In [2]:

```
# Variables
a = 0.0006; #[m**2] - area
l = 0.1; #[m] - length
# (a) using the fourier law
deltax = 0.1; #[m] - thickness of copper block
T2 = 100.; #[degC] - temp on one side of copper block
T1 = 0.; #[degC] - temp on other side of the copper block
k = 380.; #[W/mK] - thermal conductivity
# Calculations
# using the formula (q/A)*deltax = -k*(T2-T1)
g = -k*(T2-T1)/deltax;
print " a) The steady state heat flux across the copper block is q/A = %5.1e J*m**-2*sec**-1 "%(g);
# (b)
V = a*l; #[m**3] - volume
# using the overall balance equation with the accumulation and generation term
Qgen = 1.5*10**6; #[j*m**-3*sec**-1]
SIx = (g*a-Qgen*V)/a;
# Results
print " b) the flux at face 1 is %5.1e j*m**-2*sec**-1;the negative sign indicates that the \
\nheat flux is from right to left negative x direction"%(SIx);
```

In [5]:

```
from sympy import *
# Variables
x = Symbol('x')
SIx2 = -3.8*10**5; #[j*m**-2*sec**-1] - flux at x = 0.1,i.e through face2
Qgen = 1.5*10**6; #[j*m**-3*sec**-1] - uniform generation in the volume
T2 = 100+273.15; #[K] temperature at face 2
x2 = 0.1; #[m]
k = 380.; #[W/mK] - thermal conductivity
# Calculations
# using the equation der(SIx)*x = SIx+c1;where c1 is tyhe constant of integration
c1 = (Qgen*x2)-SIx2;
SIx = Qgen*x-c1;
# Results
print "SIx = ",SIx
print " where SIx is in units of J m**-2 sec**-1 and x is in units of m"
# using the equation -k*T = der(SIx)*x**2-c1*x+c2;where c2 is the constant of integration
c2 = -k*T2-(Qgen*(x2)**2)/2+c1*x2;
T = -(Qgen/k)*x**2+(c1/k)*x-(c2/k);
print "T = ",T
print " where T is in units of kelvin K"
# Answer may vary because of rouding error.
```

In [6]:

```
import math
from sympy import *
# Variables
# given
x = Symbol('x')
t = Symbol('t')
hf1 = -270.; #[J/sec] - heat flow at face 1
hf2 = -228.; #[J/sec] - heat flow at face2
Qgen = 1.5*10**6; #[J*m**-3*sec**-1] generation per unit volume per unit time
v = 6*10**-5; #[m**3] volume
Cp = 0.093; #[cal*g**-1*K**-1] heat capacity of copper
sp = 8.91; #specific gravity of copper
a = 0.0006; #[m**2] - area
# Calculation and Results
# (a) using the overall balance
acc = hf1-hf2+Qgen*v;
print "a) the rate of accumulation is %d J/sec "%(acc);
# (b)
SIx1 = hf1/a;
SIx2 = hf2/a;
x1 = 0.;
# solving for the constant of integration c1 in the equation [del(p*cp*T)/delt-der(SIx)]*x = -SIx+c1;
c1 = 0+SIx1;
x2 = 0.1;
g = (-(SIx2)+c1)/x2+Qgen;
SIx = c1-(g-Qgen)*x;
print "SI(x) = ","(b)",SIx
# solving for constant of integration c3 in the equation p*cp*T = g*t+c3
T2 = 100+273.15;
t2 = 0;
p = sp*10**3; #[kg/m**3] - density
cp = Cp*4.1840; #[J*kg**-1*K**-1]
c3 = p*cp*T2-g*t2;
T = (g*(10**-3)/(p*cp))*t+c3/(p*cp);
print "Relationship between T and t at x=0.1m is T = ",T
# solving for constant of integration c2 in the equation -k*T = der(SIx)*x**2-c1*x+c2
k = 380.; #[w/m**1*K**1]
x2 = 0.1;
c2 = k*T+(3.5*10**5)*x2**2-(4.5*10**5)*x2;
def T(t,x):
return (-(3.5*10**5)*x**2+(4.5*10**5)*x+87.7*t+1.00297*10**5)/k;
# at face 1;
x1 = 0.;
t1 = 60.; #[sec]
T1 = T(t1,x1);
print "Temperature profile as a function of x and t is T = %.2f K, at face 1"%T1
# at face 2
x2 = 0.1;
t2 = 60.; # [sec]
T2 = T(t2,x2);
print "Temperature at face 2 = %.0f K ,at face 2"%T2
```