Chapter3: Convection¶

#Boundary layer thickness

#Variable declaration
mu=10**-3        #N.s/m**2
#At distance y from surface
#ux=a+by+cy**2+dy**3
#At y=0,ux=0 therefore a=0
#i.e tao=0
#At edge of boundary layer,ie y=del
#ux=u_inf
#At y=o,c=0
#At y=del,ux=b*del+d*del**3

#Therefore, b=-3*d*del**3
#d=-u_inf/(2*del**2)
#b=3*u_inf/(2*del)

#For velocity profile,we have:
#del/x=4.64*(Nre_x)**(-1/2)

#Evaluate N re_x

x=75.0         #[mm]
x=x/1000         #[m]
u_inf=3.0             #[m/s]
rho=1000.0            #[kg/m**3] for air

#Calculation
Nre_x=u_inf*rho*x/mu        #Reynold number
#Substituting the value,we get
dell=x*4.64*(Nre_x**(-1.0/2.0))       #[m]

#Result
print"Boundary layer thickness is del=",round(dell,6),"m or ",round(dell*1000,3),"mm"
print"Wrong units in answer of book,m and mm are wrongly interchanged"

#Boundary layer thickness of plate

#Variable declaration
mu=15*10**-6    #sq m /s
v=2    #m/s
L=2    #[m] length of plate
Nre_x=3*10**5

#Calculation

xc=Nre_x*mu/v    #critical length at whihc the transition takes place
#Since xc is less than 2 m.Therefore the flow is laminar
#at any distance x,.it is calculated from
#del/x=4.64/(math.sqrt(NRe,x))
#At x=L=2 m
Nre_l=v*L/mu
del_l=4.64*L/math.sqrt(Nre_l)
del_l=del_l*1000    #[mm]

#Result
print"Boundary layer thickness at the trailing edge is",round(del_l),"mm"

#Thickness of hydrodynamic boundary layer

#Variable declaration
mu=15*10**-6    #Kinematic viscosity in [sq m /s]
x=0.4    #[m]
u_inf=3    #[m/s]


#Calculation
#At x=0.4 m,
Nre_x=u_inf*x/mu     
import math
dell=4.64*x/math.sqrt(Nre_x)    #[m]
dell=dell*1000    #[mm]
Cf_x=0.664/math.sqrt(Nre_x) 

#Result
print"Since Nre,x(",Nre_x,")is Less than 3*10**5,..the boundary layer is laminar"
print"Thickness of boundary layer at x=",x,"m",round(dell,2),"mm" 
print"Local skin friction coefficient is :",round(Cf_x,4)

#Flat plate boundary layer

import math
from scipy import integrate

#Variable declaration
mu=1.85*10**-5           #[kg/(m.s)]
P=101.325                #Pressure in [kPa]
M_avg=29.0                 #Avg molecular wt of air
R=8.31451                #Gas constant
T=300.0                    #[K]

#Calculation
rho=P*M_avg/(R*T)   #[kg/m**3]
u_inf=2.0             #Viscosity in [m/s]
#At x=20 cm =0.2 m
x=0.2                    #[m]
Nre_x=rho*u_inf*x/mu     #[Reynolds number]
del_by_x=4.64/math.sqrt(Nre_x)   #[Boundary layer]
dell=del_by_x*x              #[m]
#del=del*1000                #[mm]

#At
x=0.4           #[m]
Nre_x=(rho*u_inf*x)/mu      #<3*10**5
#Boundary layer is laminar
del_by_x=4.64/math.sqrt(Nre_x)   
del1=del_by_x*x              #[m]
#del1=del1*1000                #[mm]
d=del1-dell                      #Del 
def f(y):
    m_dot=u_inf*(1.5*(y/d)-0.5*(y/d)**3)*rho
    return(m_dot)
m_dot=integrate.quad(f,0,d)


#Result
print"Boundary layer thickness at distance 20 cm from leading edge is",round(dell,4),"m=",round(dell*1000,1),"mm"
print"Boundary layer thickness at distance 40 cm from leading edge is",round(del1,4),"m=",round(del1*1000,1),"mm"
print"Thus,Mass flow rate entering the boundary layer is",round(m_dot[0],6),"kg/s"

#Rate of heat removed from plate

#Variable declaration
mu=3.9*10**-4    #Kinematic viscosity in sq m/s
k=36.4*10**-3    #Thermal conductivity in W/(m.K)
Npr=0.69
u_inf=8    #[m/s]
L=1    #Lenght of plate in [m]
import math

#Calculation
Nre_l=u_inf*L/mu 
#Since Nre_l is less than 3*10**5 ,the flow is laminar over the entire length of plate
Nnu=0.664*math.sqrt(Nre_l)*Npr**(1.0/3.0)    #=hL/k
h=k*Nnu/L    #w/sq m.K
h=3.06      #Approximation   [W/sq m.K]
T_inf=523    #[K]
Tw=351    #[K]
W=0.3    #Width of plate [m]
A=W*L    #Area in [sq m]
Q=h*A*(T_inf-Tw)    # Rate of heat removal from one side in [W]
#from two side:
Q2=2*Q    #[W]

#Result
print"Rate of heat removal is",round(Q,1),"W"
print round(Q2,1)," W heat should be removed continously from the plate"

#Heat removed from plate

#Variable declaration
P1=101.325    #Pressure in [kPa]
mu1=30.8*10**-6    #Kinematic viscosity in[sq m /s]
k=36.4*10**-3    #[W/(m.K)]
Npr=0.69
u_inf=8    #Velocity in [m/s]
Cp=1.08    #kJ/(kg.K)
L=1.5    #Length of plate in [m]
W=0.3    #Width in [m]
A=L*W    #Area in [sq m]

#Calculation
import math
#At constant temperature: mu1/mu2=P2/P1
P2=8     #[kPa]
mu2=mu1*P1/P2    #Kinematic viscosity at P2 in [sq m/s]
Nre_l=u_inf*L/mu2    #Reynold's no.
#Since this is less than 3*10**5
Nnu=0.664*math.sqrt(Nre_l)*(Npr**(1.0/3.0))
h=Nnu*k/L  # Heat transfer coeffficient in [W/sq m.K]
h=round(h,1)

T_inf=523    #[K]
Tw=353    #[K]
Q=2*h*A*(T_inf-Tw)    #Heat removed from both sides in [W]

#Result
print"Rate of heat removed from both sides of plate is",Q,"W"

#Local heat transfer coefficient

#Variable declaration
rho=0.998    #kg/cubic m
v=20.76*10**-6    #[sq m/s]
Cp=1.009    #[kJ/kg.K]
k=0.03    #[W/m.K]
u_inf=3    #[m/s]
x=0.4    #[m]
w=1.5    #[m]

#Calculation
Nre_x=u_inf*x/v    #Reynolds no at x=0.4 m
#Since this is less than 3*10**5.The flow is laminar upto x=0.4 m
mu=rho*v  #[kg/(m.s)]
import math
Cp=1.009    #[kJ/kg.K]
Cp=Cp*1000    #[J/kg.K]
k=0.03    #W/(m.K)
Npr=Cp*mu/k
Nnu_x=0.332*(math.sqrt(Nre_x))*(Npr**(1.0/3.0))
hx=Nnu_x*k/x    #[W/(m.K)]
#Average value is twice this value
h=round(2*hx,1)    #[W/(m.K)]
A=x*w    #Area in [sq m]
Tw=407    #[k]
T_inf=293    #[K]
Q=h*A*(Tw-T_inf)     #[W]
#From both sides of the plate:
Q=2*Q    #[W]
#Result
print"The heat transferred from both sides of the plate is",round(Q),"W"

#Width of plate
import math

#Variable declaration
rho=0.998    #[kg/cubic m]
v=20.76*10**-6    #[sq m/s]
k=0.03    #[W/m.K]
Npr=0.697
x=0.4    #[m] from leading edge of the plate
u_inf=3    #[m/s]

#Calculation
Nre_x=u_inf*x/v    #Reynold numebr at x=0.40 m
#Since this is less than 3*10**5    
#therefore flow is laminar and 
Nnu_x=0.332*math.sqrt(Nre_x)*(Npr**(1.0/3.0)) 
hx=Nnu_x*k/x    #[W/sq m.K]
#Average heat tarnsfer coefficient is twice this value
h=2*hx    #[W/sq m.K]
#Given:
Q=1450    #[W]
Tw=407    #[K]
T_inf=293    #[K]
L=0.4    #[m]
#Q=h*w*L*(Tw-T_inf)
#L=Q/(h*w*(Tw-T_inf))
w=Q/(h*L*(Tw-T_inf))    #[m]

#Result
print"Width of plate is",round(w),"m"

#Heat transferred in flat plate

#Variable declaration
v=17.36*10**-6    #Viscosity for air    [sq m./s]
k=0.0275    #for air ..[W/(m.K)]
Cp=1.006    #[kJ/(kg.K)]
Npr=0.7    #for air
u_inf=2    #[m/s]
x=0.2    #[m]

#Calculation
Nre_x=u_inf*x/v    #Reynolds number at x=0.2    m
#Since this is less than 3*10**5
import math
Nnu_x=0.332*math.sqrt(Nre_x)*(Npr**(1.0/3.0))
hx=Nnu_x*k/x    #[W/(sq m.K]
#Average value of heat transfer coeff is twice this value
h=round(2*hx,1)    #[W/sq m.K)]
w=1    #width in [m]
A=x*w    #[sq m] Area of plate
Tw=333    #[K]
T_inf=300    #[K]
Q=h*A*(Tw-T_inf)    #Heat flow in [W]

#Result
print"Heat flow is :",Q,"W"
#From both sides of plate:
Q=2*Q    #[W]
print"Heat flow from both sides of plate is",Q,"W"
 

#Rate of heat transferred in turbulent flow

#Variable declaration
v=16.96*10**-6    #[sq m./s]
rho=1.128    #[kg/cubic m]
Npr=0.699    #Prandtl number
k=0.0276    #[W/m.K]
u_inf=15    #[m/s]
L=0.2    #[m]

#Calculation
Nre_l=L*u_inf/v    #Reynold's number
import math

#Since this is less than 3*10**5,the boundary layer is laminar over entire length
Nnu=0.664*math.sqrt(Nre_l)*(Npr**(1.0/3.0))
h=Nnu*k/L    #[W/sq m.K]
A=L**2    #Area in [sq m]
Tw=293    #[K]
T_inf=333    #[K]
#Rate of heat transfer from BOTH sides is:
Q=2*h*A*(T_inf-Tw)    #[W]
print"Rate of heat transfer from both sides of plate is",round(Q,1),"W\n"
#ii-With turbulent boundary layer from the leading edge:
h=k*0.0366*(Nre_l**(0.8))*(Npr**(1.0/3.0))/L       #[W/(sq m.K)]
#Heat transfer from both sides is :
Q=2*h*A*(T_inf-Tw)            #[W]
print "With turbulent boundary layer,\nRate of heat transfer from both sides of the plate=",round(Q,1)
print"\nThese calculations show that the that transfer rate is approximately doubled if boundary layer is turbulent from the leading edge \n"

#Heat transfer from plate in unit direction

#Variable declaration
mu=1.906*10**-5    #[kg/(m.s)]
k=0.02723    #W/m.K
Cp=1.007    #[kJ/(kg.K)]
rho=1.129    #[kg/cubic m]
Npr=0.70
Mavg=29
u_inf=35    #[m/s]
L=0.75    #[m]
Tm=313    #[K]
P=101.325    #[kPa]

#Calculation
Nre_l=rho*u_inf*L/mu    #Reynold's number >5*10**5
Nnu=0.0366*Nre_l**(0.8)*Npr**(1.0/3.0) 
h=Nnu*k/L    #[W/s m.K]
A=1*L    #[sq m]
Tw=333    #[K]
T_inf=293    #[K]
Q=h*A*(Tw-T_inf)     #[W]

#Result
print"Heat transfer from the plate is",round(Q,1),"W(approx)"

#Heat lost by sphere
import math

#Variable declaration
v=18.23*10**-6    #sq m/s
k=0.02814    #[W/m.K]
D=0.012    #[m]
r=0.006    #[m]
u_inf=4    #[m/s]

#Calculation
Nre=D*u_inf/v    #Reynold's number
Nnu=0.37*Nre**(0.6) 
h=Nnu*(k/D)    
A=4*math.pi*r**2    #Area of sphere in [sq m]
Tw=350    #[K]
T_inf=300    #[K]
Q=h*A*(Tw-T_inf)    #Heat lost by sphere in [W]

#Result
print"Heat lost by sphere is",round(Q,2),"W"

#Heat lost by sphere

#Variable declaration
v=15.69*10**-6    #[sq m./s]
k=0.02624    #[W/m.K]
Npr=0.708    #Prandtl number
mu=2.075*10**-5    #kg/m.s
u_inf=4    #[m/s]
mu_inf=1.8462*10**-5    #[m/s] velocity
Tw=350    #[K]
T_inf=300    #[K]
D=0.012    #[m]
r=D/2    #Radius in [m]

#Calculation
Nre=u_inf*D/v    #Reynold's numbe
Nnu=2+(0.4*Nre**(1.0/2.0)+0.06*Nre**(2.0/3.0))*Npr**(0.4)*(mu_inf/mu)**(1.0/4.0)
h=Nnu*k/D    #[W/sq m.K]
import math
A=4*math.pi*r**2    #Area in [sq m]
Q=h*A*(Tw-T_inf) 

#Result
print"\n Heat lost by the sphere is",round(Q,3),"W"

#Percent power lost in bulb
#Variable declaration
v=2.08*10**-5    #[sq m/s]
k=0.03    #W/(m.K)
Npr=0.697    #Prandtl number
D=0.06    #[m]
u_inf=0.3    #[m/s]

#Calculation
Nre=D*u_inf/v    #Reynolds number
#Average nusselt number is given by:
Nnu=0.37*(Nre**0.6) 
h=Nnu*k/D    #W/sq m.K
Tw=400    #[K]
T_inf=300    #[K]
D=0.06    #[m]
r=0.03    #[m]
import math
A=4*math.pi*r**2    #Area in [sq m]
Q=h*A*(Tw-T_inf)    #[W]
per=Q*100/100    #Percent of heat lost by forced convection

#Result
print"Heat transfer rate is",round(Q,2),"W,And percentage of power lost by convection is:",round(per,2),"%" 
 

 
#Heat lost by cylinder  
u_inf=50    #velocity in [m/s]
mu=2.14*10**-5    #[kg/(m.s)]
rho=0.966    #[kg/cubic m]
k=0.0312    #[W/(m.K)]
Npr=0.695    #Prandtl number
D=0.05    #Diameter in [m]
Nre=D*u_inf*rho/mu    #Reynold's number
Nnu=0.0266*Nre**0.805*Npr**(1.0/3.0) 
h=round(Nnu*k/D,1)     #W/sq m.K
Tw=423    #[K]
T_inf=308    #[K]
import math
#Heat loss per unit length is :
Q_by_l=h*math.pi*D*(Tw-T_inf)   #[W]

#Result
print"Heat lost per unit length of cylinder is",round(Q_by_l),"W" 

#Heat transfer in tube

#Variable declaration
v=20.92*10**-6    #sq m/s
k=3.0*10.0**-2    #W/(m.K)
Npr=0.7
u_inf=25.0    #[m/s]
d=50.0    #[mm]
d=d/1000    #[m]
Nre=u_inf*d/v    #Reynold's number
Tw=397.0    #[K]
T_inf=303.0    #[K]


#Calculation
import math
#Case 1: Circular tube
Nnu=0.0266*Nre**(0.805)*Npr**(1.0/3.0) 
h=Nnu*k/d    #[W/sq m.K]
A=math.pi*d    #Area in [sq m]
Q=h*A*(Tw-T_inf)    #[W]
Q_by_l1=h*math.pi*d*(Tw-T_inf)    #[W/m]

#Case 2:Square tube
A=50.0*50.0    #Area in [sq mm]
P=2.0*(50.0+50.0)    #Perimeter [mm]
l=4.0*A/P       #[mm]
l=l/1000        #[m]
Nnu=0.102*(Nre**0.675)*(Npr**(1.0/3.0))
h=Nnu*k/d    #W/(sq m.K)
A=4*l*l        #[sq m]

Q=h*A*(Tw-T_inf)
Q_by_l2=Q/l    #[W/m]


#Result

print"Rate of heat flow from the circular pipe is",round(Q_by_l1,1),"W/m" 
print"Rate of heat flow from the square pipe=",round(Q_by_l2,1),"W/m"
print"Hence rate of heat flow from square pipe is more than that from circular pipe"

#Heat transfer coefficient

#Variable declaration
mu=0.8    #Viscosity of flowing fluid [N.s/sq m]
rho=1.1    #Density of flowinf fluid [g/cubic cm]
rho=rho*1000    #Density in [kg/cubic m]
Cp=1.26    #Specific heat [kJ/kg.K]
Cp=Cp*10**3    # in[J/(kg.K)]
k=0.384    #[W/(m.K)]
mu_w=1.0    #Viscosity at wall temperature [N.s/sq m]
L=5.0    #[m]
vfr=300.0    #Volumetric flow rate in [cubic cm/s]
vfr=vfr*10.0**-6    #[cubic m/s]
mfr=vfr*rho    #Mass flow rate of flowinf fluid [kg/s]
Di=20.0    #Inside diameter in[mm]
Di=Di/1000    #[m]


#Calculation
import math
Area=(math.pi/4)*Di**2    #Area of cross-section [sq m]
u=vfr/Area    #Veloctiy in [m/s]
Nre=Di*u*rho/mu    #Reynold's number
#As reynold's number is less than 2100,he flow is laminar
Npr=Cp*mu/k    #Prandtl number
Nnu=1.86*(Nre*Npr*Di/L)**(1.0/3.0)*(mu/mu_w)**(0.14)
hi=Nnu*k/Di    #inside heat transfer coefficient [W/sq m.K]

#Result
print"Inside heat transfer coefficient is",round(hi),"W/(sq m.K)"
#Note:
print"NOTE:The answer given in book..ie 1225 is wrong.please redo the calculation of last line manually to check\n"

#Heat transfer coefficient in heated tube

#Variable declaration
m=5500.0    #Mass flow rate in [kg/h]
m=m/3600.0    #[kg/s]
rho=1.07    #Density of fluid in [g/cm**3]
rho=rho*1000    #[kg/m**3]
vfr=m/rho    #Volumetric flow rate in [m**3/s]
Di=40.0    #Diameter of tube [mm]
Di=Di/1000    #[m]
import math

#Calculation
A=(math.pi/4)*Di**2    #Area of cross-section in [sq m]
u=vfr/A    #Velocity of flowing fluid    [m/s]
rho=1070.0    #Density in [kg/m**3]
mu=0.004    #Viscosity in [kg/m.s]
Nre=Di*u*rho/mu
Nre=12198.0       #Approx
#Since this reynold's number is less than 10000,the flow is turbulent
Cp=2.72    #Specific heat in [kJ/kg.K]
Cp=Cp*10**3    #Specific heat in [J/kg.K]
k=0.256    #thermal conductivity in [W/m.K]
Npr=Cp*mu/k    #Prandtl number
Nnu=0.023*(Nre**0.8)*(Npr**0.4)    #Nusselt number
hi=k*Nnu/Di    #Inside heat transfer coefficient in [W/m**2.K]

#Result
print"Inside heat transfer coefficient is ",round(hi,1),"W/sq m.K"

#h of water flowing in tube

#Variable declaration
rho=984.1    #Density of water [kg/m**3]
Cp=4187.0    #Specific heat in [J/kg.K]
mu=485.0*10**-6    #Viscosity at 331 K[Pa.s]
k=0.657    #[W/(m.K)]
mu_w=920.0*10**-6    #Viscosity at 297 K [Pa.s]


#Calculation
D=16.0    #Diameter in [mm]
D=D/1000    #Diameter in [m]
u=3.0    #Velocity in [m/s]
rho=984.1    #[kg/m**3]
Nre=D*u*rho/mu    #Reynolds number
Nre=round(Nre)
Npr=Cp*mu/k    #Prandtl number

#Dittus-Boelter equation (i)
Nnu=0.023*(Nre**0.8)*(Npr**0.3)    #nusselt number
h=k*Nnu/D    #Heat transfer coefficient [W/m**2.K]

#Result
print"ANSWER-(i) \nBy Dittus-Boelter equation we get h=",round(h,1),"W/sq m.K"
#sieder-tate equation (ii)
Nnu=0.023*(Nre**0.8)*(Npr**(1.0/3.0))*((mu/mu_w)**0.14)    #Nusselt number
h=k*Nnu/D    #Heat transfer coefficient in [W/sq m.K]
print"Answer-(ii)\nBy Sieder-Tate equation we get h=",round(h,2),"W/sq m.K"
print"\nNOTE:Calculation mistake in book in part 2 ie sieder tate eqn"

#Overall heat transfer coefficient

#Variable declaration
m_dot=2250    #Mass flow arte in [kg/h]
Cp=3.35    #Specific heat in [kJ/(kg.K)]
dT=316-288.5    #Temperature drop for oil    [K]
Q=Cp*m_dot*dT    #Rate of heat transfer in [kJ/h]
Q=round(Q*1000/3600)    #[J/s] or[W]
Di=0.04    #Inside diameter [m]
Do=0.048    #Outside diamter in [m]
hi=4070    #for steam [W/sq m.K]
ho=18.26    #For oil [W/sq m.K]
Rdo=0.123    #[sq m.K/W]
Rdi=0.215    #[sq m.K/W]

#Calculation
Uo=1.0/(1.0/ho+Do/(hi*Di)+Rdo+Rdi*(Do/Di))    #[W/m**2.K]
Uo=2.3
import math
dT1=373-288.5    #[K]
dT2=373-316    #[K]
dTm=(dT1-dT2)/math.log(dT1/dT2)    #[K]
Ao=Q/(Uo*dTm)    #Heat transfer area in [m**2]

#Result
print"Heat transfer area is:",round(Ao,1),"m**2"

#Number of tubes in exchanger
import math

#Variable declaration
k_tube=111.65    #[W/m.K]
W=4500.0    #[kg/h]
rho=995.7    #[kg/sq m]
Cp=4.174    #[kJ/(kg.K)]
k=0.617    #[W/(m.K)]
v=0.659*10**-6    #Kinematic viscosity [sq m/s]
m_dot=1720.0    #kg/h
T1=293.0    #Initial temperature in [K]
T2=318.0    #Final temperature in [K]


#Calculation
dT=T2-T1    #[K]
Q=m_dot*Cp*dT    #Heat transfer rate in [kJ/h]
Q=Q*1000.0/3600.0    #[J/s] or [W]
Di=0.0225    #[m]
u=1.2    #[m/s]
#Nre=Di*u*rho/mu or
Nre=Di*u/v    #Reynolds number
#As Nre is greater than 10000,Dittus Boelter equation is applicable
Cp=Cp*10**3    #J/(kg.K)
mu=v*rho    #[kg/(m.s)]
Npr=Cp*mu/k    #Prandtl number
#Dittus-Boelter equation for heating is 
Nnu=0.023*(Nre**0.8)*(Npr**0.4)
hi=k*Nnu/Di    #Heat transfer coefficient  [W/(sq m.K)]
Do=0.025    #[m]
Dw=(Do-Di)/math.log(Do/Di)    #math.log mean diameter in [m]
ho=4650.0    #[W/sq m.K]
k=111.65    #[W/m.K]
xw=(Do-Di)/2    #[m]
Uo=1.0/(1.0/ho+Do/(hi*Di)+xw*Do/(k*Dw))    #Overall heat transfer coefficient in W/(m**2.K)
T_steam=373.0    #Temperature of condensing steam in [K]
dT1=T_steam-T1+10    #[K]
dT2=T_steam-T2+10    #[K]
dTm=(dT1-dT2)/math.log(dT1/dT2)    #[K]
Ao=Q/(Uo*dTm)#Area in [m**2]
L=4.0    #length of tube [m]
n=Ao/(math.pi*Do*L)    #number of tubes


#Result
print"No. of tubes required=",round(n) 
print"\nNOTE: there is an error in book in calculation of dT1 and dT2,\n373-293 is written as 90,instead of 80...similarly in dT2,\nSo,in compliance with the book,10 is added to both of them"

#Convective film coefficient
import math

#Variable declaration
m_dot=25000    #massflow rate of water [kg/h]
rho=992.2    #[kg/m**3]
k=0.634    #[W/m.K]
vfr=m_dot/rho    #[m**3/h]
Npr=4.31    #Prandtl numberl
Di=50    #[mm]
Di=0.05    #[m]
dT=10    #[K] as the wall is at a temperature of 10 K above the bulk temperature

#Calculation
u=round((vfr/3600)/(math.pi*(Di/2)**2),2)    #Velocity of water in [m/s]
#Nre=Di*u*rho/mu=Di*u/v    as v=mu/rho
v=0.659*10**-6     #[m**2/s]
Nre=Di*u/v    #Reynolds number
#As it is less than 10000,the flow is in the turbulent region for heat transfer and Dittus Boelter eqn is used
Nnu=0.023*(Nre**0.8)*(Npr**0.4)     #Nusselt number
hi=Nnu*k/Di    #Heat transfer coefficiet in [W/sq m.K]
q_by_l=hi*math.pi*Di*dT   #Heat transfer per unit length[kW/m]

#Result
print"Average value of convective film coefficient is hi=",round(hi)," W/sq m.K"
print"Heat transferred per unit length is Q/L=",round(q_by_l/1000,1),"kW/m"
 

#Length of tube
import math

#Variable declaration
vfr=1200.0     #Water flow rate in [l/h]
rho=0.98     #Density of water in g/[cubic cm]
m_dot=vfr*rho    #Mass flow rate of water [kg/h]
m_dot2=m_dot/3600.0    #[kg/s]
Cp=4.187*10**3     #[J/kg.K]
Di=0.025     #Diameter in [m]
mu=0.0006     #[kg/(m.s)]

#Calculation
Ai=math.pi*((Di/2)**2)    #Area of cross-section in [m**2]
Nre=(Di/mu)*(m_dot2/Ai)    #Reynolds number
k=0.63   #for metal wall in [W/(m.K)]
Npr=Cp*mu/k     #Prandtl number
#Since Nre>10000
#therefore ,Dittus boelter eqn for heating is 
Nnu=0.023*(Nre**(0.8))*(Npr**(0.4))
ho=5800.0     #Film heat coefficientW/(m**2.K)
hi=Nnu*k/Di    #Heat transfer coeffcient in [W/(sq m.K)]
Do=0.028     #[m]
Di=0.025     #[m]
xw=(Do-Di)/2     #[m]
Dw=(Do-Di)/math.log(Do/Di)     #[m]
k=50.0     #for metal wall in [W/(m.K)]
Uo=1.0/(1.0/ho+Do/(hi*Di)+xw*Do/(k*Dw))     #in [W/sq m.K]
dT=343.0-303.0   #[K]
dT1=393.0-303.0     #[K]
dT2=393.0-343.0     #[K]
dTm=(dT1-dT2)/math.log(dT1/dT2)     #[K]
Cp=Cp/1000.0   #[in [kJ/kg.K]]
Q=m_dot*Cp*dT     #Rate of heat transfer in [kJ/h]
Q=Q*1000.0/3600.0     #[J/s] or [W]
Ao=Q/(Uo*dTm)     #Heat transfer area in [sq m]
#Also,..Ao=math.pi*Do*L    ..implies that
L=Ao/(math.pi*Do)    #[m]

#Result
print"Length of tube required is",round(L),"m"

#Cooling coil
#1.For initial conditions:
import math
from scipy.optimize import fsolve

#Variable declaration
T=360            #[K]
T1=280          #[K]
T2=320          #[K]
dT1=T-T1        #[K]
dT2=T-T2        #[K]

#Calculation
#Q1=m1_dot*Cp1*(T2-T1)
Cp1=4.187           #Heat capacity 
dTlm=(dT1-dT2)/math.log(dT1/dT2) #[K]
m1_by_UA=dTlm/(Cp1*(T2-T1)) 
#For final conditions :
#m2_dot=m1_dot
#U2=U1
#A2=5*A1
def f(t):
    x=m1_by_UA*Cp1*(t-T1)-5*((dT1-(T-t))/math.log(dT1/(T-t)))
    return(x)
T=fsolve(f,350.5)

#Result
print"Outlet temperature of water is",T[0],"K"

#Outlet temperature of water
import math


#Variable declaration
mo_dot=60.0   #Mass flow rate of oilin [g/s]
mo_dot=6.0*10**-2  #[kg/s]
Cpo=2.0     #Specific heat of oil in [kJ/(kg.K)]
T1=420.0  #[K]
T2=320.0  #[K]

#Calculation
Q=mo_dot*Cpo*(T1-T2)    #Rate of heat flow in [kJ/s]
mw_dot=mo_dot   #Mass flow rate of water   #kg/s
t1=290.0 #[K]
Cpw=4.18    #[kJ/(kg.K)]
#For finding outlet temperature of water
t2=t1+Q/(mw_dot*Cpw)    #[K]
dT1=T1-t2   #[K]
dT2=T2-t1   #[K]
dTm=(dT1-dT2)/math.log(dT1/dT2)  #[K]
ho=1.6  #Oil side heat transfer coefficient in [kW/(sq m.K)]
hi=3.6  #Water side heat transfer coeff in [kW/(sq m.K)]
#Overall heat transfer coefficient is:
U=1.0/(1.0/ho+1.0/hi) #[kW/(m**2.K)]
A=Q/(U*dTm) #[sq m]
Do=25.0   #[mm]
Do=Do/1000  #[m]
L=A/(math.pi*Do)    #Length of tube in [m]

#Result
print"Outlet temperature of water is:",round(t2),"K" 
print"Area of heat transfer required is",round(A,2),"m^2"
print"Length of tube required is",round(L,2),"m"

#Inside heat transfer coefficient
import math

#Variable declaration
k=0.14  # for oil[W/m.K]
Cp=2.1 # for oil [kJ/kg.K]
Cp=Cp*10**3  #J/kg.K
mu=154  #[mN.s/sq m]
mu_w=87 #(mn.s/sq m)
L=1.5   #[m]
m_dot=0.5   #Mass flow rate of oil[kg/s]
Di=0.019 #Diameter of tube [m]
mean_T=319  #Mean temperature of oil [K]

#Calculation
mu=mu*10**-3 #[N.s/sq m] or [kg/(m.s)]
A=math.pi*(Di/2)**2   #[sq m]
G=m_dot/A   #Mass velocity in [kg/sq m.s]
Nre=Di*G/mu  #Reynolds number
#As Nre<2100,the flow is laminar
mu_w=mu_w*10**-3 #[N.s/sq m] or kg/(m.s)
#The sieder tate equation is 
hi=(k*(2.0*((m_dot*Cp)/(k*L))**(1.0/3.0)*(mu/mu_w)**(0.14)))/Di   #Heat transfer coeff in [W/sq m.K]

#Result
print"The inside heat transfer coefficient is",round(hi,2),"W/(m**2.K) " 
print"NOTE:Calculation mistake in last line.ie in the calculation of hi in book,please perform the calculation manually to check the answer"

#Film heat transfer coefficient
import math

#Variable declaration
m_dot=0.217 #Water flow rate in [kg/s]
Do=19.0   #Outside diameter in [mm]
rho=1000.0    #Density
t=1.6   #Wall thickness in [mm]
Di=Do-2*t   #i.d of tube in [mm]
Di=Di/1000.0  #[m]
Do=Do/1000.0  #[m]


#Calculation
Ai=math.pi*(Di/2)**2 #Cross-sectional area in sq m
u=m_dot/(rho*Ai)  #Water velocity through tube  [m/s]
u=1.12  #approx in book
Di=0.0157   #apprx in book
T1=301.0  #Inlet temperature of water in [K]
T2=315.0  #Outlet temperature of water in [K]
T=(T1+T2)/2 #[K]
hi=(1063.0*(1+0.00293*T)*(u**0.8))/(Di**0.20) #Inside heat transfer coefficient W/(sq m.K)
hi=5084.0     #Approximation
hio=hi*(Di/Do)  #Inside heat transfer coeff based on outside diameter  in W/(sq m.K)

#Result
print"Based on outside temperature,Inside heat transfer coefficient is",round(hio),"W/(m**2.K) or ",round(hio/1000,1),"kW/(m**2.K)"

#Area of exchanger
import math
#Variable declaration
mair_dot=0.90   #[kg/s]
T1=283.0  #[K]
T2=366.0  #[K]
dT=(T1+T2)/2    #[K]
Di=12.0   #[mm]
Di=Di/1000.0  #[m]
G=19.9  #[kg/(sq m.s)]
mu=0.0198   #[mN.s/(sq m)]
mu=mu*10**-3 #[N.s/sq m] or [kg/(m.s)]

#Calculation
Nre=Di*G/mu #Reynolds number
#It is greater than 10**4
k=0.029 #W/(m.K)
Cp=1.0    #[kJ/kg.K]
Cp1=Cp*10**3  #[J/kg.K]
Npr=Cp1*mu/k #Parndtl number
#Dittus-Boelter equation is
hi=0.023*(Nre**0.8)*(Npr**0.4)*k/Di   #[W/sq m.K]
ho=232.0  #W/sq m.K
U=1.0/(1.0/hi+1.0/ho) #Overall heat transfer coefficient  [W/m**2.K]
Q=mair_dot*Cp*(T2-T1)   #kJ/s
Q=Q*10**3    #[J/s] or [W]
T=700.0   #[K]
dT1=T-T2    #[K]
dT2=T2-T1   #[K]
dTm=(dT1-dT2)/math.log(dT1/dT2)  #[K]
#Q=U*A*dTm
A=Q/(U*dTm) #Area in sq m

#Result
print"Heat transfer area of equipment is",round(A,2),"m^2" 

#Natural and forced convection
import math
#Variable declaration
v=18.41*10**-6   #[sq m./s]
k=28.15*10**-3   #[W/m.K]
Npr=0.7 #Prandtl number
Beta=3.077*10**-3    #K**-1
g=9.81  #m/s**2
Tw=350  #[K]
T_inf=300   #[K]
dT=Tw-T_inf #[K]
L=0.3   #[m]


#Calculation
#1.Free Convection
Ngr=(g*Beta*dT*L**3)/(v**2) #Grashof number
Npr=0.7 #Prandtl number
Nnu=0.59*(Ngr*Npr)**(1.0/4.0)    #Nusselt number
h=Nnu*k/L   #Average heat transfer coefficient [W/sq m K]

#2.Forced Convestion
u_inf=4 #[m/s]
Nre_l=u_inf*L/v
Nnu=0.664*(Nre_l**(1.0/2.0))*(Npr**(1.0/3.0))     #Nusselt number
h1=Nnu*k/L   #[W/sq m.K]


#Result
print"In free convection,heat transfer coeff,h=",round(h,1),"W/(m^2.K)"
print"In forced convection,heat transfer coeff,h=",round(h1,2),"W/(m^2.K)"
print"From above it is clear that heat transfer coefficient in forced convection is much larger than that in free convection"

#Natural convection

import math
#Variable declaration
k=0.02685   #W/(m.K)
v=16.5*10**-6  #kg/(m.s)
Npr=0.7 #Prandtl number
Beta=3.25*10**-3 #K**-1
g=9.81  #m/(s**2)
Tw=333  #[k]
T_inf=283   #[K]
dT=Tw-T_inf  #[K]
L=4 #Length/height  of plate [m]

#Calculation
Ngr=(g*Beta*dT*(L**3))/(v**2)   #Grashoff number
#Let const=Ngr*Npr
const=Ngr*Npr
#Sice it is >10**9
Nnu=0.10*(const**(1.0/3.0))   #Nusselt number
h=round(Nnu*k/L,1)   #W/(sq m.K)
W=7 #width in [m]
A=L*W   #Area of heat transfer in [sq m]
Q=h*A*dT    #[W]

#Result
print"Heat transferred is",Q,"W"

#Free convection in vertical pipe
import math

#Variable declaration
v=18.97*10**-6   #m**2/s
k=28.96*10**-3   #W/(m.K)
Npr=0.696
D=100.0   #Outer diameter [mm]
D=D/1000    #[m]
Tf=333.0  #Film temperature in [K]
Tw=373.0  #[K]
T_inf=293.0   #[K]

#Calculation
dT=Tw-T_inf #[K]
Beta=1.0/Tf   #[K**-1]
g=9.81  #[m/s**2]
L=3.0 #Length of pipe [m]
Ngr=(g*Beta*dT*(L**3))/(v**2) #Grashof number
Nra=Ngr*Npr
Nnu=0.10*(Ngr*Npr)**(1.0/3.0)    #nusselt number for vertical cylinder
h=Nnu*k/L   #W/(sq m.K)
Q_by_l=h*math.pi*D*dT   #Heat loss per metre length [W/m]

#Result
print"Hence,Heat loss per metre length is",round(Q_by_l,2),"W/m"

#Heat loss per unit length
import math

#Variable declaration
k=0.630 #W/(m.K
Beta=3.04*10**-4 #K**-1
rho=1000.0    #kg/m**3
mu=8.0*10**-4    #[kg/(m.s)]
Cp=4.187    #kJ/(kg.K)
g=9.81  #[m/(s**2)]
Tw=313.0  #[K]
T_inf=298.0   #[K]
dT=Tw-T_inf #[K]
D=20.0    #[mm]
D=D/1000    #[m]

#Calculation
Ngr=9.81*(rho**2)*Beta*dT*(D**3)/(mu**2)   #Grashoff number
Cp1=Cp*1000.0 #[J/kg.K]
Npr=Cp1*mu/k #Prandtl number
#Average nusselt number is
Nnu=0.53*(Ngr*Npr)**(1.0/4.0)
h=Nnu*k/D  #[W/ sqm.K]
Q_by_l=h*math.pi*D*dT   #Heat loss per unit length [W/m]

#Result
print"Heat loss per unit length of the heater is",round(Q_by_l,1),"W/m(APPROX)"

#Free convection in pipe
import math

#Variable declaration
k=0.03406   #[W/(m/K)]
Beta=2.47*10**-3 #K**-1
Npr=0.687   #Prandtl number
v=26.54*10**-6   #m**2/s
g=9.81  #[m/s**2]
Tw=523.0  #[K]
T_inf=288.0   #[K]
dT=Tw-T_inf #[K]
D=0.3048    #[m]

#Calculation
Ngr=(g*Beta*dT*(D**3))/(v**2)     #Grashof number
Nra=Ngr*Npr 
#For Nra less than 10**9,we have for horizontal cylinder
Nnu=0.53*(Nra**(1.0/4.0))    #Nusselt number
h=Nnu*k/D   #[W/sq m.K]
Q_by_l=h*math.pi*D*dT   #W/m

#Result
print"Heat loss of heat transfer per meter lengh is",round(Q_by_l,1),"W/m"

#Free convection in plate
import math

#Variable declaration
rho=960.63  #Density in [kg/m**3]
Cp=4.216*10**3   #Specific heat in [J/(kg.K)]
D=16.0    #Diameter in [cm]
D=D/100 #[m]
k=0.68  #Thermal conductivity in [W/m.K]

#Calculation
A=(math.pi*(D/2)**2)
L=A/(math.pi*D)   #Length=A/P  in [m]
Beta=0.75*10**-3  #[K**-1]
alpha=1.68*10**-7    #[m**2/s]
g=9.81  #[m/s**2]
Tw=403.0  #[K]
T_inf=343.0   #[K]
dT=Tw-T_inf #[K]
v=0.294*10**-6   #[m**2/s]
Nra=(g*Beta*(L**3)*dT)/(v*alpha) 
#1.For Top surface
Nnu=0.15*(Nra)**(1.0/3.0)   #Nusselt number
ht=Nnu*k/L  #Heat transfer coeff for top surface in W/(m**2.K)
ht=round(ht)
#2.For bottom surface
Nnu=0.27*Nra**(1.0/4.0)  #Nusselt number
hb=Nnu*k/L  #[W/sq m.K]
hb=round(hb)
Q=(ht+hb)*A*dT  #[W]

#Result
print"The rate of heat input is",round(Q,1),"W"

#Heat transfer from disc
import math

#Variable declaration
v=2.0*10**-5   #[m**2/s]
Npr=0.7 #Prandtl number
k=0.03  #[W/m.K]
D=0.25   #Diameter in [m]
L=0.90*D    #Characteristic length,let  [m]
T1=298.0  #[K]
T2=403.0  #[K]
dT=T2-T1    #[K]

#Calculation
Tf=(T1+T2)/2  #[K]
Beta=1.0/Tf   #[K**-1]
A=math.pi*(D/2)**2   #Area in[sq m]
g=9.81  #[m/s**2]

#Case 1: Hot surface facing up
Ngr=g*Beta*dT*(L**3)/(v**2)   #Grashoff number
Nnu=0.15*((Ngr*Npr)**(1.0/3.0))    #Nusselt number
print "Nnu in the book is wrongly calculated as 66.80,\nActually it is:58.22"
h=Nnu*k/L   #[W/sq m.K]
Q=h*A*dT    #[W]

#Case 2:For hot surface facing down
Nnu=0.27*(Ngr*Npr)**(1.0/4.0)    #Grashof Number
h=Nnu*k/L   #[W/sqm.K]
Q1=h*A*dT    #[W]


#Result
print"\nHeat transferred when hot surface is facing up is",round(Q,2),"W"
print"NOTE:Taking into consideration the correct value of Nnu\n"
print"Heat transferred when hot surface is facing down is",round(Q1,2),"W"

#Rate of heat input to plate
import math
#Variable declaration
rho=960 #[kg/m**3]
Beta=0.75*10**-3 #[K**-1]
k=0.68  #[W/m.K]
alpha=1.68*10**-7    #[m**2/s]
v=2.94*10**-7    #[m**2/s]
Cp=4.216    #[kJ/kg.K]
Tw=403  #[K]
T_inf=343   #[K]
dT=Tw-T_inf     #[K]
g=9.81  #[m/s**2]
l=0.8   #[m]
W=0.08  #[m]

#Calculation
A=l*W   #Area in [m**2]
P=2*(0.8+0.08)   #Perimeter in [m]
L=A/P   #Characteristic dimension/length,L in   [m]
Nra=g*Beta*L**3*dT/(v*alpha) 
#(i) for natural convection,heat transfer from top/upper surface heated 
Nnu=0.15*(Nra**(1.0/3.0))    #Nusselt number
ht=Nnu*k/L  #[W/m**2.K]
ht=2115.3   #Approximation in book,If done manually then answer diff
#(ii)For the bottom/lower surface of the heated plate
Nnu=0.27*(Nra**(1.0/4.0))    #Nusselt number
hb=Nnu*k/L  #[W/(m**2.K)]
hb=round(hb)
#Rate of heat input is equal to rate of heat dissipation from the upper and lower surfaces of the plate
Q=(ht+hb)*A*(Tw-T_inf)   #[W]

#Result
print"Rate of heat input is equal to heat dissipation =",round(Q,1),"W"

#Two cases in disc
import math

#Variable declaration
k=0.03  #W/(m.K)
Npr=0.697   #Prandtl number
v=2.076*10**-6   #m**2/s
Beta=0.002915   #K**-1
D=25.0      #[Diameter in cm]
D=D/100 #[m]
Tf=343.0  #Film temperature in [K]

#Calculation
A=math.pi*(D/2)**2   #Area in [m**2]
P=math.pi*D #Perimeter [m]
T1=293.0 #[K]
T2=393.0  #[K]
g=9.81  #[m/s**2]
#Case (i) HOT SURFACE FACING UPWARD
L=A/P   #Characteristic length in [m]
Beta=1.0/Tf   #[K**-1]
dT=T2-T1    #[K]
Ngr=(g*Beta*dT*(L**3))/(v**2) #Grashoff number
Nra=Ngr*Npr 
Nnu=0.15*(Nra**(1.0/3.0))    #Nusselt number
h=Nnu*k/L   #[W/m**2.K]
Q1=h*A*dT    #[W]

#Case-(ii) HOT FACE FACING DOWNWARD
Nnu=0.27*(Nra**(1.0/4.0))    #Nusselt number
h=Nnu*k/L   #W/(m**2.K)
Q2=h*A*dT    #[W]

#Case-(iii)-For disc vertical 
L=0.25  #Characteristic length[m]  
D=L #dia[m]
A=math.pi*((D/2)**2) #[sq m]
Ngr=(g*Beta*dT*(L**3))/(v**2)     #Grashoff number
Npr=0.697
Nra=Ngr*Npr 
Nnu=0.10*(Nra**(1.0/3.0))    #Nusselt number
h=Nnu*k/D   #[W/(m**2.K)]
Q3=h*A*dT    #[W]


#Result
print"Heat transferred when disc is horizontal with hot surface facing upward is",round(Q1,1),"W"
print"Heat transferred when disc is horizontal with hot surface facing downward is",round(Q2,1),"W" 
print"For vertical disc,heat transferred is",round(Q3),"W"

#Total heat loss in a pipe
import math

#Variable declaration
v=23.13*10**-6     #[m**2/s]
k=0.0321     #[W/m.K]
Beta=2.68*10**-3  #[K**-1]
Tw=443.0   #[K]
T_inf=303.0    #[K]
dT=Tw-T_inf     #[K]
g=9.81   #[m/s**2]
Npr=0.688    #Prandtl number
D=100.0    #Diameter [mm]
D=D/1000    #Diameter [m]

#Calculation
Nra=(g*Beta*dT*(D**3)*Npr)/(v**2)
Nnu=0.53*(Nra**(1.0/4.0))    #Nusselt number
h=Nnu*k/D   #[W/(m**2.K)]
h=7.93      #Approximation
e=0.90  #Emissivity
sigma=5.67*10**-8     
#Q=Q_conv+Q_rad  #Total heat loss
#for total heat  loss per meter length
Q_by_l=h*math.pi*D*dT+sigma*e*math.pi*D*(Tw**4-T_inf**4)  #[W/m]

#Result
print"Total heat loss per metre length of pipe is",round(Q_by_l,1),"W/m"

#Heat loss by free convection
import math

#Result
k=0.035  #[W/(m.K)]
Npr=0.684    #Prandtl number
Beta=2.42*10**-3  #[K**-1]
v=27.8*10**-6     #[m**2/s]
Tw=533.0   #[K]
T_inf=363.0       #[K]
dT=Tw-T_inf #[K]
D=0.01   #[m]
g=9.81  #[m/s**2]

#Calculation
Nra=(g*Beta*dT*(D**3))/(v**2)
#For this <10**5,we have for sphere
A=4*math.pi*(D/2)**2 #Area of sphere in [m**2]
Nnu=(2+0.43*Nra**(1.0/4.0))#Nusslet number
h=Nnu*k/D   #W/(m**2.K)
Q=h*A*dT    #[W]

#Result
print"Rate of heat loss is",round(Q,2),"W"

#Heat loss from cube
import math
#Variable declaration
v=17.95*10**-6   #[m**2/s]
dT=353.0-293.0  #[K]
k=0.0283    #[W/m.K]
g=9.81  #[m/s**2]
Npr=0.698   #Prandtl number
Cp=1005.0 #J/(kg.K)
Tf=323.0  #Film temperature in [K]
Beta=1.0/Tf   #[K**-1]
l=1.0 #[m]

#Calculation
Nra=(g*Beta*dT*(l**3)*Npr)/(v**2)
#In textbook result of above statement is wrongly calculated,So
Nra=3.95*10**8
#For Nra <10**9,for a vertical plate,the average nusselt number is
Nnu=0.59*Nra**(1.0/4.0)  #Nusselt number
h=round(Nnu*k/l,2)   #[W/m**2.K]
A=l**2   #Area [m**2]
#Heat loss form 4 vertical faces of 1m*1m is 
Q1=4.0*(h*A*dT)   #[W]
#For top surface 
P=4.0*l   #Perimeter in [m]
L=A/P   #[m]
Nra=(Npr*g*Beta*dT*(L**3))/(v**2)
Nnu=0.15*Nra**(1.0/3.0)  #Nusselt number
h=round(Nnu*k/L,1)   #[W/m**2.K]
Q2=h*A*dT   #[W]
Q_total=Q1+Q2   #Total heat loss[W]

#Result
print"Total heat loss is",Q_total,"W"

#Plate exposed to heat

#Variable declaration
rho=0.910            #Density in [kg/m**3]
Cp=1.009*1000             #[J/kg.K]
k=0.0331             #[W/m.K]
mu=22.65*10**-6       #[N.s/m**2]

#Calculation
#Let a=smaller side
#b=bigger side
#Qa=ha*A*dT
#Qb=hb*A*dT
#Qa=1.14*Qb
#Given a*b=15*10**-4
#On solving we get:
a=0.03           #[m]
b=0.05           #[m]
A=a*b           #Area in [sq m]
Tf=388           #[K]
Beta=1.0/Tf       #[K**-1]
T1=303           #[K]
T2=473          #[K]
dT=T2-T1        #[K]
v=mu/rho        
g=9.81      #m/s**2[acceleration due to gravity ]
hb=0.59*(((g*Beta*dT*(b**3))/(v**2))*Cp*mu/k)**(1.0/4.0)*(k/b)       #[W/sq m.K]
Qb=hb*A*(dT)            #[W]

Qa=1.14*Qb              #[W]

#Result
print"Dimensions of the plate are",a,"x",b,"m=",a*100,"x",b*100,"cm"
print"Heat transfer when the bigger side held vertical is",round(Qb,2),"W"
print"Heat transfer when the small side held vertical is",round(Qa,2),"W" 

#Nucleate poolboiling
import math

#Variable declaration
Ts=373.0  #[K]
rho_l=957.9    #rho*l[kg/m**3]
Cpl=4217.0    #[J/kg.K]
mu_l=27.9*10**-5 #[kg/(m.s)]
rho_v=0.5955   #[kg/m**3]
Csf=0.013
sigma=5.89*10**-2    #[N/m]
Nprl=1.76
lamda=2257.0 #[kJ/kg]
lamda=lamda*1000  #in [J/kg]
n=1 #for water
m_dot=30.0    #Mass flow rate [kg/h]

#Calculation
m_dot=m_dot/3600    #[kg/s]
D=30.0    #Diameter of pan [cm]
D=D/100 #[m]
g=9.81  #[m/s**2]
A=math.pi*(D/2)**2   #Area in [sq m]
Q_by_A=m_dot*lamda/A   #[W/sq m]
#For nucleate boiling point we have:
dT=(lamda/Cpl)*Csf*(((Q_by_A)/(mu_l*lamda))*math.sqrt(sigma/(g*(rho_l-rho_v))))**(1.0/3.0)*(Nprl**n) #[K]
Tw=Ts+dT    #[K]

#Result
print"Temperature of the bottom surface of the pan is",round(Tw,1),"W/(m^2)"

#Peak Heat flux
import math

#Variable declaration
lamda=2257.0 #[kJ/kg]
lamda=lamda*1000  #in [J/kg]
rho_l=957.9    #rho*l[kg/m**3]
rho_v=0.5955   #[kg/m**3]
sigma=5.89*10**-2    #[N/m]
g=9.81  #[m/s**2]

#Calculation
#Peak heat flux is given by
Q_by_A_max=(math.pi/24)*(lamda*rho_v**0.5*(sigma*g*(rho_l-rho_v))**(1.0/4.0))    #W/m**2
Q_by_A_max=Q_by_A_max/(10**6)    #MW/(sq m)


#Result
print"Peak heat flux is",round(Q_by_A_max,3),"MW/sq m"

#Stable film pool boiling
import math

#Variable declaration
rho_l=957.9 #[kg/m**3]
lamda=2257.0 #[kJ/kg]
lamda=lamda*10**3  #[J/kg]
rho_v=31.54 #[kg/m**3]
Cpv=4.64    #[kJ/kg.K]
Cpv=Cpv*10**3    #[J/kg.K]
kv=58.3*10**-3#[W/(m.K)]
g=9.81  #[m/s**2]
mu_v=18.6*10**-6 #[kg/(m.s)]
e=1.0   #Emissivity
sigma=5.67*10**-8 
Ts=373.0  #[K]
Tw=628.0  #[K]

#Calculation
dT=Tw-Ts    #[K]
D=1.6*10**-3 #[m]
T=(Tw+Ts)/2 #[K]

hc=0.62*((kv**3)*rho_v*(rho_l-rho_v)*g*(lamda+0.40*Cpv*dT)/(D*mu_v*dT))**(1.0/4.0)  #Convective heat transfer coeff  [W/sq m.K]
hr=e*sigma*(Tw**4-Ts**4)/(Tw-Ts)  #Radiation heat transfer coeff in [W/sq m.K]
h=hc+(3.0/4.0)*hr   #Total heat transfer coefficient W/(sq m.K)
Q_by_l=h*math.pi*D*dT   #Heat dissipation rate per unit length in [kW/m]
Q_by_l=Q_by_l/1000  #[kW/m]
#Result
print"Stable film boiling point heat transfer coefficient is",round(h,1),"W/(sq m.K)"
print"Heat dissipated per unit length of the heater is",round(Q_by_l,1),"kW/m"

print"\nNOTE:In textbook,value of hc is wrongly calculated as 1311.4,Actually it is 1318.9,"
print"So,there is a difference in final values of 'h'"