riaology
Unit 02 · Lesson 02 of 09

How do we read property tables?

Choose the region before choosing a row.

Explore the model. Derive the result. Check its limits.

What you’ll be able to do
  • Classify before lookup.
  • Read the headings and units.
  • Interpolation is a local approximation.
Before you touch the controls

At 200 kPa, is water at 275°C a mixture or superheated vapor?

An experiment you can change

Interpolate a superheated-water table

Predict the direction of change, move one control, then explain the response using the equation.

Linear interpolation between property table rows at temperature 275 degrees Celsius200°C2,870.78 kJ/kg250°C2,971.26 kJ/kg300°C3,072.08 kJ/kg350°C3,173.89 kJ/kgChoose rows at the same pressureInterpolate a superheated-water table. Temperature (°C) versus Enthalpy (kJ/kg). Selected result 3,021.6718 kJ/kg. Saturation at 200 kPa is 120.21 °C. This entire control range is superheated. At an exact row no interpolation is needed.02001,586.9462753,173.891350Enthalpy (kJ/kg)Temperature (°C)
h=2971.3+275−25050(3072.1−2971.3)=3021.7 kJ/kgh=2971.3+\frac{275-250}{50}(3072.1-2971.3)=3021.7\,\mathrm{kJ/kg}

Saturation at 200 kPa is 120.21 °C. This entire control range is superheated. At an exact row no interpolation is needed.

Model assumptions and units

Superheated water at 200 kPa. Interpolation is linear between the displayed IAPWS-IF97 table rows; values are not interpolated across phase boundaries.

h=ha+T−TaTb−Ta(hb−ha)h=h_a+\frac{T-T_a}{T_b-T_a}(h_b-h_a)
Superheated water at 200 kPa · IAPWS-IF97
Temperature (°C)h (kJ/kg)u (kJ/kg)v (m³/kg)
2002,870.7792,654.6761.080519
2502,971.2592,731.4771.198909
3003,072.0842,808.8381.316231
3503,173.8912,887.2991.432964
01

Classify before lookup.

Compare the specified temperature with saturation temperature at the specified pressure. A lower temperature indicates compressed liquid, an equal temperature allows saturation, and a higher temperature indicates superheated vapor. At saturation, another independent property is needed to determine phase proportions.

T<Tsat: liquid;T>Tsat: vaporT<T_{sat}:\text{ liquid};\quad T>T_{sat}:\text{ vapor}
02

Read the headings and units.

Specific volume is volume per mass. Internal energy and enthalpy are also listed per mass. Subscripts f and g denote saturated liquid and vapor; fg denotes their difference. Multiply specific quantities by mass to obtain totals.

V=mv,U=mu,H=mhV=mv,\quad U=mu,\quad H=mh
03

Interpolation is a local approximation.

Between two rows at the same pressure and within the same phase, linear interpolation uses the fraction of the temperature interval traversed. It assumes the property curve is sufficiently close to a line over that interval.

h(T)=ha+T−TaTb−Ta(hb−ha)h(T)=h_a+\frac{T-T_a}{T_b-T_a}(h_b-h_a)
04

Avoid interpolation through a phase boundary.

A compressed-liquid row and a superheated row cannot be blended as if boiling were an ordinary temperature interval. First identify the region, then use the corresponding property relation. For mixtures, use quality and saturated endpoints.

A worked example

Interpolate at 275°C and 200 kPa.

  1. Saturation at 200 kPa is about 120.21 ∘C120.21\,{}^\circ\mathrm C; 275°C is superheated.
  2. The temperature is halfway between 250°C and 300°C: (275−250)/(300−250)=0.5(275-250)/(300-250)=0.5.
  3. Use h275=h250+0.5(h300−h250)h_{275}=h_{250}+0.5(h_{300}-h_{250}) with the table displayed below. The live marker shows the interpolated result.
Make the idea your own

Calculate. Compare. Test the assumptions.

Work out your answer, check it, then explore the visual solution. The answer checker uses a 1.5% rounding tolerance; for model-limit questions it checks the displayed model result.

1 · Calculate

A midpoint table lookup

At 200 kPa and 275°C, find specific enthalpy by interpolation in the displayed table.

Hint

The requested temperature lies halfway between 250 and 300°C.

Explore the visual solution
Linear interpolation between property table rows at temperature 275 degrees Celsius200°C2,870.78 kJ/kg250°C2,971.26 kJ/kg300°C3,072.08 kJ/kg350°C3,173.89 kJ/kgChoose rows at the same pressureInterpolate a superheated-water table. Temperature (°C) versus Enthalpy (kJ/kg). Selected result 3,021.6718 kJ/kg. Saturation at 200 kPa is 120.21 °C. This entire control range is superheated. At an exact row no interpolation is needed.02001,586.9462753,173.891350Enthalpy (kJ/kg)Temperature (°C)
h=2971.3+275−25050(3072.1−2971.3)=3021.7 kJ/kgh=2971.3+\frac{275-250}{50}(3072.1-2971.3)=3021.7\,\mathrm{kJ/kg}

Saturation at 200 kPa is 120.21 °C. This entire control range is superheated. At an exact row no interpolation is needed.

Model assumptions and units

Superheated water at 200 kPa. Interpolation is linear between the displayed IAPWS-IF97 table rows; values are not interpolated across phase boundaries.

h=ha+T−TaTb−Ta(hb−ha)h=h_a+\frac{T-T_a}{T_b-T_a}(h_b-h_a)

Controls explore variations. The steps below retain the original question’s values.

Superheated water at 200 kPa. Interpolation is linear between the displayed IAPWS-IF97 table rows; values are not interpolated across phase boundaries.

h=ha+T−TaTb−Ta(hb−ha)h=h_a+\frac{T-T_a}{T_b-T_a}(h_b-h_a)
2 · Compare

A quarter of an interval

At 200 kPa and 212°C, find interpolated specific enthalpy. Does averaging the endpoint values equally give the correct result?

Hint

Use (212 − 200)/(250 − 200), not 1/2.

Explore the visual solution
Linear interpolation between property table rows at temperature 212 degrees Celsius200°C2,870.78 kJ/kg250°C2,971.26 kJ/kg300°C3,072.08 kJ/kg350°C3,173.89 kJ/kgChoose rows at the same pressureInterpolate a superheated-water table. Temperature (°C) versus Enthalpy (kJ/kg). Selected result 2,894.8945 kJ/kg. Saturation at 200 kPa is 120.21 °C. This entire control range is superheated. At an exact row no interpolation is needed.02001,586.9462753,173.891350Enthalpy (kJ/kg)Temperature (°C)
h=2870.8+212−20050(2971.3−2870.8)=2894.9 kJ/kgh=2870.8+\frac{212-200}{50}(2971.3-2870.8)=2894.9\,\mathrm{kJ/kg}

Saturation at 200 kPa is 120.21 °C. This entire control range is superheated. At an exact row no interpolation is needed.

Model assumptions and units

Superheated water at 200 kPa. Interpolation is linear between the displayed IAPWS-IF97 table rows; values are not interpolated across phase boundaries.

h=ha+T−TaTb−Ta(hb−ha)h=h_a+\frac{T-T_a}{T_b-T_a}(h_b-h_a)

Controls explore variations. The steps below retain the original question’s values.

Superheated water at 200 kPa. Interpolation is linear between the displayed IAPWS-IF97 table rows; values are not interpolated across phase boundaries.

h=ha+T−TaTb−Ta(hb−ha)h=h_a+\frac{T-T_a}{T_b-T_a}(h_b-h_a)
3 · Challenge the model

An exact table row

Find enthalpy at 200 kPa and 350°C. Explain why no temperature interpolation is needed and why pressure must still match.

Hint

An exact row can be used directly, but only in the correct pressure table.

Explore the visual solution
Linear interpolation between property table rows at temperature 350 degrees Celsius200°C2,870.78 kJ/kg250°C2,971.26 kJ/kg300°C3,072.08 kJ/kg350°C3,173.89 kJ/kgChoose rows at the same pressureInterpolate a superheated-water table. Temperature (°C) versus Enthalpy (kJ/kg). Selected result 3,173.8914 kJ/kg. Saturation at 200 kPa is 120.21 °C. This entire control range is superheated. At an exact row no interpolation is needed.02001,586.9462753,173.891350Enthalpy (kJ/kg)Temperature (°C)
h=3173.9+350−3501(3173.9−3173.9)=3173.9 kJ/kgh=3173.9+\frac{350-350}{1}(3173.9-3173.9)=3173.9\,\mathrm{kJ/kg}

Saturation at 200 kPa is 120.21 °C. This entire control range is superheated. At an exact row no interpolation is needed.

Model assumptions and units

Superheated water at 200 kPa. Interpolation is linear between the displayed IAPWS-IF97 table rows; values are not interpolated across phase boundaries.

h=ha+T−TaTb−Ta(hb−ha)h=h_a+\frac{T-T_a}{T_b-T_a}(h_b-h_a)

Controls explore variations. The steps below retain the original question’s values.

Superheated water at 200 kPa. Interpolation is linear between the displayed IAPWS-IF97 table rows; values are not interpolated across phase boundaries.

h=ha+T−TaTb−Ta(hb−ha)h=h_a+\frac{T-T_a}{T_b-T_a}(h_b-h_a)
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