initial commit / add all books

23 files changed, 371 insertions, 0 deletions

diff --git a/858/CH4/EX4.1/example_1.sce b/858/CH4/EX4.1/example_1.sce
new file mode 100755
index 000000000..d87134ff3
--- /dev/null
+++ b/858/CH4/EX4.1/example_1.sce
@@ -0,0 +1,9 @@
+clc

+clear 

+printf("example 4.1 page number 125\n\n")

+

+//to find water compressibility

+delta_p=70;   //in bar

+Et=20680     //in bar

+compressibility = delta_p/Et;

+printf("compressibilty of water = %f",compressibility)

diff --git a/858/CH4/EX4.10/example_10.sce b/858/CH4/EX4.10/example_10.sce
new file mode 100755
index 000000000..ed8b4f6e7
--- /dev/null
+++ b/858/CH4/EX4.10/example_10.sce
@@ -0,0 +1,13 @@
+clc

+clear 

+printf("example 4.10 page number 139\n\n")

+

+//to find the temperature increase

+

+Q=0.001*10^5   //in J/s

+w=0.001*1000   //in kg/s

+density=1000   //in kg/m3

+cp=4.19*10^3   //in J/kg K

+

+delta_T=Q/(w*cp);

+printf("Temperature increase = %f degree celcius",delta_T)

diff --git a/858/CH4/EX4.11/example_11.sce b/858/CH4/EX4.11/example_11.sce
new file mode 100755
index 000000000..a10120041
--- /dev/null
+++ b/858/CH4/EX4.11/example_11.sce
@@ -0,0 +1,23 @@
+clc

+clear 

+printf("example 4.11 page number 142\n\n")

+

+//to find the pressure

+

+u1=0;   //in m/s

+ws=0;

+P1=0.7*10^5    //in Pa

+P3=0

+density=1000   //in kg/m3

+

+u3=((2*(P1-P3))/density)^0.5;

+printf("u3 = %f m/s",u3)

+

+ratio_area=0.5;

+u2=u3/ratio_area;

+printf("\n\nu2 = %f m/s",u2)

+

+//applying bernoulli's equation

+P2=1.7*10^5-((density*u2^2)/2)

+printf("\n\nP2 = %f Pa",P2)

+printf("\nthis flow is physically unreal")

diff --git a/858/CH4/EX4.12/example_12.sce b/858/CH4/EX4.12/example_12.sce
new file mode 100755
index 000000000..0d524f6f8
--- /dev/null
+++ b/858/CH4/EX4.12/example_12.sce
@@ -0,0 +1,19 @@
+clc

+clear 

+printf("example 4.12 page number 143\n\n")

+

+//to find the power requirements

+

+Q=3800/(24*3600)    //in m3/s

+d=0.202    //in m

+

+u=Q/((3.14/4)*d^2);    //in m/s

+delta_P=5.3*10^6    //in Pa

+density=897    //in kg/m3

+F=delta_P/density;    //in J/kg

+ws=9.8*30+F;

+mass_flow_rate= Q*density;

+power=(ws*mass_flow_rate)/0.6;

+

+printf("power required = %f kW",power/1000)

+

diff --git a/858/CH4/EX4.13/example_13.sce b/858/CH4/EX4.13/example_13.sce
new file mode 100755
index 000000000..6d0973283
--- /dev/null
+++ b/858/CH4/EX4.13/example_13.sce
@@ -0,0 +1,22 @@
+clc

+clear 

+printf("example 4.13 page number 146\n\n")

+

+//to find the tube length

+density=1000    //in kg/m3

+viscosity=1*10^-3   //in Pa s

+P=100*1000    //in Pa

+

+vdP=P/density;

+

+Q=2.5*10^-3/(24*3600)

+A=3.14*(0.0005)^2/4;

+u=Q/A;

+printf("u = %f m/s",u)

+

+Re=density*u*0.0005/viscosity;

+printf("\n\nRe = %f",Re)

+

+//F=18.86*L

+L=(-u^2+vdP)/18.86;

+printf("\n\nL = %f m",L)

diff --git a/858/CH4/EX4.14/example_14.sce b/858/CH4/EX4.14/example_14.sce
new file mode 100755
index 000000000..4cf05a660
--- /dev/null
+++ b/858/CH4/EX4.14/example_14.sce
@@ -0,0 +1,20 @@
+clc

+clear 

+printf("example 4.14 page number 151\n\n")

+

+//to find the discharge pressure

+d=0.025     //in m

+u=3        //in m/s

+density=894   //in kg/m3

+viscosity=6.2*10^4     //in Pa-s

+

+Re=(u*d*density)/viscosity;

+f=0.0045;

+L=50;

+

+delta_P=2*f*density*u^2*(L/d)

+printf("frictional head loss = %f kPa",delta_P/1000)

+

+required_P=25*density*9.8;

+total_head=delta_P+required_P;

+printf("\n\ntotal pressure head = %f bar",total_head/10^5)

diff --git a/858/CH4/EX4.15/example_15.sce b/858/CH4/EX4.15/example_15.sce
new file mode 100755
index 000000000..2aa7c554e
--- /dev/null
+++ b/858/CH4/EX4.15/example_15.sce
@@ -0,0 +1,20 @@
+clc

+clear 

+printf("example 4.15 page number 152\n\n")

+

+//to find the level difference

+

+Q=0.8*10^-3;   //in m3/s

+d=0.026    //in m

+A=(3.14*(d^2))/4   //in m2

+

+u=Q/A;    //in m/s

+density=800    //in kg/m3

+viscosity=0.0005   //in Pa-s

+

+Re=(u*density*d)/viscosity;

+f=0.079*(Re)^-0.25;

+L=60

+h_f=2*f*((u^2)/9.8)*(L/d);

+

+printf("level difference = %f m",h_f)

diff --git a/858/CH4/EX4.16/example_16.sce b/858/CH4/EX4.16/example_16.sce
new file mode 100755
index 000000000..ad8351654
--- /dev/null
+++ b/858/CH4/EX4.16/example_16.sce
@@ -0,0 +1,24 @@
+clc

+clear 

+printf("example 4.16 page number 153\n\n")

+

+//to find the engery cost

+delta_z=50;    //in m

+L=290.36    //in m

+d=0.18    //in m

+Q=0.05    //in m3/s

+

+A=(3.14*d^2)/4;    //in m2

+u=Q/A;    //in m/s

+density=1180;   //in kg/m3

+viscosity=0.0012   //in Pa-s

+Re=u*density*d/viscosity;

+

+f=0.004;

+sigma_F=2*f*u^2*L/d;

+ws=((9.8*50)+sigma_F)/0.6;

+mass_flow_rate=Q*density;    //in Kg/s

+power=mass_flow_rate*ws/1000;   //in KW

+energy_cost=power*24*0.8;

+

+printf("Energy cost = Rs %f",energy_cost)

diff --git a/858/CH4/EX4.17/example_17.sce b/858/CH4/EX4.17/example_17.sce
new file mode 100755
index 000000000..2656cc805
--- /dev/null
+++ b/858/CH4/EX4.17/example_17.sce
@@ -0,0 +1,20 @@
+clc

+clear 

+printf("example 4.17 page number 154\n\n")

+

+//to find the pressure loss

+density=998   //in kg/m3

+viscosity=0.0008  //in Pa-s

+d=0.03   //in m

+u=1.2   //in m/s

+

+Re=density*d*u/viscosity;

+

+f=0.0088;

+D=1   //in m

+N=10

+L=3.14*D*N;

+delta_P=(2*f*u^2*L)/d;   //in Pa

+delta_P_coil=delta_P*(1+(3.54*(d/D)));

+

+printf("frictional pressure drop = %f kPa",delta_P_coil)

diff --git a/858/CH4/EX4.18/example_18.sce b/858/CH4/EX4.18/example_18.sce
new file mode 100755
index 000000000..5f7d40a9f
--- /dev/null
+++ b/858/CH4/EX4.18/example_18.sce
@@ -0,0 +1,20 @@
+clc

+clear 

+printf("example 4.18 page number 154\n\n")

+

+//to find pressure drop per unit length

+

+b=0.050    //in m

+a=0.025    //in m

+d_eq=b-a  //in m

+density=1000  //in kg/m3

+u=3    //in m/s

+viscosity = 0.001

+

+Re=d_eq*u*density/viscosity;

+

+e=40*10^6   //in m

+f=0.0062;

+P_perunit_length=2*f*density*u^2/d_eq;   //in Pa/m

+

+printf("pressure per unit length = %f Pa/m",P_perunit_length)

diff --git a/858/CH4/EX4.19/example_19.sce b/858/CH4/EX4.19/example_19.sce
new file mode 100755
index 000000000..eb57ad25e
--- /dev/null
+++ b/858/CH4/EX4.19/example_19.sce
@@ -0,0 +1,10 @@
+clc

+clear 

+printf("example 4.19 page number 155\n\n")

+

+//to find the flow rate

+d = 0.3   //in m

+u = 17.63   //avg velocity in m/s

+

+q = (3.14/4)*d^2*u;

+printf("volumetric flow rate = %f cubic meter per second",q)

diff --git a/858/CH4/EX4.2/example_2.sce b/858/CH4/EX4.2/example_2.sce
new file mode 100755
index 000000000..faf242b5a
--- /dev/null
+++ b/858/CH4/EX4.2/example_2.sce
@@ -0,0 +1,5 @@
+clc

+clear 

+printf("example 4.2 page number 125\n\n")

+

+disp("this is a theoritical problem,book shall be referred for solution")

diff --git a/858/CH4/EX4.20/example_20.sce b/858/CH4/EX4.20/example_20.sce
new file mode 100755
index 000000000..2ecddb63f
--- /dev/null
+++ b/858/CH4/EX4.20/example_20.sce
@@ -0,0 +1,10 @@
+clc

+clear 

+printf("example 4.20 page number 156\n\n")

+

+//to find the size of pipe required

+d = 0.15   //in m

+u = (0.0191/0.15^2);   //in m/s

+

+q = (3.14/4)*d^2*u;

+printf("volumetric flow rate = %f cubic meter/s",q)

diff --git a/858/CH4/EX4.21/example_21.sce b/858/CH4/EX4.21/example_21.sce
new file mode 100755
index 000000000..f00ef1cb9
--- /dev/null
+++ b/858/CH4/EX4.21/example_21.sce
@@ -0,0 +1,26 @@
+clc

+clear 

+printf("example 4.21 page number 160\n\n")

+

+//to find the pressure gradient

+

+Q=0.0003    //in m3/s

+d=0.05    //in m

+A=(3.14*d^2)/4;

+

+u=Q/A;

+

+density=1000;   //in kg/m3

+viscosity=0.001;  //in Pa-s

+e=0.3;

+dp=0.00125;   //particle diameter in m

+

+Re=(dp*u*density)/(viscosity*(1-e));

+fm=(150/Re)+1.75;

+L=0.5   //in m

+delta_Pf=fm*((density*L*u^2)/dp)*((1-e)/e^3);   //in Pa

+

+//applying bernoulli's equation, we get

+delta_P=delta_Pf-(density*9.8*L);

+pressure_gradient=delta_P/(L*1000);   //in kPa/m

+printf("required pressure gradient = %f kPa/m of packed height",pressure_gradient)

diff --git a/858/CH4/EX4.22/example_22.sce b/858/CH4/EX4.22/example_22.sce
new file mode 100755
index 000000000..8508ea21f
--- /dev/null
+++ b/858/CH4/EX4.22/example_22.sce
@@ -0,0 +1,29 @@
+clc

+clear 

+printf("example 4.22 page number 163\n\n")

+

+//to find minimum fluidization velocity

+

+d=120*10^-6    //in m

+density=2500   //particle density in kg/m3

+e_min=0.45;

+density_water=1000   //in kg/m3

+viscosity=0.9*10^-3;   //in Pa-s

+umf=(d^2*(density-density_water)*9.8*e_min^3)/(150*viscosity*(1-e_min));

+printf("minimum fludization velocity = %f m/s",umf)

+

+Re_mf=(d*umf*density_water)/(viscosity*(1-e_min));

+

+

+//given that uo/umf=10

+function[f] = F(e)

+    f = e^3+1.657*e-1.675;

+endfunction

+

+//initial guess

+x = 10;

+e = fsolve(x,F);

+

+printf("\n\ne = %f",e)

+length_ratio=(1-e_min)/(1-e);

+printf("\n\nratio of heights = %f",length_ratio)

diff --git a/858/CH4/EX4.23/example_23.sce b/858/CH4/EX4.23/example_23.sce
new file mode 100755
index 000000000..a8c7ddc34
--- /dev/null
+++ b/858/CH4/EX4.23/example_23.sce
@@ -0,0 +1,17 @@
+clc

+clear 

+printf("example 4.23 page number 167\n\n")

+

+//to find the power requirements

+

+P=9807   //in Pa

+density=1000   //in kg/m3

+Q=250/(60*density)

+head=25   //in m

+

+w= head*Q*P;     //in kW

+power_delivered=w/0.65;

+power_taken=power_delivered/0.9;

+

+printf("power_delivered = %f kW",power_delivered/1000)

+printf("\n\npower taken by motor = %f kW",power_taken/1000)

diff --git a/858/CH4/EX4.3/example_3.sce b/858/CH4/EX4.3/example_3.sce
new file mode 100755
index 000000000..89b3cddc0
--- /dev/null
+++ b/858/CH4/EX4.3/example_3.sce
@@ -0,0 +1,14 @@
+clc

+clear 

+printf("example 4.3 page number 128\n\n")

+

+//to find the viscosity of oil

+

+F=0.5*9.8;   //in N

+A=3.14*0.05*0.15;   //in m2

+shear_stress=F/A;   //in Pa

+printf("shear_stress = %f Pa",shear_stress)

+

+velocity_distribution =0.1/(0.05*10^-3);

+viscosity=shear_stress/velocity_distribution;

+printf("\n\nviscosity = %f Pa-s",viscosity) 

diff --git a/858/CH4/EX4.4/example_4.sce b/858/CH4/EX4.4/example_4.sce
new file mode 100755
index 000000000..22ffeefa5
--- /dev/null
+++ b/858/CH4/EX4.4/example_4.sce
@@ -0,0 +1,4 @@
+clc

+clear 

+printf("example 4.4 page number 130\n\n")

+printf("this is a theoritical problem,book shall be referred for solution")

diff --git a/858/CH4/EX4.5/example_5.sce b/858/CH4/EX4.5/example_5.sce
new file mode 100755
index 000000000..718f55335
--- /dev/null
+++ b/858/CH4/EX4.5/example_5.sce
@@ -0,0 +1,9 @@
+clc

+clear 

+printf("example 4.5 page number 133\n\n")

+

+//to find variation of losses with velocity

+loss_ratio=3.6;     //delta_P2/delta_P1=3.6

+velocity_ratio=2;   //u2/u1=2

+n=log2(loss_ratio);  //delta_P2/delta_P1=(u2/u1)^n

+printf("power constant = %f flow is turbulent",n)

diff --git a/858/CH4/EX4.6/example_6.sce b/858/CH4/EX4.6/example_6.sce
new file mode 100755
index 000000000..77bf07c1e
--- /dev/null
+++ b/858/CH4/EX4.6/example_6.sce
@@ -0,0 +1,4 @@
+clc

+clear 

+printf("example 4.6 page number 133\n\n")

+printf("this is a theoritical problem,book shall be referred for solution")

diff --git a/858/CH4/EX4.7/example_7.sce b/858/CH4/EX4.7/example_7.sce
new file mode 100755
index 000000000..1d0b4af71
--- /dev/null
+++ b/858/CH4/EX4.7/example_7.sce
@@ -0,0 +1,4 @@
+clc

+clear 

+printf("example 4.7 page number 134")

+disp("this is a theoritical problem,book shall be referred for solution")

diff --git a/858/CH4/EX4.8/example_8.sce b/858/CH4/EX4.8/example_8.sce
new file mode 100755
index 000000000..f0d9f2fd7
--- /dev/null
+++ b/858/CH4/EX4.8/example_8.sce
@@ -0,0 +1,31 @@
+clc

+clear 

+printf("example 4.8 page number 137\n\n")

+

+//to find the boundary layer properties

+

+disp('part 1')

+x=0.05   //in m

+density=1000   //in kg/m3

+viscosity=1*10^-3    //in Pa-s

+u=1   //in m/s

+Re=(density*u*x)/viscosity;

+

+printf("Reynolds Number = %f",Re)

+

+thickness=4.65*x*(Re)^-0.5;

+printf("\nboundary layer thickness = %f m\n",thickness)

+

+disp('part 2')

+Re_x=3.2*10^5;

+x_cr=(Re_x*viscosity)/(density*u);

+printf("transition takes place at x = %f m\n",x_cr) 

+

+disp('part 3')

+x=0.5   //in m

+Re=(density*u*x)/viscosity;

+thickness=0.367*x*(Re)^-0.2;

+printf("boundary layer thickness= %f m",thickness)

+

+t_sublayer=71.5*x*(Re)^-0.9;

+printf("\nsub layer thickness= %f m",t_sublayer)

diff --git a/858/CH4/EX4.9/example_9.sce b/858/CH4/EX4.9/example_9.sce
new file mode 100755
index 000000000..3ccdea2eb
--- /dev/null
+++ b/858/CH4/EX4.9/example_9.sce
@@ -0,0 +1,18 @@
+clc

+clear 

+printf("example 4.9 page number 138\n\n")

+

+//to find the flow properties

+d1=0.05   //in m

+A1=(3.14*d1^2)/4;

+density_1=2.1   //in kg/m3

+u1=15     //in m/s

+P1=1.8;   //in bar

+P2=1.3;   //in bar

+

+w=density_1*A1*u1;

+density_2=density_1*(P2/P1);

+printf("density at section 2 = %f kg/cubic meter",density_2)

+

+u2=u1*(density_1/density_2)*(0.05/0.075)^2;

+printf("\n\nvelocity at section 2 = %f m/s",u2)