+clc;

+clear;

+T1=50;//degree farenheit

+D=0.73;//in

+vol=0.0125;//ft^3

+T2=140;//degree farenheit

+

+vis1=2.73/100000;//lb*s/ft^2 at 50 degree farenheit

+vis2=0.974/100000;//lb*s/ft^2 at 140 degree farenheit

+

+//for 50 degree farenheit

+//if flow is laminar, maximum Re=2100; Re=d*V*D/vis

+V1=2100*vis1/(1.94*D/12);

+t1=vol/(%pi*((D/12)^2)/4*V1);

+//if flow is turbulent, minimum Re=4000

+V2=4000*vis1/(1.94*D/12);

+t2=vol/(%pi*((D/12)^2)/4*V2);

+

+//for 140 degree farenheit

+//if flow is laminar, maximum Re=2100; Re=d*V*D/vis

+V3=2100*vis2/(1.94*D/12);

+t3=vol/(%pi*((D/12)^2)/4*V3);

+//if flow is turbulent, minimum Re=4000

+V4=4000*vis2/(1.94*D/12);

+t4=vol/(%pi*((D/12)^2)/4*V4);

+

+disp("For laminar flow")

+disp("seconds",t1,"The time taken to fill the glass at 50 degree F=")

+disp("seconds",t3,"The time taken to fill the glass 100 degree F=")

+disp("For turbulent flow:")

+disp("seconds",t2,"The time taken to fill the glass at 50 degree F=")

+disp("seconds",t4,"The time taken to fill the glass at 140 degree F=") \ No newline at end of file

+clc;

+clear;

+D=4;//in

+l=20;//ft

+n=4;//number of 90 degree elbows

+h=0.2;//in

+T=100;//degree F

+//energy equation between the inside of the dryer and the exit of the vent pipe

+p1=(h/12)*62.4;//lb/(ft^2)

+KLentrance=0.5;

+KLelbow=1.5;

+sw=0.0709;//lb/(ft^3)

+f=0.022;//assumption

+//hence,

+V=((p1/sw)*2*32.2/(1+(f*l/(D/12))+KLentrance+(n*KLelbow)))^0.5;//ft/sec

+Q=V*(%pi*((D/12)^2)/4);//(ft^3)/sec

+disp("(ft^3)/sec",Q,"The flowrate=")

+

+clc;

+clear;

+Pa=50;//hp

+D=1;//ft

+l=300;//ft

+f=0.02;

+z1=90;//ft

+//energy equation between the surface of the lake and the outlet of the pipe

+//p1=V1=p2=z2=0; V2=V

+//hL=f*l*(V^2)/(D *2*g)

+//hT=Pa/(sw*%pi*(D^2)*V/4)

+c1=(Pa*550)/(62.4*%pi*(D^2)/4)//561

+c2=f*l/(D*2*32.2)//0.0932

+fn=poly([c1 (-z1) 0 ((1/(2*32.2))+(c2))],"V","c");

+r=roots(fn);

+V1=r(1);//ft/sec

+V2=r(2);//ft/sec

+Q1=(%pi*(D^2)/4)*V1;//(ft^3)/sec

+Q2=(%pi*(D^2)/4)*V2;//(ft^3)/sec

+disp("(ft^3)/sec",Q2,"and","(ft^3)/sec",Q1,"The possible flowrates are=") \ No newline at end of file

+clc;

+clear;

+roughness=0.0005;//ft

+Q=2;//(ft^3)/sec

+pd=0.5;//psi; where pd=pressure drop

+l=100;//ft

+d=0.00238;//slugs/(ft^3)

+vis=3.74*(10^(-7));//lb*sec/(ft^2)

+x=Q/(%pi/4);//where x =V*(D^2)

+//energy equation with z1=z2 and V1=V2

+y=l*d*(x^2)*0.5/(pd*144);//where y=(D^5)/f

+f=0.027;//using reynolds number, roughness and moody's chart

+D=(y*f)^(1/5);//ft

+disp("ft",D,"The diameter of the pipe should be =")

+q=0.01:0.01:3;

+count=1;

+for i=0.01:0.01:3

+ dia(count)=((l*d*((i/(%pi/4))^2)*0.5/(pd*144))*f)^(1/5);

+ count=count+1;

+end

+plot2d(q,dia,rect=[0,0,3,0.25])

+xtitle("D vs Q","Q, (ft^3)/sec","D, ft") \ No newline at end of file

+clc;

+clear;

+T=60;//degree F

+kvis=1.28*(10^(-5));//(ft^2)/sec

+l=1700;//ft

+roughness=0.0005;//ft

+Q=26;//(ft^3)/sec

+n=4;//number of flanged 45 degree elbows

+z1=44;//ft

+x=Q/(%pi/4);//where x=V*(D^2)

+KLentrance=0.5;

+KLelbow=0.2;

+KLexit=1;

+//Finding f from Re, roughness and moody's chart

+f=0.01528;

+sumKL=(n*KLelbow)+KLentrance+KLexit;

+y=f*l;

+//V^2 = (x^2)/(D^4)

+//energy equation with p1=p2pV1=V2=z2=0

+z=(2*32.2*z1)/((x^2)*l);

+k=sumKL/l;

+fn=poly([(-f) (-k) 0 0 0 z],'D','c');

+r=roots(fn);

+disp("ft",r(1),"The diameter=")

+count=1;

+len=400:2000;

+for i=400:2000

+ root=roots(poly([(-f) (-(sumKL/i)) 0 0 0 ((2*32.2*z1)/((x^2)*i))],'a','c'));

+ dia(count)=root(1);

+ count=count+1;

+end

+plot2d(len,dia,rect=[0,0,2000,1.8])

+xtitle("D vs l","l, ft","D, ft")

+clc;

+clear;

+D=1;//ft

+f=0.02;

+z1=100;//ft

+z2=20;//ft

+z3=0;//ft

+l1=1000;//ft

+l2=500;//ft

+l3=400;//ft

+//assuming fluid flows into B

+//applying energy equation bwtween (1 and 3) and (1 and 2) and using the relation V1=V2+V3

+c1=z1*32.2*2/(f*l1);

+c2=(z1-z2)*32.2*2/(f*l1);

+x=(c1-c2)/(l3/l1);//160

+y=(l2/l1)/(l3/l1);//1.25

+a=c2-x;//98

+b=(a*2*(y+(l2/l1)));//539

+c=4*x+b;//1179

+d=-((y+(l2/l1))^2)+(4*y);//-2.5625

+e=-(a^2);//-9604

+fn=poly([e 0 c 0 d],'V2','c');

+r=roots(fn);

+V2=r(1);

+V1=(c2-(l2/l1)*V2)^0.5;

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

+Q1=V1*A;

+Q2=V2*A;

+Q3=Q1-Q2;

+disp("(ft^3)/sec",Q1,"Q1 (out of A)=")

+disp("(ft^3)/sec",Q2,"Q2 (into B)=")

+disp("(ft^3)/sec",Q3,"Q3 (into C)=") \ No newline at end of file

+clc;

+clear;

+D=60;//mm

+pdiff=4;//kPa

+Q=0.003;//(m^3)/sec

+d=789;//kg/(m^3)

+vis=1.19*(10^(-3));//N*sec/(m^2)

+Re=d*4*Q/(%pi*D*vis);

+//assuming B=dia/D=0.577, where dia=diameter of nozzle, and obtaining Cn from Re as 0.972

+Cn=0.972;

+B=0.577;

+dia=((4*Q/(Cn*%pi))/((2*pdiff*1000/(d*(1-(B^4))))^0.5))^0.5;

+disp("mm",dia*1000,"Diameter of the nozzle=")

+clc;

+clear;

+vis=0.4;//Ns/(m^2)

+d=900;//kg/(m^3)

+D=0.02;//m

+Q=2.0*(10^-5);//(m^3)/s

+x1=0;

+x2=10;//m

+p1=200;//kPa

+x3=5;//m

+V=Q/(%pi*(D^2)/4);//m/s

+Re=d*V*D/vis;

+disp("Hence the flow is laminar.",Re,"a) Reynolds number =")

+pdiff=128*vis*(x2-x1)*Q/(%pi*(D^4)*1000);

+//for part b0 p1=p2; Q=%pi*(pdiff-(sw*l*sin(ang)))*(D^4)/(128*vis*l)

+ang=(asin(-128*vis*Q/(%pi*d*9.81*(D^4))))*180/%pi;

+//since sin(ang) doesn= not depend on pdiff, the the pressure is constant all along the pipe

+//hence for c)

+p3=p1;//kPa

+disp("kPa.",pdiff,"The pressure drop required if the pipe is horizontal=")

+disp("degrees.",ang,"b) The angle of the hill the pipe must be on if the oil is to flow at the same rate as a) but with (p1=p2) =")

+disp("kPa",p3,"c) For conditions of part b), the pressure at x3=5 m = ") \ No newline at end of file

+clc;

+clear;

+T=[60 80 100 120 140 160];//degree F

+d=[2.07 2.06 2.05 2.04 2.03 2.02];//(slugs/(ft^3))

+vis=[0.04 0.019 0.0038 0.00044 0.000092 0.000023];//lb*sec/(ft^2)

+Q=0.5;//(ft^3)/sec

+T1=100;//degree F

+l=6;//ft

+D=3;//in

+//Q=K*pdiff; where pdiff=p1-p2

+//hence K=%pi*(D^4)/(128*vis*l)

+count=1;

+for i=1:6

+ K(i)=(%pi*((D/12)^4))/(128*vis(i)*l);

+end

+plot2d(T,K,logflag='nl')

+xtitle("K vs T","T, degree F","K, (ft^5)/(lb.sec)")

+pdiff=(128*Q*vis(3)*l)/(%pi*((D/12)^4));//when temperature is 100 degree F

+disp("lb/(ft^2)",pdiff,"The pressure drop for the given Q and T =")

+V=Q/(%pi*((D/12)^2)/4);//ft/sec

+Re=d(3)*V*(D/12)/vis(3);

+disp("hence the flow is laminar",Re,"The reynolds number=")

+stress=pdiff*(D/12)/(4*l);//lb/(ft^2)

+disp("lb/(ft^2)",stress,"The wall stress for the given Q and T =")

+Fp=(%pi/4)*((D/12)^2)*pdiff;//lb

+Fv=(2*%pi)*((D/12)/2)*l*stress;//lb

+disp("lb",Fp,"The net pressure force =")

+disp("lb",Fv,"The net viscous/shear force =") \ No newline at end of file

+clc;

+clear;

+T=20;//degree C

+d=998;//kg/(m^3)

+kvis=1.004*(10^-6);//(m^2)/s; where kvis=kinematic viscosity

+D=0.1;//m

+Q=0.04;//(m^3)/sec

+pgrad=2.59;//kPa/m; where pgrad is pressure gradient

+r=0.025;//m

+stress=D*(pgrad*1000)/(4*1);//N/(m^2)

+uf=(stress/d)^0.5;//m/sec; where uf is frictional velocity

+ts=5*kvis*1000/(uf);//mm; where ts is the thickness of the viscous sublayer

+disp("mm",ts,"The thickness of the viscous sublayer=")

+V=Q/(%pi*(D^2)/4);//m/s

+Re=V*D/kvis;

+disp("hence the flow is turbulent.",Re,"The reynolds number=")

+n=8.4;//from turbulent flow velocity profile diagram

+

+//Q=(%pi)*(R^2)*V

+R=1;//assumption

+//let Q/Vc=x

+x=integrate('((1-(r/R))^(1/n))*(2*%pi*r)','r',0,R);

+q=%pi*(R^2)*V;

+Vc=q/x;//m/s

+disp("m/s",Vc,"The approximate centerline velocity=")

+stress1=(2*stress*r)/D;//N/(m^2)

+//d(uavg)/dr=urate=-(Vc/(n*R))*((1-(r/R))^((1-n)/n)); where uavg=average velocity

+urate=-(Vc/(n*(D/2)))*((1-(r/(D/2)))^((1-n)/n));//s^(-1)

+stresslam=-(kvis*d*urate);//N/(m^2)

+stressratio=(stress1-stresslam)/stresslam;

+disp(stressratio,"The ratio of teh turbulent to laminar stress at a point midway between the centreline and the pipe wall =") \ No newline at end of file

+clc;

+clear;

+D=4;//mm

+V=50;//m/sec

+l=0.1;//m

+d=1.23;//kg/(m^3)

+vis=1.79/100000;//N*sec/(m^2)

+Re=d*V*(D/1000)/vis;

+//if flow is laminar

+f=64/Re;

+pdiff=f*l*0.5*d*(V^2)/((D/1000)*1000);//kPa

+disp("kPa",pdiff,"The pressure drop if the flow is laminar=")

+//if flow is turbulent

+//roughness=0.0015; hence f=0.028

+f1=0.028;

+pdiff1=f1*l*0.5*d*(V^2)/((D/1000)*1000);//kPa

+disp("kPa",pdiff1,"The pressure drop if flow is turbulent=")

+

+clc;

+clear;

+A=[22 28 35 35 4 4 10 18 22];

+V=[36.4 28.6 22.9 22.9 200 200 80 44.4 36.4];

+//minimum area is at location 5, hence max velocity is at 5

+c5=(1.4**1716*(460+59))^0.5;//ft/sec

+Ma5=V(5)/c5;

+//applying energy equation between locations 1 and 9

+//hL=hp=(p1-p9)/sw=pdiff/sw

+//Pa=sw*Q*hp=sw*A(5)*V(5)*hL

+KLcorner=0.2;

+KLdif=0.6;

+KLscr=4;

+hL=((KLcorner*(((V(7))^2)+((V(8))^2)+((V(2))^2)+((V(3))^2))) + (KLdif*(((V(6))^2))) + (KLcorner*((V(5))^2)) + (KLscr*((V(4))^2)))/(2*32.2);//ft

+Pa=0.0765*A(5)*V(5)*hL/550;//hp

+pdiff=0.0765*hL/144;//psi

+disp("psi",pdiff,"The value of (p1-p9)=")

+disp("hp",Pa,"The horsepower supplied to the fluid by the fan=")

+v=50:300;

+count=1;

+for i=50:300

+ power(count)=0.0765*((((KLcorner*((A(5)*i/A(7))^2)+((A(5)*i/(A(8)))^2)+((A(5)*i/A(2))^2)+((A(5)*i/A(3))^2))) + (KLdif*(((A(5)*i/A(6))^2))) + (KLcorner*((i)^2)) + (KLscr*((A(5)*i/A(4))^2)))/(2*32.2))*(A(5))*i/550;

+ count=count+1;

+end

+plot2d(v,power,rect=[0,0,300,250])

+xtitle("Pa vs V5","V5, ft/sec","Pa, hp")

+clc;

+clear;

+T=120;//degree F

+D=8;//in

+vavg=10;//ft/s

+roughness=0;

+kvis=1.89/10000;//(ft^2)/s

+Re=vavg*(D/12)/kvis;

+//from this value of Re and roughness/D=0, and using Moody's chart

+f=0.022;

+hLperl=f*(vavg^2)/(D*2*32.2/12);

+//Dh=4*A/P=4*(a^2)/(4*a)=a

+

+//Vs=(%pi*((D/12)^2)*vavg)/(4*a^2)

+//a=f*((%pi*((D/12)^2)*vavg)/(4*a^2))/(2*32.2) and Reh=((%pi*((D/12)^2)*vavg)/(4*a^2))*a/kvis

+//by trial and error

+f=0.023;

+x=(%pi*((D/12)^2)*vavg/4)^2;

+y=x*f/(2*32.2);

+a=((y/0.0512)^(1/5))*12;//in

+disp("inches",a,"The duct size(a) for the square duct if the head loss per foot remains the same for the pipe and the duct=") \ No newline at end of file

+clc;

+clear;

+T=60;//degree F

+D=0.0625;//ft

+Q=0.0267;//(ft^3)/sec

+Df=0.5;//in

+l1=15;//ft

+l2=10;//ft

+l3=5;//ft

+l4=10;//ft

+l5=10;//ft

+l6=10;//ft

+V1=Q/(%pi*(D^2)/4);//ft/sec

+V2=Q/(%pi*((Df/12)^2)/4);//ft/sec

+d=1.94;//slugs/ft

+vis=2.34/100000;//lb*sec/(ft^2)

+Re=d*V1*D/vis;

+disp("hence the flow is turbulent",Re,"The reynolds number =")

+//applying energy equation between points 1 and 2

+//when all head losses are excluded

+p1=(d*32.2*(l2+l4))+(0.5*d*((V2^2)-(V1^2)));//lb/(ft^2)

+disp("psi",p1/144,"a)The pressure at point 1 when all head losses are neglected=")

+//if major losses are included

+f=0.0215;

+hLmajor=f*(l1+l2+l3+l4+l5+l6)*(V1^2)/(D*2*32.2);

+p11=p1+(d*32.2*hLmajor);//lb/(ft^2)

+disp("psi",p11/144,"b)The pressure at point 1 when only major head losses are included=")

+//if major and minor losses are included

+KLelbow=1.5;

+KLvalve=10;

+KLfaucet=2;

+hLminor=(KLvalve+(4*KLelbow)+KLfaucet)*(V1^2)/(2*32.2);

+p12=p11+(d*32.2*hLminor);//lb/(ft^2)

+disp("psi",p12/144,"c)The pressure at point 1 when both major and minor head losses are included=")

+H=(p1/(32.2*1.94))+(V1*V1/(2*32.2));//ft

+dist=0:60;

+for i=0:15

+ press(i+1)=p1/144;

+ press1(i+1)=((d*32.2*(l2+l4))+(0.5*d*((V2^2)-(V1^2)))+(d*32.2*(f*(l1+l2+l3+l4+l5+l6-i)*(V1^2)/(D*2*32.2)))+(d*32.2*(KLvalve+(4*KLelbow)+KLfaucet)*(V1^2)/(2*32.2)))/144;

+ head(i+1)=H;

+ head1(i+1)=((press1(i+1))*144/(32.2*1.94))+((V1^2)/(2*32.2));

+end

+for i=16:25

+ press(i+1)=((d*32.2*((l2+l4)-(i-15)))+(0.5*d*((V2^2)-(V1^2))))/144;

+ press1(i+1)=((d*32.2*((l2+l4)-(i-15)))+(0.5*d*((V2^2)-(V1^2)))+(d*32.2*f*(l1+l2+l3+l4+l5+l6-i)*(V1^2)/(D*2*32.2))+(d*32.2*(KLvalve+(3*KLelbow)+KLfaucet)*(V1^2)/(2*32.2)))/144;

+ head(i+1)=H;

+ head1(i+1)=(press1(i+1)*144/(32.2*1.94))+((V1^2)/(2*32.2))+(i-l1);

+end

+for i=26:30

+ press(i+1)=((d*32.2*((l2+l4)-(25-15)))+(0.5*d*((V2^2)-(V1^2))))/144;

+ press1(i+1)=((d*32.2*((l2+l4)-(25-15)))+(0.5*d*((V2^2)-(V1^2)))+(d*32.2*(f*(l1+l2+l3+l4+l5+l6-i)*(V1^2)/(D*2*32.2)))+(d*32.2*(KLvalve+(2*KLelbow)+KLfaucet)*(V1^2)/(2*32.2)))/144;

+ head(i+1)=H;

+ head1(i+1)=(press1(i+1)*144/(32.2*1.94))+((V1^2)/(2*32.2))+l2;

+end

+for i=31:40

+ press(i+1)=((d*32.2*((l2+l4)-(i-l1-l3)))+(0.5*d*((V2^2)-(V1^2))))/144;

+ press1(i+1)=((d*32.2*((l2+l4)-(i-l1-l3)))+(0.5*d*((V2^2)-(V1^2)))+(d*32.2*(f*(l1+l2+l3+l4+l5+l6-i)*(V1^2)/(D*2*32.2)))+(32.2*d*(KLvalve+(KLelbow)+KLfaucet)*(V1^2)/(2*32.2)))/144;

+ head(i+1)=H;

+ head1(i+1)=(press1(i+1)*144/(32.2*1.94))+((V1^2)/(2*32.2))+(i-(l1+l3));

+end

+for i=41:50

+ press(i+1)=((d*32.2*((l2+l4)-(40-l1-l3)))+(0.5*d*((V2^2)-(V1^2))))/144;

+ press1(i+1)=((d*32.2*((l2+l4)-(40-l1-l3)))+(0.5*d*((V2^2)-(V1^2)))+(d*32.2*(f*(l1+l2+l3+l4+l5+l6-i)*(V1^2)/(D*2*32.2)))+(d*32.2*(KLvalve+KLfaucet)*(V1^2)/(2*32.2)))/144;

+ head(i+1)=H;

+ head1(i+1)=(press1(i+1)*144/(32.2*1.94))+((V1^2)/(2*32.2))+(l2+l4);

+end

+for i=51:60

+ press(i+1)=((d*32.2*((l2+l4)-(40-l1-l3)))+(0.5*d*((V2^2)-(V1^2))))/144;

+ press1(i+1)=((d*32.2*((l2+l4)-(40-l1-l3)))+(0.5*d*((V2^2)-(V1^2)))+(d*32.2*(f*(l1+l2+l3+l4+l5+l6-i)*(V1^2)/(D*2*32.2)))+d*32.2*((KLfaucet)*(V1^2)/(2*32.2)))/144;

+ head(i+1)=H;

+ head1(i+1)=(press1(i+1)*144/(32.2*1.94))+((V1^2)/(2*32.2))+(l2+l4);

+end

+plot(dist,press,"o-")

+plot(dist,press1,"x-")

+h1=legend(['without losses';'with losses'])

+xtitle("p vs distance long pipe from (1)","distance along pipe from (1), ft","p, psi")

+xclick(1);

+clf();

+plot(dist,head,"o-")

+plot(dist,head1,"x-")

+h2=legend(['energy line with no losses';'energy line including losses'])

+xtitle("H vs distance long pipe from (1)","distance along pipe from (1), ft","H,elevation to energy line, ft")

+

+end

+clc;

+clear;

+T=140;//degree F

+sw=53.7;//lb/(ft^3)

+vis=8/100000;//lb*sec/(ft^2)

+l=799;//miles

+D=4;//ft

+Q=117;//(ft^3)/sec

+V=9.31;//ft/sec

+//energy equation=> hp=hL=f*(l/D)*((V^2)/(2*g))

+f=0.0125;

+hp=f*(l*5280/D)*((V^2)/(2*32.2));//ft

+Pa=sw*Q*hp/550;//hp

+disp("hp",Pa,"The horsepower required to drive the system=")

+dia=2:0.01:6;

+count=1;

+for i=2:0.01:6

+ power(count)=sw*Q*(f*(l*5280/i)*(((Q/(%pi*(i^2)/4))^2)/(2*32.2)))/550;

+ count=count+1;

+end

+plot2d(dia,power,rect=[0,0,6,4000000])

+xtitle("Pa vs D","D, ft","Pa, hp") \ No newline at end of file