Let’s talk about turbines and compressors. In the previous video, we discussed what a steady flow
system is, so we going to build on that knowledge by focusing in on turbines and compressors. In
a turbine, fluid will flow through and do work against the blades, which then rotates a shaft,
and produces power. So in simple terms, a turbine will create work like the ones in
steam, gas or hydroelectric power plants. Now a compressor, or even pumps or fans,
are used to increase the pressure of a fluid. For this process, we actually need to supply it with
power, in other words, work input. This is usually supplied from an external source through a shaft.
So in simple terms, compressors use power to increase the pressure of a fluid. Pumps and
fans pretty much work in the same way, with a pump just handling liquids instead of gases.
So let’s discuss the characteristics of these 2 types of steady flow devices.
Remember that for a steady flow system, our energy balance equation is this. The rate of
energy in must equal the rate of energy out. We can expand this to include all the types of energy
associated with a steady flow system like this. Now with a turbine, we have work output,
but heat transfer is usually negligible. Potential energy is also negligible. However, the fluid
velocities in turbines tend to be high, and these fluids can experience large changes in its kinetic
energy. So kinetic energy has to be considered. However, in certain cases, it’s possible that this
change is relatively small compared to enthalpy change, in which case, it can be negligible.
So with all of this in mind, our big energy balance equation for steady flow systems usually
simplifies to this. We have mass flow, enthalpy, kinetic energy and of course work output.
Now for compressors, we already said that it required work input.
As with turbines, potential energy is negligible,
and heat transfer is also negligible unless there is internal cooling involved. For compressors, the
velocity change of the fluid isn’t significant, so kinetic energy is negligible. So for compressors,
our energy balance equation usually simplifies to this. We have work input and enthalpy.
Notice again, how for turbines we have work output, while for compressors, we have work input.
Now let’s move onto some examples to see how we can apply what we learned.
In this question, we have refrigerant-134a entering a compressor and we need to
figure out the power input to the compressor and the volume flow rate
of the refrigerant at the compressor inlet. Let’s write down what we know. The initial
temperature is -24 degrees Celsius, and final temperature is 60 degrees Celsius. We don’t know
initial pressure but we know the final pressure which is 0.8 MPa. The mass flow rate is 1.2 kg/s.
So to find the power input, we can write an energy balance equation. We can expand the equation to
account for all the types of energy that can be associated with steady flow systems. Now
since this is a compressor problem, rate of heat transfer, kinetic energy, and potential energy
is negligible. Again, this is a compressor which means there is work input but no work output. So,
our equation simplifies to this. Keep in mind that there is only one input and one exit, which
means the mass flow at the input is equal to the one at the exit side. Let’s isolate the equation
for work input. So looking at the equation, we need the initial and final enthalpy values.
We can find that from the refrigerant table. Our initial temperature is -24 degrees Celsius and
it’s a saturated vapor. So our enthalpy value is right here. For the second part of our question,
we’re going to need the specific volume value, and since we have the table up,
we can write that down as well. That’s the value right here.
Next, we need the final enthalpy value. Now it’s a superheated vapor and our pressure
is 0.8 MPa and the temperature is 60 degrees Celsius. So our enthalpy value is right here.
We can now plug our values in. Solving gives us the power input for this compressor.
The next part of the question wants to us find the volume flow rate of the refrigerant
at the inlet of the compressor. For that we can use the equation for volume flow. So that’s mass
flow times specific volume. We can plug values in straight away since we already found the
specific volume from before. Solving gives us the volume flow at the inlet of the compressor.
In this problem, we have helium being compressed and we need to figure out
how much power is required for this process. Let’s write down what we know.
The helium enters the compressor at 105 kPa, with a temperature of 295 K
and exists with a pressure of 700 kPa and a temperature of 460 K. There is a heat loss
during this compression process, which is 15 kJ/kg and the mass flow rate is 60 kg/min.
We have a slight problem, which is the heat loss and mass flow. We need these in terms we can work
with. In other words, we need these values to be flow rates that are given with respect to
seconds. Mass flow is easy to figure out. We just divide it by 60 and we get it per second.
Now we are told that 15 kJ of energy is lost per every kg of helium, and we also know that 1 kg of
helium passes through this compressor each second, so that means our heat loss is 15 kJ per second.
Now that we have the givens covered, let’s write our energy balance equation. This is a compressor
problem, so we have power input, there is only 1 inlet and one outlet, so mass flow is the same,
and we have heat loss. Let’s bring the enthalpy terms to one side.
We can factor out the mass flow. Now we can find the difference in enthalpy using this equation.
All that’s left for us to do is plug values in, but we don’t have the specific heat value, so
let’s look it up on a table for helium. So that’s the value right here. Now we can plug our values
in. So we need 871.8 kW to run this compressor. In this question, we have steam that flows into
a turbine and we need to figure out the heat transfer due to this process.
Let’s write down what we know. The steam flows in at a rate of 26 kg/s with
a negligible velocity. The initial pressure is 6 MPa and the temperature is 600 degrees Celsius.
The steam leaves the turbine at 0.5 MPa and a temperature of 200 degrees Celsius. The velocity
is now 180 m/s. During this process, the steam produces work which is measured to be 20 MW.
Let’s write our energy balance equation. This is a turbine problem, so we have enthalpy,
kinetic energy, work output, and heat loss. There is no potential energy change.
Keep in mind, it’s one inlet and one outlet, so mass flow is the same on both sides. So,
what we need are the enthalpy values. Since this is steam, we can find those values using a table.
Our initial pressure is 6 MPa and temperature is 600 degrees Celsius. So our value is right here.
Our final pressure is 0.5 MPa and the temperature is 200 degrees Celsius.
So the value we are looking for is right here. Now we can start plugging in values. One thing
to note here is that since meters squared over seconds squared is equal to J/kg,
you have to either divide this by a 1000 or multiply all the kJ terms by 1000 to
get the answer. Remember, we can’t add kJ to joules. The units have to be the same.
Solving tells us that 456 kW of energy is transferred as heat through this process.
That should cover the types of problems you will face when it comes to turbines
and compressors. Thanks for watching and best of luck with your studies!