diff options
Diffstat (limited to '914/CH12')
| -rwxr-xr-x | 914/CH12/EX12.10/ex12_10.sce | 18 | ||||
| -rwxr-xr-x | 914/CH12/EX12.11/ex12_11.sce | 16 | ||||
| -rwxr-xr-x | 914/CH12/EX12.12/ex12_12.sce | 20 | ||||
| -rwxr-xr-x | 914/CH12/EX12.13/ex12_13.sce | 16 | ||||
| -rwxr-xr-x | 914/CH12/EX12.14/ex12_14.sce | 63 | ||||
| -rwxr-xr-x | 914/CH12/EX12.15/ex12_15.sce | 20 | ||||
| -rwxr-xr-x | 914/CH12/EX12.16/ex12_16.sce | 29 | ||||
| -rwxr-xr-x | 914/CH12/EX12.17/ex12_17.sce | 62 | ||||
| -rwxr-xr-x | 914/CH12/EX12.2/ex12_2.sce | 22 | ||||
| -rwxr-xr-x | 914/CH12/EX12.3/ex12_3.sce | 18 | ||||
| -rwxr-xr-x | 914/CH12/EX12.5/ex12_5.sce | 26 |
11 files changed, 310 insertions, 0 deletions
diff --git a/914/CH12/EX12.10/ex12_10.sce b/914/CH12/EX12.10/ex12_10.sce new file mode 100755 index 000000000..a4b561f72 --- /dev/null +++ b/914/CH12/EX12.10/ex12_10.sce @@ -0,0 +1,18 @@ +clc;
+warning("off");
+printf("\n\n example12.10 - pg590");
+// given
+T=293.15; //[K]
+pp=999; //[kg/m^3] - density of water
+mu=0.01817*10^-3; //[kg/m*sec] - viscosity of air
+p=1.205; //[kg/m^3] - density of air
+d=5*10^-6; //[m] - particle diameter
+g=9.80; //[m/sec^2]
+rp=d/2;
+Ut=((2*g*(rp^2))*(pp-p))/(9*mu);
+Nre=(d*Ut*p)/(mu);
+// clearly the flow is in the stokes law region at this low reynolds number;therefore , the drag force is
+Fp=6*%pi*mu*rp*Ut;
+printf("\n\n The drag force is \n Fp = %e N",Fp);
+
+
diff --git a/914/CH12/EX12.11/ex12_11.sce b/914/CH12/EX12.11/ex12_11.sce new file mode 100755 index 000000000..e640cd2f4 --- /dev/null +++ b/914/CH12/EX12.11/ex12_11.sce @@ -0,0 +1,16 @@ +clc;
+warning("off");
+printf("\n\n example12.11 - pg591");
+// given
+T=293.15; //[K]
+pp=999; //[kg/m^3] - density of water
+mu=0.01817*10^-3; //[kg/m*sec] - viscosity of air
+p=1.205; //[kg/m^3] - density of air
+d=5*10^-6; //[m] - particle diameter
+g=9.80; //[m/sec^2]
+rp=d/2;
+Ut=((2*g*(rp^2))*(pp-p))/(9*mu);
+Nre=(d*Ut*p)/(mu);
+t=((-2*(rp^2)*pp))/(9*mu)*(log(1-0.99));
+printf("\n\n Time for the drop of water in previous example from an initial velocity of zero to 0.99*Ut is \n t = %e sec",t);
+printf("\n\n In other words, the drop accelerates almost instantaneously to its terminal velocity");
diff --git a/914/CH12/EX12.12/ex12_12.sce b/914/CH12/EX12.12/ex12_12.sce new file mode 100755 index 000000000..32ec2ccae --- /dev/null +++ b/914/CH12/EX12.12/ex12_12.sce @@ -0,0 +1,20 @@ +clc;
+warning("off");
+printf("\n\n example12.12 - pg 594");
+// given
+pp=1.13*10^4; //[kg/m^3] - density of lead particle
+p=1.22; //[kg/m^3] - density of air
+g=9.80; //[m/sec^2] - acceleration due to gravity
+d=2*10^-3; //[m] - diameter of particle
+mu=1.81*10^-5; //[kg/m*sec] - viscosity of air
+// let us assume
+Cd=0.44;
+Ut=((4*d*g*(pp-p))/(3*p*Cd))^(1/2);
+disp(Ut)
+Nre=(Ut*d*p)/(mu);
+// from fig 12,16 value of Cd is
+Cd=0.4;
+Ut=((4*d*g*(pp-p))/(3*p*Cd))^(1/2);
+Nre=(Ut*d*p)/(mu);
+// Within the readibility of the chart Cd is unchanged and therefore the above obtained Cd is the final answer
+printf("\n\n The terminal velocity is \n Ut = %f m/sec",Ut);
diff --git a/914/CH12/EX12.13/ex12_13.sce b/914/CH12/EX12.13/ex12_13.sce new file mode 100755 index 000000000..f776336a2 --- /dev/null +++ b/914/CH12/EX12.13/ex12_13.sce @@ -0,0 +1,16 @@ +clc;
+warning("off");
+printf("\n\n example12.13 - pg595");
+// given
+distance=1/12; //[ft]
+time=60; //[sec]
+Ut=distance/time;
+mu=1.68; //[lb/ft*sec] - viscosity
+pp=58; //[lb/ft^3] - density of sphere
+p=50; //[lb/ft^3] - density of polymer solution
+g=32; //[ft/sec] - acceleration due to gravity
+rp=((9*mu)*(Ut)*((2*g)^(-1))*((pp-p)^(-1)))^(1/2);
+printf("\n\n The required particle diameter would be about %f inch",rp*2*12);
+Nre=(rp*2*Ut*p)/(mu);
+disp(Nre,"Nre=");
+printf("\n\n This reynolds number is well within the stokes law region ; thus the design is reasonable");
diff --git a/914/CH12/EX12.14/ex12_14.sce b/914/CH12/EX12.14/ex12_14.sce new file mode 100755 index 000000000..87a5c2946 --- /dev/null +++ b/914/CH12/EX12.14/ex12_14.sce @@ -0,0 +1,63 @@ +clc;
+warning("off");
+printf("\n\n example12.14 - pg616");
+// given
+T=842; //[degF] - temperature
+P=14.6; //[psia] - pressure
+p=0.487; //[kg/m^3] - density of air
+mu=3.431*10^-5; //[kg/m*sec] - viscosity of air
+k=0.05379; //[W/m*K] - thermal conductivity
+Npr=0.7025; //prandtl no.
+// (a) static void fraction
+mcoal=15*2000; //[lb] - mass of coal
+pcoal=94; //[lbm/ft^3] - density of coal
+d=10; //[ft]
+L=7; //[ft]
+area=((%pi*(d^2))/4);
+Vcoal=mcoal/pcoal;
+Vtotal=area*L;
+e=(Vtotal-Vcoal)/(Vtotal);
+disp(e,"(a) The void fraction is E=");
+// (b) minimum void fraction and bed height
+d=200; //[um] - particle diameter
+Emf=1-0.356*((log10(d))-1);
+// this value seems to be a lottle low and therefore 0.58 will be used
+Emf=0.58;
+Lmf=((L)*(1-e))/(1-Emf);
+printf("\n\n (b) The bed height is \n Lmf = %f ft",Lmf);
+// (c) Minimum fluidization velocity
+P1=20; //[psia]
+P2=14.696; //[psia]
+p1=(p*P1)/(P2);
+// the archimides no. is
+g=9.78; //[m/sec^2]
+Nar=p1*g*((d*10^-6)^3)*(1506-p1)*((1/(mu)^2));
+C1=27.2;
+C2=0.0408;
+Nremf=(((C1^2)+C2*Nar)^(1/2))-C1;
+Umf=(Nremf*mu)/((d*10^-6)*p1);
+printf("\n\n (c) The minimum fluidization velocity is \n Umf = %f m/sec",Umf);
+// (d) Minimum pressure
+deltapmf=(1506-p1)*(g)*(1-Emf)*((Lmf*12*2.54)/(100))+p1*g*Lmf;
+printf("\n\n (c) The minimum pressure drop for fluidization is \n -deltapmf = %e Pa",deltapmf);
+// (e) Particle settling velocity
+Cd=0.44;
+Ut=(((8*((d*10^-6)/2)*g)*(1506-p1))/(3*p1*Cd))^(1/2);
+Nrep=(Ut*d*10^-6*p1)/(mu);
+disp(Nrep,"Nrep=");
+// clearly at the point of minimum velocity for fast fluidization , the terminal settling velocity is not in the range of Newtons law.Therefore the eq. for the transition region will be tried
+Ut=((5.923/18.5)*(((d*10^-6)*p1)/(mu))^(0.6))^(1/(2-0.6))
+printf("\n\n (e) The particle settling velocity is \n Ut = %f m/sec",Ut);
+// (f) Bed to wall heat transfer coefficient
+Nrefb=(d*10^-6)*2.5*Umf*p1*(1/mu);
+Nnufb=0.6*Npr*((Nrefb)^(0.3));
+hw=Nnufb*(k/(d*10^-6));
+printf("\n\n (f) The bed to wall heat transfer coefficient is \n hw = %f W/m^2*K",hw);
+
+
+
+
+
+
+
+
diff --git a/914/CH12/EX12.15/ex12_15.sce b/914/CH12/EX12.15/ex12_15.sce new file mode 100755 index 000000000..7f328e896 --- /dev/null +++ b/914/CH12/EX12.15/ex12_15.sce @@ -0,0 +1,20 @@ +clc;
+warning("off");
+printf("\n\n example12.5 - pg618");
+// given
+pp=249.6; //[lb/ft^3] - density of catalyst
+p=58; //[lb/ft^3] - density of liquid
+g=32.174; //[ft/sec^2]
+gc=32.174;
+Lmf=5; //[ft] - height of bed
+mu=6.72*10^-3; //[lbm/ft*sec] - viscosity of liquid
+dp=0.0157/12; //[ft] - diameter of particle
+emf=0.45;
+deltapmf=(pp-p)*(g/gc)*(1-emf)*(Lmf);
+Nar=(p*g*dp^3)*(pp-p)*(1/(mu)^2);
+C1=27.2;
+C2=0.0408;
+Nremf=(((C1^2)+C2*Nar)^(1/2))-C1;
+Umf=Nremf*(mu/(dp*p));
+printf("\n\n Minimum fluidization velocity is \n Umf = %e ft/sec",Umf);
+
diff --git a/914/CH12/EX12.16/ex12_16.sce b/914/CH12/EX12.16/ex12_16.sce new file mode 100755 index 000000000..174e4a389 --- /dev/null +++ b/914/CH12/EX12.16/ex12_16.sce @@ -0,0 +1,29 @@ +clc;
+warning("off");
+printf("\n\n example12.16 - pg624");
+// given
+d=24*10^-6; //[m] - diameter of wire
+T=415; //[K] - operating temperature of hot wire anemometer
+P=0.1; //[W] - power consumption
+L=250*d;
+Tair=385; //[K] - temperature of air in duct
+A=%pi*d*L;
+Tfilm=(T+Tair)/2;
+// properties of air at Tfilm
+p=0.8825; //[kg/m^3]
+mu=2.294*10^-5; //[kg/m*s]
+cpf=1013; //[J*kg/K]
+kf=0.03305; //[W/m*K]
+Npr=0.703;
+h=P/(A*(T-Tair));
+Nnu=(h*d)/kf;
+function y=func(x)
+ y=Nnu-0.3-((0.62*(x^(1/2))*(Npr^(1/3)))/((1+((0.4/Npr)^(2/3)))^(1/4)))*((1+((x/(2.82*(10^5)))^(5/8)))^(4/5));
+endfunction
+// on solving the above function for x by using some root solver technique like Newton raphson method , we get
+x=107.7;
+// or
+Nre=107.7;
+y=func(x);
+Um=(Nre*mu)/(d*p);
+printf("\n\n The velocity is \n Um = %f m/sec = %f ft/sec",Um,Um*3.28);
diff --git a/914/CH12/EX12.17/ex12_17.sce b/914/CH12/EX12.17/ex12_17.sce new file mode 100755 index 000000000..1b56e4e43 --- /dev/null +++ b/914/CH12/EX12.17/ex12_17.sce @@ -0,0 +1,62 @@ +clc;
+warning("off");
+printf("\n\n example12.17 - pg630");
+// given
+dt=0.75;
+St=1.5*dt;
+Sl=3*dt;
+Lw=1; //[m]
+N=12;
+Stotalarea=N*(St/12)*Lw;
+Sminarea=N*((St-dt)/12)*Lw*0.3048;
+// properties of air at 293.15 K
+p=1.204; //[kg/m^3]
+mu=1.818*10^-5; //[kg/m*s]
+cp=1005; //[J*kg/K];
+k=0.02560; //[J/s*m*K]
+Npr=(cp*mu)/k;
+U_inf=7; //[m/sec]
+Umax=U_inf*(St/(St-dt));
+w=p*Umax*Sminarea;
+C_tubes=0.05983; //[m^2/m] - circumference of the tubes
+N_tubes=96;
+Atubes=N_tubes*C_tubes*Lw;
+Tw=328.15; //[K]
+Tinf=293.15; //[K]
+Tin=293.15; //[K]
+Tout=293.15; //[K]
+u=100;
+while u>10^-1
+ T=(Tin+Tout)/2
+ Told=Tout;
+ p=-(0.208*(10^-3))+(353.044/T);
+ mu=-(9.810*(10^-6))+(1.6347*(10^-6)*(T^(1/2)));
+ cp=989.85+(0.05*T);
+ k=0.003975+7.378*(10^-5)*T;
+ Npr=(cp*mu)/k;
+ dt=0.75*0.0254;
+ Gmax=w/Sminarea;
+ Nre=(dt*Gmax)/mu;
+ h=0.27*(k/dt)*(Npr^0.36)*(Nre^0.63);
+ h=h*0.98;
+ deltaT=(h*Atubes*(Tw-Tinf))/(w*cp);
+ Tout=Tin+deltaT;
+ u=abs(Tout-Told);
+end
+T=(Tin+Tout)/2
+p=-(0.208*(10^-3))+(353.044/T);
+mu=-(9.810*(10^-6))+(1.6347*(10^-6)*(T^(1/2)));
+dt=0.75;
+dv=(4*(St*Sl-(%pi*(dt^2)*(1/4))))/(%pi*dt)*(0.09010/3.547);
+de=dv;
+Nre=(dv*24.72)/mu;
+dv=dv/(0.09010/3.547);
+ftb=1.92*(Nre^(-0.145));
+Zt=Sl;
+Ltb=8*Sl;
+deltap=(ftb*(24.72^2))/(2*p*(dv/Ltb)*((St/dv)^0.4)*((St/Zt)^0.6));
+printf("\n\n -deltap = %f kg/m*s = %f N/m^2 = %f psia",deltap,deltap,deltap*(0.1614/1113));
+
+
+
+
diff --git a/914/CH12/EX12.2/ex12_2.sce b/914/CH12/EX12.2/ex12_2.sce new file mode 100755 index 000000000..2ef8a2015 --- /dev/null +++ b/914/CH12/EX12.2/ex12_2.sce @@ -0,0 +1,22 @@ +clc;
+warning("off");
+printf("\n\n example12.2 - pg562");
+p=1.2047*0.06243; //[lb/ft^3]
+mu=(18.17*10^-6)*(0.6720); //[lb/ft*sec]
+v=mu/p;
+x=2; //[ft]
+U=6; //[ft/sec]
+Nre=(x*U)/v;
+disp("The Reynolds number is well within the laminar region",Nre,"Nre=");
+del=5*x*(Nre)^(-1/2);
+C1=0.33206;
+Cd=2*C1*(Nre)^(-1/2);
+L2=2; //[ft]
+L1=1; //[ft]
+b=1;
+F=((2*(C1)*U*b))*((mu*p*U)^(1/2))*(((L2)^(1/2))-((L1)^(1/2)));
+gc=32.174;
+F=F/gc;
+printf("\n\n The value of F properly expressed in force units is \n F=%e lbf",F);
+
+
diff --git a/914/CH12/EX12.3/ex12_3.sce b/914/CH12/EX12.3/ex12_3.sce new file mode 100755 index 000000000..a6526e687 --- /dev/null +++ b/914/CH12/EX12.3/ex12_3.sce @@ -0,0 +1,18 @@ +clc;
+warning("off");
+printf("\n\n example12.3 - pg569");
+U=3; //[m/sec]
+x1=1; //[m]
+x2=2; //[m]
+p=1/(1.001*10^-3); //[kg/m^3];
+mu=1*10^-3; //[kg/m*sec]
+Nre1=(x1*U*p)/(mu);
+Nre2=(x2*p*U)/(mu);
+tauw=(1/2)*(p*(U^2))*((2*log10(Nre1)-0.65)^(-2.3));
+B=1700;
+Cd=(0.455*(log10(Nre2))^-2.58)-(B/(Nre2));
+Lb=2.0;
+F=(1/2)*(p*(U^2))*(Lb)*(Cd);
+printf("\n\n the drag on the plate is \n F = %f kg*m/sec^2 = %f N",F,F);
+
+
diff --git a/914/CH12/EX12.5/ex12_5.sce b/914/CH12/EX12.5/ex12_5.sce new file mode 100755 index 000000000..131f37826 --- /dev/null +++ b/914/CH12/EX12.5/ex12_5.sce @@ -0,0 +1,26 @@ +clc;
+warning("off");
+printf("\n\n example12.5 - pg576");
+T=290; //[K] - temperature of flowing water
+U=3; //[m/sec] - free stream velocity
+Tfs=285; //[K] - temperature of free stream
+vr=10^-3; //[m^3/kg] - volume per unit mass
+p=1/vr; //[kg/m^3] - density of water at Tfs
+mu=1225*10^-6; //[N*sec/m^2]
+k=0.590; //[W/m*K]
+Npr=8.70;
+// (a) The length of laminar boundary
+Nre=5*10^5;
+xc=(Nre)*(mu/(p*U));
+printf("\n\n (a) The length of laminar boundary is \n xc = %f m",xc);
+// (b) Thickness of the momentum boundary layer and thermal boundary layer
+del=5*xc*((Nre)^(-1/2));
+delh=del*((Npr)^(-1/3));
+printf("\n\n (b) The thickness of momentum boundary layer is \n del = %e m\n The thickness of the hydryodynamic layer is \n delh = %e m",del,delh);
+// (c) Local heat transfer coefficient
+x=0.2042; //[ft]
+hx=((0.33206*k)/(x))*((Nre)^(1/2))*((Npr)^(1/3));
+printf("\n\n (c) The local heat transfer coefficient is \n h = %f W/m^2*K = %f Btu/hr*ft^2*degF",hx,hx*0.17611);
+// (d) Mean heat transfer coefficient
+hm=2*hx;
+printf("\n\n (d) The mean heat transfer coefficient is \n h = %f W/m^2*K = %f Btu/hr*ft^2*degF",hm,hm*0.17611);
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