# Chapter 12 : Transport past immersed bodies¶

### Example 12.2 - Page No :562¶

In [1]:
import math

# Variables
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]

# Calculation and Results
Nre = (x*U)/v;
print "The Reynolds number is well within the laminar region %.3e 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;
print " The value of F properly expressed in force units is  F = %.3e lbf"%(F);

The Reynolds number is well within the laminar region 7.391e+04 Nre
The value of F properly expressed in force units is  F = 1.204e-04 lbf


### Example 12.3 - Page No :569¶

In [2]:
import math

# Variables
U = 3.;  			 #[m/sec]
x1 = 1.;  			 #[m]
x2 = 2.;  			 #[m]

# Calculations
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*math.log10(Nre1)-0.65)**(-2.3));
B = 1700.;
Cd = (0.455*(math.log10(Nre2))**-2.58)-(B/(Nre2));
Lb = 2.0;
F = (1./2)*(p*(U**2))*(Lb)*(Cd);

Xc = round((5*10**5 * mu)/(U*p),3)
CDlaminar = round(4*.33206*(5*10**5)**(-1./2),5)
Flaminar= round(1./2*(p*U**2)*Xc*CDlaminar,3)
Cd = round(.455*((math.log10(Nre2))**-2.58),6)
Fturbulent1 = round(1./2*(p*U**2)*x2*Cd,2)
Fturbulent2 = round(1./2*(p*U**2)*Xc*.005106,3)
Factual = 1.411 + Fturbulent1 - Fturbulent2

# Results
print " the drag on the plate is  F  =  %f kg*m/sec**2  =  %.1f N"%(F,F);
print ' total drag on the plate Factual = %.2f N'%Factual
print " the shear stress is %.f N/m^2"%tauw

 the drag on the plate is  F  =  26.801111 kg*m/sec**2  =  26.8 N
total drag on the plate Factual = 26.93 N
the shear stress is 14 N/m^2


### Example 12.5 - Page No :576¶

In [8]:
# Variables
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;

# Calculation and Results
#  (a) The length of laminar boundary
Nre = 5.*10**5;
xc = (Nre)*(mu/(p*U));
print " a) The length of laminar boundary is  xc  =  %.4f m"%(xc);
#  (b) Thickness of the momentum boundary layer and thermal boundary layer
del_ = 5*xc*((Nre)**(-1./2));
del_h = del_*((Npr)**(-1./3));
print " b) The thickness of momentum boundary layer is  del_  =  %.3e m \n The \
thickness of the hydryodynamic layer is  del_h  =  %.3e m"%(del_,del_h);

# (c) Local heat transfer coefficient
x = 0.2042;  			 #[ft]
hx = ((0.33206*k)/(x))*((Nre)**(1./2))*((Npr)**(1./3));
print " c) The local heat transfer coefficient is  h  =  %.0f W/m**2*K \
=  %.0f Btu/hr*ft**2*degF"%(hx,hx*0.17611);

# (d) Mean heat transfer coefficient
hm = 2*hx;
print " d) The mean heat transfer coefficient is  h  =  %.0f W/m**2*K \
=  %.0f Btu/hr*ft**2*degF"%(hm,round(hm*0.17611,1));

# Answer may vary because of rounding error.

 a) The length of laminar boundary is  xc  =  0.2042 m
b) The thickness of momentum boundary layer is  del_  =  1.444e-03 m
The  thickness of the hydryodynamic layer is  del_h  =  7.019e-04 m
c) The local heat transfer coefficient is  h  =  1395 W/m**2*K  =  246 Btu/hr*ft**2*degF
d) The mean heat transfer coefficient is  h  =  2791 W/m**2*K   =  492 Btu/hr*ft**2*degF


### Example 12.10 - Page No :590¶

In [9]:
# Variables
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]

# Calculations
rp = d/2;
Ut = ((2*g*(rp**2))*(pp-p))/(9*mu);
Nre = (d*Ut*p)/(mu);
Fp = 6*math.pi*mu*rp*Ut;

# Results
print " The drag force is  Fp  =  %.2e N"%(Fp);

 The drag force is  Fp  =  6.40e-13 N


### Example 12.11 - Page No :591¶

In [11]:
# Variables
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]

# Calculations
rp = d/2;
Ut = ((2*g*(rp**2))*(pp-p))/(9*mu);
Nre = (d*Ut*p)/(mu);
t = ((-2*(rp**2)*pp))/(9*mu)*(math.log(1-0.99));

# Results
print " Time for the drop of water in previous Example  from an initial \
velocity of zero to 0.99*Ut is  \n  t  =  %.3e sec"%(t);
print " In other words, the drop accelerates almost instantaneously to its terminal velocity"

 Time for the drop of water in previous Example  from an initial  velocity of zero to 0.99*Ut is
t  =  3.517e-04 sec
In other words, the drop accelerates almost instantaneously to its terminal velocity


### Example 12.12 - Page No : 594¶

In [23]:
# Variables
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

# Calculations
# let us assume
Cd = 0.44;
Ut = ((4*d*g*(pp-p))/(3*p*Cd))**(1./2);
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);

# Results
# Within the readibility of the chart Cd is unchanged and therefore the above obtained Cd is the final answer

print " The terminal velocity is  Ut  =  %.2f m/sec"%(Ut);

 The terminal velocity is  Ut  =  24.60 m/sec


### Example 12.13 - Page No :595¶

In [14]:
# Variables
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

# Calculations
rp = ((9*mu)*(Ut)*((2*g)**(-1))*((pp-p)**(-1)))**(1./2);
Nre = (rp*2*Ut*p)/(mu);

# Results
print " The required particle diameter would be about %.2f inch"%(rp*2*12);
print "Nre = %.2e"%Nre
print " This reynolds number is well within the stokes law region ; thus the design is reasonable"

 The required particle diameter would be about 0.15 inch
Nre = 5.29e-04
This reynolds number is well within the stokes law region ; thus the design is reasonable


### Example 12.14 - Page No :616¶

In [22]:
# Variables
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.

# Calculation and Results
# (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 = ((math.pi*(d**2))/4);
Vcoal = mcoal/pcoal;
Vtotal = area*L;
e = (Vtotal-Vcoal)/(Vtotal);
print "(a) The void_ fraction is E = %.2f"%e

# (b) minimum void_ fraction and bed height
d = 200.;  			 #[um] - particle diameter
Emf = 1-0.356*((math.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);
print " b) The bed height is  Lmf  =  %.3f ft"%(Lmf);

# (c) Minimum fluid_ization velocity
P1 = 20.;  			 #[psia]
P2 = 14.696;  			 #[psia]
p1 = (p*P1)/(P2);

# the archimid_es 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);
print " c) The minimum fluid_ization velocity is  Umf  =  %.4f %% m/sec"%(Umf);

# (d) Minimum pressure
del_tapmf = (1506-p1)*(g)*(1-Emf)*((Lmf*12*2.54)/(100))+p1*g*Lmf;
print " d) The minimum pressure drop for fluid_ization is  -del_tapmf  =  %.3e Pa"%(del_tapmf);

# (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);
print "Nrep = %.2f"%Nrep
Ut = ((5.923/18.5)*(((d*10**-6)*p1)/(mu))**(0.6))**(1./(2-0.6))
print " e) The particle settling velocity is  Ut  =  %.5f 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));
print " f) The bed to wall heat transfer coefficient is  hw  =  %.1f W/m**2*K"%(hw);

# Answer may vary because of rounding error.

(a) The void_ fraction is E = 0.42
b) The bed height is  Lmf  =  9.675 ft
c) The minimum fluid_ization velocity is  Umf  =  0.0129 % m/sec
d) The minimum pressure drop for fluid_ization is  -del_tapmf  =  1.830e+04 Pa
Nrep = 14.18
e) The particle settling velocity is  Ut  =  0.79114 m/sec
f) The bed to wall heat transfer coefficient is  hw  =  60.6 W/m**2*K


### Example 12.15 - Page No :618¶

In [23]:
# Variables
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;

# Calculations
del_tapmf = (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));

# Results
print " Minimum fluidization velocity is  Umf  =  %.2e ft/sec"%(Umf);

 Minimum fluidization velocity is  Umf  =  1.18e-03 ft/sec


### Example 12.16 - Page No :624¶

In [25]:
# Variables
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 = math.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;

# Calculations
h = P/(A*(T-Tair));
Nnu = (h*d)/kf;
def func(x):
return 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));

# on solving the above function for x by umath.sing 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);

# Results
print " The velocity is  Um  =  %.1f m/sec  =  %d ft/sec"%(Um,Um*3.28);

 The velocity is  Um  =  116.6 m/sec  =  382 ft/sec


### Example 12.17 - Page No :630¶

In [4]:
# Variables
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]

# Calculations
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;
del_taT = (h*Atubes*(Tw-Tinf))/(w*cp);
Tout = Tin+del_taT;
u = abs(Tout-Told);

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-(math.pi*(dt**2)*(1./4))))/(math.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;
del_tap = (ftb*(24.72**2))/(2*p*(dv/Ltb)*((St/dv)**0.4)*((St/Zt)**0.6));

# Results
print " del_tap  =  %.0f kg/m*s  =  %.0f N/m**2  =  %f psia"%(del_tap,del_tap,round(del_tap*0.1614/1113,5))
print " Exit temperature : %.2f K"%T
# answer may slightly vary because of rounding error.

 del_tap  =  1113 kg/m*s  =  1113 N/m**2  =  0.161350 psia
Exit temperature : 299.87 K