Let’s talk about flow work and energy transfer. When we have a control volume, so a system where
mass can go in and out, some work is required to push the mass into or out of the system. This work
is called flow work or flow energy. In simple terms, it’s the energy required to push a fluid
into or out of a control volume. When a fluid is at rest, then it doesn’t possess any flow energy.
Flow work can be found by multiplying pressure by the volume, and if we wanted to write it per unit
mass, it would be like this and have the units kj/kg. This equation is the same whether a fluid
is pushed into or out of a control volume. Previously, we talked about the total energy
of a system using this equation. Here, u is internal energy, ke is kinetic energy and
pe is potential energy. If we were to consider the additional form of energy we talked about,
so flow energy, then the total energy of a flowing fluid can be written like this.
We use theta to represent the total energy of a flowing fluid. So all we did was add
the pressure times specific volume part in. Something very interesting happens when we write
it like this. Remember we discussed enthalpy? Well the equation for enthalpy was this. Notice how
this part is the same! That means this part can be replaced with enthalpy like this. When we write
it in terms of enthalpy, we don’t need to worry about flow work, since enthalpy takes care of it.
When we have mass that enters or leaves a system,
the energy associated with the flowing fluid of mass can be written like this. If the mass is
flowing at a certain rate, then we can write our equation like this, so rate of energy transport.
Whenever kinetic energy and potential energy are negligible,
then from our previous equation, these parts become zero. So energy transport equations also
end up simplifying to just mass times enthalpy. Now let’s get to some examples to see how we
can use the equations we talked about. To solve problems, you should keep in mind that enthalpy
can be found by multiplying specific heat by the difference in temperature. Also, things like the
ideal gas equation, equations relating mass flow and density are all going to be useful.
In this question, we have a house that’s at 1 atm and a temperature of 24 degrees Celsius.
When cold air enters the house, the warm air leaves, and we need to figure out the
rate of net energy loss due to this mass transfer. Let’s write down what we know.
The pressure is 1 atm or 101 kPa, the temperature outside is 5 degrees Celsius, the inside air is
24 degrees Celsius and the rate at which the warm air leaves the house is 150 cubic meters per hour.
We will also need the gas constant of air, which we can find from this table and we need
the specific heat of air at 24 degrees, which is 297.15 Kelvin. So that’s the value right here.
Now, how do we figure out the energy loss? The net energy loss is the difference between the
mass of energy flow that’s leaving minus the mass of energy flow that’s coming in.
How do we figure out the energy flow coming in and out? We can use the energy transport equation.
Now, for theta in and out, all we have is enthalpy since the potential
energy and kinetic energy change is negligible. So we can write our net mass equation like this.
Let’s factor out the mass flow. Next, we need to figure out enthalpy. For that, we can use
this equation so let’s replace it. Now all we need to do is plug values in, but notice how
we have everything we need except for mass flow. To figure out mass flow, we can use this equation.
To use this equation, we need the density. For that, we can use the ideal gas equation.
Since we want to find density using this equation, we need to replace specific volume with density.
Remember that density is equal to 1 over specific volume. That means specific volume is equal to 1
over density. We can replace it in our ideal gas equation like this. Let’s isolate it for density.
Now we can plug our values in. Keep in mind that we are looking at the inside of the house,
so our temperature is 24 degrees Celsius or 297.15 kelvin. Let’s solve and that’s our density.
Now we can use this value to figure out mass flow. Our answer is in kg per hour,
so let’s divide it by 3600 to get it per second. We finally got our mass flow, so let’s go back
to our original equation and start plugging values in. We get 0.945 kJ/s or 0.945 kW.
Let’s take a look at this problem where we have refrigerant 134-a entering a compressor. We need
to figure out the rates of energy transfer for mass in and out. Let’s write down what we know.
The refrigerant enters at a pressure of 1.14 MPa, leaves at 0.8 MPa, the temperature as it
leaves is 60 degrees Celsius, and the mass flow rate at which it leaves is 0.06 kg/s.
We can find the rates of energy transfer using this equation. Since potential energy
and kinetic energy are negligible, we just need to consider enthalpy so let’s replace theta with it.
With the given values, we can look up the enthalpy values using a table.
For the initial enthalpy, we are told that it’s a saturated vapor and with a pressure of 0.14 MPa,
our value is right here. For our second enthalpy value, we are told that it’s a superheated vapor
with a pressure of 0.8 MPa and a temperature of 60 degrees Celsius, so that’s the value right here.
Now that we have these values, we can plug them into our equations. Those are our answers.
That should cover the types of problems you will face when it comes to flow work
and energy transfer by mass. Thanks for watching and best wishes with your studies!