22 files changed, 513 insertions, 0 deletions
diff --git a/3819/CH4/EX4.1/Ex4_1.sce b/3819/CH4/EX4.1/Ex4_1.sce
new file mode 100644
index 000000000..c7437d2da
--- /dev/null
+++ b/3819/CH4/EX4.1/Ex4_1.sce
@@ -0,0 +1,19 @@
+// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 4-Buoyancy and Floatation
+// Problem 4.1
+
+//Given Data Set in the Problem
+w=2.5
+d=1.5
+l=6
+dens=650
+g=9.81
+//calculations
+V=w*d*l
+W_water=dens*V*g
+W_dens=1000*g
+V_disp=W_water/W_dens           //weight of wter displaced/weight density of water
+mprintf("The Volume of water displaced is %f m^3\n",V_disp)
+//Position of Centre of Buoyancy
+h=V_disp/(w*l)
+mprintf("The Centre of Buoyancy is at %f m from the base\n",h/2)
diff --git a/3819/CH4/EX4.10/Ex4_10.sce b/3819/CH4/EX4.10/Ex4_10.sce
new file mode 100644
index 000000000..b22deea62
--- /dev/null
+++ b/3819/CH4/EX4.10/Ex4_10.sce
@@ -0,0 +1,21 @@
+// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 4-Buoyancy and Floatation
+//// Problem 4.10
+
+//Given Data Set in the Problem
+dens=1000
+g=9.81
+D=4
+h=3
+sg=0.6
+
+//calculations
+d=0.6*h
+AB=d/2
+AG=h/2
+BG=AG-AB
+//For meta centric height
+I_yy=%pi/64*D^4
+V_sub=%pi/4*D^2*d
+GM=I_yy/V_sub-BG
+mprintf("The meta centric height is at %f m\n",GM)
diff --git a/3819/CH4/EX4.11/Ex4_11.sce b/3819/CH4/EX4.11/Ex4_11.sce
new file mode 100644
index 000000000..9f65f8893
--- /dev/null
+++ b/3819/CH4/EX4.11/Ex4_11.sce
@@ -0,0 +1,34 @@
+// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 4-Buoyancy and Floatation
+//// Problem 4.11
+
+//Given Data Set in the Problem
+dens=1000
+g=9.81
+D=3
+d=1.8
+V_disp=0.6
+CB1=1.95
+CG=1.2
+W_tot=3.9*1000*g
+//For meta centric height
+//Weight of water displaced=weight density of water*Volume of water displaced
+x=poly(0,"x")
+function [f]=F(x)           //solves for x=height of body above water surface
+    f=W_tot-(dens*g*(%pi/4*D^2*(1.8-x)+V_disp))
+endfunction
+x=10
+x=fsolve(x,F)
+//Let B2 is the centre of buoyancy of the cylindrical part and B of the whole body
+//for COB of the cylindrical part
+CB2=x+0.5*(1.8-x)
+//COB of the whole body is 
+V_cyl=%pi*(D/2)^2*(1.8-x)
+CB=((V_disp*CB1)+(V_cyl*CB2))/(V_disp+V_cyl)
+//For meta centric height
+BG=CB-CG
+I_yy=%pi/64*D^4
+V_sub=V_disp+V_cyl
+GM=I_yy/V_sub-BG
+mprintf("The Meta centric height is at %f m\n",GM)
+
diff --git a/3819/CH4/EX4.12/Ex4_12.sce b/3819/CH4/EX4.12/Ex4_12.sce
new file mode 100644
index 000000000..164896dc4
--- /dev/null
+++ b/3819/CH4/EX4.12/Ex4_12.sce
@@ -0,0 +1,20 @@
+// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 4-Buoyancy and Floatation
+//// Problem 4.12
+
+//Given Data Set in the Problem
+dens=1000
+g=9.81
+D=4
+d=2.4
+h=4
+SG=0.6
+AB=d/2
+AG=h/2
+BG= AG-AB
+
+//Calculaions
+I=%pi/64*D^4
+Vol=%pi/4*D^2*d
+GM=I/Vol-BG         //Meta centric height
+mprintf("The meta centric height is %f m\n",GM)
diff --git a/3819/CH4/EX4.13/Ex4_13.sce b/3819/CH4/EX4.13/Ex4_13.sce
new file mode 100644
index 000000000..d580d0901
--- /dev/null
+++ b/3819/CH4/EX4.13/Ex4_13.sce
@@ -0,0 +1,36 @@
+// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 4-Buoyancy and Floatation
+//// Problem 4.13
+
+//Given Data Set in the Problem
+dens=1000
+g=9.81
+D=10
+L=40
+l1=1
+S1=6
+dens1=6*dens
+l2=L-l1
+S2=0.6
+dens2=S2*dens
+
+//Calculations
+AG=((%pi/4*D^2*l1*6*0.5)+(%pi/4*D^2*39*S2*(l1+39/2)))/(%pi/4*D^2*l1*6+%pi/4*D^2*39*S2)
+//Finding meta centric point to know whther it can float vertically or not
+//solving func for the value of h equating weight of cylinder to weight of the water displaced
+function [f]=F(h)
+    f=(%pi/4*D^2*39/100*dens2*g+%pi/4*D^2*l1/100*dens1*g-%pi/4*D^2*h/100*dens*g)
+endfunction
+h=10;
+h=fsolve(h,F)
+AB=h/2
+BG=AG-AB
+I=%pi/64*D^4
+Vol=%pi/4*D^2*h
+GM=I/Vol-BG
+if (GM<=0) then mprintf("No,the body cannot float vertically in water\n");
+    
+end
+if GM>=0 then mprintf("Yes,the body can float vertically in water\n");
+    
+end
diff --git a/3819/CH4/EX4.14/Ex4_14.sce b/3819/CH4/EX4.14/Ex4_14.sce
new file mode 100644
index 000000000..f1c178a2b
--- /dev/null
+++ b/3819/CH4/EX4.14/Ex4_14.sce
@@ -0,0 +1,34 @@
+// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 4-Buoyancy and Floatation
+//// Problem 4.14
+
+//Given Data Set in the Problem
+dens=1000
+g=9.81
+W1=686.7*1000
+D=5
+W2=588.6*1000
+w=10.104*10^3
+L=10
+b=7
+
+//calculations
+AG1=2.5/2
+AG2=2.5+D/2
+//dist of common centre of gravity from A is 
+AG=(W1*AG1+W2*AG2)/(W1+W2)
+//Let h be the depth of immerison 
+//Total weight of ythe pontoon and the boiler =weight of the sea water displaced
+function [f]=F(h)
+    f=(W1+W2)-(w*L*b*h)
+endfunction
+h=10;
+h=fsolve(h,F)       //depth of immersion
+//also ,dist of common centre of buoyancy 
+AB=h/2
+BG=AG-AB
+//for meta centric height
+I=1/12*L*b^3
+Vol=L*b*h
+GM=I/Vol-BG
+mprintf("The meta centric height of both the pontoonm and the boiler is %f m \n",GM)
diff --git a/3819/CH4/EX4.15/Ex4_15.sce b/3819/CH4/EX4.15/Ex4_15.sce
new file mode 100644
index 000000000..fc31ae803
--- /dev/null
+++ b/3819/CH4/EX4.15/Ex4_15.sce
@@ -0,0 +1,25 @@
+// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 4-Buoyancy and Floatation
+//// Problem 4.15
+
+//Given Data Set in the Problem
+dens=1000
+g=9.81
+SG1=0.6
+SG2=0.9
+
+//calculations
+L=poly(0,"L")
+d=poly(0,"d")
+AG=L/2
+h=%pi/4*SG1*dens*g*L/(%pi/4*SG2*dens*g)
+AB=h/2
+BG=AG-AB
+//for ,meta centric height ;
+I=%pi/64*d^4
+function [f]=F(k)
+    f(1)=(%pi/64*k(1)^4)/(%pi/4*k(1)^2*(%pi*4*SG1*dens*g*k(2)/(%pi/4*SG2*dens*g)))-k(2)/6
+    f(2)=0          //k(1)=d and k(2)=L
+endfunction
+k=[100 100];
+y=fsolve(k,F);
diff --git a/3819/CH4/EX4.16/Ex4_16.sce b/3819/CH4/EX4.16/Ex4_16.sce
new file mode 100644
index 000000000..d0ad05d15
--- /dev/null
+++ b/3819/CH4/EX4.16/Ex4_16.sce
@@ -0,0 +1,49 @@
+// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 4-Buoyancy and Floatation
+//// Problem 4.16
+
+//Given Data Set in the Problem
+dens=1000
+g=9.81
+D=1
+H=2
+w=7.848*10^3
+dens1=1030
+
+
+//calculations
+//1) to show that it cannot float vertically
+function [f]=F(h)
+    f=w-dens1*g*%pi/4*D^2*h
+endfunction
+h=1
+h=fsolve(h,F)
+//distance of the centre of gravity G,from A is AG 
+AB=h/2
+AG=H/2
+BG=AG-AB
+//now ,meta centric height is equal to 
+I=%pi/64*D^4
+Vol=%pi/4*D^2*h
+GM=I/Vol-BG
+mprintf("The meta centric height is at %f m \n",GM)
+if GM<0 then mprintf("Since M lies below G,Hence,The body cannot float vertically \n")
+    else mprintf("Since M lies above G,Hence,The body can float vertically \n")
+end
+//2)
+T=poly(0,"T")
+F_d=w+T
+//equating the total downward force to weight of awter displaced
+h0=(F_d)/(dens1*g*%pi/4*D^2)
+AB=h0/2
+//Combined CG due to weight of cylinder and the rension in the chain is
+AG=(w*H/2+T*0)/(w+T)
+BG=AG-AB
+//the metacentric height is GM
+I=%pi/64*D^4
+function [g]=G(T)
+    g=(%pi/64*D^4)/(%pi/4*D^2*(w+T)/(dens1*g*%pi/4*D^2))-((w*H/2+T*0)/(w+T))+((w+T)/2/(dens1*g*%pi/4*D^2))
+endfunction
+T=1
+T=fsolve(T,G)
+mprintf("The Force necessary in the chain to keep it vertical is minimum %f N \n",T)
diff --git a/3819/CH4/EX4.17/Ex4_17.sce b/3819/CH4/EX4.17/Ex4_17.sce
new file mode 100644
index 000000000..302f8a566
--- /dev/null
+++ b/3819/CH4/EX4.17/Ex4_17.sce
@@ -0,0 +1,5 @@
+// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 4-Buoyancy and Floatation
+//// Problem 4.17
+
+//Derivation asked(Theoretrical Work)
diff --git a/3819/CH4/EX4.18/Ex4_18.sce b/3819/CH4/EX4.18/Ex4_18.sce
new file mode 100644
index 000000000..2e70e873a
--- /dev/null
+++ b/3819/CH4/EX4.18/Ex4_18.sce
@@ -0,0 +1,5 @@
+// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 4-Buoyancy and Floatation
+//// Problem 4.18
+
+//Derivation required(Theoretical Work)
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)
diff --git a/3819/CH4/EX4.2/Ex4_2.sce b/3819/CH4/EX4.2/Ex4_2.sce
new file mode 100644
index 000000000..771b3f377
--- /dev/null
+++ b/3819/CH4/EX4.2/Ex4_2.sce
@@ -0,0 +1,23 @@
+// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 4-Buoyancy and Floatation
+// Problem 4.2
+
+//Given Data Set in the Problem
+dens=1000
+g=9.81
+d=0.6
+l=5
+SG=0.7
+r=0.3
+//W=theta angle
+//calculations
+
+//Equating the Area of ADCA from using geometry,we get;
+function[f] = F(W)
+    f=0.1979-((%pi*0.3^2*(1-W/180))+0.3^2*cos(W/180*%pi)*sin(W/180*%pi))
+endfunction
+W= 10;
+W = fsolve(W,F)
+//so, h=r+r*cos(theta)
+h=r+r*cos(W/180*%pi)
+mprintf("\nThe depth of wooden log in water is %f m\n",h)
diff --git a/3819/CH4/EX4.20/Ex4_20.sce b/3819/CH4/EX4.20/Ex4_20.sce
new file mode 100644
index 000000000..b52f4b49b
--- /dev/null
+++ b/3819/CH4/EX4.20/Ex4_20.sce
@@ -0,0 +1,15 @@
+// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 4-Buoyancy and Floatation
+//// Problem 4.20
+
+//Given Data Set in the Problem
+dens=1000
+g=9.81
+W=15696
+w1=245.25
+x=8
+theta=4
+
+//Calculations
+GM=w1*x/(W*tan (theta/180*%pi))
+mprintf("The meta centric height is %f m \n",GM)
diff --git a/3819/CH4/EX4.21/Ex4_21.sce b/3819/CH4/EX4.21/Ex4_21.sce
new file mode 100644
index 000000000..d6d8db00d
--- /dev/null
+++ b/3819/CH4/EX4.21/Ex4_21.sce
@@ -0,0 +1,13 @@
+// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 4-Buoyancy and Floatation
+//// Problem 4.21
+
+//Given Data Set in the Problem
+dens=1000
+g=9.81
+K=8
+GM=70/100
+
+//calculations
+T=2*%pi*(K^2/GM/g)^0.5
+mprintf("The Time period of Oscillation is %f seconds\n",T)
diff --git a/3819/CH4/EX4.22/Ex4_22.sce b/3819/CH4/EX4.22/Ex4_22.sce
new file mode 100644
index 000000000..34a337ba5
--- /dev/null
+++ b/3819/CH4/EX4.22/Ex4_22.sce
@@ -0,0 +1,24 @@
+// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 4-Buoyancy and Floatation
+//// Problem 4.22
+
+//Given Data Set in the Problem
+dens=1000
+g=9.81
+T=10
+I=10000
+BG=1.5
+W=29430*10^3
+SG=10100
+
+//calculations
+Vol=W/SG        //vol of water displaced
+//for meta centric height 
+GM=I/Vol-(BG)
+//Using the formula to calculate th eradius of gyration 
+function [f]=F(K)
+    f=T-2*%pi*(K^2/GM/g)^0.5
+endfunction
+K=1
+K=fsolve(K,F)
+mprintf("The Radius of gyration is %f m \n",K)
diff --git a/3819/CH4/EX4.3/Ex4_3.sce b/3819/CH4/EX4.3/Ex4_3.sce
new file mode 100644
index 000000000..83ba43215
--- /dev/null
+++ b/3819/CH4/EX4.3/Ex4_3.sce
@@ -0,0 +1,16 @@
+// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 4-Buoyancy and Floatation
+// Problem 4.3
+
+//Given Data Set in the Problem
+dens=1000
+g=9.81
+w_s_air=392.4
+w_s_water=196.2
+
+//calculations
+vol_disp=w_s_water/(dens*g)
+mprintf("The colume of stone is %f m^3 \n",vol_disp)
+dens_stone=(w_s_air/g)/vol_disp         //finding stones density
+sg=dens_stone/dens
+mprintf("The  SG of stone is %f \n",sg)
diff --git a/3819/CH4/EX4.4/Ex4_4.sce b/3819/CH4/EX4.4/Ex4_4.sce
new file mode 100644
index 000000000..742b2336d
--- /dev/null
+++ b/3819/CH4/EX4.4/Ex4_4.sce
@@ -0,0 +1,18 @@
+// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 4-Buoyancy and Floatation
+// Problem 4.4
+
+//Given Data Set in the Problem
+dens=1000
+g=9.81
+v_body=1.5*1*2
+w_body=196.2
+
+//calculations
+w_disp=dens*g *v_body
+//weight of body in air=wight of water displaced + weight in water.hence
+w_air=w_body+w_disp
+mass=w_air/g
+dens_body=mass/v_body
+SG=dens_body/dens
+mprintf("The Specific Gravity of the body is %f \n",SG)
diff --git a/3819/CH4/EX4.5/Ex4_5.sce b/3819/CH4/EX4.5/Ex4_5.sce
new file mode 100644
index 000000000..914545091
--- /dev/null
+++ b/3819/CH4/EX4.5/Ex4_5.sce
@@ -0,0 +1,23 @@
+// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 4-Buoyancy and Floatation
+// Problem 4.5
+
+//Given Data Set in the Problem
+dens=1000
+g=9.81
+SG=13.6
+v_m=0.4
+v_w=0.6
+V=poly(0,"V")
+
+//calculations
+//For equilibrium of the body ,toatl buoyancy=weight of the body
+//buoyancy due to water
+F_w=dens*g*0.6*V
+//buoyancy due to mercury
+F_m=SG*dens*g*0.4*V
+
+//Total force
+F_tot=F_m+F_w
+dens_body=(F_tot/(V*g))
+mprintf("The density of the body is %f kg/m^3\n",horner(dens_body,1))
diff --git a/3819/CH4/EX4.6/Ex4_6.sce b/3819/CH4/EX4.6/Ex4_6.sce
new file mode 100644
index 000000000..a5ed04afa
--- /dev/null
+++ b/3819/CH4/EX4.6/Ex4_6.sce
@@ -0,0 +1,30 @@
+// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 4-Buoyancy and Floatation
+// Problem 4.6
+
+//Given Data Set in the Problem
+dens=1000
+g=9.81
+sg=0.8
+theta=135
+d=15
+P=9.81
+OB=50
+OD=35
+
+//calculations
+//Let h is the depth
+h=OB*sin((180-theta)*%pi/180)-(OD)          //in cms
+//volume of oil displaced
+v_disp=2/3*%pi*(d/2)^3+h*%pi*(d/2)^2
+F_buoy=sg*dens*g*v_disp*10^-6
+//taking moment about the hinge
+//P*20=(F_buoy-W_float)*(OB*cos 45)
+function[f] = F(W)
+    f = P*20-(F_buoy-W)*(OB*cos((180-theta)/180*%pi))
+endfunction
+W= 10;
+W = fsolve(W,F)
+//Weight of the float
+mprintf("The weight of the float is %f N\n",W)
+
diff --git a/3819/CH4/EX4.7/Ex4_7.sce b/3819/CH4/EX4.7/Ex4_7.sce
new file mode 100644
index 000000000..2e84b2f1d
--- /dev/null
+++ b/3819/CH4/EX4.7/Ex4_7.sce
@@ -0,0 +1,20 @@
+// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 4-Buoyancy and Floatation
+// Problem 4.7
+
+//Given Data Set in the Problem
+dens=1000
+g=9.81
+d_p=5*3*1.2
+d_i=0.8
+AG=0.6
+AB=1/2*d_i
+dens_sw=1025
+
+//Calculations
+I_yy=1/12*5*3^3        //MOI about y-y axis
+V_sub=3*d_i*5
+//hence GM is
+BG=AG-AB
+GM=I_yy/V_sub-BG
+mprintf("The meta centric height is %f m \n",GM)
diff --git a/3819/CH4/EX4.8/Ex4_8.sce b/3819/CH4/EX4.8/Ex4_8.sce
new file mode 100644
index 000000000..595dec6a8
--- /dev/null
+++ b/3819/CH4/EX4.8/Ex4_8.sce
@@ -0,0 +1,23 @@
+// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 4-Buoyancy and Floatation
+// Problem 4.8
+
+//Given Data Set in the Problem
+dens=1000
+g=9.81
+d=3*2*1
+d_i=0.8
+AG=1/2
+AB=d_i/2
+
+//calculations
+//1)Weight of the body
+w=dens*g*(3*2*d_i)
+mprintf("The Weight of the Body is %f N\n",w)
+//2)Meta centric height
+I_yy=1/12*3*2^3        //MOI about y-y axis
+V_sub=3*2*0.8
+BG=AG-AB
+//Hence meta centric height is 
+GM=I_yy/V_sub-BG
+mprintf("The meta centric height is %f m \n",GM)
diff --git a/3819/CH4/EX4.9/Ex4_9.sce b/3819/CH4/EX4.9/Ex4_9.sce
new file mode 100644
index 000000000..75dc22ab6
--- /dev/null
+++ b/3819/CH4/EX4.9/Ex4_9.sce
@@ -0,0 +1,30 @@
+// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 4-Buoyancy and Floatation
+// Problem 4.9
+
+//Given Data Set in the Problem
+dens=1000
+g=9.81
+V=2*1*0.8
+sg=0.7
+
+//calculations
+h=poly(0,"h")
+w_d=dens*g*2*1*h
+//we know thtat at equilibrium; weight of wooden piece =weight of wter displaced
+w_w=sg*dens*g*2*1*0.8
+function[f] = F(h)
+f=w_w-(dens*g*2*1*h)            //w_wood-w_displaced
+endfunction
+h=1
+h=fsolve(h,F)
+//For centre of buoyancy
+AB=h/2
+AG=0.8/2
+BG=AG-AB
+//Meta centric heinght
+I_yy=1/12*2*1^3
+v_sub=2*1*h
+//hence GM is 
+GM=I_yy/v_sub-BG
+mprintf("The Meta centric height is %f m\n",GM)