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

Exam areas

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 = \Delta U + \Delta KE + \Delta PE

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=\Delta U

. For an ideal gas the internal energy depends only on temperature, so

\Delta U=m c_v\,\Delta T

regardless of whether the volume actually stays constant. This is the trap and the gift of the first law:

c_v

multiplies

\Delta T

to give

\Delta u

in every ideal-gas process, not just the constant-volume one.

Q - W = \Delta U = m\,c_v\,(T_2 - T_1)\quad(\text{ideal gas})

Boundary work and the P–v diagram

When a system boundary moves, it does reversible boundary work

w_b=\int P\,dv

— 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.

w_b = \int_1^2 P\,dv

The common processes

Four process types cover almost every closed-system FE problem. Constant pressure (isobaric):

w_b=P\,\Delta v

and

q=\Delta h=c_p\,\Delta T

. Constant volume (isochoric):

w_b=0

, so

q=\Delta u=c_v\,\Delta T

. Constant temperature (isothermal, ideal gas):

Pv

is constant and

w_b=RT\ln(v_2/v_1)=RT\ln(P_1/P_2)

. Reversible adiabatic (isentropic, ideal gas):

Pv^{k}

is constant with

k=c_p/c_v

q=0

, and the work equals

-\Delta u

. The general polytropic path

Pv^{n}=\text{const}

contains all of these as special values of

n

w_b = \dfrac{P_2 v_2 - P_1 v_1}{1-n}\ (n\neq 1),\qquad w_b = RT\ln\!\dfrac{v_2}{v_1}\ (n=1)

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

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.

\dot{Q}_{in} - \dot{W}_{out} = \sum_e \dot{m}_e\!\left(h_e + \tfrac{V_e^2}{2} + gZ_e\right) - \sum_i \dot{m}_i\!\left(h_i + \tfrac{V_i^2}{2} + gZ_i\right)

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=h_i-h_e

(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:

h_i+V_i^2/2=h_e+V_e^2/2

. A throttling valve has no work, no heat, and negligible velocity change, giving the isenthalpic result

h_i=h_e

. A boiler or condenser (one side) is

q=h_e-h_i

. Recognizing which device you have tells you instantly which terms survive.

\text{turbine/compressor: } w = h_i - h_e;\qquad \text{throttle: } h_i = h_e

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=\Delta U

and remember

\Delta U=mc_v\Delta T

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

V^2/2

g_c=32.2\,\text{lbm·ft/(lbf·s}^2)

and converting with

778\,\text{ft·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.

Key equations

Closed-system first law

Q - W = \Delta U + \Delta KE + \Delta PE

Energy balance for a fixed mass. $Q$ positive when added, $W$ positive when done by the system; energies in kJ (or Btu).

Ideal-gas internal energy and enthalpy change

\Delta u = c_v\,\Delta T,\qquad \Delta h = c_p\,\Delta T,\qquad c_p - c_v = R

Hold for an ideal gas in any process. $c_v,c_p$ in kJ/(kg·K); $\Delta T$ in K (= $\Delta T$ in °C).

Reversible boundary work

w_b = \int_1^2 P\,dv

Area under the path on a $P$ – $v$ diagram; per unit mass in kJ/kg. Positive for expansion.

Constant-pressure work and heat

w_b = P\,(v_2 - v_1),\qquad q = \Delta h = c_p\,\Delta T

Isobaric closed process. Heat added equals the enthalpy change for an ideal gas.

Isothermal work (ideal gas)

w_b = RT\ln\!\dfrac{v_2}{v_1} = RT\ln\!\dfrac{P_1}{P_2}

Constant- $T$ ideal gas, $Pv=\text{const}$ . Since $\Delta u=0$ , $q=w_b$ . $T$ absolute (K or °R).

Isentropic process relations (ideal gas)

Pv^{k}=\text{const},\quad \dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^{\frac{k-1}{k}}=\left(\dfrac{v_1}{v_2}\right)^{k-1}

Reversible adiabatic ideal gas, $k=c_p/c_v$ . $q=0$ , so $w=-\Delta u$ .

Isentropic / polytropic work (ideal gas)

w = \dfrac{P_2 v_2 - P_1 v_1}{1-n} = \dfrac{R(T_2-T_1)}{1-n}\ \ (n\neq 1)

Use $n=k$ for isentropic, $n=1$ for isothermal (special form), $n=0$ for isobaric.

Enthalpy definition

h = u + Pv

Internal energy plus flow work; kJ/kg. Why enthalpy, not internal energy, appears in open systems.

Steady-flow energy equation

0 = \dot{Q}_{in} - \dot{W}_{out} + \sum_i \dot{m}_i\!\left(h_i + \tfrac{V_i^2}{2} + gZ_i\right) - \sum_e \dot{m}_e\!\left(h_e + \tfrac{V_e^2}{2} + gZ_e\right)

Open control volume, no storage. $\dot{m}$ in kg/s, $h$ in J/kg if SI consistent (or use kJ with care).

Adiabatic turbine / compressor

w = h_i - h_e

Per unit mass; KE and PE neglected. Turbine $w>0$ (out), compressor $w<0$ .

Nozzle / diffuser energy

h_i + \dfrac{V_i^2}{2} = h_e + \dfrac{V_e^2}{2}

No work, no heat. Enthalpy converts to kinetic energy. In USCS divide $V^2/2$ by $g_c$ and use 778 ft·lbf/Btu.

Throttling process

h_i = h_e

Valve/porous plug: no work, no heat, $\Delta KE\approx 0$ . Isenthalpic — basis of the refrigeration expansion valve.

Worked examples

Constant-pressure heating of air in a piston-cylinder

Problem. A piston-cylinder holds $0.5\,\text{kg}$ of air at $200\,\text{kPa}$ . Heat is added at constant pressure until the temperature rises by $200\,\text{K}$ . Take $R=0.287$ , $c_v=0.718$ , $c_p=1.005\ \text{kJ/(kg·K)}$ . Find the boundary work and the heat added.

Solution. Constant pressure, so the boundary work is

W=mR\,\Delta T

(using

P\,\Delta V=mR\,\Delta T

for an ideal gas):

W = (0.5)(0.287)(200) = 28.7\,\text{kJ}

. The internal energy change is

\Delta U = m c_v\,\Delta T = (0.5)(0.718)(200) = 71.8\,\text{kJ}

. First law:

Q = \Delta U + W = 71.8 + 28.7 = 100.5\,\text{kJ}

. Check: for an isobaric process

Q

should equal

\Delta H = m c_p\,\Delta T = (0.5)(1.005)(200) = 100.5\,\text{kJ}

. The two routes agree, and

c_p-c_v=R

is built into the consistency. Final:

W = 28.7\,\text{kJ}

Q = 100.5\,\text{kJ}

Q = mc_p\,\Delta T = (0.5)(1.005)(200) = 100.5\ \text{kJ}

Adiabatic steam turbine power

Problem. Steam enters an adiabatic turbine at $\dot{m}=3\,\text{kg/s}$ with $h_i=3248\,\text{kJ/kg}$ and leaves at $h_e=2378\,\text{kJ/kg}$ . 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

\dot{W}=\dot{m}(h_i-h_e)

\dot{W} = 3\,(3248 - 2378) = 3\,(870) = 2610\,\text{kW}

. Check: the enthalpy drop is positive (gas does work), so

\dot{W}>0

— consistent with a turbine. Units:

(\text{kg/s})(\text{kJ/kg}) = \text{kW}

. Final:

\dot{W} = 2610\,\text{kW} = 2.61\,\text{MW}

\dot{W} = \dot{m}\,(h_i - h_e) = 3(3248-2378) = 2610\ \text{kW}

Nozzle exit velocity in USCS (the gc trap)

Problem. Steam enters an adiabatic nozzle at $V_i=100\,\text{ft/s}$ with $h_i=1300\,\text{Btu/lbm}$ and expands to $h_e=1200\,\text{Btu/lbm}$ . Find the exit velocity. Use $g_c=32.174\,\text{lbm·ft/(lbf·s}^2)$ and $1\,\text{Btu}=778.17\,\text{ft·lbf}$ .

Solution. Nozzle energy balance:

h_i+\dfrac{V_i^2}{2g_c}=h_e+\dfrac{V_e^2}{2g_c}

, with

h

in Btu/lbm converted to ft·lbf/lbm via 778.17. Solve for

V_e

V_e^2 = V_i^2 + 2 g_c (778.17)(h_i-h_e)

V_e^2 = (100)^2 + 2(32.174)(778.17)(1300-1200) = 10{,}000 + 5{,}007{,}368 = 5{,}017{,}368\ \text{ft}^2/\text{s}^2

V_e = \sqrt{5{,}017{,}368} = 2240\,\text{ft/s}

. Check: dropping

100\,\text{Btu/lbm}

is a large enthalpy drop, so a near-sonic-scale velocity is reasonable; the inlet KE (

10{,}000

) is tiny against the converted enthalpy (

5.0\times10^6

), confirming why nozzle inlet velocity is often ignored. Final:

V_e = 2240\,\text{ft/s}

V_e = \sqrt{V_i^2 + 2 g_c\,(778.17)(h_i-h_e)} = 2240\ \text{ft/s}

Common pitfalls

•Sign of work: the handbook uses work-by-the-system positive, so a compressor's work is NEGATIVE in $Q-W=\Delta U$ . If you write the balance as 'heat in = work in + $\Delta U$ ' you have flipped a sign — pick one convention and label every term.
•Using $\Delta u=c_v\,\Delta T$ only for constant-volume processes. For an ideal gas it is true in EVERY process; likewise $\Delta h=c_p\,\Delta T$ always holds. The constant-volume label belongs to the work term ( $w_b=0$ ), not to $c_v$ .
•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 $g_c$ and the 778 factor in USCS energy balances. A kinetic-energy term $V^2/2$ in ft²/s² must be divided by $g_c$ to get ft·lbf/lbm, then divided by 778 to match Btu/lbm.
•Treating work as a property. Work and heat are path functions — $\int P\,dv$ 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 $w_b=P\,\Delta v$ 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 $\Delta h$ throws away the answer.

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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