14.2 Thermodynamic Processes and Property Diagrams (T-s, P-h, P-v)

14.2 Thermodynamic Processes and Property Diagrams (T-s, P-h, P-v)

NCEES lists "Thermodynamic processes (e.g., isothermal, adiabatic, reversible, irreversible)" and "Property and phase diagrams (e.g., T-s, P-h, P-v)" as two sub-topics. They belong together, because a process is a path and a diagram is where you read that path. On the exam these appear as "which diagram would you use" and "what does this area represent" items — cheap points if you know the two area rules.

The Named Processes

The Polytropic Family

$Pv^n = \text{constant}$

Ideal-gas relationships along a polytropic path:

$\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{(n-1)/n} = \left(\frac{v_1}{v_2}\right)^{n-1}$

Boundary work for a closed system:

$W = \int_1^2 P\,dv = \frac{P_2v_2 - P_1v_1}{1-n} \quad (n\ne1), \qquad W = P_1v_1\ln\frac{v_2}{v_1} \quad (n=1)$

Adiabatic vs. Isentropic — Not Synonyms

Every isentropic process is adiabatic, but not every adiabatic process is isentropic. A throttling valve is the definitive counterexample: it is adiabatic ($Q=0$) yet strongly irreversible, so entropy increases substantially. On a $T$-$s$ diagram, throttling moves to the right, not vertically. An exam option treating "adiabatic" as licence to use $Pv^k = $ const across a valve is wrong.

The Two Area Rules

These are the highest-yield facts in the topic:

$\boxed{\text{Area under a reversible path on } P\text{-}v = \text{boundary work} \; \textstyle\int P\,dv}$

$\boxed{\text{Area under a reversible path on } T\text{-}s = \text{heat transferred} \; \textstyle\int T\,ds}$

So on a cycle: the enclosed area on $P$-$v$ is the net work, and the enclosed area on $T$-$s$ is the net heat — and by the first law applied to a cycle, they are equal. A clockwise cycle produces net work (a power cycle); a counterclockwise cycle consumes it (refrigeration or heat pump).

Reading the Vapor Dome

Every property diagram for a pure substance has the same landmarks:

The essential trap: inside the dome, $P$ and $T$ are not independent — they are locked together by the saturation relationship. Knowing that steam is at 200 kPa and 120 °C tells you nothing more than 200 kPa alone, and you cannot fix the state. You need quality:

$x = \frac{m_{\text{vapor}}}{m_{\text{total}}}, \qquad y = y_f + x(y_g - y_f) = y_f + x\,y_{fg}$

which applies to $v$, $u$, $h$, and $s$ alike.

Which Diagram for Which Job

Why Refrigeration Engineers Use $P$-$h$

The four processes of the ideal vapor-compression cycle plot as:

1. Compressor: isentropic — a curve sloping up and to the right
2. Condenser: isobaric heat rejection — a horizontal line leftward
3. Expansion valve: isenthalpic throttling — a vertical line downward
4. Evaporator: isobaric heat absorption — a horizontal line rightward

Three of four are straight, and the two horizontal lines let you read $q_{\text{evap}}$ and $q_{\text{cond}}$ as simple horizontal distances. The coefficient of performance becomes a ratio of two lengths on the chart:

$COP_R = \frac{q_L}{w_{\text{in}}} = \frac{h_1-h_4}{h_2-h_1}$

Worked Example: Reading a State and a Process

Steam at 1.0 MPa has $h_f = 762.5$ kJ/kg, $h_g = 2{,}778.1$ kJ/kg, $s_f = 2.1381$ kJ/kg·K, $s_g = 6.5850$ kJ/kg·K. A wet mixture at this pressure has quality $x = 0.85$. Find $h$ and $s$, then determine the ideal work if it expands isentropically to 20 kPa where $s_f = 0.8320$ and $s_g = 7.9085$ kJ/kg·K, with $h_f = 251.4$ and $h_g = 2{,}609.7$ kJ/kg.

State 1: $h_1 = 762.5 + 0.85(2{,}778.1 - 762.5) = 762.5 + 0.85(2{,}015.6) = 762.5 + 1{,}713.3 = 2{,}475.8\ \text{kJ/kg}$ $s_1 = 2.1381 + 0.85(6.5850-2.1381) = 2.1381 + 0.85(4.4469) = 2.1381 + 3.7799 = 5.9180\ \text{kJ/kg}\cdot\text{K}$

State 2 — isentropic, so $s_2 = s_1 = 5.9180$ kJ/kg·K. Solve for quality at 20 kPa:

$x_2 = \frac{s_2 - s_f}{s_g - s_f} = \frac{5.9180 - 0.8320}{7.9085 - 0.8320} = \frac{5.0860}{7.0765} = 0.7187$

$h_2 = 251.4 + 0.7187(2{,}609.7 - 251.4) = 251.4 + 0.7187(2{,}358.3) = 251.4 + 1{,}694.9 = 1{,}946.3\ \text{kJ/kg}$

Ideal turbine work: $w = h_1 - h_2 = 2{,}475.8 - 1{,}946.3 = \boxed{529.5\ \text{kJ/kg}}$

Note the structure of that solution. On the $T$-$s$ diagram this is a vertical line downward from 1.0 MPa to 20 kPa, both endpoints inside the dome. The constant-entropy condition is what determines $x_2$, and $x_2 = 0.72$ means the steam got wetter during expansion — a real concern, since liquid droplets erode turbine blades and expansion below about 88% quality is avoided by reheating. Candidates who assume quality is unchanged at 0.85 get $h_2 = 2{,}256$ kJ/kg and a work of 220 kJ/kg, low by 58%.