Thermodynamics · Study · FE Mechanical · FE → PE Prep
Thermodynamics
10% of exam
Ideal gases and pure substances, the laws of thermodynamics, processes and component performance, power and refrigeration cycles, gas mixtures, psychrometrics, HVAC processes, and combustion.
5 concepts
A. Properties of ideal gases and pure substances
Properties of Ideal Gases and Pure Substances
The ideal-gas law and specific heats, the T–v and P–v phase diagrams, quality and the saturation tables, and the linear interpolation that turns a table into an answer.
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C. Laws of thermodynamics
The First Law, Energy Transfers and Processes
Conservation of energy as a closed-system balance and an open control-volume balance — heat, work, enthalpy, and the isobaric, isothermal, and adiabatic processes that fill the FE.
The first law is conservation of energy written for a thermodynamic system: whatever heat and work cross the boundary must show up as a change of the energy stored inside. On the FE Mechanical exam this single idea — set energy in minus energy out equal to the change in storage — solves a large fraction of the thermodynamics questions, from a piston compressing a gas to a steam turbine running at steady state. The whole skill is choosing the right system, getting the sign convention right, and knowing which terms vanish for the process at hand. The FE Reference Handbook (Thermodynamics, First Law of Thermodynamics) states both the closed-system and the open-system forms; this concept turns those bare equations into a procedure you can run under exam pressure.
Heat, work, and the sign convention
Heat Q is energy that crosses the boundary because of a temperature difference; work W is every other form of energy transfer. The handbook's convention — the classic engine convention — is that heat added to the system is positive and work done by the system is positive, which is why the closed-system law appears as Q−W rather than Q+W. Per unit mass, write q=Q/m and w=W/m. Keep that convention nailed down: a compressor does work on the gas, so W is negative; a turbine has the gas doing work, so W is positive.
Q−W=ΔU+ΔKE+ΔPE
The closed-system energy balance
A closed system exchanges energy but no mass. When the kinetic and potential energy of the bulk system do not change — the usual case for a gas in a piston-cylinder — the balance collapses to Q−W=ΔU. For an ideal gas the internal energy depends only on temperature, so ΔU=mcvΔT
Boundary work and the P–v diagram
When a system boundary moves, it does reversible boundary work wb=∫Pdv — which is exactly the area under the process path on a P–v diagram. That geometric reading is the fastest sanity check you have: a process that moves left (compression) has negative area and negative work; a process that moves right (expansion) has positive work. Because the area depends on the path, work is a path function, not a property, so two processes between the same end states generally transfer different amounts of work.
The common processes
Four process types cover almost every closed-system FE problem. Constant pressure (isobaric): wb=PΔv and q=Δh=cpΔT
Enthalpy and the open-system balance
Mass flowing across a boundary carries not only its internal energy but also the flow work Pv needed to push it in or out, and the combination u+Pv is defined as the enthalpy h. Bundling flow work into h is what makes the open-system (control-volume) balance compact. For steady state the storage term is zero and, summing over inlets i
Specialized steady-flow devices
Each common device drops most of the steady-flow terms. An adiabatic turbine or compressor with negligible velocity change reduces to w=hi−he (work out positive for a turbine, negative for a compressor). A nozzle or diffuser has no work and no heat, so the enthalpy drop becomes kinetic energy: hi+Vi2/2=he+Ve2/2
Exam strategy
Start every problem by drawing the boundary and labeling it closed or open — that one decision picks the equation. For a closed system write Q−W=ΔU and remember ΔU=mcvΔT for an ideal gas in any process; for an open device write the steady-flow balance and cross out the terms the device kills. Watch the sign convention religiously (work out positive), and in USCS keep the kinetic-energy term honest by dividing
Key equations
Closed-system first lawQ−W=ΔU+ΔKE+ΔPE
Energy balance for a fixed mass. Q positive when added,
Worked examples
Constant-pressure heating of air in a piston-cylinder
Problem. A piston-cylinder holds 0.5kg of air at 200kPa. Heat is added at constant pressure until the temperature rises by 200K. Take R=0.287, cv=0.718
Common pitfalls
•Sign of work: the handbook uses work-by-the-system positive, so a compressor's work is NEGATIVE in Q−W=ΔU. If you write the balance as 'heat in = work in + ΔU' you have flipped a sign — pick one convention and label every term.
•Using Δu=cvΔT
References
NCEES FE Reference Handbook — Thermodynamics (First Law of Thermodynamics; Steady-Flow Systems)
Moran, Shapiro, Boettner & Bailey, Fundamentals of Engineering Thermodynamics — energy balance and control-volume analysis
Cengel & Boles, Thermodynamics: An Engineering Approach — closed- and open-system first law
F. Power cycles
Power, Refrigeration and Heat-Pump Cycles
The second law and Carnot limit, the Rankine, Brayton, Otto and Diesel power cycles, and the vapor-compression refrigeration and heat-pump cycle with its coefficient of performance.
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I. Psychrometrics
Psychrometrics and HVAC Processes
The psychrometric chart and its variables — dry-bulb, wet-bulb, dew point, humidity ratio, relative humidity, and enthalpy — applied to HVAC heating, cooling, humidification, and mixing.
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K. Combustion and combustion products
Gas Mixtures and Combustion
Nonreacting ideal-gas mixtures by Dalton and Amagat, the air-fuel ratio, stoichiometric and excess-air combustion, the products of combustion, and the heating value of a fuel.
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regardless of whether the volume actually stays constant. This is the trap and the gift of the first law:
cv
multiplies
ΔT
to give
Δu
in every ideal-gas process, not just the constant-volume one.
Q−W=ΔU=mcv(T2−T1)(ideal gas)
wb=∫12Pdv
. Constant volume (isochoric):
wb=0
, so
q=Δu=cvΔT
. Constant temperature (isothermal, ideal gas):
Pv
is constant and
wb=RTln(v2/v1)=RTln(P1/P2)
. Reversible adiabatic (isentropic, ideal gas):
Pvk
is constant with
k=cp/cv
,
q=0
, and the work equals
−Δu
. The general polytropic path
Pvn=const
contains all of these as special values of
n
.
wb=1−nP2v2−P1v1(n=1),wb=RTlnv1v2(n=1)
and exits
e
, the rate balance reads as below — heat in and shaft work out plus the enthalpy, kinetic, and potential energy carried by each stream.
. A throttling valve has no work, no heat, and negligible velocity change, giving the isenthalpic result
hi=he
. A boiler or condenser (one side) is
q=he−hi
. Recognizing which device you have tells you instantly which terms survive.
turbine/compressor: w=hi−he;throttle: hi=he
V2/2
by
gc=32.2lbm⋅ft/(lbf⋅s2)
and converting with
778ft⋅lbf/Btu
. A units check on every term — kJ on both sides, or Btu/lbm on both sides — catches most mistakes before they cost you.
W
positive when done by the system; energies in kJ (or Btu).
Ideal-gas internal energy and enthalpy changeΔu=cvΔT,Δh=cpΔT,cp−cv=R
Hold for an ideal gas in any process. cv,cp in kJ/(kg·K); ΔT in K (= ΔT in °C).
Reversible boundary workwb=∫12Pdv
Area under the path on a P–v diagram; per unit mass in kJ/kg. Positive for expansion.
Constant-pressure work and heatwb=P(v2−v1),q=Δh=cpΔT
Isobaric closed process. Heat added equals the enthalpy change for an ideal gas.
Isothermal work (ideal gas)wb=RTlnv1v2=RTlnP2P1
Constant-T ideal gas, Pv=const. Since Δu=0, q=wb
Isentropic process relations (ideal gas)Pvk=const,T1T2=(P1P2)kk−1=(v2v1)k−1
Reversible adiabatic ideal gas, k=cp/cv. q=0, so w=−Δu
Isentropic / polytropic work (ideal gas)w=1−nP2v2−P1v1=1−nR(T2−T1)(n=1)
Use n=k for isentropic, n=1 for isothermal (special form), n=0 for isobaric.
Enthalpy definitionh=u+Pv
Internal energy plus flow work; kJ/kg. Why enthalpy, not internal energy, appears in open systems.
Steady-flow energy equation0=Q˙in−W˙out+∑im˙i(hi+2Vi2+gZi)−∑em˙e(he+2Ve2+gZe)
Open control volume, no storage. m˙ in kg/s, h in J/kg if SI consistent (or use kJ with care).
Adiabatic turbine / compressorw=hi−he
Per unit mass; KE and PE neglected. Turbine w>0 (out), compressor w<0.
Nozzle / diffuser energyhi+2Vi2=he+2Ve2
No work, no heat. Enthalpy converts to kinetic energy. In USCS divide V2/2 by gc and use 778 ft·lbf/Btu.
Throttling processhi=he
Valve/porous plug: no work, no heat, ΔKE≈0. Isenthalpic — basis of the refrigeration expansion valve.
,
cp=1.005kJ/(kg⋅K)
. Find the boundary work and the heat added.
Solution. Constant pressure, so the boundary work is W=mRΔT (using PΔV=mRΔT for an ideal gas):
W=(0.5)(0.287)(200)=28.7kJ.
The internal energy change is ΔU=mcvΔT=(0.5)(0.718)(200)=71.8kJ.
First law: Q=ΔU+W=71.8+28.7=100.5kJ.
Check: for an isobaric process Q should equal ΔH=mcpΔT=(0.5)(1.005)(200)=100.5kJ. The two routes agree, and cp−cv=R is built into the consistency. Final: W=28.7kJ, Q=100.5kJ.
Q=mcpΔT=(0.5)(1.005)(200)=100.5kJ
Adiabatic steam turbine power
Problem. Steam enters an adiabatic turbine at m˙=3kg/s with hi=3248kJ/kg and leaves at he=2378kJ/kg. Velocity and elevation changes are negligible. Find the power produced.
Solution. Steady flow, adiabatic, single stream, KE and PE negligible, so the energy balance reduces to W˙=m˙(hi−he)
W˙=m˙(hi−he)=3(3248−2378)=2610kW
Nozzle exit velocity in USCS (the gc trap)
Problem. Steam enters an adiabatic nozzle at Vi=100ft/s with hi=1300Btu/lbm and expands to he=1200Btu/lbm. Find the exit velocity. Use gc=32.174lbm⋅ft/(lbf⋅s2) and 1Btu=778.17ft⋅lbf.
Solution. Nozzle energy balance: hi+2gcVi2=he+2gcVe2
Ve=Vi2+2gc(778.17)(hi−he)=2240ft/s
only for constant-volume processes. For an ideal gas it is true in EVERY process; likewise
Δh=cpΔT
always holds. The constant-volume label belongs to the work term (
wb=0
), not to
cv
.
•Double-counting flow work. Enthalpy h=u+Pv already contains the Pv flow work, so do not add a separate flow-work term in the open-system balance.
•Forgetting gc and the 778 factor in USCS energy balances. A kinetic-energy term V2/2 in ft²/s² must be divided by gc to get ft·lbf/lbm, then divided by 778 to match Btu/lbm.
•Treating work as a property. Work and heat are path functions — ∫Pdv depends on the path, so you cannot evaluate it from end states alone unless the process type is specified.
•Mixing absolute and gauge pressure in wb=PΔv or in the ideal-gas law. Always use absolute pressure in thermodynamic property relations.
•Neglecting kinetic energy in a nozzle or diffuser. Those are the one device family where the velocity terms are the whole point — keeping only Δh throws away the answer.
.
T
absolute (K or °R).
.
.
W˙=3(3248−2378)=3(870)=2610kW
.
Check: the enthalpy drop is positive (gas does work), so
W˙>0
— consistent with a turbine. Units:
(kg/s)(kJ/kg)=kW
. Final:
W˙=2610kW=2.61MW
.
, with
h
in Btu/lbm converted to ft·lbf/lbm via 778.17.
Solve for