FIRST LAW:
One of the most fundamental and common concepts used in the Mechanical PE exam is the first law of thermodynamics. The general equation can look complex, but the concept behind the first law of thermodynamics is fairly intuitive and if fully understood can be easily recalled and used without the need to root through your reference material and (more importantly) use up precious time. In simple terms the first law just says that the total energy (E) of a defined system is the sum of all the internal and external energy going into and out of the system. Internal energy includes heat (Q) and work (W) and external energy includes pressure, kinetic, and potential. This is just another way of stating the law of conservation of energy; the total energy within a system is constant, energy can be transformed from one form to another, but cannot be created or destroyed.
Examples of energy transformation include cases like a water fall. A unit volume of water at the top of a waterfall will possess some potential energy (h) in feet. When the water falls from its initial destination it will have lost that elevation and therefore its potential energy but will have gained kinetic energy when it falls and invariably picks up speed (v). This represents a transfer of energy from one type (potential) to another (kinetic). Understanding the first law will let us know that the final kinetic energy must equal the initial potential energy since energy cannot be created or destroyed (see Bernoulli’s equation for more on this subject).
Let’s take a look at the first law in its formal, most general form:
In words, this equation just states that the heat input (Q) into a system, minus the work done BY the system (W), plus the net change in pressure (h), velocity (V2/2), and potential energy (gz) into and out of a system, equals the total net change of energy of the system. Let’s break each of these components down to understand each one individually and, more importantly, understand how these elements may be presented to you on the actual exam.
HEAT ENERGY:
Heat energy (Q) as the name implies is energy transfer due to heat transfer. If a problem gives two different sets of temperatures before or after a process or amount of time or there is a temperature difference between two portions of a control volume, chances are there will be a transfer of heat energy that needs to be taken into account in the problem solution.
Conversely, systems, processes, or cycles that are said to be insulated can be assumed to be adiabatic, that is Q=0. This is a common problem setup that can greatly simplify your solution.
WARNING: It is common to confuse heat and temperature. One implies the other but they are not the same thing. Heat (Q) is the energy in a given system due to a temperature (i.e. the energy from conduction or convection for example), Q can be zero while T is non zero such as the special case of an adiabatic process mentioned above.
WORK (W):
The concept of work in terms of the first law of thermodynamics can be summed up as energy into or out of the system that can move a mass a given distance. This definition can be somewhat confusing as many thermodynamic problems may not involve any mass movement at all. Take the classic case of a battery and motor system. If we energize the motor there is work being done. It is not a typical mass moving through a linear distance as most examples will give you, but intuitively we know that the more is consuming power (i.e. work) in order to move its rotor.
In general if a problem gives you a piston cylinder setup where the piston moves through a distance, or there is a motor running a pump or compressor or any other movement within the boundary layer of the system, there is most likely work being done on the system.
A NOTE ABOUT SIGN CONVENTIONS AND THE FIRST LAW:
The sign conventions used for thermodynamic problems can be tricky and are often a common source of costly mistakes on the exam. Take a look at our first law equation again:
Heat energy (Q) is defined as what goes into the system. If 100 btus of heat is added to the system, that is shown in the equation as positive. Work (W) is traditionally defined as work done BY the system. Therefore, if the system is producing 100 Hp of power, this is considered a net negative to the system as that energy gets dissipated out of the system. To envision this, think of the classic piston cylinder apparatus with an entrained gas shown in every thermodynamic text:
Intuitively, if you consider this piston volume to be your “system”, you can reason that the heat added to the gas will increase the systems net energy. What about the work? Is it adding to the system or decreasing the net energy in the piston? It’s pushing the piston out, thereby decreasing the internal pressure of the gas which is a net loss to the system. In keeping with the above first law definition, work done BY the system is considered positive so it would be written as positive Q minus positive W:
Again, intuitively the net equation makes sense as we are adding heat energy and since the work is being done by to system to the surroundings, subtracting work energy. The key is to make sure the units make sense. Is energy being added or subtracted to your system?
SPECIAL CASES:
We know from above that employing the first law of thermodynamics is simply a mathematical way of accounting for all the given energies in a given system. This can sometimes be a lengthy process. Therefore very often PE exam problems will be given with special criteria that greatly simplify these problems. Understanding these criteria will greatly streamline solving many thermodynamic problems.
ADIABATIC:
Adiabatic processes are ones in which no heat (Q) is transferred with the surroundings. The only energy transferred is work. Therefore:
This can greatly simplify many thermodynamics problems.
ISENTROPIC:
Isentropic processes are a subset of adiabatic processes. Isentropic processes are considered fully reversible; as such the change in entropy from one stage of the process to the next is zero. Since isentropic processes are by definition also adiabatic the following holds true:
Isentropic conditions are often assumed for steady flow compression or expansion processes. Isentropic compression or expansion is considered ideal and is used in part to calculate compressor, pump, or expander efficiencies; or, given an efficiency value in the problem statement can be used to calculate the actual performance. See the associated examples for how this is done. The relationships between isentropic work (ideal work), actual work, and efficiency are shown as follows:
For turbines:
For compressors:
For nozzles:
Note: Nozzles don’t do any work, so their measure of efficiency is done by comparison between the ideal and actual changes in kinetic energy instead.
PE exam strategy: For thermodynamic problems featuring the 1st law, look for indications of a special case (adiabatic, isobaric, etc.). And use those specialized equations from your equations sheet as a starting point.
The figure below gives a breakdown of examples of different closed and open systems as well as some basic characteristics of each:
REAL AND IDEAL GASES:
Problems involving ideal gases are another major component to the PE test. Assuming a gas is ideal allows for major simplification in the problem solution. So the question naturally arises-how do we know if a gas is ideal? Practically speaking, most thermodynamic problems involving gases on the test can probably be assumed to be ideal gas problems. However, there are never any guarantees regarding what will actually be on the exam so let’s explore the relationship between real gases and ideal gases.
IDEAL GASES
Ideal gases follow the ideal gas law which is usually written as:
or
and
Where P=pressure [psia] V=volume [ft3] m=mass [lbm] T=temperature [R] MW=molecular weight R* is the universal gas constant 1545.4 (ft-lbf)/(lbmol-R).
If a mole value is given for a gas the following form of the ideal gas equation can be used as well:
Where n=moles of gas
COMPRESSIBILITY FACTOR:
The compressibility factor of a gas is the main indicator of whether a gas can be considered real or ideal. Ideal gases have a compressibility factor of 1. The compressibility factor for real gases will deviate from this value positively or negatively. The farther away from unity (1) the value is, the less like an ideal gas the real gas will behave.
The compressibility factor (Z) for all gases is a function of its reduced temperature, Tr [R], pressure, pr [psia], and specific volume, vr [ft3/lbm or ft3/lbmole] as shown below:
Where Tc, pc, and vc are the critical temperature, pressure, and volume respectively. Critical point properties for some common gases are included below:
If at least two out of three of these reduced properties are known, a generalized compressibility chart can be used to determine the compressibility factor (Z) which can then be used with a corrected ideal gas equation per the following:
PE exam strategy: If the gas characteristics in a problem statement are either at very low temperatures or very high pressure (or both), the gas may need to be treated as a real gas. Otherwise, assume the gas is ideal.
There are many other thermodynamics topics that are important for the test. This is just a highlight of some of the more important subjects you may come across. For any questions or corrections on this material drop us a line at info@passthepeexam.com