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Summary: Problem Set for Interpolation with MATLAB
Determine the saturation temperature, specific liquid enthalpy, specific enthalpy of evaporation and specific enthalpy of dry steam at a pressure of 2.04 MPa.
| 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 |
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
The following table gives data for the specific heat as it changes with temperature for a perfect gas. 1
| Temperature [F] | Specific Heat [BTU/lbmF] |
|---|---|
| 25 | 0.118 |
| 50 | 0.120 |
| 75 | 0.123 |
| 100 | 0.125 |
| 125 | 0.128 |
| 150 | 0.131 |
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
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.
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
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.
| 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 |
>> 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
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This work is licensed by Serhat Beyenir under a Creative Commons Attribution License (CC-BY 3.0), and is an Open Educational Resource.
Last edited by Serhat Beyenir on Nov 21, 2011 2:30 pm -0600.