Let’s talk about efficiency. You probably heard things like,
this motor is 85% efficient, or this car engine is 35% efficient. In simple terms, efficiency is the
desired output divided by the required input. For example, if we take an incandescent light bulb,
only a small percentage of the electrical energy it uses produces light, most of it is dissipated
as heat. So, our desired output it light, and required input is electricity, however,
the efficiency is low since not all the energy we give turns to light. We use the Greek letter Eta
to represent efficiency. So if a motor is 85% efficient, then we would write it like this.
Diving a bit deeper, we have different types of efficiencies.
The first we need to look at is something called combustion efficiency. That’s the amount of heat
released during combustion divided by the heating value of the fuel burned.
The heating value of a fuel is the amount of heat released when a unit amount is burned
at room temperature and then the combustion products are cooled to the room temperature.
If we can get 100% combustion efficiency, then the fuel is burned completely and the gases leave
the combustion chamber at room temperature. In other words, the heat released during the process
is equal to the heating value of the fuel. Next, we have mechanical efficiency. Most of
the time, mechanical energy happens through a rotating shaft, so we its usually called shaft
work. For example, a propeller that receives shaft work from a motor. When it spins, the mechanical
energy is transferred to the fluid, excluding of course the frictional losses. The mechanical
efficiency can be found using this equation. So if we have a 95% mechanical efficiency,
that means 5% of the mechanical energy is converted to thermal energy due to friction.
Next, we look at pump efficiency. Let’s say we wanted to move some water from here to here.
To do so we use a pump, and it works by suppling mechanical energy to the fluid.
You can use pumps to increase the pressure, the velocity or the elevation of a fluid.
So to figure out the efficiency, we divide the mechanical energy increase of a fluid
by the mechanical energy input. This part here is the mechanical energy out minus the
mechanical energy input in rate form, so the rate of increase in mechanical energy of the fluid.
Now let’s think about the opposite, which is when we extract mechanical energy from
a fluid using a turbine, and then the rotating shaft can drive a generator.
So turbine efficiency can be written like this. That’s the mechanical energy output divided by
the mechanical energy decrease of the fluid. Makes sense right? We’re taking
energy away from the fluid and transferring it to a shaft. If we have a 100% efficiency for this,
that means a perfect conversion between shaft work and the mechanical energy of the fluid.
Next, let’s talk about electrical motors. Electrical energy can be converted to
rotating mechanical energy, things like circular saws, blenders, washing machines,
all have motors in them. The conversion of electrical energy to mechanical energy
is not perfect, so we have to think about motor efficiency. That can be expressed like this.
When a motor receives electrical energy input, 100% of it is not transferred out as mechanical
energy. Due to friction, some of the energy is lost as heat. In other words, supplying a specific
amount of electrical energy, let’s say 20 kw does not mean, you will get 20 kw of shaft power,
in fact, you will always get an amount less than the input. Because of this,
it’s very important to not confuse mechanical efficiency with motor efficiency.
We also have generators, so the opposite way where we are generating electricity from shaft power,
and their efficiency can be written like this. So here,
if we had a 100% efficiency, then all the shaft power would be converted to electrical energy,
but again, this is never the case. There is always some loss due to friction.
Now obviously a generator is not a generator with just a shaft, it’s usually attached to a turbine.
The same is true for a pump. It’s not really a pump if there isn’t a pump attached to the motor.
So in these cases, we can go further and think about combined efficiency.
So not just shaft to electrical, or vice versa, but the overall efficiency of the
combination of a motor and pump or turbine and generator. They can be expressed like this.
To solve problems associated with conversion efficiencies, you should be familiar with mass
flow since turbines, pumps, and such are associated with fluids that flow through
blades and so forth. If you need a refresh, please check the description. Now let’s move on to some
examples to see how we can apply what we learned. Let’s take a look at this question. We have a room
that’s cooled by circulating cold water through a heat exchanger in the room. The air is then
circulated through the heat exchanger by a 0.25 horse power fan. The efficiency of this fan, or
in other words, the motor, is 54 percent. We need to figure out how much heat is being added to the
room due to the fan running. To understand what we are looking for, we need to realize that all the
electrical energy used up by the motor, including the shaft power is dissipated as heat. So what
we are really looking for is the total electric power the motor consumes. Let’s write down what
we know. It’s important to note that the 0.25 hp is shaft output, in other words shaft work. The
efficiency is 54 %. Now we need to use the motor efficiency equation. Let’s plug in what we know.
Now we can solve for electric power output. The last step is to write our horsepower answer
in watts. Remember that 1 horse power is 746 watts. Let’s discuss this in detail. While the
shaft power is 0.25 hp, or 186.5 watts, that’s not the total power that’s supplied to the motor.
Remember, there is a loss between the electric power input and shaft output. So the total energy
the motor is supplied with is 345 W, and all of that is converted to heat in the process.
In this problem, we have a wind turbine with a diameter of 100
m (animate circle and show 100 m). There is a steady wind with a speed of 8 m/s. The wind
turbine has an efficiency of 32 % and we need to find the electric power generated by the turbine.
We also need to figure out the electric energy and revenue generated for a 24 hour period.
Let’s write down what we know. The diameter is 100m. The speed of the wind is
8 m/s. The efficiency of the wind turbine is 32% and the density of air is 1.25 kg per cubic meter.
Our first goal is to figure out the maximum potential work generated by the wind turbine.
This isn’t the amount of electricity generated, this is just what the maximum
possible amount of electricity that can be generated with the wind. Once we have that,
we can use the efficiency percentage given to figure out the actual electric power generation.
So how do we figure out the maximum power generated? We know we’re going to have to
consider mass flow of the wind. We also need to think about the type of energy the wind possesses.
The wind blows with a velocity of 8 m/s, so it has to be kinetic energy. We can figure
out the work generated by the wind turbine by multiplying the mass flow by the kinetic energy.
Let’s figure out the energy on a unit mass basis, that means our answer needs to be kj/kg,
which requires us to use a unity conversion ratio. Remember that a joule is kgm^2 per second squared,
so 1 kj would be 1000 kgm^2 per s^2, and since we want it per unit mass, we divide both sides by kg.
We end up with 1 kJ per kg is equal to 1000 m^2 per s^2. Let’s solve.
Next, we need to figure out the mass flow. Think of it as how much wind mass is flowing through
the blades per second. For that, we can use this equation. This should be familiar to you
from the mass flow video, if you need a refresh, please check the description. The cross-sectional
area can be plugged in as well. Let’s fill in the values we know. Solving gives us the mass of air
going through the turbine per second. Now that we have both values we need,
we can go back to the work equation. Let’s plug in our values and solve. Note that kJ per second
is kW so let’s change it. This answer we have is the maximum amount of work the turbine can do.
In other words, this is the maximum amount of electricity that can be produced in a perfect
world with no friction, but remember, the efficiency of this turbine is only 32 percent,
so we need to account for that. So the electric power generation can be found by multiplying our
maximum power by the efficiency percentage. Solving gives us 804.25 kilowatts. To figure
out the amount of electricity generated per day, we just need to multiply it by 24 hours. To find
the total revenue, we need to multiply this value by 9 cents per kilowatt hour. Solving gives us
the total revenue per day. This also goes to show, you can make some good money with a wind turbine.
Let’s go through one last example. Here, we have 2 water reservoirs
and water is pumped from the bottom to the top by a pump that provides 20 kW of shaft power.
We need to figure out the mechanical power that is converted to thermal energy during this process.
Let’s write down what we know. The pump power is 20 kW. The volume flow rate is 0.03 cubic meters
per second. The height difference between the reservoirs is 45 m. Although not given in the
question, we will need the density of water, which is 1000 kg/cubic meter. This is a value you should
try to remember. Now we need to find the energy required to pump the water. To do so, we need to
consider potential energy. You can think of this problem in a different way. If you think of it in
the opposite sense, so water going from the top to the bottom, it might be easier to understand.
Imagine a perfect world scenario with no friction, then how much energy can be produced by the
water when it runs down the pump, which would be acting as a turbine since water is rushing down?
This value is what we need to find. In other words, the change in the total
mechanical energy of the water is equal to the change in potential energy of the water.
Let’s go through this step by step. So we can write the total energy is equal to mass flow times
the energy on a unit mass basis. This energy we are talking about is potential energy. Potential
energy per unit mass basis is gravity times height. The mass flow can be found by multiplying
the density of water by the volume flow rate. Now we can plug our values in. We need our answer in
kilowatts, so we need to use 2 unity conversion ratios to get rid of the units we have and to get
kilowatts. 1 N is 1 kgm per second squared, and 1 kilowatt is 1000 newton meters per second. What
we found tells us that in a perfect world with no friction, this is the shaft power required to pump
the water from the bottom reservoir to the top. However, the pump provides 20 kW of shaft power,
so to figure out the mechanical power that’s converted to thermal energy, we just need to
subtract the two. Solving tells us that 6.76 kW of power is used to compensate for the friction
in the piping, in other words, this is the amount of energy that is converted to thermal energy.
That should cover the types of problems you will face when it comes to energy conversion
efficiencies. I hope this video helped, thanks for watching and best of luck with your studies!