Example 5.3

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Using the table below, ﬁnd the internal energy of steam at 215 ˚C and the temperature if the internal energy is 2600 kJ/kg (use linear interpolation).

Table 5.1 An extract from Steam Tables
Temperature [C]	Internal Energy [kJ/kg]
100	2506.7
150	2582.8
200	2658.1
250	2733.7
300	2810.4
400	2967.9
500	3131.6

First let us enter the temperature and energy values

    temperature = [100, 150, 200, 250, 300, 400, 500];
energy = [2506.7, 2582.8, 2658.1, 2733.7, 2810.4, 2967.9, 3131.6];
[temperature',energy']

returns

ans =
  1.0e+003 *

    0.1000  2.5067
    0.1500  2.5828
    0.2000  2.6581
    0.2500  2.7337
    0.3000  2.8104
    0.4000  2.9679
    0.5000  3.1316

issue the following for the ﬁrst question:

new_energy = interp1(temperature,energy,215)

returns

new_energy =
    2.6808e+003

Now, type in the following for the second question:

new_temperature = interp1(energy,temperature,2600)

returns

new_temperature =
    161.4210

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Example 5.3