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Thermodynamic properties and using the tables

To solve an energy problem (a turbine, a refrigerator, a boiler) you need the properties of the fluid at each state: pressure, temperature, volume, internal energy, enthalpy and entropy. This topic explains how those properties are organized, how to read the tables, how to interpolate and when the ideal-gas model holds.

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Pure substance, phases and state

A pure substance has the same chemical composition throughout, even in several phases (liquid water with vapor). Its state is fixed by two independent intensive properties (state postulate). Inside the saturation dome $T$ and $P$ are not independent: each pressure has its saturation temperature, so another property is needed, usually the quality $x$ , the mass fraction that is vapor.

T–v diagram (schematic): at constant pressure the liquid heats up, boils at constant T inside the dome and then the vapor is superheated. Above the critical point there is no phase change.

The usual properties are the specific volume $v$ , the internal energy $u$ , the enthalpy $h = u + P v$ (the one that appears in devices with flow) and the entropy $s$ . All are given per unit mass, and their absolute values depend on a reference state: what has physical meaning is their differences.

Saturation tables and quality

The saturation table (by temperature or by pressure) gives the properties of saturated liquid (subscript f) and saturated vapor (g), and their difference ( $f g$ ). A mixture is an average weighted by quality:

y = y_{f} + x (y_{g} - y_{f}) = y_{f} + x y_{f g} x = \frac{y - y _{f}}{y _{f g}} (y = v, u, h, s)

The second formula works the other way: if $P$ (or $T$ ) and another property are known, the quality is solved from it. If $x$ comes out below 0 it is compressed liquid; above 1, superheated vapor.

How to tell which region the state is in

Data	Compare with	Region
$T$ and $P$	$P_{s a t} (T)$	$P > P_{s a t}$ : compressed liquid · $P < P_{s a t}$ : superheated vapor
$T$ (or $P$ ) and $v$ , $u$ , $h$ or $s$	$y_{f}$ and $y_{g}$ at that $T$ (or $P$ )	$y < y_{f}$ : liquid · $y_{f} \leq y \leq y_{g}$ : mixture · $y > y_{g}$ : vapor
$T$ or $P$ with $x$	—	Mixture (with $0 \leq x \leq 1$ )

Superheated vapor and interpolation

The superheated vapor table has one block per pressure and one row per temperature. If the data falls between two rows it is interpolated linearly; if the pressure also falls between two blocks, it is interpolated in each block and then between the two results (double interpolation):

y = y_{1} + \frac{x - x _{1}}{x _{2} - x _{1}} (y_{2} - y_{1})

Interpolating assumes the property varies as a straight line between the two rows. For $h$ and $s$ the error is usually small; for $v$ , which goes almost as $1/ P$ , it can reach several tenths of a percent between widely spaced pressure blocks. The simulator gives both values: the interpolated one (what is done by hand) and the exact one from the equation of state.

Compressed liquid

Liquids are nearly incompressible: their properties depend strongly on $T$ and weakly on $P$ . When there is no compressed-liquid table, the saturated-liquid approximation at the same temperature is used, with a correction for $h$ :

v \approx v_{f} (T) u \approx u_{f} (T) s \approx s_{f} (T) h \approx h_{f} (T) + v_{f} (T) [P - P_{s a t} (T)]

Ideal gas

At low pressure and high temperature relative to the critical point, a gas follows $P v = RT$ , with $R = R_{u} / M$ ( $R_{u} = 8.314$ kJ/kmol·K). The compressibility factor $Z = P v / RT$ measures how far it departs: near 1, the model holds. In an ideal gas $u$ and $h$ depend only on $T$ , and the entropy depends on $T$ and $P$ :

h_{2} - h_{1} = \int_{T_{1}}^{T_{2}} c_{p} d T s_{2} - s_{1} = s^{\circ} (T_{2}) - s^{\circ} (T_{1}) - R ln \frac{P _{2}}{P _{1}}

Ideal-gas tables integrate $c_{p} (T)$ for you: they give $h$ , $u$ and $s^{\circ}$ as functions of $T$ . For an isentropic process the relative pressure and volume are used: $P_{2} / P_{1} = P_{r 2} / P_{r 1}$ and $v_{2} / v_{1} = v_{r 2} / v_{r 1}$ . With constant $c_{p}$ this reduces to $T_{2} / T_{1} = (P_{2} / P_{1})^{(k - 1) / k}$ , convenient but less exact when the temperature changes a lot.

The values in this topic and in the tool are computed by Simuveliq with CoolProp 8.0.0 (MIT license; Bell et al., Ind. Eng. Chem. Res. 53, 2014), with the equation of state of each fluid cited in the tool (for water, IAPWS-95 by Wagner and Pruß, 2002). They may differ in the last decimal from a textbook's tables, which use other references or roundings.

Common mistakes

Using the superheated table without first checking the region: with $T$ and $P$ you must compare with $P_{s a t} (T)$ .
Interpolating with a quality outside 0–1, or computing $x$ with $y_{g}$ instead of $y_{f g}$ in the denominator.
Mixing references: $h$ from one table and $h$ from another (or from another program) may have different zeros.
Treating steam or a refrigerant as an ideal gas near saturation: $Z$ moves away from 1.
Using temperatures in °C in $P v = RT$ or in $T_{2} / T_{1}$ : they go in kelvin.

Worked examples

Example 1Saturated water mixture

Water at 200 kPa with quality 0.7. Find

T

v

u

h

and

s

Solution

With $x$ between 0 and 1 it is a mixture, so $T = T_{s a t} (200 kPa) = 120.21$ °C.
From the saturation table by pressure: $v_{f} = 0.0010605$ , $v_{g} = 0.88568$ m³/kg; $u_{f} = 504.49$ , $u_{f g} = 2024.6$ ; $h_{f} = 504.70$ , $h_{f g} = 2201.5$ kJ/kg; $s_{f} = 1.5302$ , $s_{f g} = 5.5967$ kJ/kg·K.
$v = 0.0010605 + 0.7 (0.88568 - 0.0010605) = 0.62030$ m³/kg, $u = 504.49 + 0.7 \cdot 2024.6 = 1921.7$ kJ/kg, $h = 504.70 + 0.7 \cdot 2201.5 = 2045.8$ kJ/kg and $s = 1.5302 + 0.7 \cdot 5.5967 = 5.4479$ kJ/kg·K.

T = 120.21

°C,

v = 0.6203

m³/kg,

u = 1921.7

kJ/kg,

h = 2045.8

kJ/kg,

s = 5.4479

kJ/kg·K.

Solve it in Simuveliq

In Engineering → Thermodynamic tables, Calculator, in Examples, «Load a case» → Water: mixture at 200 kPa with x = 0.7.
The table compares the exact value from the equation of state with the interpolated one (read directly here: differences of 10⁻⁴ %), and Interpolation in the tables, step by step writes the four formulas y = yf + x·yfg. «See the rows in the table» highlights the 200 kPa row.

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Example 2Quality at a turbine exit

Steam leaves a turbine at 10 kPa with

s = 7.0

kJ/kg·K. Find the region, the quality, the temperature and the enthalpy.

Solution

At 10 kPa: $s_{f} = 0.6492$ and $s_{g} = 8.1488$ kJ/kg·K. Since $s_{f} < 7.0 < s_{g}$ , it is a mixture, at $T_{s a t} = 45.81$ °C.
$x = \frac{s - s _{f}}{s _{f g}} = \frac{7.0 - 0.6492}{7.4996} = 0.8468$ .
$h = h_{f} + x h_{f g} = 191.81 + 0.8468 \cdot 2392.05 = 2217.4$ kJ/kg. About 15% of the flow leaves as liquid droplets, which in a real turbine erode the blades.

Saturated mixture at 45.81 °C with

x = 0.8468

and

h = 2217.4

kJ/kg.

Solve it in Simuveliq

«Load a case» → Water (turbine exit: P, s): 10 kPa and s = 7.0 kJ/kg·K → quality x. The region card gives the mixture with x = 0.84682 and the step by step solves the quality from s and then computes v, u and h.

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Example 3Double interpolation in superheated steam

Find

v

h

and

s

of steam at 1.3 MPa and 350 °C. The table has blocks of 1.2 and 1.4 MPa.

Solution

Region: $P_{s a t} (350^{\circ} C) = 16529$ kPa and 1300 kPa is lower: superheated vapor.
Both blocks have the 350 °C row, so only the interpolation in $P$ remains; 1.3 MPa is exactly halfway: $v = (0.234551 + 0.200292) /2 = 0.217422$ m³/kg, $h = (3154.16 + 3150.09) /2 = 3152.13$ kJ/kg and $s = (7.21391 + 7.13788) /2 = 7.17590$ kJ/kg·K.
Comparison with the exact value: $h$ matches (3152.13) and $s$ differs by 0.02%, but the exact $v$ is 0.216104: interpolation overestimates it by 0.6%, because $v$ varies almost as $1/ P$ , which is not a straight line.

Interpolated:

v = 0.21742

m³/kg,

h = 3152.13

kJ/kg,

s = 7.1759

kJ/kg·K (exact:

v = 0.21610

Solve it in Simuveliq

«Load a case» → Water: steam at 1.3 MPa and 350 °C (double interpolation). The step by step compares P with Psat(T) = 16 529 kPa, reads the 1.2 and 1.4 MPa blocks and does the «Interpolation between the two pressure blocks»; the table above gives the difference from the exact value for each property.

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Example 4Compressed liquid versus the approximation

Water at 80 °C and 5 MPa. Compare its properties with the saturated-liquid approximation at 80 °C.

Solution

$P_{s a t} (80^{\circ} C) = 47.414$ kPa and 5000 kPa is much higher: compressed liquid. The 5 MPa table gives $v = 0.00102672$ m³/kg, $u = 333.82$ kJ/kg, $h = 338.96$ kJ/kg and $s = 1.0723$ kJ/kg·K.
Saturated liquid at 80 °C: $v_{f} = 0.0010291$ , $u_{f} = 334.96$ , $h_{f} = 335.01$ , $s_{f} = 1.0756$ . The differences are 0.2–0.4% in $v$ , $u$ and $s$ , and 1.2% in $h$ .
With the correction: $h \approx 335.01 + 0.0010291 (5000 - 47.41) = 340.11$ kJ/kg, 0.34% from the table value. The correction improves $h$ because $P v$ is the part of the enthalpy that does change with pressure.

h = 338.96

kJ/kg (table);

h_{f} = 335.01

(−1.2%) and with the correction 340.11 (+0.34%).

Solve it in Simuveliq

«Load a case» → Water: compressed liquid at 80 °C and 5 MPa: it says P is higher than Psat(T) = 47.414 kPa, «compressed liquid», and reads the 5 MPa block.
For the approximation, change Property pair to «T, x (quality)» with 80 °C and x = 0: it gives P = 47.4145 kPa, v = 0.00102905, u = 334.964, h = 335.012 and s = 1.07558.

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Example 5Isentropic compression of R-134a

In a refrigerator, R-134a leaves the evaporator as saturated vapor at −10 °C and an ideal (isentropic) compressor takes it to 0.8 MPa. Find the exit state and the work per kilogram.

Solution

Inlet: saturated vapor at −10 °C, $P = 200.6$ kPa, $h_{1} = h_{g} = 244.52$ kJ/kg and $s_{1} = s_{g} = 0.93774$ kJ/kg·K.
Exit: $P_{2} = 0.8$ MPa and $s_{2} = s_{1}$ . At 0.8 MPa, $s_{g} = 0.91843 < 0.93774$ : superheated vapor. In the 0.8 MPa block, $s_{2}$ falls between the saturation row (31.33 °C, $h = 267.32$ ) and the 40 °C row ( $s = 0.94801$ , $h = 276.45$ ).
Interpolating in $s$ : $T_{2} = 36.99$ °C and $h_{2} = 273.28$ kJ/kg (exact: 36.93 °C and 273.25 kJ/kg).
Compressor work (steady flow, adiabatic, no kinetic or potential energy changes): $w = h_{2} - h_{1} = 273.25 - 244.52 = 28.73$ kJ/kg.

Superheated vapor at 36.93 °C with

h_{2} = 273.25

kJ/kg;

w = 28.73

kJ/kg.

Solve it in Simuveliq

«Load a case» → R-134a: evaporator exit (saturated vapor at −10 °C) for state 1 (h = 244.521, s = 0.93774), and then R-134a: compressor exit (0.8 MPa, s = s₁).
The second says «superheated vapor» because s > sg = 0.91843 and interpolates in s within the 0.8 MPa block (T = 36.989 °C); the exact one is 36.926 °C with h = 273.249.

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Example 6Isentropic compression of air with Pr

Air at 300 K is compressed isentropically with a pressure ratio of 8. Find the final temperature and the work, with the ideal-gas table and with constant

c_{p}

Solution

At 300 K the table gives $h_{1} = 301.02$ kJ/kg and $P_{r 1} = 1.3882$ .
Isentropic: $P_{r 2} = 8 P_{r 1} = 11.1056$ . Looking up that $P_{r}$ in the table: $T_{2} = 539.8$ K and $h_{2} = 545.07$ kJ/kg.
Work: $w = h_{2} - h_{1} = 545.07 - 301.02 = 244.05$ kJ/kg.
With constant $c_{p}$ ( $k = 1.4$ , $c_{p} = 1.005$ ): $T_{2} = 300 \cdot 8^{0.4/1.4} = 543.4$ K and $w = 1.005 (543.4 - 300) = 244.6$ kJ/kg. The final temperature differs by 3.6 K (1.5% of the rise) and the work by only 0.2%, because $c_{p}$ changes little between 300 and 540 K; at gas-turbine temperatures the difference grows.

T_{2} = 539.8

w = 244.05

kJ/kg (with constant

c_{p}

: 543.4 K and 244.6 kJ/kg).

Solve it in Simuveliq

In Thermodynamic tables, fluid «Air (ideal gas)», Calculator: with Known value T = 300 K you read h = 301.017 kJ/kg and Pr = 1.3882.
Change the property to Pr with the value 11.1056 (the button opens this step): T = 539.800 K and h = 545.065 kJ/kg, with the interpolation in the air table beside it.

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Thermodynamic properties and using the tables

Pure substance, phases and state

Saturation tables and quality

How to tell which region the state is in

Superheated vapor and interpolation

Compressed liquid

Ideal gas

Common mistakes

Worked examples

Example 1Saturated water mixture

Solution

Solve it in Simuveliq

Example 2Quality at a turbine exit

Solution

Solve it in Simuveliq

Example 3Double interpolation in superheated steam

Solution

Solve it in Simuveliq

Example 4Compressed liquid versus the approximation

Solution

Solve it in Simuveliq

Example 5Isentropic compression of R-134a

Solution

Solve it in Simuveliq

Example 6Isentropic compression of air with Pr

Solution

Solve it in Simuveliq

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