You probably know many different types of energy. For example, kinetic, potential, electric,
nuclear and more. If we take all of these types of energy in a system, the sum of that
is called total energy, and it’s represented with a capital letter E. If we were to represent
this on a unit mass basis, then we use a simple letter e and we can write it like this. All
we are saying is that on a unit mass basis, our total energy is simply the total energy
divided by the mass. So our units would be J/kg. But most of the time, you will have
it as kj/kg. In simple terms, it’s energy per kilogram.
You should be familiar with kinetic and potential energy. Kinetic energy is ½ times mass times
velocity squared. Again, if you need to represent this on a unit mass basis, you divide it by
mass, so you end up with ½ velocity squared. In simple terms, it’s kinetic energy per
kilogram. We do this because sometimes, we need to figure out the energy per unit mass
basis. You will see how this becomes helpful when we do examples.
Potential energy is mass times the acceleration due to gravity times height. On a unit mass
basis, we divide it by mass, so we get just the acceleration due to gravity times height.
The units are the same as kinetic energy, so kilojoules per kilogram.
The total energy of a system can then be represented with this equation when we consider kinetic
and potential energy. Here, the letter U represents energy relating to the molecular structure
of a system, in other words, internal energy of a system. If we write this on a unit mass
basis, we just use simple letters. Potential energy and kinetic energy are extensively
covered in the dynamics video playlist, and because of that, we will not dive deep into
those 2 topics. Rather, we are going to look at something new called mass flow. When we
have open systems, also called control volume, it involves mass flow. So air going through
a tube, or water going through a pipe, things like that. Sometimes, we need to be able to
see how mass is flowing through a section with respect to time. We call that mass flow.
And that’s the amount of mass flowing through a cross section per unit time. It’s represented
with the letter m with a dot on top, also called m dot. You might have seen this dot
before, and it simply represents time rate, in other words, something that happens per
time. So here, it’s how much mass is flowing per second and the units would be kg/s. In
simple terms, how many kilograms of mass is flowing per second? This mass flow rate is
also related volume flow rate. That’s the fancy letter V with a dot on top. That would
have the units of cubic meter per second. So in simple terms, how much volume is flowing
per second. We can relate them using this equation. Here, Rho is the fluid density.
It’s also equal to this. The letter A represent the cross-sectional
area of flow and V average is the average flow velocity. Another way we can relate them
would be like this. Here, it’s the volume flow rate divided by specific volume. Hopefully,
you remember that specific volume is volume divided by mass, so this is just isolated
for mass. Now if we wanted to find the energy flow rate
with a fluid that flows, then we can find it using this equation. The units would be
kJ per second or kilowatts. So what we are saying is that the energy flow rate is equal
the mass flow rate multiplied by the total energy of a system on a unit mass basis. Sounds
kind of complicated, but in simple terms, it’s just how much mass is flowing per second
multiplied by the total energy of a system. You might notice that this is pretty much
the equation we talked about before, just rearranged and it’s written with respect
to time rate, so with the dot on top. One last thing we need to talk about is mechanical
energy. It’s a form of energy that can be converted to mechanical work. But what we’re
interested in is how the pressure of a flowing fluid is connected to mechanical work. For
example, when water goes through a turbine, what’s happening is that the turbine extracts
mechanical energy from the water by dropping its pressure. Pressure itself is not a form
of energy, but if we look at a pressure force that acts on a fluid through a certain distance,
then that creates work, and it’s called flow work. Flow work can be expressed like
this, so Pressure multiplied by volume which would have the units joules. We can also express
it on a unit mass basis, so that would be pressure divided by density. The units for
it would be joule per kilogram. If we were to combine everything we talked
about, we can write it like this on a unit mass basis of the fluid.
This is the flow energy; this is kinetic energy and this is potential energy. If we wanted
to write this in the rate form, we can do so like this. Remember that m dot is the mass
flow rate of the fluid. Now let’s move on to some examples to see
how to apply what we learned. In this problem, we have a turbine 120m below
the free surface of a water reservoir. The reservoir can supply water at a rate of 2400
kg/s. We need to find the power generation potential.
First, we write down what we know. The height is 120 m, and we have our water rate as mass
flow, so that’s 2400 kg/s, Lastly, the acceleration due to gravity is 9.81 m/s^2.
Our first goal is to figure out the mechanical energy. In this case, we know that the water
has potential energy because we have a height difference. We can figure it out per unit
mass. We know that on a unit mass basis, potential energy is gravity times height. This is why
we discussed per unit mass since here, it’s not possible to use a mass for the water.
Let’s plug in the values we know. Keep in mind that we know for potential energy per
unit mass, our answer is supposed to have the units kj/kg. So we need to use a unity
conversion ratio. I’ll break it down step by step. A joule is equal to kg m^2 per second
squared. So that means joules divided by kilograms gives us meter squared over seconds squared.
A 1000 joules is a kilojoule. Now that we got our answer in the units we want, let’s
move on to figuring out the power generation potential, in other words, the energy flow
rate. Here, we need to use the given mass flow rate and the equation we talked about
before. So this is the total energy we can get. Let’s plug in our values, and here,
I will write out the unity conversion onto the same equation. 1 Kw is equal to 1 kj/second.
All the units get cancelled out and we are left with just kilowatts and that is our answer.
If the units give you a hard time, you can always follow the process I did previously
and break it into steps. In this question, we have buckets around a
wheel. The water is sprayed from a nozzle at a speed of 60 m/s and has a flow rate of
120 kg/s. We need to figure out the power generation potential. As always, we write
down what we know. The velocity is 60 m/s and the mass flow rate is 120 kg/s.
So as before, we need to figure out the mechanical energy per unit mass. Here, we are given the
velocity of the water, so we know it has to be kinetic energy. Again, keep in mind that
this is per unit mass. Let’s plug the velocity in, along with a unity conversion ratio. This
is the same as before, we want our answer in kJ per kilogram.
Next, we need to figure out the power generation potential, so we can write down our equation.
Let’s plug in what we know. We need another unity conversion ratio. Solving gives us our
answer. Let’s take a look at one last example. Here,
we have river flowing towards a lake with an average velocity of 3 m/s at a rate of
500 cubic meters per second. The difference in height between the river and the lake is
90 m. We need to figure out the total mechanical energy of the river water per unit mass and
the power generation potential. Notice that we’re give the rate with 500 cubic meters
per second, so that’s Volume dot, in other words, volume flow rate. We also have velocity
of the river water, which is 3 m/s, and the height is 90 m. So first, we need to calculate
the total mechanical energy. That would be the kinetic energy of the water because we
have velocity, plus the potential energy because we have height. As usual, we will write it
per unit mass. Let’s plug in our values and our unity conversion ratio.
Now that’s our total mechanical energy. To find the power generation potential, we
use this equation, which is the energy flow equation. So the same one we’ve been using
before, but notice we don’t have the mass flow. We need to figure that out. For that,
we can use this equation. The density of water is 1000 kg per cubic meter. Let’s plug in
our values and solve. This is our mass flow, so we can use this in our energy flow equation.
Let’s plug them in and solve and those are our answers.
That should cover the types of problems you will face when it comes to mass flow, energy
flow and more. I hope this video helped, thanks for watching and best of luck with your studies!