Thermodynamics and Heat Transfer

A. Thermodynamic laws

The First Law and Energy Balances

Conservation of energy for closed and open systems: internal energy, enthalpy, the steady-flow energy equation, and the heat/work sign conventions every thermo problem rides on.

Almost every thermodynamics question on the FE Other Disciplines exam is, at heart, one bookkeeping statement applied carefully: energy is conserved. The handbook's Thermodynamics chapter opens the First Law as "the net energy crossing the system boundary equals the change in energy inside the system." Where points are lost is never the idea — it is the sign of the work term, whether you should use internal energy

u

or enthalpy

h

, and whether mass is allowed to cross the boundary. Get those three decisions right and the algebra is short.

Heat, work, and the sign convention

Heat

Q

is energy that crosses the boundary because of a temperature difference; work

W

is every other organized energy transfer. The handbook fixes the classical engineering convention: heat added TO the system is positive, and work done BY the system is positive. That is why the closed-system law reads

Q - W

and not

Q + W

— a turbine doing work on its surroundings has

W>0

, draining the system's energy. Specific (per-unit-mass) quantities are lowercase:

q = Q/m

,

w = W/m

.

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

Internal energy versus enthalpy

Internal energy

u

is the microscopic energy stored in a substance. Enthalpy bundles internal energy with the flow-work group

Pv

:

h = u + Pv

. The reason enthalpy exists is the open system — fluid pushing its way across a boundary carries both its internal energy and the

Pv

work needed to shove the upstream fluid out of the way. For an ideal gas with constant specific heats, both changes collapse to a temperature change, which is the workhorse relation for gas problems.

h = u + Pv, \qquad \Delta u = c_v\,\Delta T, \qquad \Delta h = c_p\,\Delta T

Closed systems and boundary work

When no mass crosses the boundary — a sealed piston-cylinder, a rigid tank — the system is closed and

Q - W = \Delta U

(kinetic and potential changes are usually negligible). The only work is reversible boundary (moving-boundary) work

w_b = \int P\,dv

. For a constant-pressure process this is simply

w_b = P\,\Delta v

; for a rigid (constant-volume) tank the boundary work is zero, so all heat goes into internal energy.

W_b = \int_{1}^{2} P\,dV

Open systems: mass and energy balances together

An open system (control volume) lets mass cross the boundary, so you must satisfy TWO balances. Mass in equals mass out plus storage; at steady state,

\sum \dot m_i = \sum \dot m_e

. Energy then balances the heat and shaft-work rates against the enthalpy, kinetic, and potential energy each stream carries. Note that enthalpy — not internal energy — appears, because the

Pv

flow work is already folded in.

\dot Q_{in} - \dot W_{net} = \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)

The steady-flow energy equation in practice

For a single inlet and single outlet at steady state, dividing the energy balance by

\dot m

gives the steady-flow energy equation (SFEE). Most devices then drop terms: a turbine or compressor is adiabatic with negligible

KE

/

PE

, so

w = h_i - h_e

; a throttling valve has no work and no heat, so

h_i = h_e

; a nozzle has no work or heat but significant velocity, so

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

; a boiler or condenser has no work, so

q = h_e - h_i

.

q - w = (h_e - h_i) + \frac{V_e^2 - V_i^2}{2} + g(Z_e - Z_i)

Why the device dictates which terms survive

The exam rarely asks you to keep all five terms. The skill is recognizing the device and crossing out what is negligible. Heat exchangers, throttles, turbines, pumps, nozzles, and diffusers each have a signature reduced form (the handbook tabulates them under "Special Cases of the Steady-Flow Energy Equation"). Memorize the reductions, and a paragraph-long problem becomes a one-line subtraction of enthalpies.

Exam strategy

First decide: closed (use

u

,

Q-W=\Delta U

) or open (use

h

, write the SFEE)? Then identify the device and kill the negligible terms before plugging in numbers. Watch units: enthalpies from steam tables are in

\text{kJ/kg}

, so a velocity term

V^2/2

in

\text{m}^2/\text{s}^2

must be divided by

1000

to become

\text{kJ/kg}

. Keep the sign convention straight — if your turbine power comes out negative, you almost certainly flipped inlet and exit.

Key equations

Closed-system first law

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

No mass crosses the boundary. $Q$ positive into system, $W$ positive out (done by system); usually $\Delta KE=\Delta PE=0$ . Energy in $\text{kJ}$ .

Definition of enthalpy

H = U + PV, \qquad h = u + Pv

Enthalpy combines internal energy with flow work $Pv$ ; $h$ in $\text{kJ/kg}$ , $P$ in $\text{kPa}$ , $v$ in $\text{m}^3/\text{kg}$ .

Ideal-gas energy changes

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

Constant specific heats (cold-air-standard). $c_v,c_p$ in $\text{kJ/(kg·K)}$ ; valid for ideal gases regardless of process path.

Boundary (moving-boundary) work

W_b = \int_{1}^{2} P\,dV

Reversible work of a moving boundary in a closed system; equals area under the process curve on a $P$ - $V$ diagram.

Constant-pressure boundary work

w_b = P\,\Delta v = R\,\Delta T

Charles' law process for an ideal gas; the second form uses $Pv = RT$ . $R$ is the specific gas constant in $\text{kJ/(kg·K)}$ .

Steady-flow mass balance

\sum \dot m_i = \sum \dot m_e, \qquad \dot m = \rho A V = \frac{AV}{v}

Conservation of mass for a control volume at steady state; $\dot m$ in $\text{kg/s}$ , $A$ area, $V$ velocity, $v$ specific volume.

General SFEE (per unit mass)

q - w = (h_e - h_i) + \frac{V_e^2 - V_i^2}{2} + g(Z_e - Z_i)

Single-inlet/single-exit steady flow. Divide velocity and gravity terms by $1000$ to convert $\text{J/kg}$ to $\text{kJ/kg}$ before adding to enthalpy.

Adiabatic turbine / compressor

w = h_i - h_e

Negligible heat, $KE$ , $PE$ . $w>0$ for a turbine (work out); $w<0$ for a compressor (work in).

Throttling valve

h_i = h_e

No work, no heat, negligible $KE$ ; enthalpy is conserved (isenthalpic). Used for expansion valves and the Joule-Thomson effect.

Adiabatic nozzle

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

No work or heat; enthalpy converts to kinetic energy. Solve for exit velocity $V_e=\sqrt{2(h_i-h_e)}$ with $h$ in $\text{J/kg}$ .

Heat exchanger / boiler (one stream)

\dot Q = \dot m\,(h_e - h_i)

No work; heat added equals enthalpy rise of the stream. Positive $\dot Q$ heats the fluid.

Worked examples

Constant-pressure heating of air (closed system)

Problem. A piston-cylinder holds $2\,\text{kg}$ of air at constant pressure while it is heated from $300\,\text{K}$ to $500\,\text{K}$ . Take $c_p = 1.005$ and $c_v = 0.718\,\text{kJ/(kg·K)}$ , $R = 0.287\,\text{kJ/(kg·K)}$ . Find the heat added, the boundary work, and the change in internal energy.

Solution. Internal energy change:

\Delta U = m c_v \Delta T = 2(0.718)(200) = 287\,\text{kJ}

. Boundary work (constant

P

):

W = mR\,\Delta T = 2(0.287)(200) = 114.8\,\text{kJ}

. First law:

Q = \Delta U + W = 287.2 + 114.8 = 402\,\text{kJ}

. Equivalently

Q=\Delta H = m c_p \Delta T = 2(1.005)(200)=402\,\text{kJ}

. Answers:

Q = 402\,\text{kJ}

,

W = 115\,\text{kJ}

,

\Delta U = 287\,\text{kJ}

. Sanity check:

Q-W = 402-114.8 = 287.2\,\text{kJ} = \Delta U

. \checkmark

Q = \Delta U + W = m c_v \Delta T + m R \Delta T = m c_p \Delta T

Adiabatic steam turbine (open system)

Problem. Steam enters an adiabatic turbine at $\dot m = 5\,\text{kg/s}$ with $h_i = 3248\,\text{kJ/kg}$ and leaves with $h_e = 2360\,\text{kJ/kg}$ . Kinetic and potential changes are negligible. Find the power output.

Solution. Adiabatic, negligible

KE

/

PE

:

w = h_i - h_e = 3248 - 2360 = 888\,\text{kJ/kg}

. Power:

\dot W = \dot m\, w = 5(888) = 4440\,\text{kW} = 4.44\,\text{MW}

. Answer:

\dot W = 4.44\,\text{MW}

out. Sanity check: enthalpy drops across a turbine, so

w>0

— work leaves the system, as expected. \checkmark

Steam nozzle exit velocity

Problem. Steam enters a well-insulated nozzle with negligible velocity at $h_i = 3000\,\text{kJ/kg}$ and exits at $h_e = 2762\,\text{kJ/kg}$ . Find the exit velocity.

Solution. Nozzle: no work, no heat,

V_i \approx 0

, so

h_i = h_e + V_e^2/2

.

V_e = \sqrt{2(h_i - h_e)} = \sqrt{2(3000-2762)\times 10^3\,\text{J/kg}} = \sqrt{4.76\times 10^5}

.

V_e = 690\,\text{m/s}

. Sanity check: a

238\,\text{kJ/kg}

enthalpy drop into kinetic energy gives a near-sonic-to-supersonic steam speed — entirely reasonable for a converging-diverging nozzle. \checkmark

V_e = \sqrt{2(h_i - h_e)} = \sqrt{2(238{,}000)} \approx 690\ \text{m/s}

Common pitfalls

•Sign of work: the closed-system law is $Q - W = \Delta U$ . Work done BY the system is positive; if you write $Q + W$ you have silently switched to the physics (work-done-on) convention and will get the wrong sign.
•Using $u$ in an open-system balance. Flowing streams carry enthalpy $h = u + Pv$ ; substituting internal energy omits the flow work and undercounts the energy.
•Forgetting the $1000$ factor on the kinetic term. $V^2/2$ is in $\text{J/kg}$ but tabulated $h$ is in $\text{kJ/kg}$ — divide $V^2/2$ by $1000$ before combining.
•Applying $\Delta u = c_v\Delta T$ to a liquid or two-phase mixture. That relation is for ideal gases only; for water use the steam tables.
•Setting boundary work to $P\Delta V$ for a rigid tank. A constant-volume process has $\Delta V = 0$ , so $W_b = 0$ — all the heat becomes internal energy.
•Confusing $c_p$ and $c_v$ : use $c_p$ for enthalpy and constant-pressure problems, $c_v$ for internal energy and constant-volume problems. Mixing them is a classic trap.
•Treating a throttle as isentropic. A throttle is isenthalpic ( $h_i=h_e$ ), not isentropic — entropy actually increases.

References

NCEES FE Reference Handbook — Thermodynamics (First Law, Closed and Open Systems)
NCEES FE Reference Handbook — Thermodynamics (Special Cases of the Steady-Flow Energy Equation)
Moran, Shapiro et al., Fundamentals of Engineering Thermodynamics — control-volume energy balance background

The Second Law, Entropy, and Carnot

Why energy has direction: entropy and entropy generation, reversibility, Carnot limits, heat-engine/heat-pump/refrigerator efficiencies, COP, and isentropic processes.