Chapter 10: Heat Transfer With Phase Change¶

Ex10.1: Page 643¶

In :
print "Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 10, Example 1"
#Surface temperature of polished stainless steel surface in degree celcius
T_s=106.0;
#Boiling point of water under at atmospheric pressure in degree celcius
T_b=100.0;
#Value of empirical constant
C_sf=0.0132;
#latent heat of vaporization in J/kg
h_fg=2.25e6;
#gravitational acceleration in m/s**2
g=9.81;
#Value of proportionality factor in British Gravitational system
g_c=1;
#density of saturated liquid in kg/m**3
rho_l=962.0;
#density of saturated vapor in kg/m**3
rho_v=0.60;
#specific heat of saturated liquid in J/kg K
c_l=4211.0;
#prandtl number of saturated liquid
Pr_l=1.75;
#surface tension of the liquid-to-vapor interface in N/m
sigma=58.8e-3;
# vismath.cosity of the liquid in kg/ms
mu_l=2.77e-4;
#Excess temperature in degree Celcius
delta_Tx= T_s-T_b;

print "Heat flux from the surface to the water in W/m**2"
#Heat flux in W./m2
q=(c_l*delta_Tx/(C_sf*h_fg*Pr_l))**3*mu_l*h_fg*math.sqrt((g*(rho_l-rho_v))/(g_c*sigma))
print round(q,1)

print "Critical heat flux in W/m**2"
#Heat flux in W./m2
q_maxZ=(math.pi/24.0)*math.sqrt(rho_v)*h_fg*(sigma*g*(rho_l-rho_v)*g_c)**0.25

print "At 6°C excess temperature the heat flux is less than the critical value; therefore nucleate pool boiling exists"
print "For the Teflon-coated stainless steel surface, heat flux in W/m**2"
#Heat flux in W./m2
q=29669*(C_sf/0.0058)**3
print round(q,2)
print "Thus for Teflon-coated stainless steel surface there is a remarkable increase in heat flux; however, it is still below the critical value."

# the answers in the textbook is slightly different in the textbook due to approximation
Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 10, Example 1
Heat flux from the surface to the water in W/m**2
28673.9
Critical heat flux in W/m**2
At 6°C excess temperature the heat flux is less than the critical value; therefore nucleate pool boiling exists
For the Teflon-coated stainless steel surface, heat flux in W/m**2
349736.31
Thus for Teflon-coated stainless steel surface there is a remarkable increase in heat flux; however, it is still below the critical value.

Ex10.2: Page 646¶

In :
print "Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 10, Example 2"
#density of saturated liquid in kg/m**3
rho_l=962;
#gravitational acceleration in m/s**2
g=9.8;
#latent heat of vaporization in J/kg
h_fg=2250000;
#density of saturated vapor in kg/m**3
rho_v=0.60;
#Surface temperature of polished stainless steel surface in degree celcius
T_s=400;
#Value of proportionality factor in British Gravitational system
g_c=1;
#Boiling point of water under at atmospheric pressure in degree celcius
T_b=100;
#surface tension of the liquid-to-vapor interface in N/m
sigma=58.8e-3;
#Excess temperature in degree Celcius
delta_Tx= T_s-T_b;
#Wavelength in m from eq. 10.7
lamda=2*math.pi*math.sqrt(g_c*sigma/(g*(rho_l-rho_v)));
#Thermal conductivity in W/mK
k_c=0.0249;
#Absolute vismath.cosity in Ns/m**2
mu_c=12.1e-6;
#Specific heat in J/kg K
c_pc=2034;
#Heat transfer coefficient due to conduction alone in W/m**2 K
h_c=(0.59)*(((g*(rho_l-rho_v)*rho_v*(k_c**3)*(h_fg+(0.68*c_pc*delta_Tx)))/(lamda*mu_c*delta_Tx))**0.25); # math.expression obtained assuming diameter D tending to infinity
#Emissivity
epsilon_s= 0.05; #math.since surface is polished and hence heat transfer coefficient due to radiation is negligible
print "Heat flux in W/m**2"
#Heat flux in W/m**2
q= h_c*delta_Tx
print round(q,1)
# the answers in the textbook is slightly different in the textbook due to approximation
Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 10, Example 2
Heat flux in W/m**2
44739.1

Ex10.3: Page 655¶

In :
print "Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 10, Example 3"
#Flow rate of n-butyl alcohol in kg/hr
m=161;
#Internal diameter of copper tube in meters
D=0.01;
#Tube wall temperature in degree C
T=140;
#surface tension in N/m
sigma=0.0183;
#Heat of vaporization in J/kg
h_fg=591500;
#atmospheric pressure boiling point in degree C
T_sat=117.5;
# saturation pressure corresponding to a saturation temperature of 140°C in atm
P_sat=2;
#Density of vapor in kg/m**3
rho_v=2.3;
#Vismath.cosity of vapor in kg/m s
mu_v=.0143e-3;
#Property values for n-butyl alcohol are taken from Appendix 2, Table 19
#Density in kg/m**3
rho_l=737;
#Absolute vismath.cosity in Ns/m**2
mu_l=0.39e-3;
#Specific heat in J/kg K
c_l=3429.0;
#Prandtl number
Pr_l=8.2;
#Thermal conductivity in W/m K
k_l=0.13;
#Empirical constant
C_sf=0.00305;# Value taken from table 10.1
#Mass velocity in kg/m**2 s
G=(m/3600.0)*(4/(math.pi*0.01**2));
print "Mass velocity in kg/m**2 is ",round(G,2)
#Reynolds number for liquid flow
Re_D=(G*D)/mu_l;
print "Reynolds number for liquid flow is",round(Re_D,2)
#The contribution to the heat transfer coefficient due to the two-phase annular flow is [(0.023)*(14590)**0.8*(8.2)**0.4*16.3*(1-x)**0.8*F]
#Since the vapor pressure changes by 1 atm over the temperature range from saturation temperature to 140°C,so saturation pressure in N/m**2
delta_p_sat=101300;
#Therefore the contribution to the heat transfer coefficient from nucleate boiling is
#h_b= 0.00122*[(0.163**0.79*3429**0.45*737**0.49*1**0.25)/(0.0183**0.5*0.39e-3**0.29*591300**0.24*2.3**0.24)]*(140-117.5)**0.24*(101300)**0.75*S
#or h_b= 8393S
#Now 1/Xtt will be calculated by
#1/Xtt=12.86*(x/(1-x))**0.9
#Now a table is prepared showing stepwise calculations that track the increase in quality, from x=0 to x=0.5,assuming that the steps delta x are small enough that the heat flux and other parameters are reasonably constant in that step
print "The tube length required to reach 50% quality is 1.35 m"

# as the answer is found by hit and trial thus answer is printed through table
Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 10, Example 3
Mass velocity in kg/m**2 is  569.42
Reynolds number for liquid flow is 14600.54
The tube length required to reach 50% quality is 1.35 m

Ex10.4: Page 666¶

In :
print "Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 10, Example 4"
#Outer diameter of the tube in meters
D=0.013;
#Acceleration due to gravity in m/s**2
g=9.81;
#Length of the tube in meters
L=1.5;
#Temperature of saturated vapour in Kelvin
T_sv=349.0;
#Average tube wall temperature in Kelvin
T_s=325;
#Average temperature of the condensate film in degree K
Tf=(T_sv+T_s)/2.0;
#Thermal conductivity of liquid in W/m-K
k_l=0.661;
#Vismath.cosity of liquid in N s/m**2
mu_l=4.48e-4;
#Dendity of liquid in kg/m**3
rho_l=980.9;
#Specific heat of liquid in J/kg K
c_pl=4184.0;
#Latent heat of condensation in J/kg
h_fg=2.349e6;
#Density of vapor in kg/m**3
rho_v=0.25;
#Modified latent heat of condensation in J/kg
h_fg_dash=h_fg+(3/8.0)*c_pl*(T_sv-T_s);

print "Heat transfer coefficient for tube in horizontal position in W/m**2 K"
#Heat transfer coefficient in W/m2K
h_c_bar=0.725*(((rho_l*(rho_l-rho_v)*g*h_fg_dash*k_l**3)/(D*mu_l*(T_sv-T_s)))**0.25)
print round(h_c_bar,2)
print "Heat transfer coefficient for tube in vertical position in W/m**2 K"
##Heat transfer coefficient in W/m2K
h_c_bar=0.943*(((rho_l*(rho_l-rho_v)*g*h_fg_dash*k_l**3)/(mu_l*(T_sv-T_s)))**0.25)
print round(h_c_bar,2)

# the answer is incorrect in the textbook
Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 10, Example 4
Heat transfer coefficient for tube in horizontal position in W/m**2 K
10648.3
Heat transfer coefficient for tube in vertical position in W/m**2 K
4676.7

Ex10.5: Page 667¶

In :
print "Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 10, Example 5"
#Acceleration due to gravity in m/s**2
g=9.81;
#Length of the tube in meters
L=1.5;
#Temperature of saturated vapour in Kelvin
T_sv=349.0;
#Average tube wall temperature in Kelvin
T_s=325.0;
#Average temperature of the condensate film in Kelvin
Tf=(T_sv+T_s)/2;
#Thermal conductivity of liquid in W/m-K
k_l=0.661;
#Vismath.cosity of liquid in N s/m**2
mu_l=4.48e-4;
#Dendity of liquid in kg/m**3
rho_l=980.9;
#Specific heat of liquid in J/kg K
c_pl=4184.0;
#Latent heat of condensation in J/kg
h_fg=2.349e6;
#Density of vapor in kg/m**3
rho_v=0.25;
#Modified latent heat of condensation in J/kg
h_fg_dash=h_fg+(3/8.0)*c_pl*(T_sv-T_s);

print "Reynolds number at the lower edge"
#Reynolds number
Re=(4/3.0)*(((4*k_l*L*(T_sv-T_s)*rho_l**(2/3.0)*g**(1/3.0))/(mu_l**(5/3.0)*h_fg_dash))**0.75)
print round(Re,2)
print "Since the Reynolds number at the lower edge of the tube is below 2000, the flow of the condensate is laminar"

# the answers in the textbook is slightly different in the textbook due to approximation
Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 10, Example 5
Reynolds number at the lower edge
569.05
Since the Reynolds number at the lower edge of the tube is below 2000, the flow of the condensate is laminar

Ex10.6: Page 682¶

In :
print "Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 10, Example 6"
#Length of Heat pipe in meters
L_eff=0.30;
#Temperature of the heat pipe in degree celcius
T=100.0;
#Diameter of the heat pipe in meters
D=1e-2;
#Density of water at 100 degree celcius in k/m**3
rho=958.0;
#Vismath.cosity of water in N s/m**2
mu=279.0e-6;
#surface tension of the liquid-to-vapor interface in N/m
sigma=58.9e-3;
#latent heat of vaporization in J/kg
h_fg=2.26e6;
#Inclination angle in degree
theta=30;
#Acceleration due to gravity in meter/sec**2
g=9.81;
#Wire diameter for wick in metres
d=0.0045e-2;
#So thickness of four layers of wire mesh
t=4.0*d;
#Area of the wick in m**2
Aw=math.pi*D*t;
#For phosphorus-bronze,heat pipe wick pore size in meters
r=0.002e-2;
#For phosphorus-bronze,heat pipe wick permeability in m**2
K=0.3e-10;
print "Maximum liquid flow rate in kg/sec"
#flow rate in kg/sec
m_max=((2*sigma/r)-rho*g*L_eff*0.5)*((rho*Aw*K)/(mu*L_eff))
print round(m_max,6)
print "Maximum heat transport capability in Watt"
#heat transport capability in W
q_max=m_max*h_fg
print round(q_max,1)
Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 10, Example 6
Maximum liquid flow rate in kg/sec
9e-06
Maximum heat transport capability in Watt
19.7

Ex10.7: Page 686¶

In :
print "Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 10, Example 7"
#Temperature of the brine spray used for internal refrigeration in degree celcius
T_inf=-11.0;
#Required thickness of ice layer in meters
epsilon= 0.0025;
#Water-liquid temperature in degree celcius
T1=4.4;
#Liquid-surface conductance in W/m**2 K
h_epsilon=57.0;
#Conductance between brine and ice(including metal wall) in W/m**2 K
h_not=570.0;
#Latent heat of fusion for ice in J/Kg
Lf=333700.0;
#Density for ice in Kg/m**3
rho=918.0;
#Thermal conductivity for ice in W/m K
k=2.32;
#Freezing point temperature in degree K
Tfr=0;
#Dimensionless R, T, epsilon and t are as follows
#R plus parameter
R_plus= h_epsilon/h_not;
#T plus parameter
T_plus= (T1-Tfr)/(Tfr-T_inf);
#Epsilon plus parameter
Epsilon_plus= h_not*epsilon/k;
#t plus parameter
t_plus=(Epsilon_plus/(R_plus*T_plus))-((1/(R_plus*T_plus)**2)*math.log(1+(R_plus*T_plus*Epsilon_plus/(1+R_plus*T_plus))))

print "Time taken for 0.25cm thick ice layer deposition in sec"
#time in seconds
t=t_plus*rho*Lf*k/((h_not)**2*(Tfr-T_inf))
print round(t,1)
# the answers in the textbook is incorrect
Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 10, Example 7
Time taken for 0.25cm thick ice layer deposition in sec
151.6