Electrical Efficiency

1. 1

2. ∗ Energy Efficiency is the percentage of energy consumed
by a machine or a system that was transformed into
useful energy.
∗ No technical object is 100% efficient.
∗ Energy can be lost due to heat, resistance, friction.
∗ A lamp would be considered 100% Efficient if all the
electrical energy was transformed to light energy.
∗ A heater is considered 100% Efficient if all the electrical
energy is transformed to heat energy.
∗ Efficiency depends on the purpose of the device.
Energy Efficiency
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3. Energy Transformation
3

4. Energy Efficiency
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Energy
Conversion
Device
Energy Input
Useful Energy
Output
Energy Dissipated
to the Surroundings
InputEnergyTotal
OutputEnergyUseful
Efficiency
)100(
% =

5. Vehicle Efficiency – Gasoline Engine
5
25% Of the gasoline is used to propel a car, the rest is
“lost” as heat. i.e an efficiency of 0.25
Source: Energy Sources/Applications/Alternatives

6. Energy Efficiencies

7. Comparing Efficiency of 2 Types
of Space Heating
Electricity = 24%
Fuel Oil = 53%
Natural Gas = 70%

8. ∗ An electric motor consumes 100 J of energy to obtain 90 J of
mechanical energy. Determine its efficiency.
∗ %E = Useful E x 100%
∗ Total E
= 90 J x 100
100 J
= 90%
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Electric Motor
InputEnergyTotal
OutputEnergyUseful
Efficiency
)100(
% =

9. An electric heater gives off 50000 Joules of energy
every minute. What is its Electrical Efficiency, if it
uses 120 V at 8 A?
Step 1: Useful energy output = 50000 J
Step 2: Total energy input = E = Pt = VIt
= 120V x 8A x 60 s
= 57600 Joules
Step 3: % Electrical Efficiency = (Useful E/ Total E) X 100 %
= (50000 / 57600) x 100
= ~ 87% Efficient
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Electric Heater Question

10. ∗ An electric motor gives off 48000 Joules of kinetic
energy when it turns. If it loses 12000 Joules of
energy due to friction, what is its Electrical Efficiency?
∗ Step 1: Useful Energy Output = 48000 J
∗ Step 2: Total Energy Input = Useful E + Waste E
= 48000 J +
12000 J
∗ = 60000 J
∗ Step 2: Step 3: %E = (Useful E/Total E) x 100%
∗ = (48000/60000) x 100%
∗ = 80%
Electric Motor Efficiency
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11. ∗ Observatory Text
∗ Page 99, Q. 1,2,3,4,5
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Classwork

12. ∗ What is the efficiency of a kettle that heats
200 mL of water up by 50 degrees if it uses
120 V at 10 amps for 5 minutes?
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Efficiency

13. ∗ Heat Energy: Q= mc∆t
∗ Where m is mass in grams
∗ c = specific heat capacity – 4.19J/g°C for water
∗ ∆T = Change in Temperature
∗ Energy to heat 100 mL of water by 5 °C
∗ Q = 100 g x 4.19 x 5
∗ = 2095 Joules
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Heat Energy

14. ∗ Useful Energy = Energy that performs a task
∗ Q= mc∆t
∗ N.B. Density of water = 1 g/mL
∗ Q = 200 g x 4.19 J/g°C x 50 °C
∗ = 41 900 Joules
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Calculation: Useful Energy

15. ∗ Energy supplied = Electrical Energy
∗ E = Pt
∗ E = VIt
∗ E.g. A heater uses 120 V at 8 A for 5 minutes.
∗ E = 120V x 10A x 5 x 60s/min
∗ = 360 000 Joules
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Calculation: Total Energy

16. ∗ %E = Useful Energy x 100
Total Energy
= 41 900 J x 100
360 000 J
= 11.6 %
Therefore the % Efficiency is 11.6 %.
88.4% of the energy is dissipated or wasted.
360 000 J – 41 900 J = 318 100 J are wasted.
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Calculation: % Efficiency

17. ∗ KE: Energy due to motion
∗ KE = 1/2 mv2
∗ m = mass in kg
∗ v = velocity in m/s
∗ What is KE for a 100 kg man running at 5 m/s
∗ KE = ½ (100 kg) (5 m/s) 2
∗ = ½ (100) (25)
∗ =1250 J
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Kinetic Energy