# Internal Flow¶

## Example 8.2 Page 499¶

In [1]:
import math
#Operating Conditions
m = .1;            #[kg/s] mass flow rate of water
Ti = 20+273.;      #[K] Inlet temp
To = 60+273.;      #[K] Outlet temperature
Di = .02;          #[m] Inner Diameter
Do = .04;          #[m] Outer Diameter
q = 1000000.;	   #[w/m^3]  Heat generation Rate
Tsi = 70+273.;     #[K] Inner Surface Temp
#Table A.4 Air Properties T = 313 K
cp = 4179;         #[J/kg.K] specific heat
#calculations
L = 4*m*cp*(To-Ti)/(math.pi*(Do*Do-Di*Di)*q);

#From Newtons Law of cooling, Equation 8.27, local heat convection coefficient is
h = q*(Do*Do-Di*Di)/(Di*4*(Tsi-To));
#results
print '%s %.1f %s' %("\n Length of tube needed to achieve the desired outlet temperature = ",L,"m ")
print '%s %.1f %s' %("\n Local convection coefficient at the outlet =",h," W/m^2.K");

#END

 Length of tube needed to achieve the desired outlet temperature =  17.7 m

Local convection coefficient at the outlet = 1500.0  W/m^2.K


## Example 8.3 Page 503¶

In [2]:
import math
#Operating Conditions
m = .25;            	#[kg/s] mass flow rate of water
Ti = 15+273.;       	#[K] Inlet temp
To = 57+273.;       	#[K] Outlet temperature
D = .05;          		#[m] Diameter
L = 6;          		#[m] Length of tube
Ts = 100+273.;      	#[K] outer Surface Temp

#Table A.4 Air Properties T = 309 K
cp = 4178;          	#[J/kg.K] specific heat
#calculations
Tlm = ((Ts-To)-(Ts-Ti))/(2.30*math.log10((Ts-To)/(Ts-Ti)));

h = m*cp*(To-Ti)/(math.pi*D*L*Tlm);
#results
print '%s %d %s' %("\n Average Heat transfer Convection Coefficient = ",h,"W/m^2.K");

#END

 Average Heat transfer Convection Coefficient =  754 W/m^2.K


## Example 8.4 Page 506¶

In [4]:
import math
#Operating Conditions
m = .01;            #[kg/s] mass flow rate of water
Ti = 20+273;       	#[K] Inlet temp
To = 80+273;       	#[K] Outlet temperature
D = .06;          	#[m] Diameter
q = 2000;          	#[W/m^2] Heat flux to fluid

#Table A.4 Air Properties T = 323 K
cp = 4178;          #[J/kg.K] specific heat
#Table A.4 Air Properties T = 353 K
k = .670;           #[W/m] Thermal Conductivity
u = 352*math.pow(10,-6);#[N.s/m^2] Viscosity
Pr = 2.2;           #Prandtl Number
cp = 4178;          #[J/kg.K] specific heat
#calculations
L = m*cp*(To-Ti)/(math.pi*D*q);

#Using equation 8.6
Re = m*4/(math.pi*D*u);
print '%s %.2f %s' %("\n (a) Length of tube for required heating =",L,"m")
print '%s %.2f %s' %("\n\n (b)As Reynolds Number is",Re,".The flow is laminar.");

Nu = 4.364;        #Nusselt Number
h = Nu*k/D;        #[W/m^2.K] Heat convection Coefficient

Ts = q/h+To;        #[K]
#results
print '%s %.2f %s' %("\n Surface Temperature at tube outlet = ",Ts-273,"degC");

#END

 (a) Length of tube for required heating = 6.65 m

(b)As Reynolds Number is 602.86 .The flow is laminar.

Surface Temperature at tube outlet =  121.04 degC


## Example 8.5 Page 509¶

In [5]:
import math
#Operating Conditions
um1 = .13;                  #[m/s] Blood stream
um2 = 3*math.pow(10,-3);    #[m/s] Blood stream
um3 = .7*math.pow(10,-3);   #[m/s] Blood stream
D1 = .003;                  #[m] Diameter
D2 = .02*math.pow(10,-3);   #[m] Diameter
D3 =  .008*math.pow(10,-3); #[m] Diameter
Tlm = .05;
kf = .5;                    #[W/m.K] Conductivity
#Table A. Water Properties T = 310 K
rho = 993.;                 #[kg/m^3] density
cp = 4178.;                 #[J/kg.K] specific heat
u = 695*math.pow(10,-6);    #[N.s/m^2] Viscosity
kb = .628;                  #[W/m.K] Conductivity
Pr = 4.62;                  #Prandtl Number
i=1.;
#calculations
#Using equation 8.6
Re1 = rho*um1*D1/u;
Nu = 4;
hb = Nu*kb/D1;
hf = kf/D1;
U1 = 1/(1/hb + 1/hf);
L1 = -rho*um1*D1/U1*cp*2.303*math.log10(Tlm)/4.;
xfdh1 = .05*Re1*D1;
xfdr1 = xfdh1*Pr;

Re2 = rho*um2*D2/u;
Nu = 4;
hb = Nu*kb/D2;
hf = kf/D2;
U2 = 1/(1/hb + 1/hf);
L2 = -rho*um2*D2/U2*cp*2.303*math.log10(Tlm)/4.;
xfdh2 = .05*Re2*D2;
xfdr2 = xfdh2*Pr;

Re3 = rho*um3*D3/u;
Nu = 4;
hb = Nu*kb/D3;
hf = kf/D3;
U3 = 1/(1/hb + 1/hf);
L3 = -rho*um3*D3/U3*cp*2.303*math.log10(Tlm)/4.;
xfdh3 = .05*Re3*D3;
xfdr3 = xfdh3*Pr;
#results
print ' %s' %("\n Vessel          Re   U(W/m^2.K)    L(m)   xfdh(m)  xfdr(m)")
print '%s %.3f %d %.1e %.1e %.1e' %("\n Artery          ",Re1,      U1         ,L1,   xfdh1     , xfdr1)
print '%s %.3f %d %.1e %.1e %.1e' %("\n Anteriole       ",Re2,      U2         ,L2,   xfdh2     , xfdr2)
print '%s %.3f %d %.1e %.1e %.1e' %("\n Capillary      ",Re3,U3,L3,xfdh3,xfdr3);

#END


Vessel          Re   U(W/m^2.K)    L(m)   xfdh(m)  xfdr(m)

Artery           557.223 138 8.7e+00 8.4e-02 3.9e-01

Anteriole        0.086 20849 8.9e-06 8.6e-08 4.0e-07

Capillary       0.008 52124 3.3e-07 3.2e-09 1.5e-08


## Example 8.6 Page 516¶

In [6]:
import math
#Operating Conditions
m = .05;            	#[kg/s] mass flow rate of water
Ti = 103+273.;       	#[K] Inlet temp
To = 77+273.;     		#[K] Outlet temperature
D = .15;          		#[m] Diameter
L = 5;            		#[m] length
ho = 6.;           		#[W/m^2.K] Heat transfer convective coefficient
Tsurr = 0+273.;    		#[K] Temperature of surrounding

#Table A.4 Air Properties T = 363 K
cp = 1010;          	#[J/kg.K] specific heat
#Table A.4 Air Properties T = 350 K
k = .030;           	#[W/m] Thermal Conductivity
u = 20.82/1000000.;   	#[N.s/m^2] Viscosity
Pr = .7;           		#Prandtl Number
#calculations and results
q = m*cp*(To-Ti);

Re = m*4/(math.pi*D*u);
print '%s %d %s' %("\n As Reynolds Number is",Re,". The flow is Turbulent.");

#Equation 8.6
n = 0.3;
Nu = .023*math.pow(Re,.8)*math.pow(Pr,.3);
h = Nu*k/D;
q2 = (To-Tsurr)/(1/h + 1/ho);
Ts = -q2/h+To;

print '%s %d %s' %("\n\n Heat Loss from the Duct over the Length L, q =",q," W ")
print '%s %.1f %s %.1f %s' %("\n Heat flux and suface temperature at x=L is",q2,"W/m^2 &",Ts-273,"degC respectively");

#END

 As Reynolds Number is 20384 . The flow is Turbulent.

Heat Loss from the Duct over the Length L, q = -1313  W

Heat flux and suface temperature at x=L is 304.3 W/m^2 & 50.7 degC respectively


## Example 8.7 Page 525¶

In [7]:
import math
#Operating Conditions
T1 = 125+273.;    			#[K] Chip Temperature 1
T2 = 25+273.;    			#[K] Chip Temperature 2
Ti = 5+273.;    			#[K] Inlet Temperature
D = .01;          			#[m] Diameter
L = .02;            		#[m] length
delP = 500*1000.;    		#[N/m^2] Pressure drop
#Dimensions
a = 40*math.pow(10,-6);
b = 160*math.pow(10,-6);
s = 40*math.pow(10,-6);

#Table A.5 Ethylene Glycol Properties T = 288 K
rho = 1120.2;           	#[kg/m^3]    Density
cp = 2359.;             	#[J/kg.K]    Specific Heat
u = 2.82*math.pow(10,-2);	#[N.s/m^2] Viscosity
k = 247*math.pow(10,-3); 	#[W/m.K]  Thermal Conductivity
Pr = 269;              		#Prandtl number
#Table A.5 Ethylene Glycol Properties T = 338 K
rho2 = 1085.;            	#[kg/m^3]    Density
cp2 = 2583.;             	#[J/kg.K]    Specific Heat
u2 = .427*math.pow(10,-2);	#[N.s/m^2] Viscosity
k2 = 261*math.pow(10,-3); 	#[W/m.K]  Thermal Conductivity
Pr2 = 45.2;               	#Prandtl number
#calculations
P = 2*a+2*b;              	#Perimeter of microchannel
Dh = 4*a*b/P;             	#Hydraulic Diameter

um2 = 2/73.*Dh*Dh/u2*delP/L;#[[m/s] Equation 8.22a
Re2 = um2*Dh*rho2/u2;       #Reynolds Number
xfdh2 = .05*Dh*Re2;        	#[m] From Equation 8.3
xfdr2 = xfdh2*Pr2;        	#[m] From Equation 8.23
m2 = rho2*a*b*um2;        	#[kg/s]
Nu2 = 4.44;            		#Nusselt Number from Table 8.1
h2 = Nu2*k2/Dh;        		#[W/m^2.K] Convection Coeff
Tc2 = 124+273.;        		#[K]
xc2 = m2/P*cp2/h2*2.303*math.log10((T1-Ti)/(T1-Tc2));
tc2 = xc2/um2;

um = 2/73.*Dh*Dh/u*delP/L;  #[[m/s] Equation 8.22a
Re = um*Dh*rho/u;        	#Reynolds Number
xfdh = .05*Dh*Re;        	#[m] From Equation 8.3
xfdr = xfdh*Pr;        		#[m] From Equation 8.23
m = rho2*a*b*um;        	#[kg/s]
Nu = 4.44;            		#Nusselt Number from Table 8.1
h = Nu*k/Dh;        		#[W/m^2.K] Convection Coeff
Tc = 24+273.;        		#[K]
xc = m/P*cp/h*2.303*math.log10((T2-Ti)/(T2-Tc));
tc = xc/um;

#results
print '%s %.1f %s' %("\nTemp in case 2= ",T2-273," [degC]")
print '%s %.1f %s' %("\nTemp in case 1= ",T1-273," [degC]")
print '%s %.3f %s' %("\nFlow rate in case 2 = ",um2,"[m/s]")
print '%s %.3f %s' %("\nFlow rate in case 1 = ",um,"[m/s]")
print '%s %.1f' %("\nReynolds number in case 2 = ",Re2)
print '%s %.1f' %("\nReynolds number in case 1 = ",Re)
print '%s %.1f' %("\nHydrodynamic entrance Length [m] =",xfdh)
print '%s %.1f' %("\nHydrodynamic entrance Length in case 2 [m] =",xfdh2)
print '%s %.1e' %("\nThermal entrance Length [m] = ",xfdr)
print '%s %.1e' %("\nThermal entrance Length in case 2 [m] = ",xfdr2)
print '%s %.2e' %("\nMass Flow rate [kg/s] = ",m)
print '%s %.2e' %("\nMass Flow rate in case 2 [kg/s] = ",m2)
print '%s %.2e' %("\nConvective Coeff [W/m^2.K] = ",h)
print '%s %.2e' %("\nConvective Coeff in case 2 [W/m^2.K] = ",h2)
print '%s %.2e' %("\nTransition Length [m] = ",xc)
print '%s %.2e' %("\nTransition Length in case 2 [m] = ",xc2)
print '%s %.3f' %("\nRequired Time [s] = ",tc)
print '%s %.3f' %("\nRequired Time in case 2 [s] = ",tc2)
#END

Temp in case 2=  25.0  [degC]

Temp in case 1=  125.0  [degC]

Flow rate in case 2 =  0.657 [m/s]

Flow rate in case 1 =  0.099 [m/s]

Reynolds number in case 2 =  10.7

Reynolds number in case 1 =  0.3

Hydrodynamic entrance Length [m] = 0.0

Hydrodynamic entrance Length in case 2 [m] = 0.0

Thermal entrance Length [m] =  2.2e-04

Thermal entrance Length in case 2 [m] =  1.5e-03

Mass Flow rate [kg/s] =  6.91e-07

Mass Flow rate in case 2 [kg/s] =  4.56e-06

Convective Coeff [W/m^2.K] =  1.71e+04

Convective Coeff in case 2 [W/m^2.K] =  1.81e+04

Transition Length [m] =  7.12e-04

Transition Length in case 2 [m] =  7.79e-03

Required Time [s] =  0.007

Required Time in case 2 [s] =  0.012


## Example 8.8 Page 529¶

In [8]:
import math
#Operating Conditions
m = .0003;            		#[kg/s] mass flow rate of water
T = 25+273;    				#[K] Temperature of surrounding and tube
D = .01;          			#[m] Diameter
L = 1;            			#[m] length
#calculations and results
#Table A.4 Air Properties T = 298 K
uv = 15.7*math.pow(10,-6);  #[m^2/s] Kinematic Viscosity
u = 18.36*math.pow(10,-6);  #[N.s/m^2] Viscosity
#Table A.8 Ammonia-Air Properties T = 298 K
Dab = .28*math.pow(10,-4);  #[m^2/s] Diffusion coeff
Sc = .56;

Re = m*4/(math.pi*D*u);
print '%s %d %s' %("\n As Reynolds Number is",Re,". The flow is Laminar.");

#Using Equation 8.57
Sh = 1.86*math.pow((Re*Sc*D/L),.3334);
h = Sh*Dab/D;
print '%s %.3f %s' %("\n Average mass trasnfer convection coefficient for the tube",h,"m/s");

#END

 As Reynolds Number is 2080 . The flow is Laminar.

Average mass trasnfer convection coefficient for the tube 0.012 m/s