In this video, we are going to look at mixing chambers and heat exchangers.
You probably encountered both in your daily life.
A common example of a mixing chamber is when you take a shower and you turn the knob to
get that perfect temperature.
Behind the scenes, a valve mixes hot water and cold water together to give you that perfect
temperature you desire.
Let’s draw a simple mixing chamber.
We have 2 inlets and one outlet.
Now as with all steady flow systems, we know that the rate of energy in must equal the
rate of energy out.
We express that like this.
We can expand this to account for all the types of energy associated with a steady flow
system.
Now mixing chambers tend to be very well insulated, so heat transfer between the system and the
surroundings are negligible.
There are no work interactions as well.
The kinetic energy and potential energy of the fluid is also negligible.
So our big equation simplifies to this.
Now let’s label each mass flow for the 2 inlets and outlet.
Then our energy balance equation can be written like this.
Each inlet will have its own mass flow and enthalpy, and the outlet will also have its
own mass flow and enthalpy.
Since this a steady flow system, we know that the outlet mass flow must be equal to the
2 inlet mass flows.
In simple terms, the 2 mass flows coming in is equal to the mass flow leaving.
We can express it like this.
To simplify our equation, we can go further and just substitute this value in for outlet
mass flow.
So for simple mixing chambers, our energy balance equation will simplify to this.
Now let’s talk about heat exchangers.
You probably seen these in radiators, and heating and cooling systems.
Usually, they have 2 streams, and one of the fluids in the stream will transfer heat to
another stream.
In simple terms, two fluids will exchange heat without mixing.
A simple system would look like this.
Here, some fluid is flowing through this blue pipe.
Initially, it has a temperature of 10 degrees Celsius.
Now this red pipe brings in another fluid into the heat exchanger.
This fluid is hot, so let’s say 60 degrees Celsius.
As this fluid travels through the exchanger, the fluid in the blue pipe absorbs this heat.
Through this process, the fluid in the red pipe decreases in temperature, while the fluid
in the blue pipe increases in temperature.
This is the basics of a heat exchanger.
Now let’s discuss the energy balance of heat exchangers.
So, we start off with our energy balance equation, and we can expand it like this.
If we select the whole heat exchanger to be the system, then there is no heat transfer
between the system and the surroundings.
There is no work, and potential and kinetic energy is negligible.
So our energy balance equation simplifies to just mass flow and enthalpy.
Now if we look at our heat exchanger, we have 2 separate inputs, and 2 separate exists.
Let’s focus on just the blue pipe.
It has one input and one output.
The inlet has its own mass flow and enthalpy value and the same goes for the outlet.
Now let’s focus on the red pipe.
That also has one inlet and one outlet with its own mass flow and enthalpy.
So for our energy balance equation, what are the inputs?
We have m dot 1 and h1 and mdot3 and h3.
Those are the inputs of each of the pipes.
Now what are the outputs?
We have m dot 2 and h2 and m dot 4 and h4.
That’s our energy balance equation, but we can simplify this further.
Notice how for the blue pipe, we have only 1 inlet and one outlet, that means the mass
flow of the input must be equal to the mass flow of the outlet for this blue pipe.
The same is true for the red pipe.
That also has one inlet and one outlet, which means the mass flow at the inlet must be equal
to the mass flow at the outlet.
We can represent each of the mass flows that are equal to each other as one single mass
flow.
I’ll label them red and blue.
Now let’s bring the blue terms to one side and red terms to the other.
We can now factor the mass flow out and that gives us the simplified energy balance equation
for a heat exchanger if we consider the whole thing to be the system.
Remember that when you have a difference in enthalpy, you can use this equation to figure
it out, where all we need are the temperatures and specific heat of the fluid.
Now what if we consider just one of the fluids to be our system?
Let’s say we pick the blue pipe to be the system.
In that case, we have one input and one output and we would also have a transfer of heat
into the fluid.
It’s a heat input because the heat from fluid of the red pipe is being transferred
to the fluid in the blue pipe.
Again, it’s one input and one output, so mass flow is the same and can be represented
as one variable.
If we had selected the red pipe as the system, then there would be a heat loss, in other
words, heat output.
All of this will make more sense with examples, so let’s get started.
In this problem, we have water that’s being heated up to 60 degrees Celsius with superheated
steam.
We need to figure out the mass flow rate of the steam required to increase the temperature
of the water.
Let’s write down what we know.
The initial temperature of the water is 20 degrees Celsius and the pressure is 300 kPa.
This water is then mixed with superheated steam, which is at 300 kPa and a temperature
of 300 degrees Celsius.
The cold water enters the chamber at a rate of 1.8 kg/s, and the mixture leaves the mixing
chamber with a temperature of 60 degrees Celsius.
Let’s start off by writing our energy balance equation.
We have energy in is equal to energy out.
Now this is a mixing chamber problem.
So all we have is mass flow and enthalpy.
Notice how we have 2 mass flows entering, and 1 exiting.
So that means the mass flow times the enthalpy of this top line plus the mass flow times
the enthalpy of the bottom line is equal to the mass flow times enthalpy of the exiting
line.
Now forgetting about enthalpy, we know that there are 2 entrances and 1 exit, which means
the mass flow of the 2 entrances must equal the mass flow of the exit.
In other words, we can replace mass flow 3 with this.
The next step is to figure out these enthalpy values, and we can do that by looking at temperature
tables.
Let’s start off by finding the first enthalpy value.
So our pressure is 300 kPa, and temperature is 20 degrees Celsius.
But notice how the saturation temperature for water at 300 kPa is 133 degrees Celsius.
Our temperature is lower than the saturation temperature, which means this is a compressed
liquid.
So when it comes to compressed liquids, we can find our enthalpy values by thinking of
it as a saturated liquid at a given temperature.
If you need a refresh on this, please take a look at the pure substances and property
tables video, where I explain how pressure and temperature determine the states of liquids
and gases.
So let’s look at the temperature table for 20 degrees Celsius.
Here is our enthalpy value.
Next, we have 60 degrees Celsius, which is the exit temperature.
This is our third enthalpy value.
Now we need to figure out the 2nd enthalpy value.
For that, we need to look at a superheated water table.
The pressure of the steam is 300 kPa, and the temperature is 300 degrees Celsius.
So our enthalpy value is right here.
We now have all the values we need, so we can start plugging them in.
Solving gives us the mass flow rate of the steam.
In this problem, we have two streams of r-134a being mixed together.
We need to figure out the temperature and the quality of the exit stream.
Let’s write down what we know.
For the first stream, we have a pressure of 1 MPa and a temperature of 20 degrees Celsius.
For the 2nd stream, we have a pressure of 1 MPa and a temperature of 80 degrees Celsius.
We are also told that the mass flow rate of the first stream is twice that of the 2nd
stream.
So we can start off with the energy balance equation.
Since this is a mixing chamber, we have mass flow and enthalpy to worry about.
We have 2 inlets and 1 outlet.
We can replace mass flow 3 with the sum of mass flow 1 and 2 just like we did in the
previous question since those 2 flows added together must equal the mass flow leaving
through the outlet.
Now in the question, we are told that mass flow 1 is equal to 2 times mass flow 2.
We can replace that as well.
Let’s simplify.
We can one step further by dividing all the terms by mass flow 2.
Now all we need to do is figure out the enthalpy values.
For that, we can use a temperature table for r134a.
So for our first inlet, the temperature is 20 degrees Celsius with a pressure of 1 MPa.
Notice that for a temperature of 20 degrees Celsius, our saturated pressure should be
572 kPa.
Since our pressure is higher than the saturation pressure, this is a compressed liquid.
So we can treat it like a saturated liquid at the given temperature.
Here is the enthalpy value we are looking for.
Next, for the 2nd inlet, we have a temperature of 80 degrees Celsius with a pressure of 1
MPa.
But now, if we look at the saturated pressure, it’s higher than our pressure.
That means this is a superheated vapor.
Let’s look at the superheated vapor table for r134a.
Our pressure is 1 MPa, and the temperature is 80 degrees Celsius.
This is our enthalpy value.
Now that we have these values, we can plug it in to solve for enthalpy of the outlet.
Now that we know the enthalpy of the outlet, how do we use that to figure out the temperature?
We first need to determine what type of mixture this is.
So let’s look at a pressure table for R134a, and our pressure is 1 MPa.
Note these values here, which are the saturated liquid and saturated vapor enthalpy values.
Notice how our outlet enthalpy value is greater than the saturated liquid enthalpy value but
less than the saturated vapor enthalpy value.
That means our refrigerant at the outlet is now a saturated mixture.
So we just need to look for the saturation temperature at 1 MPa.
Here is the value we are looking for and that’s our temperature.
The last part of the question wants us to figure out the quality.
Quality is just the ratio of vapor to the liquid in a mixture.
We can find that using this equation.
Remember this part is the difference between saturated vapor and saturated liquid enthalpy.
Now we can plug our values in.
Solving gives us the quality.
Let’s take a look at this question where we are told that oil is cooled using water
with the use of a heat exchanger.
We need to figure out rate of heat transfer in the heat exchanger and the exit temperature
of the water that’s used to cool the oil down.
Let’s write down what we know.
The initial temperature of the oil is 150 degrees Celsius and the final temperature
is 40 degrees Celsius.
The mass flow rate of the oil is 2 kg/s.
The water that is used to cool the oil enters with a temperature of 22 degrees Celsius and
a mass flow rate of 1.5 kg/s.
We are also given the specific heats of both the oil and water.
Let’s start by writing our energy balance equation.
This is a heat exchanger problem, and we are going to pick the oil tube as our system.
What that means is, we will have heat output since heat is lost from the oil and transfers
into the water.
So we have mass flow times enthalpy and that’s equal to the heat loss plus the exit side
mass flow times enthalpy.
Let’s bring the mass flow terms to one side.
Remember that for our oil tube, it’s one inlet and one outlet, so mass flow will be
equal.
We can factor it out.
Now we have a difference of enthalpy, and we can replace that with this equation.
Now we can plug our values in since we have everything we need.
This tells us the heat lost by the oil, which is then absorbed by the water.
We can now do the same thing we did before, but this time, we will look at the water tube
as our system.
Now we have heat input since heat is transferred from the oil to the water.
We also have mass flow and enthalpy at the inlet and the outlet.
Isolating as before, and simplifying gives us pretty much the same equation.
We can use this to figure out the temperature of the water.
Let’s plug our values in.
Again, there is now a heat transfer into the water, which is the heat transferred by the
oil.
Solving gives us the final temperature of the water.
Let’s take a look at this problem where we have refrigerant 134a flowing through a
pipe and it’s cooled by air.
We need to figure out the mass flow rate of the refrigerant.
Let’s write down what we know.
The initial and final pressure of the refrigerant is 1 MPa.
The initial temperature is 90 degrees Celsius and the final temperature is 30 degrees Celsius.
The air that’s cooling the refrigerant enters with a pressure of 100 kPa and has a temperature
of 27 degrees Celsius.
It leaves with a pressure of 95 kPa and a temperature of 60 degrees Celsius.
We are also told that the volume flow rate of the air is 600 cubic meters per minute.
To solve this problem, we are going to take the whole heat exchanger as the system.
That means there is no heat transfer between the heat exchanger and the surroundings, so
the only thing we need to worry about is mass flow and enthalpy.
So for inlet, we have the mass flow of the refrigerant and the air.
The same is true for the outlet.
Now if we look at just the refrigerant tube, we have one inlet and one outlet.
That means the mass flow for both will be the same.
The same is true for the air.
So the mass flow for the refrigerant, we will label m dot r and the mass flow for the air
will be m dot a.
Let’s bring the air terms to one side and refrigerant terms to the other.
We can factor out the mass flows.
Now we can isolate it for the mass flow of the refrigerant since that’s what we are
trying to find.
First, we have to figure out the mass flow rate of the air.
We can do that by using this equation.
Remember, mass flow can be found by dividing volume flow by the specific volume.
So now we don’t have specific volume, but we can find that using the ideal gas equation
since with the given parameters, the air acts as an ideal gas.
Let’s isolate for specific volume.
We have the temperature and pressure, we just need the gas constant.
So let’s look at our table, and here is the gas constant for air.
With that, we can plug our values in.
Remember, we need the temperature in kelvin to use the ideal gas equation, so 27 degrees
Celsius is 300.15 K. Let’s solve and we get our specific volume.
Now we can go back to our mass flow equation and plug values in.
Let’s solve and we get our mass flow per minute.
So looking back at our original equation, we figured out the mass flow of the air.
Next, we have to think about the enthalpy values.
For the refrigerant enthalpies, we can easily find them from a refrigerant table.
But there are no tables for air, which means we need to substitute this part with this
equation.
The specific heat of air can be found from a table, and that’s the value right here.
We have the initial temperature and final temperature.
So all we need are the enthalpy values for r134a.
Let’s look at our table.
The pressure is 1 MPa and the temperature is 90 degrees Celsius so it’s a superheated
vapor.
Here is our inlet enthalpy value.
Now we need the outlet enthalpy value.
The pressure is the same, but the temperature is now 30 degrees Celsius.
Looking at the pressure table, we see that our saturation temperature is 39.37 degrees
Celsius.
Our temperature is lower than the saturation temperature, which means this is now a compressed
liquid.
So we need to treat it as a saturated liquid at the given temperature.
So let’s look at our temperature table for 30 degrees Celsius.
Here is the saturated liquid enthalpy.
We finally have everything we need to solve this problem.
Let’s start plugging in values.
Solving tells us that 100 kg of refrigerant passes through each minute which is our mass
flow rate.
That should cover the types of problems you will face when it comes to mixing chambers
and heat exchangers.
Thanks for watching and best of luck with your studies!