The table has a row for 100 kPa and a row for 125 kPa. Your system runs at 115. Every thermodynamics student meets interpolation this way, and it stays the most common practical use of it anywhere in engineering.
The quick answer
Work out how far between the two rows you are as a fraction, then apply that same fraction to every property column. One fraction, reused down the whole row.
At 115 kPa between rows at 100 and 125, the fraction is 15 ÷ 25 = 0.6. Add 60 percent of each column's change to the lower row's value.
Single interpolation, step by step
Three rows from a saturated water table, indexed by pressure:
| P (kPa) | Tsat (°C) | vg (m3/kg) | hg (kJ/kg) |
|---|---|---|---|
| 100 | 99.61 | 1.6941 | 2675.0 |
| 125 | 105.97 | 1.3749 | 2684.9 |
| 150 | 111.35 | 1.1594 | 2693.1 |
You need the saturated vapour properties at 115 kPa. The rows either side are 100 and 125, so start with the fraction:
t = (115 − 100) ÷ (125 − 100) = 15 ÷ 25 = 0.6115 is 60 percent of the way from 100 to 125. That single number now does all the work. For each column, find how much the value changes between the two rows, take 60 percent of it, and add it to the 100 kPa value.
| Property | Change across the rows | 60 percent of it | Answer at 115 kPa |
|---|---|---|---|
| Tsat | 105.97 − 99.61 = 6.36 | 3.816 | 103.43 °C |
| vg | 1.3749 − 1.6941 = −0.3192 | −0.1915 | 1.5026 m3/kg |
| hg | 2684.9 − 2675.0 = 9.9 | 5.94 | 2680.9 kJ/kg |
Notice vg going down while the other two go up. Specific volume falls as pressure rises, so a negative change there is correct. If all three of your columns move the same direction, check the table again.
Working out t once and reusing it is the step people skip. Recomputing it per property wastes time and gives three chances to make the same mistake.
How much accuracy are you losing?
You can measure it directly, because the table above has a middle row you can pretend not to have. Interpolate across the wide gap from 100 to 150 and compare against the real 125 kPa entry.
Start with specific volume. Halfway between 100 and 150 is 125, so just average the two:
(1.6941 + 1.1594) ÷ 2 = 1.4268, against the tabulated 1.3749That is 3.8 percent too high. Specific volume against pressure curves downward sharply, roughly like 1 over P, so a straight line drawn across it always sits above the real curve. Straight-line interpolation will always overestimate this particular property.
Now the same test on enthalpy:
(2675.0 + 2693.1) ÷ 2 = 2684.05, against the tabulated 2684.9That is 0.03 percent too low, and it errs in the opposite direction. Enthalpy of saturated vapour bends the other way in this range, so the straight line sits slightly below.
Two properties, same table, same gap, and one is more than a hundred times less accurate than the other. The conclusion is not that interpolation is unreliable. It is that error depends on how hard the particular column bends, so specific volume needs tighter rows than enthalpy does. When you need vg to be right, interpolate against 1 ÷ P rather than P, since volume is close to straight against that.
Double interpolation in a superheated table
Superheated tables have pressure down the side and temperature across the top. A state that falls between rows and between columns needs two rounds of the same arithmetic.
Find enthalpy at 1.1 MPa and 225 °C from these four entries:
| h (kJ/kg) | 200 °C | 250 °C |
|---|---|---|
| 1.0 MPa | 2828.3 | 2943.1 |
| 1.2 MPa | 2816.1 | 2935.6 |
Round one. Fix the pressure and interpolate in temperature. 225 °C is exactly halfway between 200 and 250, so both of these are simple averages:
At 1.0 MPa: (2828.3 + 2943.1) ÷ 2 = 2885.70At 1.2 MPa: (2816.1 + 2935.6) ÷ 2 = 2875.85
You now have enthalpy at 225 °C for two different pressures. Your target pressure, 1.1 MPa, sits halfway between them.
Round two. Interpolate those two results in pressure:
(2885.70 + 2875.85) ÷ 2 = 2880.78 kJ/kgCheck it by reversing the order. Interpolating in pressure first gives 2822.20 at 200 °C and 2939.35 at 250 °C, and averaging those gives 2880.78 again. Order independence is a property of bilinear interpolation, and it costs almost nothing to use as a check on a step where sign errors are easy.
The one that looks like interpolation and is not
Inside the vapour dome, finding the state from an enthalpy is a quality calculation, not a table lookup. Mixing these up is the classic exam mistake.
At 200 kPa, hf is 504.71 kJ/kg and hfg is 2201.6 kJ/kg. Given h = 2000 kJ/kg, the quality is:
x = (2000 − 504.71) ÷ 2201.6 = 1495.29 ÷ 2201.6 = 0.6792That arithmetic is identical to interpolation, and it is interpolation, between the saturated liquid state and the saturated vapour state. The difference is that the fraction has a physical meaning here: 67.92 percent of the mass is vapour. There is no approximation involved and no curvature error, because enthalpy really does vary in a straight line with quality at fixed pressure.
Rules that save you marks and money
Use adjacent rows. Skipping a row you already have throws away accuracy for nothing.
Never interpolate across the saturation line. Properties jump there. A pair of rows with one in the compressed liquid region and one in the two-phase region gives you a meaningless number.
Report sensible precision. Quoting an interpolated enthalpy to six figures claims accuracy the method does not have. Three or four significant figures is honest.
Spot-check the curvature. The test above, interpolating across a gap you can verify, takes thirty seconds and tells you whether the column you are reading is well behaved.
The same routine works for psychrometric charts, refrigerant tables, compressibility charts, air property tables and turbine maps. Learn it once on steam and it carries over. To skip the arithmetic, put the two rows into the interpolation calculator, and for a whole column of states the Excel guide shows how to automate the row lookup.