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Physics Lesson 5.5.3 - The Meaning of Efficiency
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Welcome to our Physics lesson on The Meaning of Efficiency, this is the third lesson of our suite of physics lessons covering the topic of Power and Efficiency, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.
The Meaning of Efficiency
In general, work is done by machines, which take their energy from a power source. Even the human body can be though as a machine which takes its energy from food.
The quality of machines depend on how complicated they are or how much they meet our demands. These are general features, which are difficult to measure. However, there is a very simple physical quantity, which allows us to measure numerically the quality of a given machine. It is called "Efficiency" (in short, e) and is calculated by the following formula:
Equation 4
e = Poutput/Pinput × 100%Efficiency is a dimensionless quantity that is calculated as a percentage of the output power to the total (or input) one.
We have
and
Therefore, when multiplying both numerator and denominator in the equation (4) by time t, we obtain
= Poutput × t/Pinput × t × 100%
In this way, we obtain another equation for the efficiency of a machine:
Equation 5
e = W/Etotal × 100%Example 2
A crane can lift a 200 kg load at 12 m above the ground in 6 s. During this process, the crane draws 30 000 J of electrical energy from the power source. Calculate the efficiency of this crane. For convenience, take g = 10 N/kg.
Solution 2
As stated in the Physics tutorial "Gravitational Potential Energy. The Law of Mechanical Energy Conservation", the work done on an object by a source of energy when raising it from the ground level to a certain height h, contributes in the increase of its gravitational potential energy from GPE1 = 0 to GPE2 = m × g × h. Therefore, we can write
= GPE2 - GPE1
= m × g × h - 0
= m × g × h
= 200 kg × 10 N/kg × 12 m
= 24 000 J
Now, there are two ways for calculating the efficiency of the crane.
Method I - Using the equation (4). For this, we have to calculate the output and input power by divining both work and total energy by time. Thus,
= 24000 J/6 s
= 4000 W
and
= 30000 J/6 s
= 5000 W
Therefore,
= 4000 W/5000 W × 100%
= 0.8 × 100%
= 80%
Method 2 - Using the equation (5), i.e. calculating the efficiency directly, without dividing anything by time. Thus,
= 4000 W/5000 W × 100%
= 0.8 × 100%
= 80%
Interpretation of result: Having an efficiency of 80 percent, means that only 80% of the energy provided by the source is converted into work done by the machine. The rest (20 %) has turned into other forms of energy such as heat, light, sound energy etc. This part is considered as wasted energy although it may be useful for the operation processes of machine such as to turn on the buttons, to make the gears rotate, etc. The fact that it is not used to do work, i.e. to lift the object, makes us consider this part of total energy as wasted.
Another Formula for Calculating the Mechanical Power
Let's transform the equation of (output) power as follows:
= F⃗ × ∆x⃗/t
= F⃗ × ∆x⃗/t
Thus,
Equation 6
P = F⃗ × v⃗where v⃗ is the velocity of the object in the direction of force F⃗.
The equation (6) enables us formulate another alternating definition for power:
"Power is the scalar product of the force applied by a source and the velocity this object gains due to the action of the force."
Example 3
An object starts moving from rest and it accelerates constantly reaching a velocity of 8 m/s after 10 s. The output power delivered by the source is 120 W and the efficiency of the process is 40%.
- What is the mass of the object?
- What is the total energy supplied by the source for this process?
Solution 3
Clues:
v⃗0 = 0
v⃗ = 8 m/s
t = 10 s
Poutput = 120 W
e = 40 %
- m = ?
- Etotal = ?
- The first thing to do is finding the average velocity of the process as its value will be used in the equation (6) in order to work out the average force of the source later. Thus,
= 0 + 8 m/s/2
= 4 m/s
Now, let's calculate the average force produced by the source. We have
< F⃗ > = Poutput/< v⃗ >
120 W/4 m/s
=30 N
Now, using the kinematic equation
we calculate the acceleration, and then use this value to find the object's mass by applying the Newton's Second Law of Motion. Thus,
= 0.8 m/s2
Hence, given that
then,
= 30 N/0.8 m/s2
= 37.5 kg
- Using the equation
we calculate the work done by the source on the object. Thus,
= 120 W × 10 s
= 1200 J
Therefore, giving that
we obtain after the substitutions
Etotal = 1200 J × 100%/40%
= 1200 J × 2.5
= 3000 J
Remark! Constructors of electrical appliances often write the power of appliance (in watts) on its back or lateral side. Power is stamped on a white plate amongst other parameters such as operating voltage, frequency, noise level etc. This value represents the output power delivered by the appliance. Since Energy = Power × Time, people (wrongfully) try to calculate the electrical energy consumed by the appliance by multiplying the power shown at the plate by the time (in seconds). Then, using the conversion factor 1 kw - h = 3 600 000 J (we discuss this in detail in our Physics tutorials on Electricity), they find a value in kw-h which in fact doesn't correspond to the value measured by the electric board meter. This occurs because the board meter counts the total energy consumed by the appliance while the abovementioned calculations give the output (useful) energy (i.e. the part of total energy that is used for doing work).
You have reached the end of Physics lesson 5.5.3 The Meaning of Efficiency. There are 3 lessons in this physics tutorial covering Power and Efficiency, you can access all the lessons from this tutorial below.
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