# Introduction to Convection¶

## Example 6.2 Page 356¶

In :
import math
#Operating Conditions

h = .05;          			#[W/m^2.K] Heat Convection coefficient
D = .02;          			#[m] Diameter of cylinder
Cas = 5*math.pow(10,-6);    #[kmol/m^3] Surface molar Conc
Casurr = 0;       			#[kmol/m^3] Surrounding molar Conc
Ma = 128;         			#[Kg/kmol] Molecular weight
#calculations
#From Eqn 6.15
Na = h*(math.pi*D)*(Cas-Casurr);
na = Ma*Na;
#results
print '%s %.2e %s' %("\n\n Mass sublimation Rate is =",na," kg/s.m ");
#END


Mass sublimation Rate is = 2.01e-06  kg/s.m


## Example 6.3 Page 357¶

In :
import math
#Operating Conditions

Dab = .288*math.pow(10,-4);        	#[m^2/s] Table A.8 water vapor-air (319K)
pas = .1;            				#[atm] Partial pressure at surface
pasurr = .02;            			#[atm] Partial pressure at infinity
y0 = .003;            				#[m] Tangent at y = 0 intercepts y axis at 3 mm
#calculations
#From Measured Vapor Pressure Distribution
delp = (0 - pas)/(y0 - 0);            #[atm/m]
hmx = -Dab*delp/(pas - pasurr);       #[m/s]
#results
print '%s %.4f %s' %("\n\n Convection Mass Transfer coefficient at prescribed location =",hmx," m/s");
#END


Convection Mass Transfer coefficient at prescribed location = 0.0120  m/s


## Example 6.4 Page 362¶

In :
import math
#Operating Conditions
v = 1;        				#[m/s]  Velocity of water
L = 0.6;      				#[m] Plate length
Tw1 = 300.;    				#[K]
Tw2 = 350.;    				#[K]
#Coefficients [W/m^1.5 . K]
Clam1 = 395;
Cturb1 = 2330;
Clam2 = 477;
Cturb2 = 3600;

#Water Properties at T = 300K
p1 = 997;    				#[kg/m^3]  Density
u1 = 855*math.pow(10,-6);   #[N.s/m^2] Viscosity
#Water Properties at T = 350K
p2 = 974;    				#[kg/m^3]  Density
u2 = 365*math.pow(10,-6);   #[N.s/m^2] Viscosity

Rec = 5*math.pow(10,5);     #Transititon Reynolds Number
xc1 = Rec*u1/(p1*v); 		#[m]Transition length at 300K
xc2 = Rec*u2/(p2*v); 		#[m]Transition length at 350K
#calculations
#Integrating eqn 6.14
#At 300 K
h1 = (Clam1*math.pow(xc1,.5) /.5 + Cturb1*(math.pow(L,.8)-math.pow(xc1,.8))/.8)/L;

#At 350 K
h2 = (Clam2*math.pow(xc2,.5) /.5 + Cturb2*(math.pow(L,.8)-math.pow(xc2,.8))/.8)/L;
#results
print '%s %.2f %s %.2f %s' %("\n\n Average Convection Coefficient over the entire plate for the two temperatures at 300K =",h1," W/m^2.K and at 350K =",h2," W/m^2.K");
#END


Average Convection Coefficient over the entire plate for the two temperatures at 300K = 1622.45  W/m^2.K and at 350K = 3707.93  W/m^2.K


## Example 6.5 Page 372¶

In :
#Operating Conditions
v = 160;        				#[m/s]  Velocity of air
L = 0.04;      					#[m] Blade length
Tsurr = 1150+273.;    			#[K]
Ts = 800+273.;    				#[K] Surface Temp
q = 95000;        				#[W/m^2] Original heat flux
#calculations
#Case 1
Ts1 = 700+273.;   	 			#[K] Surface Temp
q1 = q*(Tsurr-Ts1)/(Tsurr-Ts);

#Case 2
L2 = .08;            			#[m] Length
q2 = q*L/L2;        			#[W/m^2] Heat flux
#results

print '%s %d %s' %("\n\n (a) Heat Flux to blade when surface temp is reduced =",q1/1000. ," KW/m^2")
print '%s %.2f %s' %("\n (b) Heat flux to a larger turbine blade = ",q2/1000. ,"KW/m^2");
#END


(a) Heat Flux to blade when surface temp is reduced = 122  KW/m^2

(b) Heat flux to a larger turbine blade =  47.50 KW/m^2


## Example 6.6 Page 379¶

In :
import math
#Operating Conditions
v = 100;        			#[m/s]  Velocity of air
Tsurr = 20+273.;    		#[K] Surrounding Air Temperature
L1 = 1;      				#[m] solid length
Ts = 80+273.;    			#[K] Surface Temp
qx = 10000;        			#[W/m^2] heat flux at a point x
Txy = 60+273.;        		#[K] Temp in boundary layer above the point

#Table A.4 Air Properties at T = 323K
v = 18.2*math.pow(10,-6);   #[m^2/s] Viscosity
k = 28*math.pow(10,-3);    	#[W/m.K] Conductivity
Pr = 0.7;          			#Prandttl Number
#Table A.6 Saturated Water Vapor at T = 323K
pasat = 0.082;     			#[kg/m^3]
Ma = 18;           			#[kg/kmol] Molecular mass of water vapor
#Table A.8 Water Vapor-air at T = 323K
Dab = .26*math.pow(10,-4);	#[m^2/s]
#calculations
#Case 1
Casurr = 0;
Cas = pasat/Ma;        		#[kmol/m^3] Molar conc of saturated water vapor at surface
Caxy = Cas + (Casurr - Cas)*(Txy - Ts)/(Tsurr - Ts);

#Case 2
L2 = 2.;
hm = L1/L2 * Dab/k * qx/(Ts-Tsurr);
Na = hm*(Cas - Casurr);
#results

print '%s %.4f %s' %("\n (a) Water vapor Concentration above the point =",Caxy,"Kmol/m^3 \n")
print '%s %.2e %s' %("(b) Molar flux to a larger surface = ",Na,"Kmol/s.m^2");
#END

 (a) Water vapor Concentration above the point = 0.0030 Kmol/m^3

(b) Molar flux to a larger surface =  3.53e-04 Kmol/s.m^2


## Example 6.7 Page 383¶

In :
import math
#Operating Conditions
Tsurr = 40+273.;    		#[K] Surrounding Air Temperature
#Volatile Wetting Agent A
hfg = 100;        			#[kJ/kg]
Ma = 200;         			#[kg/kmol] Molecular mass
pasat = 5000;     			#[N/m^2] Saturate pressure
Dab = .2*math.pow(10,-4);   #[m^2/s] Diffusion coefficient

#Table A.4 Air Properties at T = 300K
p = 1.16;                	#[kg/m^3] Density
cp = 1.007;              	#[kJ/kg.K] Specific Heat
alpha = 22.5*math.pow(10,-6)#[m^2/s]
R = 8.314;               	#[kJ/kmol] Universal Gas Constt
#calculations
#Applying Eqn 6.65 and setting pasurr = 0
# Ts^2 - Tsurr*Ts + B = 0     , where the coefficient B is
B = Ma*hfg*pasat*math.pow(10,-3) /(R*p*cp*math.pow((alpha/Dab),(2./3.)));
Ts = (Tsurr + math.sqrt(Tsurr*Tsurr - 4*B))/2. ;
#results
print '%s %.1f %s' %("\n Steady State Surface Temperature of Beverage =",Ts-273.,"degC");
#END

 Steady State Surface Temperature of Beverage = 5.9 degC