| << Chapter < Page | Chapter >> Page > |
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 |
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
Notification Switch
Would you like to follow the 'A brief introduction to engineering computation with matlab' conversation and receive update notifications?
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|