Property tables, enthalpy and interpolation

Tomás spent his first week at Northgate shadowing Graham on the morning rounds, writing down pressures and temperatures from gauges all over the plant: boiler outlets, the turbine exhaust, the deaerator, the feedwater pump discharge. Elena then asked him a simple question: how much energy does each kilogram of water carry at each of those points? Answering it means turning two measured properties into the property we actually need, which is usually enthalpy. For water and refrigerants the relations between properties are too complicated for simple equations, so engineers use tables, or software built on the same equations. This lesson is about reading those tables quickly and correctly.

Enthalpy

In the analysis of flowing fluids, internal energy almost always appears together with the product of pressure and specific volume. The combination is so common that it is defined as a property of its own, enthalpy:

Enthalpy is a property because \(u\), \(P\) and \(v\) are properties. Lesson 7 will show where the \(Pv\) term comes from: it is the flow work needed to push each kilogram of fluid into or out of a control volume. For now, note two things. First, in a constant-pressure process in a closed system, the heat transferred equals the change in enthalpy, which is why enthalpy is sometimes loosely called heat content. Second, enthalpy is tabulated alongside internal energy, and for steam it is the property you will use most.

Steam tables list values of \(u\), \(h\) and \(s\) relative to an arbitrary reference state, because thermodynamics deals only with changes in energy. The usual reference sets the internal energy and entropy of saturated liquid water at the triple point (0.01 °C) to zero. Refrigerant tables often use a different reference, so values from different sources can differ by a constant. Changes come out the same whatever the reference, but never mix values from two tables with different references in one calculation.

The saturation tables

Saturated water tables come in two forms: one listed by temperature and one by pressure, so that whichever property is known, the other can be read without interpolation. Each row gives the saturation pressure or temperature and the properties of saturated liquid (subscript \(f\)) and saturated vapour (subscript \(g\)), and often the difference (subscript \(fg\)). The enthalpy of vaporization \(h_{fg}\) is the latent heat introduced in Lesson 3. For mixtures, the quality relations give any property: \(y = y_f + x\,y_{fg}\).

The superheated vapour table

In the superheated region, pressure and temperature are independent, so the table is laid out in blocks, one for each pressure, with rows for a series of temperatures. Each block begins with the saturation temperature at that pressure, in parentheses, and the properties of saturated vapour, followed by rows at higher temperatures. Compared with saturated vapour at the same pressure, superheated vapour has a higher temperature, a larger specific volume, and higher internal energy, enthalpy and entropy.

These values describe the steam in the district main. If it leaves the turbine at 1 MPa and 250 °C, it carries 2943.1 kJ/kg; if the turbine is bypassed and the main is fed through a pressure-reducing valve from the 4 MPa header, its temperature after the valve will be higher, as Lesson 8 will show.

Compressed liquid

Compressed-liquid tables exist but are rarely needed, because a liquid's properties depend much more on temperature than on pressure. A common and accurate approximation treats a compressed liquid as a saturated liquid at the same temperature: \(v \approx v_f(T)\), \(u \approx u_f(T)\), \(s \approx s_f(T)\). For enthalpy, which contains the \(Pv\) term, a small correction improves the estimate:

For feedwater at 5 MPa and 80 °C, \(h_f = 335.0\ \text{kJ/kg}\) and the correction adds \(0.001029(5000 - 47.4) = 5.1\ \text{kJ/kg}\), giving 340.1 kJ/kg. The full compressed-liquid table gives about 339 kJ/kg, so the uncorrected value is within about 1 percent and the corrected one within a fraction of a percent. At the very high pressures of supercritical boilers the correction matters more, but at district-energy pressures the simple approximation is usually enough.

Finding the region first

Before reading any table, decide which region the state lies in. Elena's rule for Tomás is to look up the saturation values for whichever property is given, then compare. The flowchart below summarizes the logic for a state given by pressure and one other property.

The same logic works with temperature given instead of pressure: compare the pressure with \(P_{\text{sat}}(T)\). If the given pressure is higher, the substance is a compressed liquid (it is squeezed above its boiling pressure); if lower, it is superheated. A frequent slip is to reverse this comparison, so it is worth checking against intuition: water at 1 MPa and 150 °C is below its boiling point of 179.9 °C, so it must be liquid.

Interpolation

Tables cannot list every value, so we interpolate. Over the short intervals in a steam table, properties vary nearly linearly, and linear interpolation is accurate enough for almost all engineering work. For a quantity \(y\) known at \(x_0\) and \(x_1\), its value at \(x\) between them is

Sometimes neither the pressure nor the temperature appears in the table, and we must interpolate twice: first in temperature within each of two pressure blocks, then in pressure between the two results. This double interpolation is tedious by hand, which is why most practising engineers now use property software or a spreadsheet add-in. You should still be able to do it, both for examinations and to check that software output is reasonable.

The program below automates both interpolations, classifies several states, solves an inverse problem (finding the temperature that gives a known enthalpy, which arises whenever an energy balance gives \(h\) and we need \(T\)), and applies the compressed-liquid correction.

# Locating a state and interpolating in steam tables
# Superheated steam: (P kPa, T C) -> (v m3/kg, h kJ/kg, s kJ/kg.K), rounded from IAPWS-based tables
SUPER = {
    (800, 250): (0.29321, 2950.4, 7.0402),  (800, 300): (0.32416, 3057.3, 7.2348),
    (1000, 250): (0.23275, 2943.1, 6.9265), (1000, 300): (0.25799, 3051.6, 7.1246),
    (1200, 250): (0.19241, 2935.6, 6.8313), (1200, 300): (0.21386, 3046.3, 7.0335),
}
# Saturation: P kPa -> (Tsat C, vf, vg, hf, hg)
SAT = {800: (170.41, 0.001115, 0.24035, 720.9, 2768.3),
       1000: (179.88, 0.001127, 0.19436, 762.5, 2777.1),
       1200: (187.96, 0.001138, 0.16326, 798.3, 2783.8)}

def lerp(x, x0, x1, y0, y1):
    return y0 + (x - x0) / (x1 - x0) * (y1 - y0)

def region(P, T):
    Ts = SAT[P][0]
    if T < Ts - 1e-9: return "compressed liquid"
    if T > Ts + 1e-9: return "superheated"
    return "saturated (need a second property)"

for P, T in ((1000, 150), (1000, 179.88), (1000, 280), (800, 300)):
    print(f"{P:5d} kPa, {T:7.2f} C: {region(P, T)}")

# Single interpolation: 1 MPa, 280 C
v = [lerp(280, 250, 300, a, b) for a, b in zip(SUPER[(1000, 250)], SUPER[(1000, 300)])]
print(f"\n1 MPa, 280 C: v = {v[0]:.5f} m3/kg, h = {v[1]:.1f} kJ/kg, s = {v[2]:.4f} kJ/kg.K")

# Double interpolation: 0.9 MPa, 265 C
def at(P, T):
    return [lerp(T, 250, 300, a, b) for a, b in zip(SUPER[(P, 250)], SUPER[(P, 300)])]
lo, hi = at(800, 265), at(1000, 265)
mid = [lerp(900, 800, 1000, a, b) for a, b in zip(lo, hi)]
print(f"0.9 MPa, 265 C: v = {mid[0]:.5f} m3/kg, h = {mid[1]:.1f} kJ/kg, s = {mid[2]:.4f} kJ/kg.K")

# Inverse problem: 1 MPa steam with h = 3000 kJ/kg -> temperature
h250, h300 = SUPER[(1000, 250)][1], SUPER[(1000, 300)][1]
print(f"1 MPa, h = 3000 kJ/kg: T = {lerp(3000, h250, h300, 250, 300):.1f} C")

# Compressed liquid: feedwater at 5 MPa and 80 C, approximated with saturated liquid at 80 C
Psat80, vf80, hf80 = 47.42, 0.001029, 335.0
print(f"\nfeedwater 5 MPa, 80 C: h ~ hf = {hf80:.1f}; with pressure correction "
      f"{hf80 + vf80 * (5000 - Psat80):.1f} kJ/kg")

 1000 kPa,  150.00 C: compressed liquid
 1000 kPa,  179.88 C: saturated (need a second property)
 1000 kPa,  280.00 C: superheated
  800 kPa,  300.00 C: superheated

1 MPa, 280 C: v = 0.24789 m3/kg, h = 3008.2 kJ/kg, s = 7.0454 kJ/kg.K
0.9 MPa, 265 C: v = 0.27141 m3/kg, h = 2979.1 kJ/kg, s = 7.0423 kJ/kg.K
1 MPa, h = 3000 kJ/kg: T = 276.2 C

feedwater 5 MPa, 80 C: h ~ hf = 335.0; with pressure correction 340.1 kJ/kg

Putting it together: Tomás's morning readings

Software and the IAPWS formulations

Modern steam tables are generated from equations published by IAPWS. The scientific formulation, IAPWS-95, is a very accurate equation of state for water over the whole fluid range. The industrial formulation, IAPWS-IF97, divides the range into regions with faster equations and is the basis of most commercial software and of contract guarantees for turbines and boilers. Spreadsheet add-ins, online calculators and engineering programs all use one of these. Results from different sources should agree to four or five significant figures; if they do not, suspect a unit or reference-state problem rather than the equations.

Software makes double interpolation obsolete, but it does not remove the need for judgement. It will happily return properties for a state that cannot exist in your system, or interpret 1 MPa as gauge pressure if that is what you typed. The habit of locating the state on a T-v or P-v diagram, and checking that the result is in the right region and of the right size, is what separates a reliable engineer from a fast typist.