initial commit / add all books

12 files changed, 159 insertions, 0 deletions

diff --git a/896/CH6/EX6.1/1.sce b/896/CH6/EX6.1/1.sce
new file mode 100755
index 000000000..af1dd8839
--- /dev/null
+++ b/896/CH6/EX6.1/1.sce
@@ -0,0 +1,25 @@
+clc

+//Example 6.1

+//calculate the drop in pressure per unit length in a pipe

+q=50//gal/min flow rate

+d=3.068//in inner diameter

+a=(%pi)*(3.068/12)^2/4//ft^2

+//1 ft^3 = 7.48 gal

+//1 min = 60 sec

+v_avg=q/a/60/7.48//ft/s

+mew=50//cP

+//1 cP = 0.000672 lbm/ft/s

+rho=62.3//lbm/ft^3

+R=(d/12)*v_avg*rho/(mew*0.000672)//dimentionless reynold's no.

+if (R<2000)

+  printf("Laminar flow\n");

+else

+  printf("Turbulent flow\n");

+end

+dx=3000//ft length of pipe

+//1 gal = 231 in^3

+//1 cP.ft^3 = 0.0000209 lbf.s

+dp=(q/60)*(128/%pi)*(mew/d^4)*dx*231*0.0000209/12//lbf/in^2 or psi

+//let D represent d/dx

+Dp=(dp/dx)*100//psi/ft

+printf("The pressure gradient in the pipe is %f psi/100ft",Dp);
\ No newline at end of file
diff --git a/896/CH6/EX6.10/10.sce b/896/CH6/EX6.10/10.sce
new file mode 100755
index 000000000..0a5757a4e
--- /dev/null
+++ b/896/CH6/EX6.10/10.sce
@@ -0,0 +1,10 @@
+clc

+//Example 6.10

+//Calculate the pressure difference created due to expansion and contraction

+rho=62.3//lbm/ft^3

+K=1.5//dimentionless

+v=13//ft/s

+//1 ft = 12 in

+//1 lbf.s^2 = 32.2 lbm.ft

+dp=rho*K*(v^2/2)/32.2/144//lbf/in^2

+printf("The pressure drop due to expansion and contraction is %f lbf/in^2",dp);
\ No newline at end of file
diff --git a/896/CH6/EX6.11/11.sce b/896/CH6/EX6.11/11.sce
new file mode 100755
index 000000000..18878274e
--- /dev/null
+++ b/896/CH6/EX6.11/11.sce
@@ -0,0 +1,10 @@
+clc

+//Example 6.11

+//Calculate the pressure drop in the pipe due to fittings

+dx=3000//ft actual length of pipe

+dx1=281//ft equivalent length of fittings

+p=484//psi

+dx_total=dx+dx1//ft

+dp_total=p*(dx_total/dx)//psi

+dp_vnf=dp_total-p//psi pressure drop fue to valves and fittings

+printf("The pressure drop due to valves and fittings is %f psi",dp_vnf);
\ No newline at end of file
diff --git a/896/CH6/EX6.12/12.sce b/896/CH6/EX6.12/12.sce
new file mode 100755
index 000000000..f2f8b4497
--- /dev/null
+++ b/896/CH6/EX6.12/12.sce
@@ -0,0 +1,10 @@
+clc

+//Example 6.12

+//Calculate pressure drop due to valves and fittings

+K=27.56//deimentionless

+rho=62.3//lbm/ft^3

+v=13//ft/s

+//1 ft = 12 in

+//1 lbf.s^2 = 32.2 lbm.ft

+dp=rho*K*(v^2/2)/32.2/144//psi

+printf("THe pressure drop due to valves and fittings is %d psi",dp);
\ No newline at end of file
diff --git a/896/CH6/EX6.13/13.sce b/896/CH6/EX6.13/13.sce
new file mode 100755
index 000000000..a725dc3c8
--- /dev/null
+++ b/896/CH6/EX6.13/13.sce
@@ -0,0 +1,15 @@
+clc

+//Example 6.13

+//Calculate the gasoline leakage rate through a seal

+p=100//lbf/in^2

+l=1//in length od seal in direction of leak

+mew=0.6//cP

+d=0.25//in diameter of valve stem

+t=0.0001//in thickness of valva stem

+//1 cP = 0.0000209 lbf.s/ft^2

+//1 ft = 12 in

+q=(p/l)*(1/12/mew)*(%pi)*d*t^3/0.0000209*144*3600//in^3/hr

+printf("The volumetric leakage rate of gasoline is %f in^3/hr\n",q);

+rho=0.026//lbm/in^3

+m=q*rho//lbm/hr

+printf("The mass leakage rate of gasoline is %f lbm/hr",m);
\ No newline at end of file
diff --git a/896/CH6/EX6.2/2.sce b/896/CH6/EX6.2/2.sce
new file mode 100755
index 000000000..f68b7dce8
--- /dev/null
+++ b/896/CH6/EX6.2/2.sce
@@ -0,0 +1,11 @@
+clc

+//Example 6.2

+//calculate viscosity of fluid using a viscometer

+rho=1050//Kg/m^3

+g=9.81//m/s^2

+dz=0.12//m change in height

+d=0.001//m inner diameter of capillary of viscometer

+q=10^(-8)//m^3/s

+dx=0.1//m length of capillary

+mew=(rho*g*dz*(%pi)*d^4)*1000/128/(q*dx)//cP

+printf("The viscosity of the fluid is %f cP",mew);
\ No newline at end of file
diff --git a/896/CH6/EX6.3/3.sce b/896/CH6/EX6.3/3.sce
new file mode 100755
index 000000000..0a9883518
--- /dev/null
+++ b/896/CH6/EX6.3/3.sce
@@ -0,0 +1,7 @@
+clc

+//Example 6.3

+//Calculate the fanning friction factor

+R=10^5//dimentionless reynold's no.

+ratio_ED=0.0002//dimentionless

+f=0.001375*(1+(20000*ratio_ED+10^6/R)^(1/3))//dimentionless

+printf("The fanning friction factor is %f",f);
\ No newline at end of file
diff --git a/896/CH6/EX6.4/4.sce b/896/CH6/EX6.4/4.sce
new file mode 100755
index 000000000..45e3262e6
--- /dev/null
+++ b/896/CH6/EX6.4/4.sce
@@ -0,0 +1,14 @@
+clc

+//Example 6.4

+//Calculate the gauge pressure in the tank

+q=300//gal/min flow rate

+d=3.068//in inner diameter

+a=(%pi)*(3.068/12)^2/4//ft^2

+//1 ft^3 = 7.48 gal

+//1 min = 60 sec

+v_avg=q/a/60/7.48//ft/s

+f=0.0091//dimentionless fanning friction factor

+dx=3000//ft

+rho=62.3//lbm/ft^3

+dp=4*f*(dx/(d/12))*rho*(v_avg^2/2)/32.2/144//lbf/in^2 or psi

+printf("The gauge pressure in the tank is %f psi",dp);
\ No newline at end of file
diff --git a/896/CH6/EX6.5/5.sce b/896/CH6/EX6.5/5.sce
new file mode 100755
index 000000000..0d6847b94
--- /dev/null
+++ b/896/CH6/EX6.5/5.sce
@@ -0,0 +1,13 @@
+clc

+//Example 6.5

+//Calculate volumetric flow rate of gasoline through a pipe

+d=0.1//m internal diameter of pipe

+A=%pi*d^2/4//m^2

+dx=100//m length of pipe

+f=0.005//dimentionless fanning friction factor

+dz=10//m difference in water level

+g=9.81//m/s^2

+v=((2*g*dz/4/f)*d/dx)^0.5//m/s

+printf("The velocity of gasoline through pipe is %f m/s\n",v);

+q=A*v//m^3/s

+printf("The volumteric flow arte od gasoline through the pipe is %f m^3/s",q);
\ No newline at end of file
diff --git a/896/CH6/EX6.6/6.sce b/896/CH6/EX6.6/6.sce
new file mode 100755
index 000000000..acc9a7d97
--- /dev/null
+++ b/896/CH6/EX6.6/6.sce
@@ -0,0 +1,21 @@
+clc

+//Example 6.6

+//Calculate pressure difference across the duct

+p=14.75//lbf/in^2

+M=29//lbm/lbmol

+R=10.73//lbf.ft^3/(in^2.lbmol.R)

+T=500//R Rankine temperature scale

+rho=p*M/(R*T)//lbm/ft^3

+q=500//ft^3/min

+d=1//ft

+A=(%pi)*d^2/4//ft^2

+v=(q/60)/A//ft/s

+mew=0.017//cP

+//1cP = 0.000672 lbm/ft/s

+R=d*v*rho/(mew*0.000672)//dimentionless reynold's no.

+f=0.00465//fanning friction factor

+dx=800//ft lenght of duct

+//1 ft = 12 in

+//1 lbf.s^2 = 32.2 lbm.ft

+dP=rho*(4*f*(dx/d)*(v^2/2))/32.2/144//lbf/in^2

+printf("The pressure drop across the duct is %f lbf/in^2",dP);
\ No newline at end of file
diff --git a/896/CH6/EX6.8/8.sce b/896/CH6/EX6.8/8.sce
new file mode 100755
index 000000000..1cf291b69
--- /dev/null
+++ b/896/CH6/EX6.8/8.sce
@@ -0,0 +1,13 @@
+clc

+//Example 6.8

+//Calculate the pump power required

+q=200//gal/min

+rho=62.3//lbm/ft^3

+//1 ft^3 = 7.48 gal

+m=(q/60)*rho/7.48//lbm/s

+dx=2000//ft

+dp=3.87//psi/100ft

+F=(dp/100)*dx/rho*32.2*144//ft

+//1 hp = 550 lbf.ft/s

+Po=F*m/550//hp

+printf("The pump power required is %f hp",Po);
\ No newline at end of file
diff --git a/896/CH6/EX6.9/9.sce b/896/CH6/EX6.9/9.sce
new file mode 100755
index 000000000..bcf6b7a3c
--- /dev/null
+++ b/896/CH6/EX6.9/9.sce
@@ -0,0 +1,10 @@
+clc

+//Example 6.9

+//Calculate the drop in pressure per unit length in a pipe

+dp=0.1//psi

+dx=800//ft

+//let D represent d/dx

+//1 psi = 6895 Pa

+//1 m = 3.28 ft

+Dp=(dp/dx)*6895*3.28//Pa/m

+printf("The drop in pressure per unit length in the pipe is %f Pa/m",Dp);
\ No newline at end of file