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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 /551/CH13/EX13.30 | |
| 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 '551/CH13/EX13.30')
| -rwxr-xr-x | 551/CH13/EX13.30/30.sce | 84 |
1 files changed, 84 insertions, 0 deletions
diff --git a/551/CH13/EX13.30/30.sce b/551/CH13/EX13.30/30.sce new file mode 100755 index 000000000..f8823db0f --- /dev/null +++ b/551/CH13/EX13.30/30.sce @@ -0,0 +1,84 @@ +clc
+v=10:1:100;
+function p=f(v)
+ p=1/v^1.4;
+endfunction
+plot(v,f)
+
+v=[10 20]
+p=[f(10) f(10)]
+plot(v,p,'r')
+
+v=20:1:100;
+function p=fa(v)
+ p=2.6515/v^1.4;
+endfunction
+plot(v,fa,'g')
+
+v=[100 100]
+p=[f(100) fa(100)]
+plot(v,p,'--p')
+
+v=[15 15]
+p=[f(15) 0.040]
+plot(v,p,'--')
+
+v=[20 20]
+p=[f(20) 0.040]
+plot(v,p,'--r')
+
+xtitle("p-v diagram", "v", "p")
+legend("1-2b","2b-3", "3-4", "4-1", "2a-3a", "2-3")
+
+//The air-standard Otto, Dual and Diesel cycles are drawn on common p-v and T-s diagrams for the same maximum pressure and maximum temperature, for the purpose of comparison.
+// Otto 1-2-3-4-1
+// Dual 1-2a-3a-3-4-1
+// Diesel 1-2b-3-4-1
+
+
+xset('window', 1)
+
+s=10:1:50;
+function T=fb(s)
+ T=s^2
+endfunction
+plot(s,fb)
+
+s=10:1:50;
+function T=fc(s)
+ T=(s+30)^2
+endfunction
+plot(s,fc,'r')
+
+s=[12 12];
+T=[fb(12) fc(12)];
+plot(s,T,'--p')
+
+s=[45 45];
+T=[fb(45) fc(45)]
+plot(s,T,'m')
+
+s=10:1:27;
+T=5*(s)^2;
+plot(s,T,'g')
+
+s=10:1:20;
+T=7*s^2;
+plot(s,T,'--r')
+
+xtitle("T-s diagram", "s", "T")
+legend("1-4", "2b-3", "1-2b", "3-4", "2-3", "2a-3a")
+
+// The construction of cycles on T-s diagram proves that for the given conditions the heat rejected is same for all the three cycles (area under process line 4-1).
+// η=1-(Heat rejected)/(Heat supplied)=1-constant/Qs
+
+// The cycle with greater heat addition will be more efficient.
+// From the T-s diagram
+
+// Qs(diesel) = Area under 2b-3
+// Qs(dual) = Area under 2a-3a-3
+// Qs(otto) = Area under 2-3.
+
+// Qs(diesel) > Qs(dual) > Qs(otto)
+
+disp("Thus, ηdiesel > ηdual > ηotto")
\ No newline at end of file |