initial commit / add all books

8 files changed, 145 insertions, 0 deletions

diff --git a/1052/CH19/EX19.1/191.sce b/1052/CH19/EX19.1/191.sce
new file mode 100755
index 000000000..c6d260c9d
--- /dev/null
+++ b/1052/CH19/EX19.1/191.sce
@@ -0,0 +1,26 @@
+clc;

+//Example 19.1

+//page no. 246

+printf("Example 19.1 page no 246\n\n");

+//we have to find pressure at different point in a oil tank

+//apply manometer equation between point 1 and 2

+//since rho1=rho2,z1=z2

+//it gives P1=P2

+//applying manometer equation between points 2 and 3

+rho_oil=0.8*1000//density of oil

+//since rho3=rho_oil=rho2

+rho3=rho_oil

+z_32=.4//height difference between point 2 and 3

+g=9.807//grav. acc.

+P7=0//pressure at point 7,on gauge basis

+z_76=0.8//height difference between point 6 and 7

+rho_hg=13600//density of mercury

+P6=P7 + rho_hg*g*z_76//pressure at point 6

+P5=P6//pressure at point 5

+rho_air=1.2//density of air

+z_54=1//height difference between point 5 and 4

+P4=P5 + rho_air*g*z_54//pressure at point 4 

+P3=P4//pressure at point 3

+P2=P3 + rho_oil*g*z_32//pressure at point 2

+P1=P2//air pressure in the oil tank

+printf("\n pressure P1=%f Pag",P1); 

diff --git a/1052/CH19/EX19.2/192.sce b/1052/CH19/EX19.2/192.sce
new file mode 100755
index 000000000..72818b8da
--- /dev/null
+++ b/1052/CH19/EX19.2/192.sce
@@ -0,0 +1,17 @@
+clc;

+//Example 19.2

+//page no 250

+printf("Example 19.2 page no 250\n\n");

+//pitot tube is located at the center line of a horizontal pipe transporting air

+rho=0.075//density of gas ,lb/ft^2

+h=0.0166667//height difference,ft

+g=32.2//gravitational acc. lb/ft^2

+rho_m=62.4//density of medium which is air

+v=sqrt(2*g*h*(rho_m-rho)/rho)//velocity

+printf("\n velocity v=%f ft/s",v);

+v_max=v//because at that point where the reading was taken is the centerline

+printf("\n maximum veocity v_max=%f ft/s",v_max);

+//since the flowing fluid is air at a high velocity the flow has a high probability of being turbilent .from chapter 14,assume

+//v_av/v_max=0.815

+v_av=v_max*0.815

+printf("\n average velocity v_av=%f ft/s",v_av);

diff --git a/1052/CH19/EX19.3/193.sce b/1052/CH19/EX19.3/193.sce
new file mode 100755
index 000000000..1ce97930c
--- /dev/null
+++ b/1052/CH19/EX19.3/193.sce
@@ -0,0 +1,12 @@
+clc;

+//Example 19.3

+//page no 251

+printf("Example 19.3 page no 251\n\n");

+//refer to example 19.3 

+S=0.785//cross sectional area,ft^2

+v_av=24.4//average velocity,ft/s

+q=v_av*S*60//flow rate,factor 60 for minute

+printf("\n flow rate q=%f ft^3 min",q);

+rho=0.075//density 

+m_dot=q*rho*60//mass flow rate 

+printf("\n m_dot mass flow rate=%f lb/hr",m_dot);

diff --git a/1052/CH19/EX19.4/194.sce b/1052/CH19/EX19.4/194.sce
new file mode 100755
index 000000000..c5b9ed0a5
--- /dev/null
+++ b/1052/CH19/EX19.4/194.sce
@@ -0,0 +1,19 @@
+clc;

+//Example 19.4

+//page no 251

+printf("Example 19.4 page no\n\n")

+//water flow ina circular pipe,a pitot tube is used to measure the water velocity 

+h=0.07//manometer height,m

+rho=1000//density of water,kg/m^3

+rho_m=13600//density of mercury,kg/m^3

+g=9.807

+v=sqrt(2*g*h*(rho_m-rho)/rho)

+printf("\n water velocity v=%f m/s ",v);

+D=0.0779//pipe inside diameter,by using table A.5 in the appendix for a 3 inch schedule 40 pipe

+S=(%pi/4)*D^2

+printf("/n cross sectional area S=%f m^2",S);

+q=v*S//flow rate

+printf("\n flow rate q=%f m^3/s",q);

+meu=0.001//viscosity of water,kg/m.s

+R_e=rho*v*D/meu//reynolds number

+printf("\n reynolds no R_e=%f ",R_e);

diff --git a/1052/CH19/EX19.5/195.sce b/1052/CH19/EX19.5/195.sce
new file mode 100755
index 000000000..360836ac7
--- /dev/null
+++ b/1052/CH19/EX19.5/195.sce
@@ -0,0 +1,23 @@
+clc;

+//Example 19.5

+//page no 254

+printf("Example 19.5 page no 254\n\n");

+//a venturi meter has gasoline flowing through it.

+h=0.035//height of venturi meter

+D1=0.06//upsteeam diameter,m

+D2=0.02//throat diameter,m

+rho_m=13600//density of mercury

+rho=680//density of gasoline

+g=9.807

+v2=sqrt((2*g*h*(rho_m-rho)/rho)/1-D2^4/D1^4)//velocity of gasoline at the the throat

+printf("\n velocity at throat v2=%f m/s",v2);

+q=(%pi/4)*D2^2*v2//flow rate

+printf("\n flow rate q =%f m^3/s",q);

+P1=101325//upstream pressure,Pa

+P2=P1-g*h*(rho_m-rho)//pressure at throat P2

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

+P_d=P1-P2//pressure difference

+P_l=.1*P_d//pressure loss is 10 %

+printf("\n pressure loss P_l=%f Pa",P_l);

+W_l=q*P_l//power loss

+printf("\n power loss W_l=%f W",W_l);

diff --git a/1052/CH19/EX19.6/196.sce b/1052/CH19/EX19.6/196.sce
new file mode 100755
index 000000000..f5a22b9e0
--- /dev/null
+++ b/1052/CH19/EX19.6/196.sce
@@ -0,0 +1,16 @@
+clc;

+//Example 19.6

+//page no. 255

+printf("\n Example 19.6 page no. 255\n\n");

+//refer to example 19.5

+//if gasoline has vapor pressure of 50000Pa ,we have to calculate flow rate at whhich cavitation to occur

+P1=101325//upstream pressure,Pa

+P2=50000//given vapor pressure,Pa

+D1=0.06//upstream diameter,m

+D2=0.02//throat diameter,m

+rho=680//density of gasoline

+v2=sqrt((2*(P1-P2))/rho*(1-D2^4/D1^4))//velocity

+printf("\n velocity v2=%f m/s",v2); 

+q=(%pi/4)*D2^2*v2//flow rate

+printf("\n flow rate q=%f m^3/s",q);

+ 

diff --git a/1052/CH19/EX19.7/197.sce b/1052/CH19/EX19.7/197.sce
new file mode 100755
index 000000000..186d2730b
--- /dev/null
+++ b/1052/CH19/EX19.7/197.sce
@@ -0,0 +1,23 @@
+clc; 

+//Example 19.7

+//page no 258

+printf("Example 19.7 page no 258\n\n");

+//an  orifice meter is equipped with flange top is installed to measure the flow rate of air in a circular duct

+D1=0.25//diameter of circular duct,m

+D2=0.19//orifice diamter,m

+v2=4/(%pi*D2^2)//velocity through orifice

+printf("\n velocity through orifice v2=%f m/s",v2);

+C_o=1// assuming orifice discharge coefficient

+rho=1.23//density of air,kg/m^3

+P=rho*v2^2*[1-(D2^4/D1^4)]/2//pressure 

+printf("\n pressure P=%f Pa",P);

+meu=1.8e-5// absolute viscosity

+R_e=rho*v2*D2/(meu)//reynolds no.

+printf("\n Reynolds no. R_e=%f ",R_e);

+C_ac=0.62//actual discharge cefficient,from fig.19.8

+P_ac=P/(C_ac)^2//actual pressure drop 

+P_rec=14*(D2/D1) + 80*((D2/D1)^2)//equation for percentage pressure recovery

+P_loss=100-P_rec//precentage pressure loss

+P_l=round((P_loss/100)*P_ac)//actual pressure  drop after recovery

+printf("\n actual pressure drop P_l=%f Pa",P_l);

+

diff --git a/1052/CH19/EX19.9/199.sce b/1052/CH19/EX19.9/199.sce
new file mode 100755
index 000000000..5af570f48
--- /dev/null
+++ b/1052/CH19/EX19.9/199.sce
@@ -0,0 +1,9 @@
+clc;

+//Example 19.9

+//page no 259

+printf("\n Example 19.9 page no 259\n\n");

+//air at ambient condition is flowing in a pipe

+rho=0.075//density of air ,lb/ft^3

+m_dot=0.5//mass flow rate ,lb/s

+q=m_dot/rho//volumatric flow rate

+printf("\n volumatric flow rate q=%f ft^3/s",q);