In [5]:

```
from __future__ import division
import math
#Variables
T1=300; #air temperature,K
v=1.5; #air velocity, m/s
t=0.5; #thickness, m
u=1.853*math.pow(10,-5); #dynamic viscosity,kg/(m*s)
v1=1.566*math.pow(10,-5); #kinematic viscosity,m**2/s
#Calculations
Rex=v*t/v1; #reynolds no. is low enough to permit the use of laminar flow analysis.
b=4.92*t/(math.sqrt(Rex))*100; # b.l. thickness, cm
#in this case b/x=1.124/50=0.0225 so laminar flow is valid.
v2=0.8604*math.sqrt(v1*v/t);
#since v2 grows larger as x grows smaller, the condition v2<u is not satisfied very near the leading edge.
#in this case del/thickness is 0.0225.
x=0.8604*math.sqrt(v1*v/t); #velocity,m/s
y=x/t;
#Results
print "Boundary layer thickness is :",round(b,3),"cm\n"
print "Since velocity grows larger as thickness grows smaller, the condition x<<u is not satisfied very near the leading edge. therefore the BI approximation themselves breakdown."
```

In [7]:

```
from __future__ import division
import math
#Variables
l=0.5; #total length of surface,m
Cf=0.00607; #overall friction coefficient
#Calculations
tw=1.183*(2.25)*Cf/2; # wall shear, kg/(m*s**2)
a=0.5; #ratio of wall shear at x=l and average wall shear
#tw(x)=twavg where 0.664/(x**0.5)=1.328/(47,)893, x=1/8 m thus the wall shear stress plummets to twavg one fourth of the way from the leading edge and drops only to one half of twavg in the remaining 75 percent plate.x<600*1.566*10**(-5)/1.5=0.0063 m.
# preceding analysis should be good over almost 99 percent of the 0.5 m length of the surface.
#Results
print "Overall friction coefficient is :",Cf,"\n"
print "Wall shear is :",tw,"kg/(m*s^2)\n"
print "The preceding analysis should be good over almost 99 percent of the 0.5m length of the surface."
```

In [7]:

```
from __future__ import division
import math
#Variables
l=0.06; #length of heater, m
p=15; #pressure of heater, atm
T1=440; #temperature of heater, K
v=2; #free stream velocity,m/s
T2=460; #constant temperature of heater, K
T3=450; #mean temperature of heater, K
k=0.674; #thermal conductivity ,W/(m.K)
#Calculations
q=2*(0.332)*(k/l)*math.sqrt(v*l/(1.72*10**-7))*(T2-T1)/1000; #(from Fig 6.10)formula for heat flux is q=2*(0.664)*k/l*(Rel**0.5)*(T2-T1), in kW/m^2
#Results
print "Heat flux is :",round(q,3),"kW/m^2\n"
```

In [5]:

```
from __future__ import division
import math
#Variables
T1=293; #air temperature,K
v=15; #air velocity,m/s
T2=383; #temperature of plate,K
l=0.5; #length of plate,m
w=0.5; #width of plate,m
Pr=0.707; #prandtl no.
k=0.02885; #thermal conductivity of ,W/(m.K)
#Calculations
Rel=v*l/(0.0000194); #reynolds no.
Nul=0.664*(Rel)**0.5*Pr**(1/3); #nusset no.
h1=367.8*(k)/l; #average convection coefficient, W/(m**2*K)
Q=h1*l**(2)*(T2-T1); #heat transferred,W
h2=h1/2 #convection coefficient at trailing , W/(m**2*K)
a1=4.92*l/math.sqrt(Rel)*1000 #hydrodynamic boundary layer,m
a2=a1/(Pr)**(1/3); #thermal boundary layer,mm
#Results
print "Average heat trensfer coefficient is :",round(h1,3),"W/m^2/K\n"
print "Total heat transferred is",round(Q,3),"W\n"
print "Convection coefficient at trailing is :",round(h2,3),"W/(m^2*K)\n"
print "Hydrodynamic boundary layer is : ",round(a1,3),"m\n"
print "Thermal boundary layer is : ",round(a2,3),"mm\n"
```

In [4]:

```
from __future__ import division
import math
#Variables
T1=288; # air temperature,K
v=1.8; # air velocity,m/s
l=0.6; # length of panel, m
Q=420; # power per unit area, m**2
T2=378; # maximum temperature of surface, K
k=0.0278; #thermal conductivity of ,W/(m.K)
Pr=0.709; #Prandtl no.
#Calculations
T3=Q*l/(k)/(0.453*(l*v/(1.794*10**-5))**(0.50)*(Pr)**(1/3));#maximum temperature difference
Twmax=T1+T3; #Twmax comes out to be 106.5 C, this is very close to 105 C,if 105 is at all conservative, Q = 420 should be safe.
T4=0.453/0.6795*T3; #average temperature difference,K
Twavg=T1+T4; #average wall temperature,K
Twa=Twavg-273; #average wall temp. in °C
#Results
print "Average wall temperature is :",round(Twa,3),"C\n"
```

In [8]:

```
from __future__ import division
import math
#Variables
v=15; #air velocity,m/s
T2=383; # temperature of plate,K
l=0.5; # length of plate,m
w=0.5; # width of plate,m
Pr=0.707; # prandtl no.
rho=1.05; #Air desnity, kg/m^3
#Calculations
Rel=v*l/(0.0000194); #reynolds no.
Nul=0.664*math.sqrt(Rel)*Pr**(1/3); # nusset no.
Cf=2*Nul/(Rel*Pr**(1/3)); #friction coefficient
s=Cf*0.5*rho*v**2; #drag shear, kg/(m*s**2)
f=s*0.5**2; #drag force, N
#Results
print "Drag force on heat transfer surface is :",round(f,5),"N\n"
```

In [9]:

```
from __future__ import division
import math
#Variables
T1=297; # river water temp.,K
T2=283; # ocean water temp., K
n=5; # no. of knots
k=0.5927; # thermal conductivity,W/(m*K)
a=998.8; #density of water, kg/m**3
Cp=4187; # heat capacity, J/kg/K
Pr=7.66; #Prandtle no.
x=1; #distance from forward edge,m
v=1.085*10**-6; # kinematic viscosity, m**2/s
u=2.572; # velocity of knot,m/s
#Calculations
T3=(T1+T2)/2; # avg. temp.,K
Rex=u/v # reynolds no.
Cf=0.455/(math.log(0.06*Rex))**2 # friction coefficient
h=k/x*0.032*(Rex)**(0.8)*Pr**(0.43); # heat transfer coefficient,W/(m**2*K)
h1=a*Cp*u*Cf/2/(1+12.8*(Pr**0.68-1)*math.sqrt(Cf/2)); #heat transfer coefficient,W/(m**2*K)
#Results
print "Friction coefficient is :",round(Cf,5),"\n"
print "Convective heat transfer coefficient at a distance of 1 m fom the forward edge is :",round(h,3),"W/(m^2*K)\n"
print "Heat transfer coefficient by another method is :",round(h1,3),"W/(m^2*K)\n"
print "The two values of h differ by about 18 percent, which is within the uncertainity \n"
```

In [23]:

```
from __future__ import division
import math
#Variables
l=2; # length of plate,m
p=1000; # power density,W/m**2
u=10; # air velocity,m/s
T1=290; # wind tunnel temp.,K
p2=1; # pressure,atm
v=1.578*10**-5; # kinematic viscosity, m**2/s
k=0.02623; # thermal conductivity,W/(m*K)
Pr=0.713; # prandtl no.
Rel=u*l/v; # reynolds no. at 10 m/s
Nul=1845; # nusselt no.
Re=400000; #Reynolds no. at the end of turbulent transition engine
#Calculations
h=Nul*k/l; #convection coefficient,W/(m**2*K)
Tavg=T1+p/h; #Average temperature of plate, K
#to take better account of the transition region, we can use churchill eqn.
x=Rel*Pr**(2/3)/(math.sqrt(1+(0.0468/Pr)**(2/3)));
x1=1.875*x*Re/Rel;
Nul1=0.45+0.6774*math.sqrt(x)*math.sqrt(1+((x/12500)**(3/5))/(1+(x1/x)**3.5)**0.4);
H=Nul1*k/l; #convection coefficient,W/(m**2*K)
Tw=290+1000/H; #average temperature of plate,K
#Results
print "Average temperature of plate is :",round(Tavg,2)," K\n"
print "Average temperature of plate is :",round(Tw,2)," K , thus in this case, the average heat transfer coefficient is 33 percent higher when the transition regime is included.\n"
```