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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /542/CH16/EX16.3/Example_16_3.sci | |
| download | Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.gz Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.bz2 Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.zip | |
initial commit / add all books
Diffstat (limited to '542/CH16/EX16.3/Example_16_3.sci')
| -rwxr-xr-x | 542/CH16/EX16.3/Example_16_3.sci | 60 |
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diff --git a/542/CH16/EX16.3/Example_16_3.sci b/542/CH16/EX16.3/Example_16_3.sci new file mode 100755 index 000000000..cc88b6783 --- /dev/null +++ b/542/CH16/EX16.3/Example_16_3.sci @@ -0,0 +1,60 @@ +clear;
+clc;
+printf("\n example 16.3");
+//(a) Air
+//G kg/s dry air enter with 0.006G kg/s water vapour and hence the heat content of this stream=
+//[(1.00G) + (0.006G × 2.01)](385 − 273) = 113.35G kW
+
+//(b) Wet solid
+//0.125 kg/s enter containing 0.40 kg water/kg wet solid, assuming the moisture is expressed on a wet basis.
+flowWater = (0.125*0.40); //in kg/sec
+flowDrySolid = (0.125-0.050); //in kg/sec
+//Hence heat content of this stream
+q = [(0.050*4.18)+(0.075*0.88)]*(295-273);
+printf("\n The heat content of this stream = %.2f kW",q);
+
+//Heat out
+//(a) Air
+//Heat in exit air = [(1.00 G) + (0.006 G × 2.01)](310 − 273) = 37.45G kW.
+fd = 0.075; //mass flow rate of dry solids in kg/sec
+w = 0.05*0.075/(1+0.05); //water in the dried solids leaving in kg/secs
+we = (0.050 - w); //The water evaporated into gas stream in kg/secs
+
+//Assuming evaporation takes place at 295 K,then:
+qout = 0.0464*[2.01*(310-295)+2449+4.18*(295-273)];
+printf("\n Heat in the water vapour = %.1f kW",qout);
+
+//the total heat in this stream = (119.30 + 37.45G) kW.
+//(b) Dried solids
+
+
+ //The dried solids contain 0.0036 kg/s water and hence heat conten t of this stream is:
+ q1 = [(0.075*0.88)+(0.0036*4.18)/(305-273)];
+ printf("\n The dried solids contain 0.0036 kg/s water and hence heat content of this stream is: %.2f kW",q1);
+
+
+//(c) Losses
+//These amount to 20 kJ/kg dry air or 20m kW.
+//Heat Balance
+G = poly([0],'G');
+G1 = roots(113.35*G + 6.05 - 119.30 - 37.45*G - 2.59 - 20*G);
+printf("\n G = %.2f kg/secs",G1);
+printf("\n Water in the outlet stream %.4f kg/secs",0.006*2.07+0.0464);
+printf("\n The humidity H = %.4f kg/kg dry air",0.0588/2.07);
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