In [1]:

```
from __future__ import division
import math
#Variables
T1=313; #fluid temp.,K
T2=287; #air temp.,K
H=0.4; #height of sides,m
Pr=0.711; #prandtl no.
b=1/T2; # b=1/v*d(R*T/p)/dt=1/To characterisation constant of thermal expansion of solid, K**-1
g=9.8; #gravity constant
nu=1.566*10**-5; #dynamic viscocity, m^3/s
#Calculations
RaL=g*b*(T1-T2)*H**3/(nu*2.203*10**-5); #Rayleigh no.
Nu=0.678*RaL**(0.25)*(Pr/(0.952+Pr))**(1/4); # nusselt no.
h=Nu*0.02614/H # average heat transfer coefficient, W/m**2/K
q=h*(T1-T2) # average heat transfer,W/m**2
c=3.936*((0.952+Pr)/Pr**2)**(1/4)*(1/(RaL/Pr)**0.25);#boundary layer thickness.,m
#Results
print "Average heat transfer coefficient is : ",round(h,3),"W/m^2/K\n"
print "Average heat transfer is :",round(q,3),"W/m^2\n"
print "Boundary layer thickness is :",round(c,4),"m\n"
print "Thus the BL thickness at the end of the plate is only 4 percent of the height, or 1.72 cm thick.this is thicker thsan typical forced convection BL but it is still reasonably thin."
```

In [3]:

```
from __future__ import division
import math
#Variables
T1=323; #wall temp.,K
T2=293; #air temp.,K
H=0.3; #height of wall, m
v2=2.318*10**-5; #molecular diffusivity, m**2/s
Pr=0.71; #prandtl no.
#Calculations
v1=16.45*10**-6; # molecular diffusivity, m**2/s
b=1/T2; # b=1/v*d(R*T/p)/dt=1/To characterisation constant of thermal expansion of solid, K**-1
Ral=9.8*b*(T1-T2)*H**3/((1.566*10**-5)*(2.203*10**-5));# Rayleigh no.
Nu=0.678*Ral**(0.25)*(Pr/(0.952+Pr))**(1/4); # nusselt no.
h=Nu*0.0267/H # average heat transfer coefficient, W/m**2/K
Nu1=0.68+0.67*((Ral)**(1/4)/(1+(0.492/Pr)**(9/16))**(4/9));#churchill correlation
h1=Nu1*(0.0267/0.3)-.11; #average heat transfer coefficient, W/m**2/K
#Results
print "Correlation average heat transfer coefficient is :",round(h1,3),"W/m^2/K\n"
print "The prediction is therefore within 5 percent of corelation .we should use the latter result in preference to the theoritical one, although the difference is slight."
```

In [4]:

```
from __future__ import division
import math
from numpy import mat
from numpy import array
#Variables
T1=400; #hot oil temp.,K
D=0.005; #diameter of line carrying oil, m
T2=300; #temp. of air around the tube,K
Tav=350; #average BI temp.,K
#Calculations & Results
#we evaluate properties at this temp. and write g as ge*(g-level), where ge is g at the earth surface and the g-level is the fraction of ge in the space vehicle.
b=1/T2; # b=1/v*d(R*T/p)/dt=1/To characterisation constant of thermal expansion of solid, K**-1
v1=2.062*10**-5; # molecular diffusivity, m**2/s
v2=2.92*10**-5; #molecular diffusivity, m**2/s
Pr=0.706; #prandtl no.
g=array(([10**-6, 10**-5, 10**-4, 10**-2]));
i=0;
while i<4:
Ral=0;
Nu=0;
h=0;
Q=0;
Ral=(9.8*b*((T1-T2))*(D**(3))/(v1*v2))*g.item(i); # Rayleigh no.
Nu=(0.6+0.387*(Ral/(1+(0.559/Pr)**(9/16))**(16/9))**(1/6))**2;
#Nu(i)=(0.6+0.387*((Ral)/(1+(0.559/Pr)**(9/16))**(16/9))**1/6)**2; churchill correlation.
print "Nusselt no. are : ",round(Nu,2),"\n"
h=Nu*0.0297/D; # convective heat transfer coefficient,W/(m**2*K)
print "Convective heat transfer coefficient are : ",round(h,2),"W/(m^2*K)\n"
Q=math.pi*D*h*(T1-T2); #heat transfer,W/m
print "Heat transfer is :",round(Q,2),"W/m of tube\n"
i=i+1;
```

In [5]:

```
from __future__ import division
import math
#Variables
T2=300; #air temp.,K
P=15; #delivered power,W
D=0.17; #diameter of heater,m
v1=1.566*10**-5; #molecular diffusivity, m**2/s
b=1/T2; #b=1/v*d(R*T/p)/dt=1/To characterisation constant of thermal expansion of solid, K**-1
Pr=0.71; #prandtl no.
v2=2.203*10**-5; #molecular diffusivity, m**2/s
v3=3.231*10**-5; #molecular diffusivity at a b except at 365 K., m**2/s
v4=2.277*10**-5; #molecular diffusivity at a b except at 365 K., m**2/s
k1=0.02614; #thermal conductivity
k2=0.0314; #thermal conductivity
#Calculations
#we have no formula for this situation, so the problem calls for some guesswork.following the lead of churchill and chau, we replace RaD with RaD1/NuD in eq.
#(NuD)**(6/5)=0.82*(RaD1)**(1/5)*Pr**0.034
delT=1.18*P/(3.14*D**(2)/4)*(D/k1)/((9.8*b*661*D**(4)/(0.02164*v1*v2))**(1/6)*Pr**(0.028));
#in the preceding computation, all the properties were evaluated at T2.mow we must return the calculation,reevaluating all properties except b at 365 K.
delTc=1.18*661*(D/k2)/((9.8*b*661*D**(4)/(k2*v3*v4))**(1/6)*(0.99));
TS=T2+delTc;
TS1=TS-271.54
#Results
print "Average surface temp. is :",round(TS1,4),"K\n"
print "That is rather hot.obviously, the cooling process is quite ineffective in this case."
```

In [6]:

```
from __future__ import division
import math
#Variables
T2=363; # temp. of strip,K
T1=373; #saturated temp.,K
H=0.3; #height of strip,m
Pr=1.86; #prandtl no.
Hfg=2257; #latent heat. kj/kg
ja=4.211*10/Hfg; #jakob no.
a1=961.9; #density of water,kg/m**3
a2=0.6; #density of air,kg/m**3
k=0.677; #thermal conductivity,W/(m*K)
#Calculations
Hfg1=Hfg*(1+(0.683-0.228/Pr)*ja); #corrected latent heat,kj/kg
delta=(4*k*(T1-T2)*(2.99*10**(-4))*0.3/(a1*(a1-a2)*9.806*Hfg1*1000))**(0.25)*1000;
Nul=4/3*H/delta; #average nusselt no.
q=Nul*k*(T1-T2)/H; # heat flow on an area about half the size of a desktop,W/m**2
Q=q*H; #overall heat transfer per meter,kW/m
m=Q/(Hfg1); #mass rate of condensation per meter,kg/(m*s)
#Results
print "Overall heat transfer per meter is :",round(Q,4),"kW/m^2\n"
print "Film thickness at the bottom is :",round(delta,4),"mm\n"
print "Mass rate of condensation per meter. is : ",round(m,4),"kg/(m*s)\n"
```