# 5.1 Problem set

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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/m 2 ] Saturation Temperature [C] h f [kJ/kg] h fg [kJ/kg] h g [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. (Data available for download ). Thermodynamics and Heat Power by Kurt C. Rolle, Pearson Prentice Hall. © 2005, (p.19)

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.

MATLAB solution is as follows: >>temperature=[25;50;75;100;125;150] temperature =25 5075 100125 150>>specific_heat=[.118;.120;.123;.125;.128;.131] specific_heat =0.1180 0.12000.1230 0.12500.1280 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 0.118435.0000 0.1188 40.0000 0.119245.0000 0.1196 50.0000 0.120055.0000 0.1206 60.0000 0.121265.0000 0.1218 70.0000 0.122475.0000 0.1230 80.0000 0.123485.0000 0.1238 90.0000 0.124295.0000 0.1246 100.0000 0.1250105.0000 0.1256 110.0000 0.1262115.0000 0.1268 120.0000 0.1274125.0000 0.1280 130.0000 0.1286135.0000 0.1292 140.0000 0.1298145.0000 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 (Data available for download ) 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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