In [2]:

```
import math
# Variables
Ta = 20; #Temperature of air in degree C
Tp = 134; #Temperature of heated plate in degree C
v = 3; #Velocity of flow in m/s
L = 2; #Length of plate in m
W = 1.5; #Width of plate in m
x = 0.4; #Distance of plane from the plate in m
k = (15.06*10**-6); #Kinematic vismath.cosity in m**2/s
# Calculations
Re = (v*x)/k; #Reynold number
q = ((5*x)/math.sqrt(Re))*1000; #Thickness of boundary layer in mm
Cfx = (0.664/math.sqrt(Re))/10**-3; #Local skin friction coefficient *10**-3
# Results
print 'Thickness of boundary layer is %3.1f mm Local skin friction coefficient is %3.2f*10**-3'%(q,Cfx)
```

In [1]:

```
# Variables
Ta = 20; #Temperature of air in degree C
Tp = 134; #Temperature of heated plate in degree C
v = 3; #Velocity of flow in m/s
L = 2; #Length of plate in m
W = 1.5; #Width of plate in m
x = 0.4; #Distance of plane from the plate in m
k = (15.06*10**-6); #Kinematic vismath.cosity in m**2/s
# Calculations
Tf = (Ta+Tp)/2; #Film temperature in degree C
pw = 0.998; #Density of air at 77 degree C
Cp = 1009; #Specific heat of air at 77 degree C
kw = (20.76*10**-6); #Kinematic viscosity of air at 77 degree C
k = 0.03; #Thermal conductivity of air at 77 degree C
Pr = 0.697; #prantl number of air at 77 degree C
Re = (v*x)/kw; #Reynolds number
Nu = (0.332*Re**0.5*Pr**(1./3)); #Nusselts number
h = (Nu*k)/x; #Heat transfer coefficient in W/m**2.K
h1 = (h*2); #Average value of heat transfer coefficient in W/m**2.K
Q = h1*x*W*(Tp-Ta); #Heat flow in W
Q1 = (2*Q); #Heat flow from both sides of the plate in W
# Results
print 'Heat flow from both sides of the plate is %3.0f W'%(round(Q1,-1))
```

In [4]:

```
import math
# Variables
Ta = 20; #Temperature of air in degree C
v = 3; #Velocity of flow in m/s
L = 2; #Length of plate in m
W = 1; #Width of plate in m
x1 = 0.3; #Initial point of the boundary layer in m
x2 = 0.8; #Final point of the boundary layer in m
p = 1.17; #Density of air at 20 degree C in kg/m**3
k = (15*10**-6); #Kinematic vismath.cosity in m**2/s
Re = (5*10**5); #Reynolds number at the transition frm laminar to turbulant
# Calculations
x = (k*Re)/v; #Critical length in m
Rel = (v*L)/k; #Reynolds number
q = (4.64*L)/math.sqrt(Rel)*1000; #Boundary layer thickness at the trailing edge of plate in mm
ts = 1.292*(0.5*p*v**2)*math.sqrt(1./Rel); #Average shear stress in N/m**2
F = (2*L*ts); #Drag force on the two sides of the plate in N
q80 = (4.64*x2)/math.sqrt((v*x2)/k); #Boundray layer thickness at x = 0.8 m
q30 = (4.64*x1)/math.sqrt((v*x1)/k); #Boundray layer thickness at x = 0.3 m
m = ((5./8)*p*v*(q80-q30))/10**-3; #Mass flow of air in kg/s
# Results
print 'Boundary layer thickness at the trailing edge of plate is % 3.2f mm \n \
Drag force on the two sides of the plate is %3.4f N \n \
Mass flow of air is %3.1f*10**-3 kg/s'%(q,F,m)
```

In [5]:

```
import math
# Variables
P = 8.; #Pressure of air in kN/m**2
Ta = 250.; #Temperature of air in degree C
L = 1.; #Length of the palte in m
W = 0.3; #Width of the plate in m
v = 8.; #Velocity of air in m/s
Tp = 78.; #Temperature of plate in degree C
# Calculations
Tf = (Ta+Tp)/2; #Film temperature in degree C
Cp = 1018; #Specific heat of air at 164 degree C and 1 atm pressure
kw = (30.8*10**-6); #Kinematic viscosity of air at 164 degree C and 1 atm pressure
k = 0.0364; #Thermal conductivity of air at 164 degree C and 1 atm pressure
Pr = 0.69; #prant number of air at 164 degree C and 1 atm pressure
k1 = kw*(101330/(P*1000)); #Kinematic viscosity of air at 164 degree C and 8kN/m**2 pressure
Re = (v*L)/k1; #Reynolds number
h = 0.662*(k/L)*math.sqrt(Re)*Pr**(1./3); #Heat transfer coefficient in W/m.K
Q = 2*h*L*W*(Ta-Tp); #Rate of heat removal in W
# Results
print 'Rate of heat removal is %3.1f W'%(Q)
#note : answer is slightly different because of rounding off error.
```

In [8]:

```
import math
# Variables
P = 8.; #Pressure of air in kN/m**2
Ta = 250.; #Temperature of air in degree C
L = 1.; #Length of the palte in m
W = 0.3; #Width of the plate in m
v = 8.; #Velocity of air in m/s
Tp = 78.; #Temperature of plate in degree C
R = 287.; #Universal gas constant in J/kg.K
# Calculations
Tf = (Ta+Tp)/2; #Film temperature in degree C
Cp = 1018; #Specific heat of air at 164 degree C and 1 atm pressure
kw = (30.8*10**-6); #Kinematic viscosity of air at 164 degree C and 1 atm pressure
k = 0.0364; #Thermal conductivity of air at 164 degree C and 1 atm pressure
Pr = 0.69; #prant number of air at 164 degree C and 1 atm pressure
k1 = kw*(101330/(P*1000)); #Kinematic viscosity of air at 164 degree C and 8kN/m**2 pressure
Re = (v*L)/k1; #Reynolds number
h = 0.662*(k/L)*math.sqrt(Re)*Pr**(1./3); #Heat transfer coefficient in W/m.K
Q = 2*h*L*W*(Ta-Tp); #Rate of heat removal in W
p = (P*1000)/(R*(Tf+273)); #Density in kg/m**3
St = (h/(p*Cp*v)); #Stanton number
Cfx2 = (St*Pr**(2/3)); #Colburn factor
ts = (Cfx2*p*v**2); #Average shear stress in N/m**2
D = (0.0186*W*L); #Drag force on one side of plate in N
D2 = (2*D)/10**-3; #Total drag force on both sides of plate in N
# Results
print 'The drag force exerted on the plate is %3.2f*10**-3 N'%(D2)
```

In [8]:

```
# Variables
L = 1; #Length of the palte in m
W = 1; #Width of the plate in m
Ts = 10; #Temperature of free strem air in degree C
v = 80; #Velocity of free stream air in m/s
# Calculations
k = 0.025; #Thermal conductivity of air at 10 degree C
Pr = 0.72; #prant number of air at 10 degree C
v1 = (14.15*10**-6); #Kinematic viscosity of air at 10 degree C
Re = (v*L)/v1; #Reynolds number
q = 0.381*L*Re**(-1./5); #Thickness of the boundary layer at the trailing edge of the plate in m
Nu = (0.037*Re**(4./5)*Pr**(1./3)); #Nusselts number
h = (Nu*k)/L; #Mean value of the heat transfer coefficient in W/m**2.K
# Results
print 'Thickness of the boundary layer at the trailing edge of the plate is %3.4f m \n \
Mean value of the heat transfer coefficient is %3.0f W/m**2.K'%(q,h)
```

In [9]:

```
# Variables
Ta = 0.; #Temperature of air stream in degree C
Tp = 90.; #Temperature of heated plate in degree C
v = 75.; #Speed of air in m/s
L = 0.45; #Length of the palte in m
W = 0.6; #Width of the plate in m
Re = (5.*10**5); #Reynolds number at the transition from laminar to turbulant
# Calculations
Tf = (Ta+Tp)/2; #Film temperature in degree C
k = 0.028; #Thermal conductivity of air at 10 degree C
Pr = 0.698; #prant number of air at 10 degree C
v1 = (17.45*10**-6); #Kinematic viscosity of air at 10 degree C
x = (Re*v1)/v; #Critical length in m
Rel = (v*L)/v1; #Reynolds number
Cfl = ((0.074/Rel**(1./5))-(1740/Rel))/10**-3; #Average value of friction coefficient *10**-3
Nu = ((0.037*Rel**(4./5))-870)*Pr**(1./3); #Nussults number
h = (Nu*k)/L; #Heat transfer coefficient in W/m**2.K
Q = (2*h*L*W*Tp); #Rate of energy dissipation in W
# Results
print 'Average value of friction coefficient is %3.2f*10**-3 \n \
Heat transfer coefficient is %3.0f W/m**2.K \n \
Rate of energy dissipation is %i W'%(Cfl,h,Q)
# note : book answer is wrong
```

In [11]:

```
# Variables
D = 0.3; #Diameter of cylinder in m
L = 1.7; #Height of cylinder in m
Ts = 30.; #Surface temperature in degree C
v = 10.; #Speed of wind in m/s
Ta = 10.; #Temperature of air in degree C
# Calculations
Tf = (Ta+Ts)/2; #Film temperature in degree C
k = 0.0259; #Thermal conductivity of air at 20 degree C
Pr = 0.707; #prant number of air at 20 degree C
v1 = (15*10**-6); #Kinematic viscosity of air at 20 degree C
Re = (v*D)/v1; #Reynolds number
Nu = 0.027*Re**0.805*Pr**(1./3) #Nusselts number
h = (Nu*k)/D; #Heat transfer coefficent in W/m**2.K
Q = (h*3.14*D*L*(Ts-Ta)); #Rate of heat loss in W
# Results
print 'Rate of heat loss is %3.1f W'%(Q)
```

In [12]:

```
# Variables
Ta = 27; #Temperature of air stream in degree C
v = 0.3; #Velodity of air in m/s
Q = 100; #Poer of electric bulb in W
Te = 127; #Temperature of electric bulb in degree C
D = 0.06; #Diameter of sphere in m
# Calculations
Tf = (Ta+Te)/2; #Film temperature in degree C
k = 0.03; #Thermal conductivity of air at 77 degree C
Pr = 0.697; #prant number of air at 77 degree C
v1 = (2.08*10**-5); #Kinematic viscosity of air at 77 degree C
Re = (v*D)/v1; #Reynolds number
h = (k*0.37*Re**0.6)/D; #Heat transfer coefficient in W/m**2.K
Q = (h*3.14*D**2*(Te-Ta)); #Heat transfer rate in W
Qp = (Q*100)/100; #Percentage of heat lost by forced convection
# Results
print "Heat transfer rate is %3.2f W \n \
Percentage of power lost due to convection is %3.2f percent"%(Q,Qp)
```

In [14]:

```
import math
# Variables
D = 0.015; #Diamter of copper bus bar in m
Ta = 20; #Temperature of air stream in degree C
v = 1; #Velocity of air in m/s
Ts = 80; #Surface temperature in degree C
p = 0.0175; #Resistivity of copper in ohm mm**2/m
# Calculations
Tf = (Ta+Ts)/2; #Film temperature in degree C
k = 0.02815; #Thermal conductivity of air at 50 degree C
Pr = 0.703; #prant number of air at 50 degree C
v1 = (18.9*10**-6); #Kinematic viscosity of air at 50 degree C
Re = (v*D)/v1; #Reynolds number
Nu = 0.3+(((0.62*math.sqrt(Re)*Pr**(1./3))/(1+(0.4/Pr)**(2./3))**(1./4))*(1+(Re/28200.)**(5./8))**(4./5)); #Nusselts number
h = (Nu*k)/D; #Heat transfer coefficent in W/m**2.K
I = 1000*3.14*D*math.sqrt((h*(Ts-Ta)*D)/(4*p)); #Current in A
# Results
print 'Heat transfer coefficient between the bus bar and cooling air is %3.2f W/m**2.K \n \
Maximum admissible current intensity for the bus bar is %3.0f A'%(h,I)
```

In [15]:

```
# Variables
Ta = 30; #Temperature of air stream in degree C
v = 25; #Velocity of stream in m/s
x = 0.05; #Side of a square in m
D = 0.05; #Diameter of circular cylinder in m
Ts = 124; #Surface temperature in degree C
# Calculations
Tf = (Ta+Ts)/2; #Film temperature in degree C
k = 0.03; #Thermal conductivity of air at 77 degree C
Pr = 0.7; #prantL number of air at 77 degree C
v1 = (20.92*10**-6); #Kinematic viscosity of air at 77 degree C
Re = (v*D)/v1; #Reynolds number
Nu1 = 0.027*Re**0.805*Pr**(1./3); #Nussults number for circulat tube
h1 = (Nu1*k)/D; #Heat tansfer coefficient for circular tube in W/m**2.K
Nu2 = 0.102*Re**0.675*Pr**(1./3); #Nussults number for square tube
h2 = (Nu2*k)/D; #Heat transfer coefficient for square tube in W/m**2.K
# Results
print 'Heat transfer coefficient for circular tube is %3.1f W/m**2.K \n \
Heat transfer coefficient for square tube is %3.2f W/m**2.K'%(h1,h2)
```

In [16]:

```
# Variables
n = 7; #Number of rows of tube
Ta = 15; #Temperature of air in degree C
v = 6; #Velocity of air in m/s
ST = 0.0205; #Transverse pitch in m
SD = 0.0205; #Longitudinal pitch in m
D = 0.0164; #Outside diameter of the tube in m
Ts = 70; #Surface temperature in degree C
# Calculations
Tf = (Ta+Ts)/2; #Film temperature in degree C
k = 0.0274; #Thermal conductivity of air at 42.5 degree C
Pr = 0.705; #prant number of air at 42.5 degree C
v1 = (17.4*10**-6); #Kinematic viscosity of air at 42.5 degree C
p = 1.217; #Density in kg/m**3
vmax = (v*ST)/(ST-D); #Maximum velocity in m/s
Re = (vmax*D)/v1; #Reynolds number
Nu = (1.13*0.518*Re**0.556*Pr**(1./3))*0.97; #Nusselts number
h = (Nu*k)/D; #Heat transfer coefficent in W/m**2.K
f = 0.4; #From Fig. 7.10 on page no 303
g = 1.04; #From Fig. 7.10 on page no 303
dp = (n*f*p*vmax**2*g)/2; #Pressure drop in N/m**2
# Results
print 'Heat transfer coefficent is %3.2f W/m**2.K \n \
Pressure drop is %3.0f N/m**2'%(h,dp)
```

In [1]:

```
# Variables
n = 7; #Number of rows of tube
Ta = 15; #Temperature of air in degree C
v = 6; #Velocity of air in m/s
ST = 0.0205; #Transverse pitch in m
SD = 0.0205; #Longitudinal pitch in m
D = 0.0164; #Outside diameter of the tube in m
Ts = 70; #Surface temperature in degree C
# Calculations
Tf = (Ta+Ts)/2; #Film temperature in degree C
k = 0.0253; #Thermal conductivity of air at 15 degree C
Pr = 0.710; #prant number of air at 15 degree C
v1 = (14.82*10**-6); #Kinematic viscosity of air at 15 degree C
p = 1.217; #Density in kg/m**3
Pr1 = 0.701; #prant number of air at 70 degree C
vmax = (v*ST)/(ST-D); #Maximum velocity in m/s
Re = (vmax*D)/v1; #Reynolds number
Nu = 0.35*Re**0.6*(Pr/Pr1)**0.25; #
h = (Nu*k)/D; #Heat transfer coefficient in W/m**2.K
# Results
print ' Heat transfer coefficient is %3.1f W/m**2 K'%(h)
```

In [18]:

```
# Variables
m = 0.314; #Mass flow rate of air in m**3/s
n1 = 7; #Number of tubes in the direction of flow
n2 = 8; #Number of tubes perpendicular to the direction of flow
L = 1.25; #Length of each tube in m
D = 0.019; #Outer diameter in m
ST = 0.0286; #Transverse pitch in m
SD = 0.038; #Longitudinal pitch in m
Ta = 200; #Temperature of air in degree C
Ts = 96; #Surface temperature in degree C
# Calculations
Tf = (Ta+Ts)/2; #Film temperature in degree C
k = 0.039; #Thermal conductivity of air at 15 degree C
Pr = 0.688; #prantl number of air at 15 degree C
v1 = (3*10**-5); #Kinematic vismath.cosity of air at 15 degree C
vmax = (m/((ST*n2*L)-(D*n2*L))); #Maximum velocity in m/s
Re = (vmax*D)/v1; #Reynolds number
Nu = (0.299*Re**0.602*Pr**(1./3)); #Nusselts number
X = 0.96; #From Table 7.5 on page no.302
Nux = (X*Nu); #Average nusselts number
h = (Nux*k)/D; #Convective heat transfer coefficient in W/m**2.K
# Results
print 'Convective heat transfer coefficient is %3.2f W/m**2.K'%(h)
```

In [19]:

```
# Variables
D = 0.2; #Diameter of pipeline in m
#velocity profile is given by u = 96r-190r**2 m/s
#Temperature profile is given by T = 100(1-2r) degree C
# Calculations
vmax = (64*(D/2))-(95*(D/2)**2); #Mean velocity in m/s
T = (2/(vmax*(D/2)**2))*(((9600*(D/2)**3)/3)-((38200*(D/2)**4)/4)+((38000*(D/2)**5)/5)); #Average temperature of the fluid in degree C
# Results
print 'Average temperature of the fluid is %3.2f degree C'%(T)
```

In [1]:

```
# Variables
Di = 0.025; #I.D of the tube in m
Do = 0.04; #O.D of the tube in m
m = 5.; #Mass flow rate of water in kg/m
T = [20.,70.]; #Temperature at entry and exit of water in degree C
Q = 10.**7; #Heat in W/m**3
Ts = 80.; #Surface temperature in degree C
Cp = 4179.; #Specific heat of water in J/kg.K
# Calculations
Tb = (T[0]+T[1])/2; #Film temperature in degree C
L = ((4*(m/60)*Cp*(T[1]-T[0]))/(3.14*(Do**2-Di**2)*Q)); #Length of tube in m
qs = ((Q*(Do**2-Di**2))/(4*Di)); #Heat flux at the surface in W/m**2
h = (qs/(Ts-T[1])); #Heat transfer coefficient at the outlet in W/m**2.K
# Results
print 'Length of tube is %3.3f m \nHeat transfer coefficient at the outlet is %3.0f W/m**2.K'%(L,h)
```

In [21]:

```
# Variables
k = 0.175; #Thermal conductivity in W/m.K
Di = 0.006; #I.D of the tube in m
L = 8; #Length of the tube in m
dT = 50; #Mean temperature difference in degree C
# Calculations
h = (3.66*k)/Di; #Heat transfer coefficient in W/m**2.K
Q = (h*3.14*Di*L*dT); #Heat transfer rate in W
# Results
print 'Heat transfer coefficient is %3.2f W/m**2.K Heat transfer rate is %3.0f W'%(h,Q)
```

In [22]:

```
# Variables
Ti = 25; #Initial temperature of water in degree C
D = 0.05; #Diamter of the tube in m
Re = 1600; #Reynolds number
q = 800; #Heat flux in W/m
Tf = 50; #Final temperature of water in degree C
# Calculations
k = 0.61; #Thermal conductivity of water at 25 degree C in W/m.K
u = (915*10**-6); #Dynamic viscosity in N.s/m**2
m = (Re*3.14*D*u)/4; #Mass flow rate of water in kg/s
h = (4.364*k)/D; #Heat transfer coefficient in W/m**2.K
qs = (q/(3.14*D)); #Constant heat flux in W/m**2
Cp = 4178; #Specific heat of water in J/kg.K
L = ((m*Cp*(Tf-Ti))/q); #Length of the tube in m
# Results
print 'Average heat transfer coefficient is %3.2f W/m**2.K \n \
Length of the tube is %3.3f m'%(h,L)
```

In [2]:

```
# Variables
Di = 0.015; #I.D of the tube in m
Tb = 60; #Temperature of the tube in degree C
m = 10; #Flow rate of water in ml/s
Ti = 20; #Temperature of water at entry in degree C
x = 1; #Dismath.tance form the plane in m
Tx = 34; #Temperature of water at 1 m dismath.tance in degree C
# Calculations
Tbm = (Ti+Tx)/2; #Mean value of bulk temperature in degree C
pw = 997; #Density of air at 27 degree C in kg/m**3
Cp = 4180; #Specific heat of air at 27 degree C in J/kg.K
u = (855*10**-6); #Dynamic vismath.cosity of air at 27 degree C in N.s/m**2
k = 0.613; #Thermal conductivity of air at 27 degree C in W/m.K
Pr = 5.83; #prantl number of air at 27 degree C
us = (464*10**-6); #Dynamic vismath.cosity of air at 60 degree C in Ns/m**2
um = (m*10**-6)/((3.14/4)*Di**2); #Mean speed in m/s
Re = (pw*um*Di)/u; #Reynolds number
Nu = 3.66+((0.0668*(Di/x)*Re*Pr)/(1+(0.04*((Di/x)*Re*Pr)**(2./3)))); #Nusselts number in Haussen correlation
Nux = (1.86*((Re*Pr)/(x/Di))**(1./3)*(u/us)**0.14); #Nusselsts number in Sieder - Tate correlation
# Results
print 'Nusselts number in Haussen correlation is %3.2f \n \
Nusselsts number in Sieder - Tate correlation is %3.3f'%(Nu,Nux)
```

In [2]:

```
# Variables
Tw = 50; #Temperature of water in degree C
Di = 0.005; #Inner diameter of the tube in m
L = 0.5; #Length of the tube in m
v = 1; #Mean velocity in m/s
Ts = 30; #Surface temperature in degree C
# Calculations
Tf = (Tw+Ts)/2; #Film temperature in degree C
k = 0.039; #Thermal conductivity of air at 15 degree C
Pr = 0.688; #prant number of air at 15 degree C
p = 990; #Density of air at 50 degree C in kg/m**3
Cp = 4178; #Specific heat of air at 50 degree C in J/kg.K
v1 = (5.67*10**-7); #Kinematic viscosity of air at 50 degree C
v2 = (6.57*10**-7); #Kinematic viscosity of air at 40 degree C
Re = (v*Di)/v1; #Reynolds number
h = ((0.316/8)*((v*Di*10)/v2)**(-0.25)*p*Cp*v*(4.34)**(-2./3)); #Heat transfer coefficient umath.sing the Colburn anamath.logy in W/m**2.K
# Results
print 'Heat transfer coefficient using the Colburn analogy is %3.0f W/m**2.K'%(h)
```

In [27]:

```
# Variables
Ti = 50; #Temperature of water at inlet in degree C
D = 0.015; #Diameter of tube in m
L = 3; #Length of the tube in m
v = 1; #Velocity of flow in m/s
Tb = 90; #Temperature of tube wall in degree C
Tf = 64; #Exit temperature of water in degree C
# Calculations
Tm = (Ti+Tf)/2; #Bulk mean temperature in degree C
p = 990; #Density of air at 57 degree C in kg/m**3
Cp = 4184; #Specific heat of air at 57 degree C in J/kg.K
u = (0.517*10**-6); #Kinematic viscosity of air at 57 degree C in m**2/s
k = 0.65; #Thermal conductivity of air at 57 degree C in W/m.K
Pr = 3.15; #prantl number of air at 57 degree C
Re = (v*D)/u; #Reynolds number
Nu = (0.023*Re**(4./5)*Pr**0.4); #Nusselts number
h = (Nu*k)/D; #Heat transfer coefficient in W/m**2.K
Q = (h*3.14*D*L*(Tb-Tm))/1000; #Rate of heat transfered in kW
# Results
print 'Heat transfer coefficient is %3.0f W/m**2.K \nRate of heat transfered is %3.2f kW'%(h,Q)
```

In [28]:

```
# Variables
D = 0.022; #Diamter of the tube in m
v = 2; #Average velocity in m/s
Tw = 95; #Temperature of tube wall in degree C
T = [15,60]; #Initial and final temperature of water in degree C
# Calculations
Tm = (T[0]+T[1])/2; #Bulk mean temperature in degree C
p = 990; #Density of air at 37.5 degree C in kg/m**3
Cp = 4160; #Specific heat of air at 37.5 degree C in J/kg.K
u = (0.69*10**-3); #Dynamic viscosity of air at 37.5 degree C in Ns/m**2
k = 0.63; #Thermal conductivity of air at 37.5 degree C in W/m.K
us = (0.3*10**-3); #Dynamic viscosity of air at 37.5 degree C in Ns/m**2
Re = (p*v*D)/u; #Reynolds number
Pr = (u*Cp)/k; #Prantl number
Nu = (0.027*Re**(4./5)*Pr**(1./3)*(u/us)**0.14); #Nusselts number
h = (Nu*k)/D; #Heat transfer coefficient in W/m**2.K
# Results
print 'Heat transfer coefficient is %3.0f W/m**2.K'%(h)
```

In [29]:

```
# Variables
D = 0.05; #Diamter of the tube in m
T = 147; #Average temperature in degree C
v = 0.8; #Flow vwlocity in m/s
Tw = 200; #Wall temperature in degree C
L = 2; #Length of the tube in m
# Calculations
p = 812.1; #Density in kg/m**3 of oil at 147 degree C
Cp = 2427; #Specific heat of oil at 147 degree C in J/kg.K
u = (6.94*10**-6); #Kinematic viscosity of oil at 147 degree C in m**2/s
k = 0.133; #Thermal conductivity of oil at 147 degree C in W/m.K
Pr = 103; #prantl number of oil at 147 degree C
Re = (v*D)/u; #Reynolds number
Nu = (0.036*Re**0.8*Pr**(1./3)*(D/L)**0.055); #Nussults number
h = (Nu*k)/D; #Average heat transfer coefficient in W/m**2.K
# Results
print 'Average heat transfer coefficient is %3.1f W/m**2.K'%(h)
```

In [31]:

```
# Variables
D = [0.4,0.8]; #Dimensions of the trunk duct in m
Ta = 20; #Temperature of air in degree C
v = 7; #Velocity of air in m/s
v1 = (15.06*10**-6); #Kinematic viscosity in m**2/s
a = (7.71*10**-2); #Thermal diffusivity in m**2/h
k = 0.0259; #Thermal conductivity in W/m.K
# Calculations
Dh = (4*(D[0]*D[1]))/(2*(D[0]+D[1])); #Value of Dh in m
Re = (v*Dh)/v1; #Reynolds number
Pr = (v1/a)*3600; #Prantl number
Nu = (0.023*Re**(4./5)*Pr**0.4); #Nussults number
h = (Nu*k)/Dh; #Heat transfer coefficient in W/m**2.K
Q = (h*(2*(D[0]+D[1]))); #Heat leakage per unit length per unit difference in W
# Results
print 'Heat leakage per unit length per unit difference is %3.2f W'%(Q)
```

In [33]:

```
# Variables
Di = 0.03125; #I.D of the annulus in m
Do = 0.05; #O.D of the annulus in m
Ts = 50; #Outer surface temperature in degree C
Ti = 16; #Temeperature at which air enters in degree C
Tf = 32; #Temperature at which air exits in degree C
v = 30; #Flow rate in m/s
# Calculations
Tb = (Ti+Tf)/2; #Mean bulk temperature of air in degree C
p = 1.614; #Density in kg/m**3 of air at 24 degree C
Cp = 1007; #Specific heat of air at 24 degree C in J/kg.K
u = (15.9*10**-6); #Kinematic viscosity of air at 24 degree C in m**2/s
k = 0.0263; #Thermal conductivity of air at 24 degree C in W/m.K
Pr = 0.707; #prantl number of air at 24 degree C
Dh = (4*(3.14/4)*(Do**2-Di**2))/(3.14*(Do+Di)); #Hydraulic diameter in m
Re = (v*Dh)/u; #Reynolds number
Nu = (0.023*Re**0.8*Pr**0.4); #Nussults number
h = (Nu*k)/Dh; #Heat transfer coefficient in W/m**2.K
# Results
print 'Heat transfer coefficient is %3.1f W/m**2.K'%(h)
```

In [34]:

```
# Variables
T = [120,149]; #Initail and final temperatures in degree C
m = 2.3; #Mass flow rate in kg/s
D = 0.025; #Diameter of the tube in m
Ts = 200; #Surface temperature in degree C
# Calculations
Tb = (T[0]+T[1])/2; #Bulk mean temperature in degree C
p = 916; #Density in kg/m**3 of air at 134.5 degree C
Cp = 1356.6; #Specific heat of air at 134.5 degree C in J/kg.K
u = (0.594*10**-6); #Kinematic viscosity of air at 134.5 degree C in m**2/s
k = 84.9; #Thermal conductivity of air at 134.5 degree C in W/m.K
Pr = 0.0087; #prantl number of air at 134.5 degree C
Q = (m*Cp*(T[1]-T[0]))/1000; #Total heat transfer in kW
v = (m/(p*(3.14/4)*D**2)); #Velocity of flow in m/s
Re = (v*D)/u; #Reynolds number
Pe = (Pr*Re); #Peclet number
Nu = (4.82+(0.0185*Pe**0.827)); #Nussults number
h = (Nu*k)/D; #Heat transfer coefficient in W/m**2.K
L = ((Q*1000)/(h*3.14*D*(Ts-Tb))); #Minimum length of the tube in m if the wall temperature is not to exceed 200 degree C
# Results
print 'Minimum length of the tube if the wall temperature is not to exceed 200 degree C is %3.3f m'%(L)
```