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| author | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
|---|---|---|
| committer | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
| commit | 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (patch) | |
| tree | dbb9e3ddb5fc829e7c5c7e6be99b2c4ba356132c /3733/CH34/EX34.7 | |
| parent | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff) | |
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initial commit / add all books
Diffstat (limited to '3733/CH34/EX34.7')
| -rw-r--r-- | 3733/CH34/EX34.7/Ex34_7.sce | 26 |
1 files changed, 26 insertions, 0 deletions
diff --git a/3733/CH34/EX34.7/Ex34_7.sce b/3733/CH34/EX34.7/Ex34_7.sce new file mode 100644 index 000000000..56bd38035 --- /dev/null +++ b/3733/CH34/EX34.7/Ex34_7.sce @@ -0,0 +1,26 @@ +// Example 34_7
+clc;funcprot(0);
+//Given data
+L_p=10;// MW
+// I=5*10^6(18+12L+0.5L^2)
+L_1=5;// MW
+L_2=7;// MW
+
+// Calculation
+//(a)
+// n=(L/I)=(1/(5*10^6((18/L)+12+0.5L)));
+// The efficiency will be maximum when ((18/L)+12+0.5L)),differentiating we get
+L_m=sqrt((-18)/(-0.5));// MW
+L=L_m*10^3*3600;// kJ/hr
+I_6=5*10^6*(18+(12*L_m)+(0.5*L_m^2));// kJ/hr
+n_max=(L/I_6)*100;// Maximum efficiency in %
+printf('\n(a)The load at which the efficiency of the plant will be maximum=%0.0f MW \n The maximum efficiency=%0.0f percentage',L_m,n_max);
+//(b)
+I_5=5*10^6*(18+(12*L_1)+(0.5*L_1^2));// kJ/hr
+I_7=5*10^6*(18+(12*L_2)+(0.5*L_2^2));// kJ/hr
+dI=I_7-I_5;// Increase in output to the plant per hour in kJ/hr
+L=(L_1+L_2)/2;// MW
+IR=5*10^6*(12+L);// kJ/hr
+Ti=IR*(L_2-L_1);// Total increase in input in kJ/hr
+printf('\n(b)Total increase in input=%0.2e kJ/hr',Ti);
+
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