Thermodynamics · Chapter 12

Thermodynamic Property Relations

Property relations turn measurable pressure, volume, and temperature behavior into consistent estimates of energy and entropy. The emphasis is selecting independent state variables, respecting phase limits, and auditing derived data.

Learning objectives

Test a differential for exactness and reconstruct a state property.
Use Maxwell relations to replace inaccessible derivatives with measurable ones.
Apply Clapeyron only along a phase-equilibrium curve.
Select general property relations and real-gas departure data.
Audit throttling, entropy generation, uncertainty, and applicability.

12–1

A Little Math: Partial Derivatives and Associated Relations

State surfaces and exactness

For a simple compressible pure substance, two independent intensive properties locate a single-phase equilibrium state. A total differential states both the direction moved on that surface and which variable is held fixed.

Exact differential test

\[dz=M\,dx+N\,dy,\qquad \left(\frac{\partial M}{\partial y}\right)_x=\left(\frac{\partial N}{\partial x}\right)_y\]

Physical meaning

Exactness means every path between two states produces the same property change. Heat and work fail this test because they describe boundary interactions rather than stored state.

EXAMPLE 1Reconstructing a property from measured slopes

A dimensionless state function z has the measured differential \(dz=(2xy+3)dx+(x^2+2y)dy\), where x and y are dimensionless. Verify exactness and find its dimensionless change from \((1,2)\) to \((2,3)\).

Recognition strategy

Two differential coefficients are supplied, so test equality of cross derivatives before integrating; this distinguishes a state property from a path-dependent quantity.

Given information

The state function z, coefficients, variables x and y, and both endpoint states are dimensionless.
The target is the exact dimensionless state-function change.

Model and interpretation

This is a fixed-mass mathematical state change, not a flow process or cycle. No heat/work balance, pressure, temperature, property table, KE, or PE is involved; exactness makes the path immaterial.

Step-by-step solution

1. Plan. Differentiate the coefficient of \(dx\): \(\partial(2xy+3)/\partial y=2x\), while the derivative of the \(dy\) coefficient with respect to \(x\) is also \(2x\).

2. Formulate. Because the cross derivatives agree throughout the domain, integrate \(2xy+3\) with respect to \(x\) to obtain \(z=x^2y+3x+g(y)\).

3. Calculate. Differentiate that expression with respect to \(y\), compare with \(x^2+2y\), and obtain \(g(y)=y^2+C\).

4. Resolve. Evaluate the endpoint values: \(z_2-C=27\) and \(z_1-C=9\), so the arbitrary reference constant cancels.

Verification

Units: Because x, y, and z are explicitly dimensionless, both coefficients and every term in z are dimensionless; subtracting endpoint values therefore gives a dimensionless change.

Signs and direction: The positive result follows directly from z rising from 9 to 27; reversing the endpoints would change only the algebraic sign and give minus 18.

Plausibility: Direct endpoint subtraction gives 27 minus 9 equals 18, and integrating first in x then in y gives the same value as the reversed rectangular path, confirming path independence.

Final result: \(\Delta z=18\), dimensionless.

Engineering interpretation

Exactness is the decisive verification: any integration path gives the same 18-unit change. Experimental slope uncertainty could break cross-derivative agreement, so release correlations need residual and uncertainty checks.

12–2

The Maxwell Relations

Exact differentials of internal energy, enthalpy, Helmholtz energy, and Gibbs energy produce four derivative identities. For example, \(dg=v\,dP-s\,dT\) gives \((\partial s/\partial P)_T=-(\partial v/\partial T)_P\). These identities apply to equilibrium states of a simple compressible substance, not to arbitrary nonequilibrium paths.

Two frequently useful Maxwell relations

\[\left(\frac{\partial s}{\partial v}\right)_T=\left(\frac{\partial P}{\partial T}\right)_v,\qquad \left(\frac{\partial s}{\partial P}\right)_T=-\left(\frac{\partial v}{\partial T}\right)_P\]

12–3

The Clapeyron Equation

Along coexistence, both phases have equal Gibbs energy. Differentiating that equality gives the exact equilibrium slope below; it must not be applied off the saturation line. Near the critical point, both numerator and denominator demand high-quality data.

Phase-boundary slope

\[\left(\frac{dP}{dT}\right)_{sat}=\frac{h_{fg}}{T v_{fg}}\]

EXAMPLE 2Latent heat from a saturation slope

At 350 K, a supplied saturation dataset gives \(v_{fg}=0.950\ \mathrm{m^3/kg}\) and \(dP_{sat}/dT=3.80\ \mathrm{kPa/K}\). Estimate the latent enthalpy and state the model limits.

Recognition strategy

The cue is phase-equilibrium slope together with a specific-volume jump; Clapeyron directly replaces an unmeasured latent enthalpy with measurable state data.

Given information

Absolute saturation temperature is 350 K; pressure slope and volume change are supplied authoritative dataset values.
The target is latent enthalpy per kilogram at this equilibrium state.

Model and interpretation

Choose a two-phase equilibrium control mass at one saturation state. Clapeyron is exact for equilibrium data; interpolation of the supplied local slope is approximate. No net process, KE, PE, or heat/work sign is needed.

Step-by-step solution

1. Plan. Identify both phases at the same saturation pressure and temperature; quality is not required because only the phase-endpoint difference enters.

2. Formulate. Rearrange the Clapeyron equation to \(h_{fg}=T v_{fg}(dP_{sat}/dT)\), retaining absolute temperature.

3. Calculate. Substitute \(350\times0.950\times3.80=1263.5\); \(\mathrm{kPa\!\cdot\!m^3/kg}=\mathrm{kJ/kg}\).

4. Resolve. Round consistently with the supplied data and note that a finite-difference saturation slope would add discretization uncertainty.

Verification

Units: The product K times cubic metres per kilogram times kilopascals per kelvin reduces to kilopascal cubic metres per kilogram, exactly kilojoules per kilogram.

Signs and direction: The supplied saturation slope and vapor-minus-liquid volume are both positive, so Clapeyron requires positive vaporization enthalpy rather than a negative latent heat.

Plausibility: Back-substitution gives \(1263.5/[350(0.950)]=3.80\ \text{kPa/K}\), exactly recovering the supplied equilibrium-curve slope.

Final result: \(h_{fg}\approx1.26\times10^3\ \mathrm{kJ/kg}\).

Engineering interpretation

The positive saturation slope and volume increase require positive vaporization enthalpy. For release work, the local slope and phase properties must come from the same authoritative equation of state.

12–4

General Relations for du, dh, ds, cv, and cp

General simple-compressible relations

\[du=c_v\,dT+\left[T\left(\frac{\partial P}{\partial T}\right)_v-P\right]dv\]\[dh=c_p\,dT+\left[v-T\left(\frac{\partial v}{\partial T}\right)_P\right]dP\]

These are property relations, not energy balances. For an ideal gas the bracket in \(du\) vanishes and \(u=u(T)\); similarly \(h=h(T)\). A real fluid generally retains pressure or volume dependence. Use \(u\) for stored microscopic energy in a closed mass and \(h=u+Pv\) for flowing streams.

12–5

The Joule–Thomson Coefficient

Isenthalpic temperature response

\[\mu_{JT}=\left(\frac{\partial T}{\partial P}\right)_h=\frac{T(\partial v/\partial T)_P-v}{c_p}\]

A positive coefficient combined with a pressure decrease predicts cooling. The coefficient varies with state, so a large drop requires integration or iteration with an equation of state. An ideal gas has zero coefficient; a real gas can cross an inversion curve and warm instead.

EXAMPLE 3Real-gas throttle and exergy audit

A steady insulated valve reduces a real gas from 12.0 MPa to 2.0 MPa. A validated property package supplies \(\mu_{JT}=0.220\ \mathrm{K/MPa}\), \(\Delta s=0.0120\ \mathrm{kJ/(kg\!\cdot\!K)}\), and \(\dot m=0.800\ \mathrm{kg/s}\). Estimate outlet temperature change and exergy destruction at \(T_0=298\ \mathrm{K}\).

Recognition strategy

A valve with a large pressure drop signals isenthalpic throttling, not turbine work. The supplied Joule–Thomson coefficient predicts direction, while entropy quantifies irreversibility.

Given information

All pressures are absolute; the inlet temperature range is covered by the supplied local coefficient.
Mass flow, entropy rise, and reference environment are explicit validated-property givens.

Model and interpretation

The valve is a steady open control volume, adiabatic, with no shaft work and negligible KE and PE changes. Thus \(h_2=h_1\); the local constant-coefficient temperature estimate is approximate, while entropy and exergy audits use supplied real-gas states.

Step-by-step solution

1. Plan. Apply the local isenthalpic relation \(\Delta T\approx\mu_{JT}(P_2-P_1)\); this is not an isentropic or reversible expansion.

2. Formulate. Compute \(P_2-P_1=-10.0\ \mathrm{MPa}\), giving \(\Delta T=0.220(-10.0)=-2.20\ \mathrm{K}\).

3. Calculate. For an adiabatic steady valve, the entropy balance gives \(\dot S_{gen}=\dot m(s_2-s_1)=0.00960\ \mathrm{kW/K}\).

4. Resolve. Use Gouy–Stodola: \(\dot X_d=T_0\dot S_{gen}=298(0.00960)=2.86\ \mathrm{kW}\), which is nonnegative.

Verification

Units: K/MPa times MPa gives kelvins; kg/s times kJ/(kg K) gives kW/K; multiplying that entropy rate by the 298 K environment gives kW.

Signs and direction: Pressure falls by 10.0 MPa, so positive mu_JT produces a negative 2.20 K temperature change; positive entropy rise produces positive exergy destruction.

Plausibility: The steady adiabatic valve closes energy with h2=h1, while Sgen=0.00960 kW/K and Xd=2.86 kW are nonnegative as required by the second law.

Final result: \(\Delta T\approx-2.20\ \mathrm{K}\); \(\dot X_d=2.86\ \mathrm{kW}\).

Engineering interpretation

Cooling is modest while useful-work potential is destroyed by irreversibility. A final valve selection needs full inlet/outlet property iteration, two-phase screening, vendor capacity data, noise review, and uncertainty analysis.

12–6

The Δh, Δu, and Δs of Real Gases

Departure functions compare a real-fluid property with the ideal-gas value at the same absolute temperature and pressure. Keep the chart or software sign convention explicit, interpolate in reduced coordinates only within range, and use consistent critical-property and acentric-factor data. Phase identification comes first; quality is meaningful only inside a saturated two-phase region.

Departure bookkeeping

\[\Delta h=\Delta h^{ig}+h_2^R-h_1^R,\qquad \Delta s=\Delta s^{ig}+s_2^R-s_1^R\]

Problem-solving strategy

Recognize the property-relation problem family and identify the requested property or performance measure.
Draw a closed-system or control-volume boundary and label every state with independent properties.
Choose a property source and model before selecting equations; interpolate only between valid tabulated states.
Write conservation and constitutive relations symbolically, count unknowns, and iterate when properties depend on the answer.
Audit dimensions, signs, limiting behavior, conservation, entropy generation, and model applicability.

Chapter recap

Choose two independent state properties and verify phase before differentiating or interpolating.
Use exactness and Maxwell identities to connect measurable and inaccessible derivatives.
Apply Clapeyron only on coexistence and use real-gas departures consistently.
Before moving on, be able to declare boundaries, sources, signs, approximations, and entropy/exergy audits.

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