diff options
Diffstat (limited to '3819/CH1')
32 files changed, 588 insertions, 0 deletions
diff --git a/3819/CH1/EX1.1/Ex1_1.sce b/3819/CH1/EX1.1/Ex1_1.sce new file mode 100644 index 000000000..c69f8ad26 --- /dev/null +++ b/3819/CH1/EX1.1/Ex1_1.sce @@ -0,0 +1,13 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.1
+W=7
+V=1/1000
+g=9.81
+d_water=1000
+w=W/V
+mprintf("The Specific weight of the liquid is %f \n",w)
+d=w/g
+mprintf("The density of the liquid is %f \n",d)
+SG=d/d_water
+mprintf("The Specific Gravity o fthe liquid is %f \n",SG)
diff --git a/3819/CH1/EX1.10/Ex1_10.sce b/3819/CH1/EX1.10/Ex1_10.sce new file mode 100644 index 000000000..0edc8dda9 --- /dev/null +++ b/3819/CH1/EX1.10/Ex1_10.sce @@ -0,0 +1,14 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.10
+
+
+//Given Data Set in the Problem
+density=981
+ss=0.2452
+vel_grad=0.2
+
+//Calculations
+visc=ss/(vel_grad)
+kin_visc=visc/density
+mprintf("The Kinematic viscosity of the oil is %f stokes\n",kin_visc*10^4)
diff --git a/3819/CH1/EX1.11/Ex1_11.sce b/3819/CH1/EX1.11/Ex1_11.sce new file mode 100644 index 000000000..f4d652dba --- /dev/null +++ b/3819/CH1/EX1.11/Ex1_11.sce @@ -0,0 +1,15 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.11
+
+
+//Given Data Set in the Problem
+visc=0.05/10
+kin_visc=0.035/(10^4)
+dens_water=1000
+
+//Calculations
+dens_oil=visc/kin_visc
+SG=dens_oil/dens_water
+mprintf("The Specifc Gravity of Oil is %f \n",SG)
+
diff --git a/3819/CH1/EX1.12/Ex1_12.sce b/3819/CH1/EX1.12/Ex1_12.sce new file mode 100644 index 000000000..944a5687d --- /dev/null +++ b/3819/CH1/EX1.12/Ex1_12.sce @@ -0,0 +1,15 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.12
+
+
+//Given Data Set in the Problem
+kin_visc=6*10^-4
+SG=1.9
+dens_water=1000
+
+//Calculations
+dens_liquid=SG*dens_water
+visc=dens_liquid*kin_visc //Kinematic viscosity=Dynamic Viscosity/density of liquid
+mprintf("The Dynamic viscosity of th liquid is %f poise \n",visc*10)
+
diff --git a/3819/CH1/EX1.13/Ex1_13.sce b/3819/CH1/EX1.13/Ex1_13.sce new file mode 100644 index 000000000..c8ff859a7 --- /dev/null +++ b/3819/CH1/EX1.13/Ex1_13.sce @@ -0,0 +1,15 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.13
+
+
+//Given Data Set in the Problem
+y=poly(0,"y")
+u=3*y/4-y^2
+visc=8.5/10
+
+//Calculations
+du_dy=(horner(derivat(u),0.15))
+
+ss=visc*du_dy
+mprintf("The shear stress at y=0.15 m is %f N/m^2 \n",ss)
diff --git a/3819/CH1/EX1.14/Ex1_14.sce b/3819/CH1/EX1.14/Ex1_14.sce new file mode 100644 index 000000000..5d27e410b --- /dev/null +++ b/3819/CH1/EX1.14/Ex1_14.sce @@ -0,0 +1,24 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.15
+
+
+//Given Data Set in the Problem(SI Units)
+visc=6/10
+D=0.4
+N=190
+L=90/1000
+t=1.5/1000
+
+
+//Calculations
+u_tangent=%pi*D*N/60
+du=u_tangent-0
+dy=t
+ss=visc*du/dy
+Area=%pi*D*L
+Force=ss*Area //Force =shear stress *Area
+T=Force*D/2 //Torque =Force*(D/2)
+Power_lost=(2*%pi/60)*N*T //Power lost =(2*pi/60)*Torgue*Speed of the shaft
+mprintf("The Power lost in the bearing of the sleeve is %f W",Power_lost)
+
diff --git a/3819/CH1/EX1.15/Ex1_15.sce b/3819/CH1/EX1.15/Ex1_15.sce new file mode 100644 index 000000000..b813781a2 --- /dev/null +++ b/3819/CH1/EX1.15/Ex1_15.sce @@ -0,0 +1,18 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.15
+
+
+//Given Data Set in the Problem(SI Units)
+dist=20/100
+u_vertex=120/100
+visc=8.5/10
+y=poly(0,"y")
+a=poly(0,"a")
+b=poly(0,"b")
+c=poly(0,"c")
+c=2
+a=2
+b-2
+u=a*y^2+b*y+c
+s=poly(0,'s')
diff --git a/3819/CH1/EX1.16/Ex1_16.sce b/3819/CH1/EX1.16/Ex1_16.sce new file mode 100644 index 000000000..99a0ebafe --- /dev/null +++ b/3819/CH1/EX1.16/Ex1_16.sce @@ -0,0 +1,19 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.16
+
+
+//Given Data Set in the Problem(SI Units)
+F1=40
+F2=200
+u1=50/100
+
+//Calculations
+//We know, Shear stress=Force/Area=viscosity*(Velocity Gradient)
+//ie, F/A=viscosity*(u/y)
+//F/u=Viscosity*(A/y)
+//F1/u1=F2/u2=constant
+u2=F2*u1/F1
+mprintf("The Speed of the sleeve when a force of 200N is applied is %f cm/s",u2*100)
+
+
diff --git a/3819/CH1/EX1.17/Ex1_17.sce b/3819/CH1/EX1.17/Ex1_17.sce new file mode 100644 index 000000000..e297a278c --- /dev/null +++ b/3819/CH1/EX1.17/Ex1_17.sce @@ -0,0 +1,23 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.17
+
+
+//Given Data Set in the Problem(SI Units)
+d=15/100
+d_outer=15.10/100
+l=25/100
+T=12
+N=100
+
+
+//Calculations
+u_tang=%pi*d*N/60
+Area_surface=%pi*d*l
+du=u_tang-0
+dy=(d_outer-d)/2
+ //We know, Shear stress=Force/Area=viscosity*(Velocity Gradient)
+ //also, Torque=Force*Diameter/2.......or.. Force=(Torque*2)/Diameter
+ //hence, 2*Torque/(diameter*area)=Viscosity*(Vel. gradient)
+visc=2*T/(d*Area_surface*du/dy)
+mprintf("The Viscosity of the liquid is %f poise",visc*10)
diff --git a/3819/CH1/EX1.18/Ex1_18.sce b/3819/CH1/EX1.18/Ex1_18.sce new file mode 100644 index 000000000..5afdfdf65 --- /dev/null +++ b/3819/CH1/EX1.18/Ex1_18.sce @@ -0,0 +1,28 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.18
+
+
+//Given Data Set in the Problem(SI Units)
+Area=0.5
+du=0.6
+visc=0.81
+y=2.4/100
+dy=2.4/2/100
+
+//Calculations
+//Case 1:When the thin plate is in the middle
+ss=visc*(du/dy)
+F_upper=ss*Area
+F_lower=ss*Area
+F=F_upper+F_lower
+mprintf("The Total shear force on the thin plate in the middle of the two plates is %f N \n",F)
+
+//Case 2: When the palte is at a distanvce of 0.8 cm from one plate
+dy_upper=y-0.8/100
+dy_lower=0.8/100
+F_upper2=visc*du/dy_upper*Area
+F_lower2=visc*du/dy_lower*Area
+F2=F_upper2+F_lower2
+mprintf("The Total shear force on the thin plate at a distance 0.8 cm from one plate is %f N \n",F2)
+
diff --git a/3819/CH1/EX1.19/Ex1_19.sce b/3819/CH1/EX1.19/Ex1_19.sce new file mode 100644 index 000000000..d32092169 --- /dev/null +++ b/3819/CH1/EX1.19/Ex1_19.sce @@ -0,0 +1,28 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.19
+
+
+//Given Data Set in the Problem(SI Units)
+gap=2.2/100
+visc=2
+SG=0.9
+g=9.81
+W_dens=SG*1000*g
+Vol=1.2*1.2*0.2/100
+Area=1.2*1.2
+t=0.2/100
+vel=0.15
+W=40
+
+//Calculations
+dis_from_plate=(gap-t)/2 //Distance of the plate from each of the two plates
+ss=visc*(vel/dis_from_plate)
+Force_left=ss*1.2*1.2
+Force_right=ss*1.2*1.2
+F=Force_left+Force_right //Sum of Force on the right + left side of the plate
+Upthrust=W_dens*Vol //Calculates Buoyant force on the plate
+F_down=W-Upthrust //net downward force on the plate except shear forces
+F_ToLift=F+F_down //som total of all forces on the plate
+mprintf("The Force required to lift the plate is %f N \n",F_ToLift)
+
diff --git a/3819/CH1/EX1.2/Ex1_2.sce b/3819/CH1/EX1.2/Ex1_2.sce new file mode 100644 index 000000000..cc0354758 --- /dev/null +++ b/3819/CH1/EX1.2/Ex1_2.sce @@ -0,0 +1,20 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.21
+
+//Given Data Set in the Problem
+V=1/1000
+SG=0.7
+d_water=1000
+g=9.81
+
+//Calculations
+// Density of Petrol
+d=SG*d_water
+mprintf("The Density of Petrol is %f \n",d)
+//Specific Weight of Petrol
+w=d*g
+mprintf("The Specific weight of Petrol is %f \n",w)
+// Weight of 1 litre of Petrol
+W=w*V
+mprintf("The Weight of 1 litre of Petrol is %f \n",W)
diff --git a/3819/CH1/EX1.20/Ex1_20.sce b/3819/CH1/EX1.20/Ex1_20.sce new file mode 100644 index 000000000..688b9bda8 --- /dev/null +++ b/3819/CH1/EX1.20/Ex1_20.sce @@ -0,0 +1,20 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.20
+
+
+//Given Data Set in the Problem(SI Units)
+w=16
+t=25
+T=273+t
+p=0.25*10^6
+g=9.81
+
+//Calculations
+//1)Density
+density=w/g
+mprintf("The Density of the gas is %f kg/m^3 \n",density)
+
+//2)Gas consatnt
+R=p/(density*T)
+mprintf("The gas constant is %f Nm/kg-K \n",R)
diff --git a/3819/CH1/EX1.21/Ex1_21.sce b/3819/CH1/EX1.21/Ex1_21.sce new file mode 100644 index 000000000..30a706dd0 --- /dev/null +++ b/3819/CH1/EX1.21/Ex1_21.sce @@ -0,0 +1,26 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.21
+
+
+//Given Data Set in the Problem(SI Units)
+V1=0.6
+t=50
+T1=273+t
+P1=0.3*10^6
+V2=0.3
+k=1.4
+
+//Calculations
+//1) Isothermal
+//Using pv=constant
+P2=P1*V1/V2
+mprintf("The Final Pressure for isothermal conditions is %f N/mm^2 \n",P2*10^-6)
+
+//2) Adiabatic
+//Using PV^k=constant or P1V1^k=P2 V2^k
+P2=P1*(V1/V2)^k
+mprintf("The Final Pressure for Adiabatic conditions is %f N/mm^2 \n",P2*10^-6)
+//Using T V^(k-1) = constant
+T2=T1*(V1/V2)^(k-1)
+mprintf("The Final Temperature for Adiabatic conditions is %f C",T2-273)
diff --git a/3819/CH1/EX1.22/Ex1_22.sce b/3819/CH1/EX1.22/Ex1_22.sce new file mode 100644 index 000000000..b40bf6fb9 --- /dev/null +++ b/3819/CH1/EX1.22/Ex1_22.sce @@ -0,0 +1,16 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.22
+
+//Given Data Set in the Problem(SI Units)
+
+m=5
+t=10
+T=273+10
+V=0.4
+M=28
+R=8314 //Universal Gas constant in N-m/(kg-mole K)
+
+//Calculations
+p=((m/M)*R*T)/V
+mprintf("The pressure exerted by the 5kg Nitrogen gas is %f N/mm^2 \n",p*10^-6);
diff --git a/3819/CH1/EX1.23/Ex1_23.sce b/3819/CH1/EX1.23/Ex1_23.sce new file mode 100644 index 000000000..217a7fe5f --- /dev/null +++ b/3819/CH1/EX1.23/Ex1_23.sce @@ -0,0 +1,16 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.23
+
+//Given Data Set in the Problem(SI Units)
+
+p_i=70
+p_f=130
+dp=p_f-p_i
+dV_V=0.15/100 //Using dV/V=-dP/P
+
+//Calculations
+//Using K=dP/(-dV/V)
+K=dp/(dV_V)
+mprintf("The Bulk modulus of elasticity of the liquid is %f N.cm^2",K);
+
diff --git a/3819/CH1/EX1.24/Ex1_24.sce b/3819/CH1/EX1.24/Ex1_24.sce new file mode 100644 index 000000000..5ea092e15 --- /dev/null +++ b/3819/CH1/EX1.24/Ex1_24.sce @@ -0,0 +1,16 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.24
+
+//Given Data Set in the Problem(SI Units)
+V_i=0.0125
+V_f=0.0124
+p_i=80
+p_f=150
+
+//Caclulations
+dV=V_i-V_f
+dV_V=-dV/V_i
+dp=p_f-p_i
+K=dp/(-dV_V) //Using K=dP/(-dV/V)=Bulk modulus of elasticity
+mprintf("The bulk modulus of elasticity of the liquid is %f N/cm^2",K);
diff --git a/3819/CH1/EX1.25/Ex1_25.sce b/3819/CH1/EX1.25/Ex1_25.sce new file mode 100644 index 000000000..63c6f78f7 --- /dev/null +++ b/3819/CH1/EX1.25/Ex1_25.sce @@ -0,0 +1,12 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.25
+
+//Given Data Set in the Problem(SI Units)
+st=0.0725 //Surface tension
+p=0.02*10^4
+
+//Calculations
+//Using pressure =(4*Surface tension)/(diameter of the droplet)
+d=4*st/p
+mprintf("The diameter of the droplet is %f mm",d*10^3);
diff --git a/3819/CH1/EX1.26/Ex1_26.sce b/3819/CH1/EX1.26/Ex1_26.sce new file mode 100644 index 000000000..6876533ce --- /dev/null +++ b/3819/CH1/EX1.26/Ex1_26.sce @@ -0,0 +1,13 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.26
+
+//Given Data Set in the Problem(SI Units)
+d=40*10^-3
+p=2.5
+
+
+//Calculations
+//Using Pressure =8*Surface tension/diameter of the soap bubble
+st=p*d/8
+mprintf("The Surface tension inside the soap bubble is %f N/m",st)
diff --git a/3819/CH1/EX1.27/Ex1_27.sce b/3819/CH1/EX1.27/Ex1_27.sce new file mode 100644 index 000000000..6fa16b3b2 --- /dev/null +++ b/3819/CH1/EX1.27/Ex1_27.sce @@ -0,0 +1,15 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.27
+
+//Given Data Set in the Problem(SI Units)
+d=0.04*10^-3
+p_outside=10.32*10^4
+st=0.0725
+
+//Calculations
+//Using pressure =(4*Surface tension)/(diameter of the droplet)
+p=4*st/d
+//But this pressure obtained is p_inside-p_outside thus,
+p_inside=p_outside+p
+mprintf("The pressure inside the droplet is %f n/cm^2",p_inside*10^-4);
diff --git a/3819/CH1/EX1.28/Ex1_28.sce b/3819/CH1/EX1.28/Ex1_28.sce new file mode 100644 index 000000000..514384741 --- /dev/null +++ b/3819/CH1/EX1.28/Ex1_28.sce @@ -0,0 +1,23 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.28
+
+//Given Data Set in the Problem
+d=2.5*10^-3
+st_w=0.0725
+st_m=0.52
+SG_m=13.6
+dens_w=1000
+dens_m=13.6*1000
+g=9.81
+
+//Calculations
+//Using rise=4*surface tension/(density *g *diameter of capillary)
+//CAPILLARY RISE FOR WATER (theta =0,cos 0=1)
+h=4*st_w/(dens_w*g*d)
+mprintf("The rise for water is %f cm \n",h*100)
+
+//CAPILLARY RISE FOR MERCURY
+//Using rise=4*surface tension/(density *g *diameter of capillary)
+h=4*st_m*cos(%pi*130/180)/(dens_m*g*d)
+mprintf("The rise for mercury is %f cm",h*100)
diff --git a/3819/CH1/EX1.29/Ex1_29.sce b/3819/CH1/EX1.29/Ex1_29.sce new file mode 100644 index 000000000..8a4038cd6 --- /dev/null +++ b/3819/CH1/EX1.29/Ex1_29.sce @@ -0,0 +1,23 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.29
+
+//Given Data Set in the Problem
+d=4*10^-3
+st_w=0.073575
+st_m=0.51
+SG_m=13.6
+dens_w=998
+dens_m=13.6*1000
+g=9.81
+
+//Calculations
+//CAPILLARY RISE FOR WATER (theta =0,cos 0=1)
+//Using rise=4*surface tension/(density *g *diameter of capillary)
+h=4*st_w/(dens_w*g*d)
+mprintf("The rise for water is %f mm \n",h*1000)
+
+//CAPILLARY RISE FOR MERCURY
+//Using rise=4*surface tension/(density *g *diameter of capillary)
+h=4*st_m*cos(%pi*130/180)/(dens_m*g*d)
+mprintf("The rise for mercury is %f m",h*100)
diff --git a/3819/CH1/EX1.3/Ex1_3.sce b/3819/CH1/EX1.3/Ex1_3.sce new file mode 100644 index 000000000..d7be0dab3 --- /dev/null +++ b/3819/CH1/EX1.3/Ex1_3.sce @@ -0,0 +1,12 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.3
+
+y=poly(0,"y")
+u=2/3*y-y^2 // Defining the Velocity Function
+a=derivat(u) //Taking Derivative of the Velocity
+visc=8.63/10 //Converting Dynamic Viscosity from poise to N s/m^2
+ss1=visc*horner(a,0) //Shear stress=(Dynamic viscosity *Velocity Gradient) at y=0
+mprintf("The shear stress at y=0 is %f N/m^2 \n",ss1)
+ss2=visc*horner(a,0.15) //Shear stress=(Dynamic viscosity *Velocity Gradient) at y=0.15
+mprintf("The shear stress at y=0.15 is %f ",ss2)
diff --git a/3819/CH1/EX1.30/Ex1_30.sce b/3819/CH1/EX1.30/Ex1_30.sce new file mode 100644 index 000000000..c158d5e4d --- /dev/null +++ b/3819/CH1/EX1.30/Ex1_30.sce @@ -0,0 +1,16 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.30
+
+//Given Data Set in the Problem
+h=0.2*10^-3
+st=0.0725
+dens=1000
+g=9.81
+
+//Calculations
+//Using rise=4*surface tension/(density *g *diameter of capillary)
+d=4*st/(dens*g*h)
+mprintf("The diameter oif the capillary for the rise of 0.2 mm is %f cm",d*100)
+
+
diff --git a/3819/CH1/EX1.31/Ex1_31.sce b/3819/CH1/EX1.31/Ex1_31.sce new file mode 100644 index 000000000..6c46b9692 --- /dev/null +++ b/3819/CH1/EX1.31/Ex1_31.sce @@ -0,0 +1,15 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.31
+
+//Given Data Set in the Problem
+h=2*10^-3
+st=0.073575
+theta=0
+dens=1000
+g=9.81
+
+//Calculations
+//Using rise=4*surface tension/(density *g *diameter of capillary)
+d=4*st/(dens*g*h)
+mprintf("The diameter of the capillary is %f cm",d*100)
diff --git a/3819/CH1/EX1.32/Ex1_32.sce b/3819/CH1/EX1.32/Ex1_32.sce new file mode 100644 index 000000000..bdc9c1d1d --- /dev/null +++ b/3819/CH1/EX1.32/Ex1_32.sce @@ -0,0 +1,24 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.32
+
+//Given Data Set in the Problem
+visc=5/10
+D=0.5
+N=200
+L=100/10^3
+t=1*10^-3
+
+//Calculations
+//Using , tangential velocity=(pi*D*N)/60
+u_tang=%pi*D*N/60
+du=u_tang-0
+dy=t
+du_dy=du/dy
+ss=visc*(du_dy) //Shear stress =viscosity*Velocity gradient
+Area=%pi*D*L
+F_shear=ss*Area
+T=F_shear*D/2 //Torque=Shear force*D/2
+Power_lost=T*(2*%pi*N/60) //Power lost =Torque*(2*pi*N/60)
+mprintf("ThePower lost by the sleeve of 100m in oil is %f kW",Power_lost*10^-3)
+
diff --git a/3819/CH1/EX1.4/Ex1_4.sce b/3819/CH1/EX1.4/Ex1_4.sce new file mode 100644 index 000000000..45977bf74 --- /dev/null +++ b/3819/CH1/EX1.4/Ex1_4.sce @@ -0,0 +1,20 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.4
+
+//Given Data Set in the Problem
+dy=0.025/1000
+v=60/100
+ss=2
+
+//Calculations
+//To find the Viscosity
+//Shear Stress=Viscosity * Velocity gradient
+du=(60-0)/100
+vel_grad=du/dy //Defining velocity gradient across the plate
+visc=ss/vel_grad
+visc_poise=visc*10 //Converting viscosity to poise from Ns/m^2
+mprintf("The Viscosity between the plates is %f poise",visc_poise)
+
+
+
diff --git a/3819/CH1/EX1.5/Ex1_5.sce b/3819/CH1/EX1.5/Ex1_5.sce new file mode 100644 index 000000000..9af4e26f1 --- /dev/null +++ b/3819/CH1/EX1.5/Ex1_5.sce @@ -0,0 +1,20 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.5
+
+//Given Data Set in the Problem
+Area=1500000/(1000)^2 //Area in m^2
+du=0.4
+dy=0.15/1000 //Distance between the plates In metres
+visc=1/10 //In SI Units of Ns/m^2
+
+//Calulations
+//Force required to maintain that speed
+ss=visc*(du/dy) //ss is the shear stress
+Force=ss*Area //Force required= Shear stress * Area
+mprintf("The Force required to maintain the speed is %f N\n",Force)
+
+//Power required
+Power=Force*du //Power =(Force)*(Speed at which the plate has to be kept moving)
+mprintf("The Power required to maintain the speed is %f W\n ",Power)
+
diff --git a/3819/CH1/EX1.6/Ex1_6.sce b/3819/CH1/EX1.6/Ex1_6.sce new file mode 100644 index 000000000..abc3f1a19 --- /dev/null +++ b/3819/CH1/EX1.6/Ex1_6.sce @@ -0,0 +1,15 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.6
+
+//Given Data Set in the Problem
+visc=1/10 //In SI Units
+D=10/100 //In SI Units
+dy=1.5/1000 //Distance between shaft and journal bearing
+N=150 //In RPM
+
+//Calculations
+//Intensity of the shear due to the Oil
+du= (%pi*D*N)/60 //du=(πDN)/60....The tangential velocity which causes shaer
+ss=visc*(du/dy)
+mprintf("The Shear stress due to the oil is %f N/m^2 \n",ss)
diff --git a/3819/CH1/EX1.7/Ex1_7.sce b/3819/CH1/EX1.7/Ex1_7.sce new file mode 100644 index 000000000..65bd40066 --- /dev/null +++ b/3819/CH1/EX1.7/Ex1_7.sce @@ -0,0 +1,17 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.7
+
+//Given Data Set in the Problem
+Area=0.8*0.8
+theta=%pi/6
+W=300
+du=0.3
+dy=1.5/1000
+
+//Calculations
+W_alongPlane=W*cos(%pi/2-theta)
+Shear_Force=W_alongPlane
+ss=Shear_Force/Area
+visc=ss/(du/dy) //Shear Stress+Viscosity * Velocity Gradient
+mprintf("The Dynamic Viscosity of the Oil is %f poise",visc*10)
diff --git a/3819/CH1/EX1.8/Ex1_8.sce b/3819/CH1/EX1.8/Ex1_8.sce new file mode 100644 index 000000000..e39a2f019 --- /dev/null +++ b/3819/CH1/EX1.8/Ex1_8.sce @@ -0,0 +1,13 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.8
+
+//Given data set in the problem
+dy=1.25/100
+visc=14/10
+u=2.5
+
+//Calculations
+ss=visc*((u-0)/dy) //shear stress=viscosity*(velocity gradient across the oil)
+mprintf("The shear stress between the plates is %f N/m^2",ss)
+
diff --git a/3819/CH1/EX1.9/Ex1_9.sce b/3819/CH1/EX1.9/Ex1_9.sce new file mode 100644 index 000000000..8f33b7da3 --- /dev/null +++ b/3819/CH1/EX1.9/Ex1_9.sce @@ -0,0 +1,24 @@ +// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
+// Chapter 1-Properties of Fluid
+// Problem 1.9
+
+
+//Given Data Set in the Problem
+Area=(60*60)/(100*100)
+dy=12.5/1000
+u=2.5
+du=u-0
+Force=98.1
+ss=Force/Area
+
+//Calculations
+//1)Dynamic viscosity of Oil in poise
+ //Shear Stress=(Force/Area)=viscosity*Velocity gradient
+Dyn_visc=ss/(du/dy)
+mprintf("The Dynamic Viscosity o fthe oil is %f poise \n",Dyn_visc*10)
+
+//2) Kinematic viscosity of th eoil in stokes in SG of Oil is 0.95
+SG=0.95
+density_oil=SG*1000
+Kin_visc=Dyn_visc/density_oil
+mprintf("The Kinematic viscosity of the oil is %f stokes",Kin_visc*10^4)
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