Let’s talk about steady flow systems. When we have a device that operates under the same conditions
for a long time, we call them steady-flow devices. You’ve seen a lot of them already, like turbines,
compressors, even your garden hose nozzle are all steady-flow devices because once they start up,
the fluid flows through the system steadily. That process is called a steady-flow process.
In other words, there is no change with time. So the volume, mass and total energy content of
the control volume stays the same. Also, the mass that enters the system is equal to the
mass that leaves the system. So to simplify things, imagine we have a system like this.
Here, we have some mass flow entering the system. We will say 10 kg/s. On the other side,
we have 2 exits. One has an exit flow of 7 kg/s, and the other 3 kg/s. So that means the mass flow
that’s entering is not equal to a single exit on the right side. However, if we just look at the
mass flow that’s entering, that won’t change with time, the same for each of the individual exists.
The entering one remains at 10 kg/s, the top one at 7 kg/s, and the bottom one at 3 kg/s.
Now if we sum the mass flow coming in and then sum the mass flow leaving, then those two will be
equal. We can represent that like this. So again, while individual inlets and outlets might have
different mass flows, the total coming in must equal the total leaving. And each of these inlets
and outlets will have a steady mass flow, meaning the rate of mass flow won’t change with time.
For a system that has a single stream, so that means one entrance and one exit,
we can write a mass balance equation like this. Each mass flow can be found by multiplying
the density by the average velocity by the cross-sectional area normal to the flow direction.
During a steady flow system, the amount of energy entering a control volume must equal the amount
of energy leaving the system. We can write that like this. So this could be heat, work,
or mass. We can expand this equation to cover all of the types of energy we just mentioned.
Remember from the previous video that this theta symbol represents the energy of a flowing fluid.
So we can expand it further like this. Let’s go through each term. This is the
rate of heat transfer between the system and the surroundings. This is power. This is mass flow.
This is enthalpy, this is kinetic energy and this is potential energy.
There are a lot of different types of steady flow devices.
Some of them include nozzles and diffusers, turbines and compressors,
throttling valves, mixing chambers, heat exchangers and pipes and ducts. In this video,
we are looking at just nozzles and diffusers and in the future videos, we will cover the rest.
A nozzle is a device that increases the velocity of a fluid by reducing the pressure.
A diffuser on the other hand increases pressure by reducing the velocity of the fluid.
You probably experienced this with garden hose nozzles, but nozzles and diffusers are also in
jet engines, rockets and more. There are some general characteristics for nozzles and diffusers
that you should keep in mind. Usually, the rate of heat transfer between the fluid that’s flowing
through the system with the surroundings is small, so small that it’s negligible, which means Q dot
is usually zero. The work involved when it comes to nozzles and diffusers are also negligible.
Also, change in potential energy is negligible. However, the fluid in nozzles and diffusers tend
to experience large changes in velocity, so kinetic energy will not be zero.
Keeping this in mind, our big energy balance equation will usually simplify to look like this.
So we have enthalpy change and velocity change. Remember that enthalpy change can
be found by using this equation, so we multiply specific heat by the change in temperature.
Keeping all of this in mind, let’s go through some examples to see how we can apply what we learned.
In this question, we have a jet engine that decreases the kinetic energy of the air
entering the compressor. We need to figure out the velocity of the air as it exits the
diffuser. Let’s write down what we know. The initial pressure is 100 kPa and initial
temperature is 30 degrees Celsius. The air enters at a velocity of 350 m/s.
At the exit side, the pressure is 200 kPa and the temperature is 90 degrees Celsius.
We will start off by writing down our energy balance equation for a steady flow process.
So, rate of energy in is equal to the rate of energy out. We can express all of the
energy that’s associated with the system like this. Now since this is a diffuser problem,
there is no work, heat transfer is negligible and potential energy change is also negligible.
That means we just need to consider enthalpy and kinetic energy so our equation simplifies to this.
The mass flow is the same since it’s one inlet and one exit, so we can cancel them out to further
simplify our equation. We can bring enthalpy to one side and kinetic energy to the other side.
Enthalpy can be found using this equation so we can replace it. Next, let’s multiply everything by
2 to get rid of the fractions. We have everything we need except for the specific heat value.
To find it, we can use a table, but we need the average temperature to use it. So we just add up
the initial and final temperature and divide it by 2. We get 60 degrees or 333 Kelvin.
Now we can look at our table, so at 333 Kelvin, our value would be between 1.005 and 1.008, so we
will extrapolate and use 1.007. Now we can go back to our equation and starting plugging in values.
When we simplify, we get square meters over seconds squared, which is equal to J/kg.
Problem is, our left side value is in kJ/kg. So we need to multiply it by a 1000 to get it in J/kg.
Now we can take the square root, and that gives us 40.7 m/s. Remember again that J/kg is equal
to meter squared over second squared, so taking the square root gives us m/s.
Let’s take a look at this problem where we have refrigerant 134a entering a nozzle and
we need to figure out the exit velocity and the ratio of the inlet to exit area.
First, we will write down what we know. The initial pressure is 700 kPa,
the initial temperature is 120 degrees Celsius, and the initial velocity is 20 m/s.
The final pressure is 400 kPa and the temperature is 30 degrees Celsius.
So as usual, we start off with our energy balance equation. Since this is a nozzle, our work is
zero, heat transfer is negligible and potential energy change is also negligible. All we need to
consider is enthalpy and kinetic energy. Since in the nozzle, we have one inlet and one outlet, the
mass flow is the same, so we can cancel it out. Since this is r-134a, we can actually figure out
the enthalpy values based on the given information by looking them up on the superheated refrigerant
table. For initial enthalpy, the pressure is 700 kPa and temperature is 120 degrees Celsius.
So the value we’re looking for is right here. For the next part of our question, we will need the
specific volume, so I am going to write that down since we have the table up. That’s the value right
here. For final enthalpy, the pressure is 400 kPa and the temperature is 30 degrees Celsius.
So the value we need is right here. Again, I am going to write down the specific volume value
as well. Now that we have these values, we can plug them in into our equation. Remember that
square meters over seconds squared is j/kg, so we need to multiply the kilojoules value by a 1000
to make sure our units are the same. Solving by taking the square root gives us 409.9 m/s.
The next part of the question asks us to calculate the ratio of the inlet to the exit area of the
nozzle. For that, we just need to use the mass balance equation. We can expand further by writing
down the equation for mass flow. Then we need to eliminate density since we don’t have that value.
Remember that density is 1 over specific volume, so let’s replace it. Now, we will
isolate this equation so that we have a ratio of A1 over A2. Don’t get the specific volume
and velocity confused even though the letters look similar. So now we can plug our values in
because remember, we found the specific volume values before. Don’t forget that we found the
final velocity in the previous step of the question. Let’s solve and we get our ratio.
In this question, we have steam entering a nozzle and we need to figure out the mass flow rate,
the exit velocity of the steam and the exit area. Let’s start off by writing down what we know. The
initial pressure is 4 MPa, the temperature is 400 degrees Celsius, and the velocity is 60 m/s.
When the steam exists, it leaves with a pressure of 2 MPa and the temperature
is 300 degrees Celsius. The inlet area of the nozzle is 50 square centimeters and we
are told that there is a heat loss of 75 kJ/s. For this problem, we are going to need specific
volume values and enthalpy values, so let’s figure those out using steam tables. The initial pressure
is 4 MPa and temperature is 400 degrees Celsius so our specific volume value is right here,
and our enthalpy value is here. Next, we have a pressure of 2 MPa and a temperature
of 300 degrees Celsius, so our specific volume value is here and this is the enthalpy value.
The first step requires us to calculate mass flow rate. Since we have one inlet and one outlet,
our mass flow in is equal to mass flow out. We can find the mass flow rate using
the mass balance equation. Since we don’t have density, we can replace it with specific volume.
Now we can plug our values in. Solving gives us the mass flow rate.
The next part of the question wants us to find the exit velocity of the steam.
For that, we need to write our energy balance equation.
In this problem, we do need to consider heat loss since we are told there is some loss of heat
when the steam exists the nozzle. However, work done and potential energy are negligible.
We can actually just plug values in right now since we have all the values we need.
Solving gives us a velocity of 589.4 m/s. One thing to note here is that each of the kilojoule
terms must be multiplied by 1000 so that it’s in joules just like we did in the previous questions,
or you won’t get the right answer. The last step requires us to find the
exit area of the nozzle. For that, we can use the equation to find mass flow. We don’t have density
so we can replace it with specific volume. Let’s plug our values in. Solving gives us the area.
That should cover the types of problems you will face when it comes to steady flow
with respect to nozzles and diffusers. Thanks for watching and best of luck with your studies.