# Chapter 13 : Unsteady-state transport¶

### Example 13.1 - Page No :651¶

In [2]:
import math

# Variables
# given
h = 12.;  			 #[W/m**2*K] - heat transfer coefficeint
k = 400.;  			 #[W/m*K] - thermal conductivity

# Calculation and Results
# (a) for sphere
r = 5.*10**-2;  			 #[m] - radius of copper sphere
Lc = ((4*math.pi*((r)**3))/3)/(4*math.pi*((r)**2));
Nbi = h*Lc*(1./k);
print " a) The biot no. is  Nbi  =  %.0e"%(Nbi);

# (b) for cyclinder
r = 0.05;  			 #[m] - radius of cyclinder
L = 0.3;  			 #[m] - height of cyclinder
Lc = (math.pi*((r)**2)*L)/(2*math.pi*r*L);
Nbi = h*Lc*(1./k);
print " b) The biot no. is  Nbi  =  %.1e"%(Nbi);

# (c) for a long square rod
L = .4;  			 #[m] - length of copper rod
r = 0.05;  			 #[m] - radius of a cyclinder havimg same cross sectional area as that of square
x = ((math.pi*r**2)**(1./2));
Lc = ((x**2)*L)/(4*x*L);
Nbi = h*Lc*(1./k);
print " c) The biot no. is  Nbi  =  %.3e"%(Nbi);
a) The biot no. is  Nbi  =  5e-04
b) The biot no. is  Nbi  =  7.5e-04
c) The biot no. is  Nbi  =  6.647e-04

### Example 13.6 - Page No :684¶

In [4]:
# Variables
# given
d = 1*0.0254;  		 #[m]
Lr = d/2;  			 #[m];
Lz = (1.2/2)*(0.0254);
x = Lz;
r = Lr;
k = 0.481;
h = 20.;
mr = k/(h*Lr);
mz = k/(h*Lz);
nr = r/Lr;
nz = x/Lz;
t = 1.2;  			 #[sec]

# Calculations
alpha = 1.454*10**-4;
Xr = (alpha*t)/(Lr**2);
Xz = (alpha*t)/(Lz**2);

# using the above value of m,n,X the value for Ycz and Ycr from fig 13.14 is
Ycr = 0.42;
Ycz = 0.75;
Yc = Ycr*Ycz;
T_infinity = 400.;  			 #[K]
To = 295.;
Tc = T_infinity-(Yc*(T_infinity-To));

# Results
print " The temperature t the centre is  Tc  =  %.0f K"%(Tc);

# Answer is vary because of rounding error.
The temperature t the centre is  Tc  =  367 K

### Example 13.7 - Page No :688¶

In [3]:
from numpy import *
# Variables
# given
T_x0 = 300.;  			 #[K]
Tw = 400.;  			 #[K]
L = 0.013;  			 #[m]
alpha = 2.476*(10**-5);  			 #[m**/sec]
h = 600.;  			 #[W/m**2*K]
pcp = 3.393*(10**6);  			 #[J/m**3*K]
L = 0.013;  			 #[m]
del_tax = L/10.;
betaa = 0.5;
del_tat = 0.03;

# Calculations
del_tat = betaa*((del_tax)**2)*(1./alpha);
T_infinity = 400.;  			 #[K]

# to be sure that the solution is stable, it is customary to truncate this number
del_tat = 0.03;  			 #[sec]
# betaa = alpha*del_tat*((1./del_tax)**2);
Told = zeros(11)
for i in range(11):
Told[i] = 300.;

a = ((2*h*del_tat)/(pcp*del_tax));
b = ((2*alpha*del_tat)/(pcp*((del_tax)**2)));

Tnew = zeros(11)
for j in range(11):
Tnew[0] = (T_infinity*0.08162)+(Told[0]*(1-0.08162-0.8791))+(Told[1]*0.8791)
for k in range(9):
Tnew[k+1] = (betaa*Told[k+2])+((1.-2*betaa)*(Told[k+1]))+(betaa*Told[k]);
Tnew[10] = ((2*betaa)*(Told[9]))
Told = Tnew;
# Results
print "Told values : " ,(Told);
Told values :  [ 325.54820838  319.78194857  315.05971328  311.28295197  308.32959437
306.07276601  304.39590474  303.20406441  302.43143939  302.04512688
302.04512688]

### Example 13.9 - Page No :700¶

In [7]:
# Variables
p = 2050.;  			 #[kg/m**3] - density of soil
cp = 1840.;  			 #[J/kg*K] - heat cpapacity of soil
k = 0.52;  			 #[W/m*K] - thermal conductivity of soil
alpha = 0.138*10**-6;  			 #[m**2/sec]
t = 4*30*24*3600;  			 #[sec] - no. of seconds in 4 months
Tx = -5.;  			 #[degC]
Tinf = -20.;  			 #[degC]
T0 = 20.;  			 #[degC]

# from the fig 13.24 the dimensionless dismath.tance Z is
Z = 0.46;

# Calculations
# then the depth is
x = 2*((alpha*t)**(1./2))*Z

# Results
print " the depth is  x  =  %.1f m  =  %.1f ft"%(x,x*3.6/1.10);
the depth is  x  =  1.1 m  =  3.6 ft

### Example 13.10 - Page No :701¶

In [8]:
# Variables
d = 0.01;  			 #[m] - diameter of cyclindrical porous plug
D = 2.*10**-9;  			 #[m**2/sec] - diffusion coefficient
t = 60.*60;  			 #[sec]
r = d/2.;
m = 0.;
Ca_inf = 0.;
Ca_0 = 10.;
X = (D*t)/((r)**2);
# from fig 13.14 the ordinate is
Y = 0.7;

# Calculations
Ca_c = Ca_inf-Y*(Ca_inf-Ca_0);

# Results
print " the concentration of KCL at the centre after 60 min is  Ca  =  %.2f kg/m**3"%(Ca_c);
the concentration of KCL at the centre after 60 min is  Ca  =  7.00 kg/m**3