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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 /3819/CH4/EX4.19 | |
| parent | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff) | |
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initial commit / add all books
Diffstat (limited to '3819/CH4/EX4.19')
| -rw-r--r-- | 3819/CH4/EX4.19/Ex4_19.sce | 30 |
1 files changed, 30 insertions, 0 deletions
diff --git a/3819/CH4/EX4.19/Ex4_19.sce b/3819/CH4/EX4.19/Ex4_19.sce new file mode 100644 index 000000000..733aa216b --- /dev/null +++ b/3819/CH4/EX4.19/Ex4_19.sce @@ -0,0 +1,30 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 4-Buoyancy and Floatation
+//// Problem 4.19
+
+//Given Data Set in the Problem
+dens=1000
+g=9.81
+L=70
+b=10
+w=19620*10^3
+theta =6
+sw=10104
+w1=343.35*10^3
+x=6
+COB=2.25
+H=2.25
+
+//calculations
+//1)
+//Meta centric height
+GM=w1*x/w/tan(theta /180*%pi)
+mprintf("The meta centric height is %f m \n",GM)
+
+//2)
+//Position of centre of gravity
+I=0.75*(1/12*L*10^3) //MOI
+Vol=w/sw //vol of water displaced
+//from equation for meta centric height ,we get,
+BG=I/Vol-GM
+mprintf("The distance of G from the free water surface is %f m \n",H-BG)
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