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author | priyanka | 2015-06-24 15:03:17 +0530 |
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committer | priyanka | 2015-06-24 15:03:17 +0530 |
commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
tree | ab291cffc65280e58ac82470ba63fbcca7805165 /1835/CH12/EX12.1/Ex12_1.sce | |
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-rwxr-xr-x | 1835/CH12/EX12.1/Ex12_1.sce | 18 |
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diff --git a/1835/CH12/EX12.1/Ex12_1.sce b/1835/CH12/EX12.1/Ex12_1.sce new file mode 100755 index 000000000..cfaf5994f --- /dev/null +++ b/1835/CH12/EX12.1/Ex12_1.sce @@ -0,0 +1,18 @@ +//CHAPTER 12 ILLUSRTATION 1 PAGE NO 310
+//TITLE:Balancing of reciprocating of masses
+clc
+clear
+pi=3.141
+N=250// speed of the reciprocating engine in rpm
+s=18// length of stroke in mm
+mR=120// mass of reciprocating parts in kg
+m=70// mass of revolving parts in kg
+r=.09// radius of revolution of revolving parts in m
+b=.15// distance at which balancing mass located in m
+c=2/3// portion of reciprocating mass balanced
+teeta=30// crank angle from inner dead centre in degrees
+//===============================
+B=r*(m+c*mR)/b// balance mass required in kg
+w=2*pi*N/60// angular speed in rad/s
+F=mR*w^2*r*(((1-c)^2*(cosd(teeta))^2)+(c^2*(sind(teeta))^2))^.5// residual unbalanced forces in N
+printf('Magnitude of balance mass required= %.0f kg\n Residual unbalanced forces= %.3f N',B,F)
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