Deriving the Steady-Flow Energy Equation (SFEE)

The Grand Principle: Energy Accounting for Open Systems

Let's start with the most basic law we have: energy cannot be created or destroyed. For any system, this means:

(Total Energy Entering the System) - (Total Energy Leaving the System) = (Change in the Total Energy of the System)

This is the first law of thermodynamics in plain English. Now, let's think about this in terms of rates—how much energy is flowing per second (which we measure in Joules per second, or Watts).

(Rate of Energy In) - (Rate of Energy Out) = (Rate of Change of Energy Inside)

Now, let's imagine our "system" is a control volume. This is just a defined space, like a jet engine, a pump, or a section of pipe. Energy can cross the boundary of this control volume in three ways:

1. Heat Transfer ($\dot{Q}$): Energy crossing the boundary because of a temperature difference (like heating a pipe).
2. Work Transfer ($\dot{W}$): Energy crossing the boundary as work (like a turbine shaft spinning or a pump's impeller pushing fluid).
3. Mass Transfer ($\dot{m}$): Energy carried by the fluid itself as it flows into or out of the control volume.

Let's visualize this with a general control volume that has an inlet and an outlet.

So, our rate equation becomes:

$\frac{d{E}_{CV}}{dt}={\dot{E}}_{in}-{\dot{E}}_{out}$

Where:

• $\frac{d{E}_{CV}}{dt}$ is the rate of change of total energy inside the control volume.
• ${\dot{E}}_{in}$ is the total rate of all energy flowing in.
• ${\dot{E}}_{out}$ is the total rate of all energy flowing out.

Let's break down the "in" and "out" terms using our standard sign conventions (heat in is positive, work out is positive):

$\frac{d{E}_{CV}}{dt}=({\dot{Q}}_{in}+{\dot{W}}_{in}+\sum {\dot{E}}_{mass,in})-({\dot{Q}}_{out}+{\dot{W}}_{out}+\sum {\dot{E}}_{mass,out})$

By convention, we group the net heat and net work terms: $\dot{Q}={\dot{Q}}_{in}-{\dot{Q}}_{out}$ and $\dot{W}={\dot{W}}_{out}-{\dot{W}}_{in}$. This gives us:

$\frac{d{E}_{CV}}{dt}=\dot{Q}-\dot{W}+\sum {\dot{m}}_{in}{\theta }_{in}-\sum {\dot{m}}_{out}{\theta }_{out}$

Here, $\theta$ (theta) represents the total specific energy (energy per kg) carried by the flowing mass.

This equation is the most general form of the First Law for a control volume. It says the energy inside the volume changes if there's a net transfer of heat or work, or if the energy of the mass coming in is different from the energy of the mass going out.

From Flow Work to Enthalpy: The Key Simplification

This is where our work from the last lesson pays off. We have a work term, $\dot{W}$, and we have energy being carried by mass. But remember, part of the total work is the flow work needed to push the mass across the boundary.

Instead of keeping track of shaft work and flow work separately, we perform that brilliant thermodynamic trick: we bundle the flow work ($Pv$) with the internal energy ($u$) to create enthalpy ($h$).

This next video does an excellent job of showing exactly how the math works out. It starts from the general equation we just discussed and elegantly incorporates the flow work terms to arrive at the final, enthalpy-based equation.

As the video showed, the term $\dot{W}$ in our equation is split into two parts: ${\dot{W}}_{CV}$ (shaft work, electrical work, etc.) and ${\dot{W}}_{flow}$. The flow work terms are then grouped with the mass flow terms.

What's left in the $\dot{W}$ term is just the useful work (like shaft work). The messy business of "pushing work" is now neatly packaged inside the enthalpy term!

So, the total energy that a kilogram of fluid brings into the control volume is its enthalpy ($h$), plus its kinetic energy ($\frac{V^2}{2}$), plus its potential energy ($gz$).

Our general energy balance equation now looks like this:

$\frac{d{E}_{CV}}{dt}=\dot{Q}-{\dot{W}}_{CV}+{\sum }_{in}\dot{m}\left(h+\frac{V^2}{2}+gz\right)-{\sum }_{out}\dot{m}\left(h+\frac{V^2}{2}+gz\right)$

This is it! This is the full general energy equation for a control volume. It accounts for everything.

The Steady-Flow Assumption: Making Life Simpler

That equation is powerful, but it's also complicated because it handles situations where things are changing over time (like filling a scuba tank). However, many engineering devices (turbines, pumps, jet engines) operate for long periods under the same conditions. We call this steady flow.

"Steady" simply means "unchanging with time."

1. Steady Flow: The properties of the fluid at any point inside the control volume (like pressure, temperature, velocity) are not changing. This means the total energy inside the control volume is constant.
   $\frac{d{E}_{CV}}{dt}=0$
2. Steady Mass Flow: The total mass flowing in equals the total mass flowing out. Mass isn't piling up or being depleted.
   $\sum {\dot{m}}_{in}=\sum {\dot{m}}_{out}$
   For the common case of a device with one inlet and one outlet, this is even simpler: ${\dot{m}}_{in}={\dot{m}}_{out}=\dot{m}$.

Applying the steady-flow condition ($\frac{d{E}_{CV}}{dt}=0$) to our general equation gives:

$0=\dot{Q}-\dot{W}+{\sum }_{in}\dot{m}\left(h+\frac{V^2}{2}+gz\right)-{\sum }_{out}\dot{m}\left(h+\frac{V^2}{2}+gz\right)$

(Note: From now on, $\dot{W}$ refers to the useful shaft/electrical work, not flow work).

Rearranging this to a more intuitive "Energy In = Energy Out" format, we get the Steady-Flow Energy Equation (SFEE):

$\dot{Q}+{\sum }_{in}\dot{m}\left(h+\frac{V^2}{2}+gz\right)=\dot{W}+{\sum }_{out}\dot{m}\left(h+\frac{V^2}{2}+gz\right)$

For the most common case of a single inlet (1) and a single outlet (2), it becomes:

$\dot{Q}+\dot{m}\left(h_1+\frac{V_1^2}{2}+g{z}_1\right)=\dot{W}+\dot{m}\left(h_2+\frac{V_2^2}{2}+g{z}_2\right)$

This is one of the most important and widely used equations in engineering thermodynamics. It's our energy accounting tool for any steady-flow device.

Let's pause and appreciate what we built. We started with "energy is conserved," accounted for all the ways energy can move, used enthalpy as a clever shortcut, and applied a "steady state" condition to get a beautifully organized equation.

Worked Numerical Problems

Let's apply the SFEE to a couple of classic engineering problems.

Problem 1: Power Output of a Steam Turbine

Steam enters a turbine at a mass flow rate of 3 kg/s. The inlet conditions are: pressure = 2 MPa, temperature = 400°C (at which $h_1=3248.4$ kJ/kg), velocity $V_1=50$ m/s, and elevation $z_1=10$ m. The steam exits the turbine at: pressure = 15 kPa, $h_2=2500$ kJ/kg, velocity $V_2=180$ m/s, and elevation $z_2=6$ m. The turbine is not perfectly insulated, and it loses heat to the surroundings at a rate of 15 kW. What is the power output of the turbine?

Problem 2: Pumping Water

Water is pumped from a large lake to a storage tank 20 m higher. The pump consumes 5 kW of electrical power. The water is pumped at a rate of 50 L/s. If the water temperature doesn't change and heat transfer is negligible, what is the pressure increase across the pump? (Assume density of water $\rho =1000$ kg/m³, 1 L = 0.001 m³).

Conclusion

Excellent work today. This was a big, foundational lesson. We have successfully derived the master key for analyzing open systems.

Our Key Takeaways:

• The General Energy Balance: We started from the first law ($E_{in}-E_{out}=\Delta E_{system}$) and applied it to a control volume in terms of rates.
• Enthalpy is Key: We saw how bundling internal energy and flow work into enthalpy ($h=u+Pv$) cleans up the math and clarifies that the main work term ($\dot{W}$) is for useful output/input.
• The Steady-Flow Energy Equation (SFEE): By assuming conditions are stable over time, we arrived at the workhorse equation for many engineering devices:
  $\dot{Q}+\dot{m}\left(h_1+\frac{V_1^2}{2}+g{z}_1\right)=\dot{W}+\dot{m}\left(h_2+\frac{V_2^2}{2}+g{z}_2\right)$
  This is simply a statement that for a steady process, energy in equals energy out.

Preview of the Next Lesson

Now that we have this powerful tool, our next step is to learn how to use it efficiently. Most real-world devices don't involve all these energy terms. For a nozzle, work is zero. For a simple heater, work and kinetic energy changes might be zero.

In our next session, we will take our shiny new SFEE and apply it to a whole range of common engineering devices: nozzles, diffusers, turbines, compressors, and more. We will become experts at simplifying the SFEE to quickly analyze how each of these machines works.