13 files changed, 387 insertions, 0 deletions
diff --git a/1052/CH22/EX22.10/2210.sce b/1052/CH22/EX22.10/2210.sce
new file mode 100755
index 000000000..9e61199d9
--- /dev/null
+++ b/1052/CH22/EX22.10/2210.sce
@@ -0,0 +1,32 @@
+clc;
+//Example 22.10
+//page no 306
+printf("\n Example 22.10 page no 306\n\n");
+//benzene is pumped from a large tank to a delivery station 
+//refer fig 22.3
+q=0.003//vol. flow rate,m^3/s
+//tank is at atmosphric pressure
+D=0.03//diameter of suction and discharge line,m
+v_2=q/((%pi/4)*D^2)//discharge velocity,m/s
+//since all diameters are same likewise velocities are same
+v_3=v_2
+g=9.807//grav. acc.
+D_h=(v_3^2)/(2*g)//dynamic head
+printf("\n dynamic head D_h=%f m",D_h);
+z1=0//height at point 1,tank level
+z2=1.8//height at point 3
+//applying bernoulli's eq. between the top of the tank(open to theatomsphere)and the inlet to the pump(station3)
+rho=865//density of benzene,kg/m^3
+P3=101325-(z2+D_h)*(rho*g)//ptressure at point 3
+printf("\n pressure at point 3 P3=%f Pa",P3);
+P_v=26200//vapor pressure of benzene,Pa
+NPSH =  (P3 - P_v)/(rho*g) +  D_h
+printf("\n  NPSH=%f m",NPSH)
+//the manufacturer NPSH is 8 m, which is greater than the calculated NPSH  of 7.06m,therfore, the suction point of pump must be lowered 
+//calculation of new pressure
+NPSH_m=8//NPSH by manufacturer
+P3_n_ab=8*(rho*g)-D_h*(rho*g) + P_v
+printf("\n new pressure at point 3 P3_n_ab=%f Pa absolute",P3_n_ab);
+P3_n_bz=-1.77//pressure in terms of benzene height,m
+z3=-P3_n_bz -D_h//desired height of point 3
+printf("\n height z3=%f m",z3);  
diff --git a/1052/CH22/EX22.11/2211.sce b/1052/CH22/EX22.11/2211.sce
new file mode 100755
index 000000000..b7b3cd228
--- /dev/null
+++ b/1052/CH22/EX22.11/2211.sce
@@ -0,0 +1,41 @@
+clc;
+//Example 22.11
+//page no 308
+printf("Example 22.11 page no 308\n\n");
+//a storage tank on top of a building pumps 60 deg F water through an open pipe to it from a reservoir 
+q=1.36//vol. flow rate ,ft^3/s
+D=0.333//diameter of pipe,ft
+S=%pi/4*D^2//cross sectional area,ft^2
+v2=q/S//flow velocity,ft/s
+rho=62.37//density of water,lb/ft^3
+meu=1.129*6.72e-4//viscosity of water
+R_e=D*v2*rho/meu//reynolds no.
+printf("\n reynolds no. R_e=%f",R_e );
+//from R_e we can conclude that flow is turbulent
+k=0.0018//roughness factor
+K_r=k/D//relative roughness
+f=0.0046//friction factor
+L=525//length of pipe,ft
+g_c=32.174//grav. acc
+h_fp=(4*f*L*v2^2)/(D*2*g_c)//frictional loss due to the length of pipe
+printf("\n frictional loss h_fp=%f ft.lbf/lb",h_fp);
+//friction due to the fitings from table 18.1
+K_ff_gate=2*0.11//loss coeff. due to gates
+K_ff_elbows=5*0.64//loss coeff. due to elbows
+//friction due to the sudden contraction is obtained from eq. 18.10 .
+//note that D1/D2=0,since the upstream diameter is singnificantly larger than the downward diameter 
+K_c=0.42//coeff. of sudden contraction
+K_e=1//coeff. of sudden expansion
+K_s=K_ff_gate +K_ff_elbows +K_e +K_c//sum of loss coeff.
+h_f=K_s*v2^2/(2*g_c)//friction losses due to fitting,expansion,contraction
+h_f_total=h_fp + h_f//total frictional losses
+printf("\n total frictional loss h_f_total=%f ft.lbf/lb",h_f_total);
+v1=0
+P_drop=0//pressure drop
+z1=0//reservoir water level
+z2=200//height of reservoir
+W_s=(v2^2-v1^2)/(2*g_c)  + (z2-z1)  + h_f_total//power requirement
+m_dot=q*rho//mass flow rate,lb/s
+neta=0.6//efficiency of pump
+W_dot_s=m_dot*W_s/(550*neta)//actual horsepower requirement
+printf("\n W_dot_s=%f hp",W_dot_s);
diff --git a/1052/CH22/EX22.12/2212.sce b/1052/CH22/EX22.12/2212.sce
new file mode 100755
index 000000000..cf2e4cc60
--- /dev/null
+++ b/1052/CH22/EX22.12/2212.sce
@@ -0,0 +1,36 @@
+clc;
+//Example 22.12
+//Page no 311
+printf("Example 22.12 page no 311\n\n")
+//turpentine is being moved  from a large storage tank to a blender through a 700 ft pipeline
+rho=62.4//density
+SG=0.872//specific gravity of terpentine
+rho_t=SG*rho//density of turpentine
+v=12.67//av. velocity of the turpentine in the line,ft/s
+z1=20//height of top surface in the storage tank above floor level,ft
+z2=90//height of discharge end of pipe,ft
+neta=0.74//efficiency of pump
+W_s=401.9//average energy delivered by pump,ft/lbf/lb
+g_c=32.174//grav.acc
+L=700//length of pipeline
+//from bernoulli eq.
+h_f= neta*W_s - v^2/(2*g_c) - (z2-z1)//frictional loss  if there is no pressure drop
+printf("\n frictional loss h_f =%f ft.lbf/lb",h_f);  
+k_c=0.4//coeff. of contraction
+k_e=0.9//coeff. of expansion
+k_f=0.2//coeff. of bends and valve
+//making equation(1) from the friction coeff. due to fittings between f and D,f=0.0293*D 
+//making another equation(2) from Reynolds number in terms D ,R_e=582250*D
+//from trial and error method we get D
+D=0.184//diameter
+S=%pi*D^2/4//cross sectional area
+S=0.0266
+q=v*S//volumetric flow rate 
+printf("\n q=%f ft^3/s ",q);
+m_dot=rho_t*q//mass flow rate
+bhp =m_dot*W_s/(550*neta)//brake horse power
+printf("\n brake horse power bhp=%f hp",bhp);
+
+
+
+
diff --git a/1052/CH22/EX22.13/2213.sce b/1052/CH22/EX22.13/2213.sce
new file mode 100755
index 000000000..afbd77822
--- /dev/null
+++ b/1052/CH22/EX22.13/2213.sce
@@ -0,0 +1,39 @@
+clc;
+//Example 22.13
+//page no 313
+printf("Example 22.13 page no 313\n\n");
+//hydrogen flows through a horizontal pipe
+//properties of hydrogen at 20 deg C from table A.3 in the appendix
+rho=0.0838//density of hydrogen,kg/m^3
+meu=9.05e-6//viscosity,kg/m.s
+D=0.08//diameter of pipe,m
+L=1//unit length of pipe,m
+q=0.0004//vol. flow rate ,m^3/s
+S=.000503//cross sectional area
+v=q/S//flow velocity,m/s
+m_dot=rho*q//mass flow rate,kg/s
+R_e=(D*v*rho/meu)//reynolds no.
+printf("\n R_e reynolds no=%f ",R_e);
+//since R_e is 593<2100, flow is laminar
+//since the tube is horizontal z1=z2,calculation of pressure gradient(P/L)
+P_grad= 128*meu*q/(%pi*D^4)//pressure gradient
+printf("\n Pressure gradient P_grad=%f Pa/m",P_grad)
+v_max=2*v//m/s
+//calculation of fanning friction factor
+//since the flow is laminar 
+f=16/R_e//fanning friction factor
+printf("\n fanning friction factor f=%f ",f);
+f_d=4*f//darcy friction factor
+printf("\n darcy friction factor f_d=%f ",f_d);
+g=9.807//grav. acc.
+h_f=f_d*(L/D)*(v^2/(2*g))//friction loss
+printf("\n friction loss h_f=%f m",h_f);
+W_f = m_dot*g*h_f//friction power loss
+printf("\n friction power loss W_f=%f W",W_f); 
+
+
+
+
+
+
+ 
diff --git a/1052/CH22/EX22.14/2214.sce b/1052/CH22/EX22.14/2214.sce
new file mode 100755
index 000000000..b52d046fa
--- /dev/null
+++ b/1052/CH22/EX22.14/2214.sce
@@ -0,0 +1,43 @@
+clc;
+//Example 22.14
+//page no 315
+printf("\Example 22.14 page no 315\n\n");
+//gasoline is pump through a horizontal cast iron pipe
+L=30//length of pipe
+D=0.2//diameter of pipe,m
+S=(%pi/4)*D^2//cross sectional area
+q=0.3//vol. flow rate ,m^3/s
+v=q/S//flow velocity,m/s
+rho=680//density of gasoline,kg/m^3
+meu=2.92e-4//viscosity of gasoline,kg/m.s
+R_e=D*v*rho/meu//reynolds no.
+printf("\n reynolds no R_e=%f ",R_e);
+//since R_e is >4000 flow is turbulent
+k=0.00026//roughness factor from table 14.1 for cast iron,m
+K_r=k/D//relative roughness
+f=0.00525//fanning friction factor from fig 14.2
+//Note that the flow corresponds to complete turbulence in the rough pipe
+g=9.807//gravitational acceleration
+//h_f=4*f*(L/D)*(v^2/(2*g))//head loss
+h_f=14.647
+//applying bernoulli equation to the fluid in the pipe
+//in this case the pipe is horizontal (z1=z2) with constant diameter (v1=v2) and no shaft head (h_s=0)
+//first convert the friction head to a pressure difference
+P_diff=rho*g*h_f//pressure difference 
+P_diff= 97.68*10^3//after round off
+W_s_id=q*P_diff//ideal shaft work
+printf("\n ideal shaft work W_s_id=%f W ",W_s_id);
+neta=0.8//efficiency of pump
+W_s_ac=W_s_id/neta//actual shaft work 
+printf("\n actual shaft work W_s_ac=%f W",W_s_ac);
+f_s=0.009//friction factor smooth
+f_r=0.021//friction factor roughnes
+k=f_r/f_s
+f_inc=100*(k-1)//percentage increment in f due to roughness 
+printf("\n f_inc=%f ",f_inc);
+
+
+
+
+
+
diff --git a/1052/CH22/EX22.15/2215.sce b/1052/CH22/EX22.15/2215.sce
new file mode 100755
index 000000000..063cbd53b
--- /dev/null
+++ b/1052/CH22/EX22.15/2215.sce
@@ -0,0 +1,24 @@
+clc;
+//Example 22.15
+//page no 316
+printf("\n Example 22.15 page no 316\n\n")
+//liquid benzene flows through a smooth horizontal iron pipe 
+D=2.3//diameter of pipe,m
+L=146.304//length of pipe,m
+S=(%pi/4)*D^2//cross sectional area,m^2
+q=4000//vol. flow rate,gal/min
+v=q/(S*264.17*60)//flow velocity
+printf("\n flow velocity v=%f m/s",v);
+rho=899//density of benzene
+meu=0.0008//viscosity of benzene,kg/m.s
+R_e = D*v*rho/meu//reynolds no
+printf("\n reynolds no R_e=%f ",R_e);
+//since the reynolds number falls in the turbulent regime,determine the fanning friction factor from fig. 14.2
+f=0.0032//fanning friction factor
+// calculation of pressure drop with the assumption of no height and velocity change , and no pump work 
+//since only frictional losses are to be considered
+//applying eq. 14.3
+P_drop = 4*f*(L/D)*(v^2/2)*rho//pressure drop
+printf("\n pressure drop P_drop=%f Pa",P_drop);
+W_dot_f=q*P_drop/(264.17*60)//friction power loss
+printf("\n friction power loss W_dot_f=%f W",W_dot_f);
diff --git a/1052/CH22/EX22.16/2216.sce b/1052/CH22/EX22.16/2216.sce
new file mode 100755
index 000000000..4ee19b488
--- /dev/null
+++ b/1052/CH22/EX22.16/2216.sce
@@ -0,0 +1,22 @@
+clc;
+//Example 22.16
+//page no 317
+printf("\n Example 22.16 page no 317\n\n");
+//a power plant employs steam to generate power 
+//adiabatic conditions
+z1=0//steam vertical position at inlet,ft
+z2=-20//steam vertical position at outlet,ft
+v1=120//steam velocity at inlet,ft/s
+v2=330//steam velocity at outlet,ft/s
+H1=1505.4//steam enthalpy at inlet 
+H2=940//steam enthalpy at outlet
+Q=0//for adiabatic conditions
+g_c=32.174//grav .acc
+//applying energy equation
+W_s=-(z2/778) - v2^2/(2*g_c*778) - H2  +z1 + v1^2/(2*g_c*778) + H1//work extracted from system
+printf("\n work extracted from the system W_s=%f Btu/lb ",W_s);
+m_dot=450000//mass flow rate ,lb/h
+W_dot_s=m_dot*W_s//total power generated by the turbine 
+printf("\n W_dot_s =%f Btu/h",W_dot_s);//approx calculation in book 
+W_hp=W_dot_s*3.927e-4//power generated in horsepower hp
+printf("\n power generated W_hp=%f hp",W_hp);//approx calculation in book
diff --git a/1052/CH22/EX22.4/224.sce b/1052/CH22/EX22.4/224.sce
new file mode 100755
index 000000000..027a6e4d4
--- /dev/null
+++ b/1052/CH22/EX22.4/224.sce
@@ -0,0 +1,15 @@
+clc;
+//Example 22.4
+//page no 298
+printf("Example 22.4 page no 298\n\n");
+//a centrifugal pump operating at 1800 rpm ,we have to find the impeller diameter needed to develop a head of 200 ft
+h=200//height,ft
+g=32.2//gravitational acc. ft/s^2
+v=sqrt(2*g*h)//velocity needed to develop a head of 200 ft
+printf("\n velocity v=%f ft/s",v);
+N=1800//pump operating at this rotational speed,in rpm
+c=v*60/N//the number of feet that the impeller travels in one rotations
+//this c represents the circumference of the impeller  since it is equal to one rotation 
+printf("\n circumference c=%f ft/rotation",c);
+D=c/%pi//diameter of the impeller
+printf("\n diameter D=%f ft",D);
diff --git a/1052/CH22/EX22.5/225.sce b/1052/CH22/EX22.5/225.sce
new file mode 100755
index 000000000..a628e68ae
--- /dev/null
+++ b/1052/CH22/EX22.5/225.sce
@@ -0,0 +1,47 @@
+clc;
+//Example 22.5
+//page no 299
+printf("Example 22.5 page no 299\n\n");
+//water for a processing plant is required to be stored in a reservoir
+//assume the properties of water at 20 deg C are
+rho=998//density,kg/m^3
+meu=0.001//viscosity,N.s/m^2
+L=120//length of pipe,m
+D=0.15//diameter  of pipe,m
+S=(%pi/4)*D^2//cross sectional area of pipe
+//given:
+q=1.2/60//volumetric flow rate,m^3/s
+v=q/S//flow velocity,m/s
+R_e=D*v*rho/meu//reynolds no
+printf("\n reynolds no R_e=%f ",R_e);
+//from value of R_e ,flow is clearly turbulent
+k=0.0005//roughness factor for galvanized iron
+K_r=k/D//relative roughness
+f=0.0053//fricion factor from fig. 14.2
+h_f=4*f*(L/D)*(v^2/2)//friction loss of energy
+printf("\n h_f frictional  loss=%f J ",h_f);
+//for right elbows (from table 18.1),the estimated value of resistance coff. K for one regular 90 deg elbows is 0.5
+K=4//resstance coeff.
+V_h=v^2/2//velociy head
+e_l=K*V_h//the total loss from the elbows
+printf("\n e_l total elbow loss=%f J/kg",e_l);
+//the energy to move 1 kg of water against a head of 22m of water is
+z=22//height,m
+g=9.81//grav. acc,m/s^2
+PE=z*g
+printf("\n potential energy PE=%f J/kg",PE);
+TE = h_f + e_l  + PE//total requirement per kg
+printf("\n total energy TE=%f J/kg",TE);
+W_dot_s= TE*q*rho//theoretical power requirement 
+printf("\n theoritical power W_dot_s=%f J/s",W_dot_s); 
+h=TE/g//head equivalent to the energy requirement 
+printf("\n equivalent head h=%f m ",h);
+
+
+
+
+
+
+
+
+ 
diff --git a/1052/CH22/EX22.6/226.sce b/1052/CH22/EX22.6/226.sce
new file mode 100755
index 000000000..6f995b617
--- /dev/null
+++ b/1052/CH22/EX22.6/226.sce
@@ -0,0 +1,17 @@
+clc;
+//Example 22.6
+//page no 301
+printf("Example 22.6 page no 301\n\n");
+//oil is flowing through a standard 3/2 inch steel pipe containing a 1 inch square edged orifice
+v_gal=400//orifice velocity of oil in gal/hr
+v_o=400*144/(0.785*3600*7.48)//orifice velocity in ft/hr
+D_o=1/12//diameter of orifice
+rho=0.87*62.4//density of oil
+meu=20.6*0.000672//viscosity of oil
+R_e=D_o*v_o*rho/meu
+printf("\n reynolds no =%f ",R_e);
+D_r=0.62//ratio of orifice plate to pipe diametersD_o/D1 = 1/1.61
+C_d=0.76//discharge coeff. fro fig 19.8
+g=32.2//grav. acc. ft/s^2
+h=(v_o^2/(2*g*(C_d)^2))*(1-D_r^4)//height of oil in gauge reading
+printf("\n gauge reading h=%f ft ",h);
diff --git a/1052/CH22/EX22.7/227.sce b/1052/CH22/EX22.7/227.sce
new file mode 100755
index 000000000..f94cadbaf
--- /dev/null
+++ b/1052/CH22/EX22.7/227.sce
@@ -0,0 +1,23 @@
+clc;
+//Example 22.7
+//page no 302
+printf("Example 22.7 page no 302\n\n ");
+//natural gas consisting of essentially pure methane flows through a long straight standard 10 inch steel pipe into which is inserted a square edged orifice 2.50 inches in diameter ,with pressure taps ,each 5 inch from the orifice plate
+//manometer is attached across the orifice reads 1.60 in H20
+D_o=2.50//diameter of orifice
+D_1=10.15//diameter of plate
+D_r=D_o/D_1//ratio of diameters
+//assuming the reynolds no R_e in the orifice to be over 30,000
+C_o=0.61//coeff. of discharge from R_e value
+g=32.2//garv. acc ft/s^2
+rho_m=62.4//density of medium (water)
+rho=0.054//density of methane gas,lb/ft^3
+h=1.60//manometer reading height,in
+meu=12*0.011*0.000672//viscosity 
+v_o= C_o*sqrt((2*g*h*rho_m)/(12*rho))// orifice velocity
+printf("\n orifice veloctiy v_o=%f ft/s",v_o);
+R_e_o=D_o*v_o*rho/meu//reynolds no in the orifice
+printf("\n R_e_o reynolds no =%f ",R_e_o);
+//from R_e_o value C_o=0.61 is permissible
+m_dot=round(v_o*(%pi/4)*(D_o^2)*rho*(3600/144))//mass flow rate 
+printf("\n mass flow rate m_dot=%f lb/hr",m_dot);
diff --git a/1052/CH22/EX22.8/228.sce b/1052/CH22/EX22.8/228.sce
new file mode 100755
index 000000000..3983a3e13
--- /dev/null
+++ b/1052/CH22/EX22.8/228.sce
@@ -0,0 +1,27 @@
+clc;
+//Example 22.8
+//page no 303
+printf("Example 22.8 page no 303\n\n");
+//refer to fig 22.1
+D1=.1//upstream diameter(at station 1),m
+D2=.06//downstream diameter(station 2),m
+S2=(%pi/4)*D2^2//cross sectional area at point 2
+rho=1.22//density of air  from ideal gas law
+rho_m=827//density of medium,kg.m^3
+g=9.8//gravitational acc.
+h=0.08//manometer head,m
+//from bernoulli equation
+v2=sqrt(2*g*h*((rho_m/rho)-1))//velocity at point 2
+v1=v2*(D2/D1)^2//velocity at point 1
+q=v2*S2//volumatric flow rate
+printf("\n vol.flow rate q=%f m^3/s",q);
+//calculation of mach number from equation 15.1
+T=293//temperature in k
+c=20*sqrt(T)//speed of light at this temperature,m/s
+M_a=v2/c//mach no.
+printf("\n mach no. M_a =%f ",M_a);
+//noting that M_a=0.095 < 0.3 , we can conclude that flow is incompressible//given 
+P1=130000 //absolute pressure at point 1,pa
+//by using bernoulli eq for P2
+P2=P1-rho*v2^2*(1-(D2/D1)^4)/2//pressure at point 2
+printf("\n pressure at point 2=%f Pa",P2);
diff --git a/1052/CH22/EX22.9/229.sce b/1052/CH22/EX22.9/229.sce
new file mode 100755
index 000000000..3d5747b7d
--- /dev/null
+++ b/1052/CH22/EX22.9/229.sce
@@ -0,0 +1,21 @@
+clc;
+//Example 22.9
+//page no 305 
+printf("\n Example 22.9 page no 305\n\n");
+//water is flowing from an elevated reservoir through a conduit to a turbine at a lower level and out of the turbine through a similar conduit  
+//refer to fig 22.2
+//since the diameter of the conduit is the same at location 1 and 2 ,kinetic energy effects can be  neglected and bernoulli eq. takes the form
+//P/rho  + z(g/g_c)  -h_s + h_f  = 0
+P1=30///pressure at point 1,psia
+z1=300//height of point 1,ft
+P2=18//pressure at point 2,psia
+z2=-10//height of point 2,ft
+rho=62.4//density
+m_dot=3600//mass flow rate,tons/hr
+W_dot =1000//output at the shaft of turbine,hp
+neta=0.9//efficiency of turbine
+h_s=W_dot*550*3600/(neta*m_dot*2000)//
+printf("\n h_s =%f ft.lbf/lb",h_s);
+//put this value in bernoulli eq.
+h_f=(P2-P1)*144/rho + (z2-z1) -h_s//frictional loss
+printf("\n frictional loss h_f=%f ft.lbf/lb",h_f)