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# Interpolation with MATLAB | Problem Set

Module by: Serhat Beyenir. E-mail the author

Summary: Problem Set for Interpolation with MATLAB

## Exercise 1

Determine the saturation temperature, specific liquid enthalpy, specific enthalpy of evaporation and specific enthalpy of dry steam at a pressure of 2.04 MPa.

Table 1: An extract from steam tables
Pressure [MN/m2] Saturation Temperature [C] hf [kJ/kg] hfg [kJ/kg] hg [kJ/kg]
2.1 214.9 920.0 1878.2 2798.2
2.0 212.4 908.6 1888.6 2797.2

### Solution

MATLAB solution is as follows;

>> pressure=[2.1 2.0];
>> sat_temp=[214.9 212.4];
>> h_f=[920 908.6];
>> h_fg=[1878.2 1888.6];
>> h_g=[2798.2 2797.2];

>> sat_temp_new=interp1(pressure,sat_temp,2.04)

sat_temp_new =

213.4000

>> h_f_new=interp1(pressure,h_f,2.04)

h_f_new =

913.1600

>> h_fg_new=interp1(pressure,h_fg,2.04)

h_fg_new =

1.8844e+003

>> h_g_new=interp1(pressure,h_g,2.04)

h_g_new =

2.7976e+003



## Exercise 2

The following table gives data for the specific heat as it changes with temperature for a perfect gas. 1

Table 2: Change of specific heat with temperature
Temperature [F] Specific Heat [BTU/lbmF]
25 0.118
50 0.120
75 0.123
100 0.125
125 0.128
150 0.131
Using interp1 function calculate the specific heat for 30 F, 70 F and 145 F.

### Solution

MATLAB solution is as follows:

>> temperature=[25;50;75;100;125;150]

temperature =

25
50
75
100
125
150

>> specific_heat=[.118;.120;.123;.125;.128;.131]

specific_heat =

0.1180
0.1200
0.1230
0.1250
0.1280
0.1310

>> specific_heatAt30=interp1(temperature,specific_heat,30)

specific_heatAt30 =

0.1184

>> specific_heatAt70=interp1(temperature,specific_heat,70)

specific_heatAt70 =

0.1224

>> specific_heatAt145=interp1(temperature,specific_heat,145)

specific_heatAt145 =

0.1304



## Exercise 3

For the problem above, create a more detailed table in which temperature varies between 25 and 150 with 5 F increments and corresponding specific heat values.

### Solution

MATLAB solution is as follows:

>> new_temperature=25:5:150;
>> new_specific_heat=interp1(temperature,specific_heat,new_temperature);
>> [new_temperature',new_specific_heat']
ans =

25.0000    0.1180
30.0000    0.1184
35.0000    0.1188
40.0000    0.1192
45.0000    0.1196
50.0000    0.1200
55.0000    0.1206
60.0000    0.1212
65.0000    0.1218
70.0000    0.1224
75.0000    0.1230
80.0000    0.1234
85.0000    0.1238
90.0000    0.1242
95.0000    0.1246
100.0000    0.1250
105.0000    0.1256
110.0000    0.1262
115.0000    0.1268
120.0000    0.1274
125.0000    0.1280
130.0000    0.1286
135.0000    0.1292
140.0000    0.1298
145.0000    0.1304
150.0000    0.1310



## Exercise 4

During a 12-hour shift a fuel tank has varying levels due to consumption and transfer pump automatically cutting in and out to maintain a safe fuel level. The following table of fuel tank level versus time is missing readings for 5 and 9 AM. Using linear interpolation, estimate the fuel level at those times.

Table 3: Fuel tank level versus time
Time [hours, AM] Tank level [m]
1:00 1.5
2:00 1.7
3:00 2.3
4:00 2.9
5:00 ?
6:00 2.6
7:00 2.5
8:00 2.3
9:00 ?
10:00 2.0
11:00 1.8
12:00 1.3

### Solution

>> time=[1 2 3 4 6 7 8 10 11 12];
>> tank_level=[1.5 1.7 2.3 2.9 2.6 2.5 2.3 2.0 1.8 1.3];

>> tank_level_at_5=interp1(time,tank_level,5)

tank_level_at_5 =

2.7500

>> tank_level_at_9=interp1(time,tank_level,9)

tank_level_at_9 =

2.1500


## Footnotes

1. Thermodynamics and Heat Power by Kurt C. Rolle, Pearson Prentice Hall. © 2005, (p.19)

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