Up: Return

Linear interpolation


Linear interpolation is a way to fill in the ``holes'' in tables. As an example, if you want to find the saturated pressure of water at a temperature of 40C you can look in Table B.1.1, (p.674), for 40C in the first column. The corresponding desired pressure is then in the next column; in this case, 7.384 kPa. But what if you want to find the saturated pressure at 38C instead of 40C?

A temperature of 38C is not in the table. You could of course just ignore the difference between 38C and 40C, and still take the saturated pressure to be 7.384 kPa. But that is not acceptable in this class; it is too inaccurate. To get an accurate value, you must use linear interpolation. (Though taking the closest value, 40C, is of course better than nothing in case you forgot how to do linear interpolation during an exam.)

Let's introduce a few symbols. Let g be your given value, 38C in this example. Let g1 and g2 be the two closest approximations to g in the table. A look at Table B.1.1 shows that the two closest values you can find in the table are 35C and 40C, so in our example g1=35C and g2=40C. (The desired value is in between those two, hence the ``in'' in ``interpolation.'')

Also, let d be our desired value, in our example the saturated pressure. Let d1 and d2 be the approximate desired values corresponding to g1 and g2. In our example, Table B.1.1 gives the saturated pressure at g1=35C to be d1=5.628 kPa and the saturated pressure at g2=40C to be d2=7.384 kPa. Both d1 and d2 are approximations to our desired pressure, but neither is accurate enough.

The formula for linear interpolation is:

So, in our example, the desired saturated pressure d at 38C is:

A nonsaturated example

You need two variables to read off the compressed liquid or superheated vapor tables. In the next example, we will find the specific volume of steam at a given temperature of 100C and a given pressure of 20 kPa.

Steam (superheated water vapor) is found in Table B.1.3. We have no difficulty finding the given 100C in that table, but we cannot find the given pressure of 20 kPa. The closest pressures in the table are 10 kPa and 50 Kpa.

So in the linear interpolation formula from the previous section,

we set the given value g equal to 20 kPa, and the closest table values g1 and g2 to 10 kPa and 50 kPa.

The desired quantity d is now the specific volume at 100C and 20 kPa. We set the value d1 to the specific volume at g1=10 kPa (and 100C,) so d1=17.19561 m3/kg according to the table, and d2 to the specific volume at g2=50 kPa (and 100C,) so d2=3.41833 m3/kg.

Our formula then gives the specific volume at 20 kPa and 100C as:

Other problems

You might ask what happens to the last example if neither the given pressure nor the given temperature is in the table. For example, to find the specific volume at 20 kPa and 110C, neither 20 kpa nor 110C are in Table B.1.4. I do not think we would do this to you during the exam. But the answer would be to do three linear interpolations: first interpolate a specific volume at 110C and 10 kPa (fill in the 110C ``hole'' in the 10 kPa data), next interpolate a specific volume at 110C and 50 kPa (fill in the 110C ``hole'' in the 50 kPa data), and finally linear interpolate those 110C values in the same way as we did for 100C in the previous section.

Another problem arises if you try to interpolate the specific volume of steam at 11 kPa and 50C. You can use the B.1.3 entry for 50C and g1=10 kPa, giving d1=14.86920 m3/kg. But unfortunately, the 50 kPa data start at 81.33C; no 50C steam at 50 kPa exists. The key to find a second table entry, to give you g2 and d2, is to recognize that superheated steam ends at saturation, which is in table B.1.1. You can find the desired second table entry there; in particular, B.1.1 at 50C gives a second pressure g2=12.350 and specific volume d2=12.0318. Which means that the formula

gives the specific volume of steam at 11 kPa and 50C as:

Up: Return