Up until now, we talked about steady flow devices, meaning everything pretty much flows at the same
rate so it’s easy to do calculations. Now, we are going to look at unsteady-flow systems.
In this case, we have to account for the changes within a control volume. An example would be
filling up an empty container from a supply line. Unsteady-flow processes start and end over
a specific time span. So for example, the time it takes to fill up an empty
container. Because these processes happen over a time span, we have to consider time intervals
when we solve problems. These processes may can also involve moving boundaries.
The mass balance for any system can be expressed like this. So the amount
of mass in minus the amount of mass leaving the system gives us the change in mass of the system.
When it comes to control volumes, we can write it like this. Remember, a control volume is just a
system where mass can enter or exit. Here, the subscript i is for inlet, e for exit, 1 is for
the initial state and 2 is for the final state. In simple terms, m1 and m2 tells us the mass of the
substance at the beginning of a process or at the end of a process. M subscript i and m subscript
e tells us the amount of mass that enters or exits through an inlet or an outlet of a system.
Usually, you will find that some of these terms will end up being zero. If no mass enters the
control volume during the process, then this is zero, if no mass leaves, then this is
zero and if there is nothing inside the control volume at the initial stage, then this is zero.
The energy balance for a control volume can be written like this. All we are saying is
that energy in minus the energy out gives us the change in energy of the system.
Solving problems can be difficult with unsteady flow processes because the mass
at the inlets and outlets can change over time. To make our lives easier, we can solve problems
by assuming it to be a uniform-flow process. We have to make a few assumptions to do so.
First, we assume that the fluid flow at the inlet or the exit is uniform and steady. That means the
properties of the fluid won’t change with time over the cross section of the inlet or outlet.
If they do change, we average it out and treat it as a constant for the entire process. Keeping this
in mind, we can write the full energy balance equation for a uniform flow system like this.
Here, this is heat input and this is work input. You should be familiar with this
theta symbol from the flow work video, where it represents the energy of a fluid stream.
The lowercase e represents the energy of the nonflowing fluid inside the control volume
per unit mass. Here, the letter h represents enthalpy and u represents internal energy.
Ke and pe represents kinetic and potential energy. Whenever kinetic and potential energy change is
negligible, we can simplify it by canceling out these terms, so we end up with just enthalpy
and internal energy. This means our big equation can be simplified like this.
All of this might be a bit confusing now, but when we cover a few examples, it should give
you a better idea. So let’s get started. Let’s take a look at this problem where
we have a tank filled with steam. There is a regulator attached to the tank, in other words,
a valve that lets out steam. When the steam in the tank is heated, some steam escapes
through the regulator which keeps the pressure constant, but the temperature rises. We need to
figure out the amount of heat transferred when the temperature of the steam is 500 degrees Celsius.
Let’s start off by writing down what we know. The volume of the tank is 0.2 cubic meters.
Initially, the temperature of the steam is 300 degrees Celsius and the pressure is 2 MPa.
The final temperature is 500 degrees Celsius. We will take the tank to be our system. Let’s
start off with our mass balance equation for a control volume. We know that mass doesn’t enter
the system, so this term is zero. We are left with just mass that leaves the system on the left side.
Let’s simplify and do a bit of tidying up. Now we need to think about the energy balance
of our system. Let’s expand this equation to account for all the types of energy we have.
Now in our problem, there are no work interactions,
kinetic and potential energies are negligible, and no mass enters the system.
There is heat input but no heat output. So our equation simplifies to this.
Now remember we talked about mass balance? Well, we can plug in the value for exit mass from our
mass balance equation. Let’s isolate this equation for heat input since that’s what we need to find.
Now we just need to find these values so we can plug them in. To do so, we need a few values from
our property table for superheated steam. Let’s look at our initial conditions, so we have 2
MPa and a temperature of 300 degrees Celsius. Here are the values we need for this problem, so that’s
specific volume, internal energy and enthalpy. Next, we need to look at the final state,
so pressure is the same, but the temperature is now 500 degrees Celsius.
Using these, we can figure out the initial and final mass that’s in the tank.
Remember, all we need to do is divide the volume by the specific volume.
Next, we can figure out the amount of mass left over after some vapor leaves the tank.
We have the masses, so now all we need is the enthalpy. Since the enthalpy value is
changing during the start and end of the process, we just need the average.
We now have everything we need, so let’s start plugging in values.
Solving tells us that 606 kJ of heat was transferred to the tank.
Let’s take a look at this problem where we have a tank that contains refrigerant-134a.
The tank is connected to a supply line. When the valve is opened, R-134a enters the tank.
After the valve is closed, it’s found that it contains saturated liquid at 1.2 MPa. We need
to figure out the mass of the refrigerant that entered the tank and the amount of heat transfer.
Let’s start by writing down what we know. The volume of the tank is 0.06 cubic meters.
The initial pressure is 0.8 MPa. The supply line carries refrigerant at 1.2
MPa at a temperature of 36 degrees Celsius. The final pressure of the tank is 1.2 MPa.
The tank will be our system. Let’s start with a mass balance equation for our control volume.
In this problem, mass doesn’t leave the system, but it does enter, so we are left with just this.
Next, we need to think about energy balance. Now we have to make an assumption,
which is to assume that heat was transferred to the tank. If we get a positive answer,
then our assumption will be right. So we don’t have heat output, just heat input. There are also
no work interactions and kinetic and potential energy changes are negligible. We also have
mass coming into the tank, but none leaving. So we end up with this simplified equation.
Now we need to figure out a few values from the refrigerant property tables before
we can solve our equation. So first, we can look for initial case values.
The pressure is 0.8 MPa and it’s a saturated vapor. Here are the values we need.
Next, we look at the second state. The pressure is 1.2 MPa and it’s now a saturated liquid.
Lastly, we need this value, which is the enthalpy of the supply line. The temperature is 36 degrees
Celsius and the pressure is 1.2 MPa. Now we look and we see that the saturation pressure
is 912 kPa, which is lower than the pressure given to us. That means this is a compressed
liquid, and we can treat it as a saturated liquid. So the enthalpy value we want is right here.
Now that we have all of these, we can first figure out the initial and final mass.
That’s the initial mass in the tank. Next, the final mass.
The first part of our question wants us to figure out mass of the refrigerant that entered the tank.
So we can use our mass balance equation to figure it out. Subtracting the initial mass
from the final mass gives us the total amount of mass that entered the tank.
The second part of the question wants us to figure out the heat transfer. So
we can use our energy balance equation to figure it out. Let’s plug in our values.
Solving for our heat input gives us a positive value, which means our assumption was right.
Let’s take a look at this question where we have a tank containing helium. When a valve is opened,
helium escapes, and we need to figure out the final temperature and pressure in the tank.
We can start off by writing down what we know. The volume of the tank is 0.15 cubic meters.
The initial pressure is 3 MPa and the temperature is 130 degrees Celsius.
We are also told that one half of the initial mass escapes when the valve is opened. That means the
final mass is equal to half of the initial mass. Let’s start off with a mass balance equation
for our control volume. Since mass doesn’t enter our tank, we can eliminate this term.
Let’s do a bit of tidying up. Now remember, we know that the final mass is equal to half
the initial mass, so let’s plug that in. Let’s simplify and we get our mass balance equation.
Next, we need to think energy balance. We don’t have heat input or heat loss since the tank is
insulated, there are no work interactions, kinetic energy and potential energy are negligible, and
mass doesn’t enter the tank. So we are left with this as our simplified energy balance equation.
We can use the same value as before for the final mass value and the simplified mass balance value
for the exit mass. This way, we can simplify by dividing everything by initial mass. Next,
we will multiply by 2 to get rid of the fractions and bring all the terms to one side.
Now the question wants us to figure out the final temperature of the tank,
which means we need to have temperature as variables in our equation. We can do that by
using the relationship between enthalpy, internal energy and temperature. Since the enthalpy will be
changing from the start of the process to the end of the process, we will need to take the average.
So for the enthalpy, we can replace it with this. For internal energy, we can replace it with this.
If this is unfamiliar to you, please watch the video about internal energy and enthalpy.
Now, we are going to divide everything by specific heat at constant volume, cv. Remember that cp over
cv gives us the specific heat ratio, so we can replace that. We can find the k value by looking
at a table that gives us specific heat ratio values. Here is the value we are looking for.
We are given the initial temperature of the tank, but that needs to be in kelvin for us
to use it in this equation with the specific heat ratio. Let’s plug our values in and solve for T2.
Lastly, we need to figure out the final pressure. We can do that using the ideal gas equation and
by comparing the initial state to the final state using a ratio. We can represent mass 2
with mass 1. Let’s eliminate the common variables so that we are left with just the simplified form.
Now we can plug our values in and solving tells us that the final pressure was 954 kPa.
That should cover the types of problems you will face when it comes to unsteady flow processes.
Thanks for watching and best of luck with your studies!