Chapter 5: Heat Transfer by Forced Convection

Example 5.1(a) , Page no:209

In [1]:
import math
from __future__ import division

#Variable declaration
D = 0.015 ; #m
Q = 0.05 ; #m^3/h
H = 1000 ; #W/m^ 2
Tb = 40 ; #degree C
k = 0.634 ; #W/m K

#calculations
v = 0.659*10**-6 ; #m^2/ s
Vbar = 4*Q /((3.14)*D**2) ;
ReD = Vbar *D/v;
h = 4.364* k/D; #W/m^2 K

#result
print"(a) Local heat transfer coefficient is",round(h,4),"W/m^2 K";

(a) Local heat transfer coefficient is 184.4517 W/m^2 K

Example 5.1(b) , Page no:209

In [2]:
import math
from __future__ import division

#Variable declaration
D = 0.015 ; #m
Q = 0.05 ; #m^3/h
H = 1000 ; #W/m^2
Tb = 40 ; #degree C
k = 0.634 ; #W/m K

#calculations
v = 0.659*10**-6 ; #m^2/s
Vbar = 4*Q /((3.14)*D**2) ;
ReD = Vbar *D/v;
h = 4.364* k/D;
Tw = H/h + Tb; #the local wal to bulk mean temperature difference

#result
print"(b) Wall Temperature Tw =",round(Tw,4),"degree C";

(b) Wall Temperature Tw = 45.4215 degree C

Example 5.2 , Page no:213

In [3]:
import math
from __future__ import division

#Variable declaration
Pr1=0.01;
Pr2=0.1;
Pr3=100;

#Calculation
T1 = 0.04305* Pr1 /0.0575; #For Pr = 0.01
T2 = 0.04305* Pr2 /0.0575; #For Pr = 0.1
T3 = 0.04305* Pr3 /0.0575 ; #For Pr = 100

#result
print"Lth/Le at Pr =0.01 is",round(T1,4);
print"Lth/Le at Pr = 1 is",round(T2,4);
print"Lth/Le at Pr = 100 is",round(T3,4);

Lth/Le at Pr =0.01 is 0.0075
Lth/Le at Pr = 1 is 0.0749
Lth/Le at Pr = 100 is 74.8696

Example 5.3(i) , Page no:215

In [4]:
import math
from __future__ import division

#Variable declaration
D = 0.015 ; #m
V = 1 ; #m/s
Tw = 90 ; #degree C
Tmi = 50 ; #degree C
Tmo = 65 ; #degree C
k = 0.656 ; #W/m K
rho = 984.4 ; #kg/m^3
Pr = 3.12 ;
rhoin = 988.1 ; #kg/m^3

#calculations
v = 0.497 * 10**-6 ; #m^2/s
Cp = 4178 ; #J/kg K
mdot =3.14*(D**2)* rhoin *V/4 ; #kg/s
Re = 4* mdot /(3.14*D* rho *v) ;
f = 0.079*( Re)** -0.25 ;
Nu = (f /2) *(Re -1000) *Pr /(1+12.7*( f /2) **(1/2) *(( Pr**(2/3) ) -1));
h = Nu*k/D;
L = mdot *Cp *( Tmo -Tmi)*( math.log ((Tw - Tmi )/(Tw - Tmo )) /(((Tw -Tmi) -(Tw - Tmo ))*h*D*3.14)); #the energy equation

#result
print"The length of tube if the exit water temperature is 65 degree C =",round(L,4),"m";

The length of tube if the exit water temperature is 65 degree C = 1.0876 m

Example 5.3(ii) , Page no:215

In [5]:
import math
from __future__ import division

#Variable declaration
D = 0.015 ; #m
V = 1 ; #m/s
Tw = 90 ; #degree C
Tmi = 50 ; #degree C
Tmo = 65 ; #degree C
k = 0.656 ; #W/m K
rho = 984.4 ; #kg/m^3
Cp = 4178 ; #J/kg K
Pr = 3.12 ;
rhoin = 988.1 ; #kg/m^3
Tmo = 70 ; #degree C
Tb = 60 ; #degree C
k1 = 0.659 ; #W/m K
rho1 = 983.2 ;#kg/m^3
Cp1 = 4179 ;#J/kg K
Pr1 = 2.98 ;
f1 = 0.005928;
Nud = 154.97; #the Gnielinski Eqn
Tmo1 = 73.4 ; #degree C

#calculations
v = 0.497 * 10**-6 ; #m^2/s
mdot =3.14*(D**2)* rhoin *V/4 ; #kg/s
Re = 4* mdot /(3.14*D* rho *v) ;
f = 0.079*( Re)** -0.25 ;
Nu = (f /2) *(Re -1000) *Pr /(1+12.7*( f /2)**(1/2) *(( Pr**(2/3) ) -1) ); #W/m^2 K
h = Nu*k/D;
L = mdot *Cp *( Tmo -Tmi)* math.log ((Tw - Tmi )/(Tw - Tmo ))/(((Tw -Tmi) -(Tw - Tmo ))*h*D*3.14); #the energy equation
v1 = 0.478 * 10**-6 ; #m^2/s
Re1 = 4* mdot /(3.14*D* rho1 *v1);
h = Nud *k1/D ; #W/m^2 K

#result
print"Trial and error method";
print"Trial 1";
print"Assumed value of Tmo =70 degree C";
print"Value of Tmo obtained =73.4 degree C";
print"Trial 2";
print"Assume Tmo =73.4 degree C";
print"Value of Tmo obtained = 73.6 degree C which is in reasonably close agreement with assumed value";

Trial and error method
Trial 1
Assumed value of Tmo =70 degree C
Value of Tmo obtained =73.4 degree C
Trial 2
Assume Tmo =73.4 degree C
Value of Tmo obtained = 73.6 degree C which is in reasonably close agreement with assumed value

Example 5.4 , Page no:219

In [6]:
import math
from __future__ import division

#Variable declaration
Di = 0.05 ; #m
m = 300 ; #kg/min
m1 = m/60 ; #kg/sec
rho = 846.7 ; #kg/m^3
k = 68.34 ; #W/m K
c = 1274; #J/kg K
Pr = 0.00468 ;

#calculations
v = 0.2937*10**-6 ; #m^2/s
ReD = 4* m1 /(3.14*Di* rho *v);
NuD = 6.3 + 0.0167*( ReD**0.85) *( Pr**0.93) ; #Assuming both temperature and velocity profile are fully developed over the length of tube
h = NuD *k/ Di ;
L = 300/60*1274*(500 -400) /(h*3.14* Di *30); #Equating the heat transferred through the wall of the tube to the change of enthalpy pf sodium

#result
print"Length of tube over which the temperature rise occurs =",round(L,4),"m";

Length of tube over which the temperature rise occurs = 6.8659 m

Example 5.5 , Page no:231

In [7]:
import math
from __future__ import division

#Variable declaration
V = 15 ; #m/s
s =0.2 ; #m
rho = 1.128 ; #kg/m^3
k = 0.0276; #W/m K
Pr = 0.699;

#calculations
Tm = (20+60) /2; ##degree C
v = 16.96*10**-6; #m^2/s
A=s**2;
ReL = V *0.2/ v;
Cf = 1.328/( ReL )**0.5; #the boundary layer may be assumed to be laminar over the entire length.
Fd = 2* Cf *1/2* rho*A*V**2;
Nul = 0.664*( Pr**(1/3) )*( ReL**(1/2) );
h = Nul *k/s;
q = 2*A*h *(60 -20) ; #rate of heat transfer q is
Cf1 = 0.074*( ReL )**( -0.2) ; #boundary layer from leading edge, the drag coefficient is
Fd1 = 2* Cf1 *1/2* rho *A*V**2;
Nul1 = 0.0366*(0.699**(1/3) )*( ReL**(0.8) );
h1 = Nul1 *k/s; #W/m^2 K
q1 = 2*A*h1 *(60 -20) ;

#result
print"For Laminar Boundary Layer";
print"Rate of Heat transfer =",round(q,4),"W";
print"Drag force =",round(Fd,4),"N";
print"For Turbulent Boundary Layer from the leading edge";
print"Rate of Heat transfer =",round(q1,4),"W";
print"Drag force =",round(Fd1,4),"N";

For Laminar Boundary Layer
Rate of Heat transfer = 109.447 W
Drag force = 0.0321 N
For Turbulent Boundary Layer from the leading edge
Rate of Heat transfer = 226.3735 W
Drag force = 0.067 N

Example 5.6(i) , Page no:235

In [8]:
import math
from __future__ import division

#Variable declaration
D = 0.075 ; #m
V = 1.2 ; #m/s
Tair = 20 ; #degree C
Tsurface = 100 ; #degree C
k = 0.0290 ; #W/m K
Pr = 0.696 ;

#calculations
Tm = ( Tair + Tsurface ) /2;
v = 18.97*10**-6 ; #m^2/s
ReD = V*D/v;
Nu = 0.3 +((0.62*( ReD**(1/2) )*( Pr**(1/3) ))/((1+((0.4/ Pr)**(2/3) ))**(1/4)))*((1+(( ReD/282000)**(5/8)))**(4/5));
h = Nu*k/D ; #W/m^2 K
flux = h*( Tsurface - Tair ); #W/m^2
q = flux *3.14*D *1; #W/m

#result
print"Heat transfer rate per unit length =",round(q,4),"W/m";

Heat transfer rate per unit length = 258.8849 W/m

Example 5.6(ii) , Page no:235

In [9]:
import math
from __future__ import division

#Variable declaration
D = 0.075 ; #m
V = 1.2 ; #m/s
Tair = 20 ; #degree C
Tsurface = 100 ; #degree C
k = 0.0290 ; #W/m K
Pr = 0.696 ;
k = 0.0290 ;
Pr = 0.696 ;
Tassumd = 130 ; #degree C
Tm = 75 ; #degree C
k1 = 0.0301 ; #W/m K
Pr1 = 0.693 ;
Nu1 = 33.99;
Tavgcalc = 129.9 ; #degree C

#calculations
Tm = ( Tair + Tsurface ) /2;
v = 18.97*10**-6 ; #m^2/s
ReD = V*D/v;
Nu = 0.3 +((0.62*( ReD**0.5) *( Pr**(1/3) )) /((1+((0.4/Pr)**(2/3) ))**(1/4)))*(1+( ReD /282000)**(5/8) )**(5/8);
h = Nu*k/D ; #W/m^2 K
flux = h*( Tsurface - Tair ); #W/m^2
Tavg = 1500/ flux *( Tsurface - Tair );
v1 = 20.56*10** -6 ; #m^2/s
ReD1 = V*D/v1;
h = Nu1*k1/D;
Tdiff = 1500/ h; #degree C

#result
print"Assumed average wall temperature =",round(Tassumd,4),"degree C";
print"Calculated average wall Temperature =",round(Tavgcalc,4),"degree C";
print"Hence,Average wall Temperature =",round(Tavgcalc,4),"degree C";

Assumed average wall temperature = 130.0 degree C
Calculated average wall Temperature = 129.9 degree C
Hence,Average wall Temperature = 129.9 degree C

Example 5.7(i) , Page no:241

In [10]:
import math
from __future__ import division

#Variable declaration
D = 0.0125 ; #m
ST = 1.5* D ;
SL = 1.5* D ;
Vinf = 2 ; #v
N = 5;
Tw = 70; #degree C
Tmi = 30; #degree C
L = 1; #m
rho = 1.165 ; #kg/m^3
Cp = 1.005 ; #kJ/kg K
k = 0.0267 ; #W/m K
Pr = 0.701;
X = 1; #tube arrangement is square

#calculations
v = 16.00 *10**-6 ; #m^2/s
Vmax = ST /(SL -D)* Vinf ; #m/s
Re = Vmax *D/v ;
f = 0.37/4;
deltaP = 4*f*N*X*( rho * Vmax**2) /2 ; #N/m^2

#result
print"(i) Pressure drop of air across the bank is",round(deltaP,4),"N/m^2";

(i) Pressure drop of air across the bank is 38.7945 N/m^2

Example 5.7(ii) , Page no:241

In [11]:
import math
from __future__ import division

#Variable declaration
D = 0.0125 ; #m
ST = 1.5* D ;
SL = 1.5* D ;
Vinf = 2 ; #m/s
N = 5;
Tw = 70; #degree C
Tmi = 30; #degree C
L = 1; #m
rho = 1.165 ; #kg/m^3
v = 16.00 *10**-6 ; #[m^2/s]
Cp = 1.005*1000 ; #J/kg K
k = 0.0267 ; #W/m K
Pr = 0.701;
X = 1; #tube arrangement is square
Pr1 = 0.694 ; #At 70 degree C
C1 = 0.27;
m = 0.63;
C2 = 0.93;

#calculations
#sub all value in the following expression
# q=h*(3.14*D*L)*50*((Tw-Tmi)-(Tw-Tmo))/log((Tw-Tmi)/(Tw-Tmo))-mdot*Cp*(Tmo-Tmi), we get
Tmo=70-40/(math.exp (190.8604/439.064));


#Results
print"Tmo =",round(Tmo,2),"oC"
print"(ii) Exit temperature of air =",round(Tmo,4),"oC";

Tmo = 44.1 oC
(ii) Exit temperature of air = 44.1016 oC

Example 5.7(iii) , Page no:241

In [12]:
import math
from __future__ import division

#Variable declaration
D = 0.0125 ; #m
ST = 1.5* D ;
SL = 1.5* D ;
Vinf = 2 ; #m/s
N = 5;
Tw = 70; #degree C
Tmi = 30; #degree C
L = 1; #m
rho = 1.165 ; #kg/m^3
k = 0.0267 ; #W/m K
Pr = 0.701;
X = 1; #tube arrangement is square
Pr1 = 0.694 ; #At 70 degree C
C1 = 0.27;
m = 0.63;
C2 = 0.93;
Tmo = 44.10; #in oC

#calculations
q=439.064*(40-(70-Tmo)); #Heat transfer rate per unit length to air

#result
print"(iii) Heat transfer rate per unit length to air =",round(q,2),"W   (roundoff error)";

(iii) Heat transfer rate per unit length to air = 6190.8 W   (roundoff error)