Thermodynamics and Propulsion  
Subsections
[VWB&S, 6.1, 6.2]

The first law of thermodynamics can be written as a rate equation:
where
To derive the first law as a rate equation for a control volume we proceed as with the mass conservation equation. The physical idea is that any rate of change of energy in the control volume must be caused by the rates of energy flow into or out of the volume. The heat transfer and the work are already included and the only other contribution must be associated with the mass flow in and out, which carries energy with it. Figure 2.10 shows two schematics of this idea. The desired form of the equation will be
[Simple]
[More General]

The fluid that enters or leaves has an amount of energy per unit mass given by
where is the fluid velocity relative to some coordinate system, and we have neglected chemical energy. In addition, whenever fluid enters or leaves a control volume there is a work term associated with the entry or exit. We saw this in Section 2.3, example 1, and the present derivation is essentially an application of the ideas presented there. Flow exiting at station ``e'' must push back the surrounding fluid, doing work on it. Flow entering the volume at station ``i'' is pushed on by, and receives work from the surrounding air. The rate of flow work at exit is given by the product of the pressure times the exit area times the rate at which the external flow is ``pushed back.'' The latter, however, is equal to the volume per unit mass times the rate of mass flow. Put another way, in a time , the work done on the surroundings by the flow at the exit station is
The net rate of flow work is
Including all possible energy flows (heat, shaft work, shear work, piston work, etc.), the first law can then be written as:
where includes the sign associated with the energy flow. If heat is added or work is done on the system then the sign is positive, if work or heat are extracted from the system then the sign is negative. NOTE: this is consistent with , where is the work done by the system on the environment, thus work is flowing out of the system.
We can then combine the specific internal energy term, , in and the specific flow work term, , to make the enthalpy appear:
For most of the applications in this course, there will be no shear work and no piston work. Hence, the first law for a control volume will be most often used as:
In the special case of a steadystate flow,
Muddy Points
What is shaft work? (MP 2.5)
What distinguishes shaft work from other works? (MP 2.6)
Definition of a control volume (MP 2.7)
Suppose that our steady flow control volume is a set of streamlines describing the flow up to the nose of a blunt object, as in Figure 2.11.

The streamlines are stationary in space, so there is no external work done on the fluid as it flows. If there is also no heat transferred to the flow (adiabatic), then the steady flow energy equation becomes
The quantity that is conserved is defined as the stagnation temperature,
or
 
It is also convenient to define the stagnation enthalpy,
An area of common confusion is the frame dependence of stagnation quantities. The stagnation temperature and stagnation pressure are the conditions the fluid would reach if it were brought to zero speed relative to some reference frame, via a steady adiabatic process with no external work (for stagnation temperature) or a steady, adiabatic, reversible process with no external work (for stagnation pressure). Depending on the speed of the reference frame the stagnation quantities will take on different values.
For example, consider a high speed reentry vehicle traveling through the still atmosphere, which is at temperature, . Let's place our reference frame on the vehicle and stagnate a fluid particle on the nose of the vehicle (carrying it along with the vehicle and thus essentially giving it kinetic energy). The stagnation temperature of the air in the vehicle frame is
The confusion comes about because is usually referred to as the static temperature. In common language this has a similar meaning as ``stagnation,'' but in fluid mechanics and thermodynamics static is used to label the thermodynamic properties of the gas ( , , etc.), and these are not frame dependent.
Thus in our reentry vehicle example, looking at the still atmosphere from the vehicle frame we see a stagnation temperature hotter than the atmospheric (static) temperature. If we look at the same still atmosphere from a stationary frame, the stagnation temperature is the same as the static temperature.
For the case shown below, a jet engine is sitting motionless on the ground prior to takeoff. Air is entrained into the engine by the compressor. The inlet can be assumed to be frictionless and adiabatic.
Considering the state of the gas within the inlet, prior to passage into the compressor, as state (1), and working in the reference frame of the motionless airplane:
The stagnation temperature of the atmosphere, , is equal to since it is moving the same speed as the reference frame (the motionless airplane). The steady flow energy equation tells us that if there is no heat or shaft work (the case for our adiabatic inlet) the stagnation enthalpy (and thus stagnation temperature for constant ) remains unchanged. Thus
If then since the flow is moving at station 1 and therefore some of the total energy is composed of kinetic energy (at the expense of internal energy, thus lowering )
Equal, by the same argument as 1.
Less than, by the same argument as 2.
The form of the ``Steady Flow Energy Equation'' (SFEE) that we will most commonly use is Equation 2.11 written in terms of stagnation quantities, and neglecting chemical and potential energies,
The steady flow energy equation finds much use in the analysis of power and propulsion devices and other fluid machinery. Note the prominent role of enthalpy.
Muddy Points
What is the difference between enthalpy and stagnation enthalpy? (MP 2.8)
Using what we have just learned we can attack the tank filling problem solved in Section 2.3.3 from an alternate point of view using the control volume form of the first law. In this problem the shaft work is zero, and the heat transfer, kinetic energy changes, and potential energy changes are neglected. In addition there is no exit mass flow.
The control volume form of the first law is therefore
The equation of mass conservation is
Combining we have
Integrating from the initial time to the final time (the incoming enthalpy is constant) and using gives the result as before.
A liquid bipropellant rocket consists of a thrust chamber and nozzle and some means for forcing the liquid propellants into the chamber where they react, converting chemical energy to thermal energy.
Once the rocket is operating we can assume that all of the flow processes are steady, so it is appropriate to use the steady flow energy equation. Also, for now we will assume that the gas behaves as a perfect gas with constant specific heats, though in general this is a poor approximation. There is no external work, and we assume that the flow is adiabatic. We define our control volume as going between location , in the chamber, and location , at the exit, and then write the First Law as
Consider for example the PW4084 pictured in Figure 2.15. The engine is designed to produce about 84,000 lbs of thrust at takeoff. The engine is a twospool design. The fan and low pressure compressor are driven by the low pressure turbine. The high pressure compressor is driven by the high pressure turbine. We wish to find the total shaft work required to drive the compression system.
Douglas Quattrochi 20060806