



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.
An extract from steam tables
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)
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.
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 12hour 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.
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 
>>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
Questions & Answers
find the 15th term of the geometric sequince whose first is 18 and last term of 387
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hmm well what is the answer
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how do they get the third part x = (32)5/4
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sure. what is your question?
ninjadapaul
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ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X6)^2
so it's 20 divided by X6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x6
ninjadapaul
oops. ignore that.
ninjadapaul
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ninjadapaul
is it a question of log
Abhi
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I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
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Cesar
I'm interested in nanotube
Uday
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what is system testing?
AMJAD
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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AMJAD
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AMJAD
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Stotaw
In this morden time nanotechnology used in many field .
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Azam
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Prasenjit
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maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
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Damian
silver nanoparticles could handle the job?
Damian
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I'm interested in Nanotube
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Source:
OpenStax, A brief introduction to engineering computation with matlab. OpenStax CNX. Nov 17, 2015 Download for free at http://legacy.cnx.org/content/col11371/1.11
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