Let’s take a look at pipes and ducts, another steady flow device.
We are all very familiar with both types. They carry the water we use to drink and shower,
the refrigerant in our air conditioners and fridges and the hot air during the winter
months from our furnaces, just to name a few. Since they can be considered steady flow devices,
we know that the rate of energy in must equal the rate of energy out. If we expand this equation
to cover all the types of energy associated with steady flow systems, it looks like this.
We can simplify this equation depending on the purpose of the pipe or duct.
If we have a pipe or duct with a diameter that is constant, then the velocities of the
fluids involved are relatively small, meaning kinetic energy is negligible. Potential energy
is also negligible if there is no significant elevation change of the fluid. If the pipe is
well insulated, there is no heat loss or gain. If there is no heating section, or a pump and such,
then work interactions are negligible. In that case, the most basic pipe or duct would involve
only mass flow and enthalpy. Since pipes or ducts usually only have one opening and one exit,
the mass flow must be the same, so it can be represented as a single variable. Now
let’s go through specific cases where we have to consider the types of energy we just crossed out.
If the pipe is not well insulated, we can have heat input or heat loss. If there is a heating
element inside the pipe heating up the fluid, then we would have work input. If there is a
large elevation change, then potential energy has to be considered. If we have ducts with variable
cross-sectional areas, then kinetic energy of the fluid might have to be considered.
So depending on the type of pipe or duct, we will need to consider what type of energy is involved.
To tackle some of the problems, you should keep in mind that if you have a difference in enthalpy,
you can use this equation to figure it out. So all you need are the initial and finals temperatures
along with the specific heat of the fluid. If a question involves heating elements,
keep in mind that voltage, current and power can be related using this equation.
So you can find the power by multiplying the voltage by the current.
Let’s go through some examples to see how we can apply what we just learned.
Let’s take a look at this problem where we have an electrical heater that’s used to heat up air. We
need to figure out the current supplied to the heater. Let’s write down what we know.
The initial temperature of the air is 15 degrees Celsius and it’s heated to 30 degrees Celsius. The
initial pressure and final pressure stay the same at 100 kPa and the volume flow rate of the air is
0.3 cubic meters per second. Lastly, the voltage of the heater is 110 volts. Let’s start off with
our energy balance equation. The air is traveling through a duct and it’s heated. So for inputs,
we have work input, so that’s done by the heater, and we have mass flow and enthalpy.
For output, we have only mass flow and enthalpy since there is no heat loss. Since this is a duct
with one inlet and outlet, the mass flow will be the same on the input side and output side.
Let’s isolate the equation for work input since that’s what we are trying to find.
Now we can factor out the mass flow. Notice how we have a difference in enthalpy,
so we can use this equation to figure it out. We don’t have mass flow and the specific heat of air.
Let’s figure out the mass flow first. For that, we can use this equation. Remember
we can figure out mass flow by dividing volume flow by the specific volume. The problem is,
we don’t have the specific volume of the air. We can figure that out using the ideal gas equation
since at the given parameters, the air acts as an ideal gas. To use this equation, we need the
gas constant of air, which we can find from a table. So here is the value we are looking for.
Now let’s plug our values in. Remember that we need our temperature in kelvin to use the ideal
gas equation, so 15 degrees Celsius is 288.15 kelvin. Let’s solve for the specific volume. Now
that we have the specific volume, we can go back to the mass flow equation and plug our values in.
Solving gives us the mass flow. Next, we need to find the specific heat of air.
We can find that from a table. Here is the value we are looking for.
We now have all the pieces, so let’s start plugging values in.
We get 5.47 kJ/s or 5470 watts. How do we use this to figure out the current of the heater? Well,
power is equal to current times voltage. The work input is our power. We have the voltage, and all
we need to do is solve for current. Solving gives us the current that’s supplied to the heater.
In this question, we are told that r-134a enters and leaves a condenser. We need to figure out the
heat transfer from the refrigerant per unit mass. Let’s write down what we know. The refrigerant
enters the condenser at 900 kPa with a temperature of 60 degrees Celsius, and leaves as a saturated
liquid at the same pressure. So all we need to do is figure out the heat loss during this process.
Let’s start with our energy balance equation. Since the refrigerant is flowing through a pipe,
we have mass flow and enthalpy for the input, and on the output side, we have a heat output along
with mass flow and enthalpy. Since the question wants us to find this heat output per unit mass,
we need to divide everything by mass flow. Now we can use refrigerant tables to figure out the
enthalpy values. The initial pressure is 900 kPa and the temperature is 60 degrees Celsius.
Looking at our pressure table, we see that for a pressure of 900 kPa, our saturation temperature
is 35 degrees Celsius, but our temperature is higher than that. So initially, our refrigerant
is a superheated vapor which means we need to look at a superheated refrigerant table.
So our pressure is 900 kPa and temperature is 60 degrees Celsius. Here is our enthalpy value.
Now for our second enthalpy value, we are told that the refrigerant is a saturated liquid
so we just need to find the saturated liquid enthalpy value
at 900 kPa. Here is the value we are looking for.
Now that we have these values, we can plug them into our equation and solve for the heat loss.
In this question, we have water that’s heated in a tube and we need to figure out the mass
flow rate of the water. Let’s write down what we know. The initial temperature of the water is 20
degrees Celsius and after it’s heated up, it has a temperature of 75 degrees Celsius. The heater has
a power rating of 7 kW, which is our work input. We will start off with our energy balance
equation. For inputs, we have mass flow and enthalpy along with work input. For output,
we only have mass flow and enthalpy since there is no heat loss or work output. Let’s bring the
mass flow terms to one side. We can now factor out mass flow. Since we have a difference of enthalpy,
we can use this equation to figure it out. Now all we need is the specific heat of water.
Let’s take a look at our table and here is the value we are looking for.
Now let’s plug our values in. Let’s solve and that’s the mass flow of the water.
That should cover the types of problems you will face when it comes and steady flow systems
and pipes and ducts. Thanks for watching and best of luck with your studies!