems
• We need to express power in terms of voltage and
current, recall that:
W
voltage V 
[joules/coulomb, J/C]
Q
Q
current
I
[coulomb/sec, C/s]
t
• Combining them the expression become:
P
W W Q
 *  VI [watts, W]
t
Q t
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Power in
Electrical Systems
• Applying Ohm’s law (V = IR and I = V/R) we can
also express power as:
P  VI   IR  I  I R [watts, W]
2
2
V
V
 
P  VI  V   
R R
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[watts, W]
Power Calculation Example
Calculate the power to the heater using all three
electrical formulas.
V 120V
I 
 10 A
P  VI   IR  I  I 2 R [watts, W]
R 12
AND
V  V
P  VI  V   
R R
2
[watts, W]
V 2 (120) 2 V
P

 1200 [watts, W]
R
12
P  VI  120V *10 A  1200 [watts, W]
P  I 2 R = 102 A*12 = 1200 [watts, W]
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Power
• Power is defined as the rate of doing work or as the
rate of energy transfer.
− The greater the power rating of a light, the more light
energy it can produce each second.
− The greater the power rating of a heater, the more heat
energy it can produce.
− The greater the power rating of a motor, the more
mechanical work it can do per second.
• Power is related to energy; it is the capacity to do
work.
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Example Problem 1
A resistor draws 3 amps from a 12V battery. How
much power does the battery deliver to the resistor?
I=3A
E=12V
P  VI  12V *3 A  36 [watts, W]
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Energy
• For power, which is the rate of doing work, to
produce an energy conversion of any form, it must
be used over a period of time.
• The energy (W), lost or gained, of a system is
therefore determined by:
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Energy
• We can rearrange our formula for power to solve for energy:
Energy (W-hr)  power (W) * time (hr)
• The unit of energy is joules (J), but is also expressed as watthours (Wh) or kilowatt-hours (kWh).
power (W) * time (hr)
Energy (kW-hr) 
1000
− 1kWh is the energy dissipated by a 100 W lightbulb in 10 hours
100 (W) *10 (hr)
Energy (kW-hr) 
 1kW-hr
1000
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Energy Cost
• Determining the cost of using power we can use the
following:
Cost  Power * time *Cost Per Unit
OR
Cost  Energy * Cost Per Unit
• For example:
− The residential energy cost from BGE is
14.8 cents per kWh.
Household kilowatthour Meter
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Example Problem 2
Suppose you are at home and use 3 100-W lamps for 3 hours and
An Xbox 500W for 2.5 hours. The TV consumes 180 W. At
$0.148 per kilowatt-hour, how much will this cost you?
First, determine the total energy required to operate the equipment:
power (W) * time (hr)
1000
(3*100W *3hrs)+(500W*2.5hrs)+(180*2.5)W
Energy (kW-hr) 
1000
Energy (kW-hr) = 2.6 kW-hr
Energy (kW-hr) 
Finally, determine the total cost required to use the energy above:
Cost  Energy * Cost Per Unit
Cost  2.6 kW-hr * $0.148/kW-hr
Cost  $0.3848 = 38 cents
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Efficiency
• In the process of converting energy, energy losses
inevitably occur.
• The measure of output energy (or power) to input
energy (or power) is called efficiency.
Thermal
energy out
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Efficiency
• Poor efficiency in energy transfers results in wasted energy.
− An inefficient piece of equipment generates more heat. As heat must be
removed to guarantee a proper function, it means more $$.
• Output energy level must always be less than the applied
(input) energy due to losses and storage within the system.
• The best one can hope for is that Wout and Win are relatively
close in magnitude.
• Conservation of energy requires that:
− Energy input = energy output + energy lost or stored by the system
Energy flow through a system.
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Efficiency
• Efficiency is usually expressed in percent and
denoted by the symbol .
Pout

*100%
Pin
Wout

*100%
Win
•
Since Pin = Pout + Plosses, efficiency can also be
expressed as:
Pout

*100%
Pout  Plosses
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Efficiency
• To find the total efficiency of a system:
− Obtain product of individual efficiencies of all subsystems:
Total = 1 * 2 * 3 * ∙∙∙ x
Basic components of a generating system.
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Efficiency Example
• Suppose a power amplifier delivers 400 W to its
speaker system. If the power loss is 509 W, what is
the efficiency?
η=
Pout
* 100%
Pout + Plosses
400W

*100%
400W  509W
  44%
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Example Problem 3
A 120 V dc motor drives a pump through a gearbox. The output power to the
pump is 1100 W. Gearbox efficiency is 75%. The input power to the motor is
1600W.
What is Overall efficiency, HP output and efficiency of the motor?
a) HPout = 1100W * (
1HP
) = 1.475 HP
746W
Po
) * 100%
Pin
1100W
T = (
) * 100% = 68.75%
1600W
c) T = (
b) T = m * 2
η
ηm = T *100%
η2
ηm =
0.6875
*100% = 91.7%
0.75
alternate c) T = m * 2
ηT = 0.917 * 0.75 * 100% = 68.75%
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QUESTIONS?
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