#Example Number 1.1
# HOW MUCH HEAT IS TRANSFERRED THROUGH THE PLATE
#VARIABLE DECLARATION
k = 370 # [W/m] at 250 degree celsius
dt = 100-400 #[degree celsius] temperature difference
dx = 3*10**(-2) #[m] thickness of plate
#CALCULATION
q = -k*dt/dx #[MW/square meter]
#RESULTS
print "Rate of heat transfer per unit area is", q/1000000 ,"MW/sq meter"
#Example Number 1.2
# CALCULATE THE HEAT TRANSFER
#Variable declaration
Twall = 250 #[degree celsius] wall temperature
Tair = 20 #[degree celsius] air temperature
h = 25 #[W/square meter] heat transfer coefficient
l = 75*10**(-2) #[m] length of plate
b = 50*10**(-2) #[m] width of plate
area = l*b #[square meter] area of plate
dt = 250-20 #[degree celsius]
#Calculation
q = h*area*dt # [W] from newton's law of cooling
#Result
print"Rate of heat transfer is",round(q/1000,3),"kW"
#Example Number 1.3
# Calculate the inside plate temperature.
#variable declaration
Qconv = 2156 # [W] from previous problem
Qrad = 300 # [W] given
dx = 0.02 # [m] plate thicknesss
l = 0.75 # [m] length of plate
w = 0.5 # [m] width of plate
k = 43 #[W/m] from table 1.1
area = l*w #[square meter] area of plate
#Calculation
Qcond = Qconv+Qrad # [W]
dt = Qcond*dx/(k*area) # [degree celsius] temperature difference
Ti = 250+dt # inside temperature
#Results
print"Inside plate temperature is",round(Ti,2),"degree C"
#Example Number 1.4
# Calculate electric power to be supplied to the wire
#Variable declaration
d = 1*10**(-3) #[m] diameter of wire
l = 10*10**(-2) #[m] length of wire
Sarea = 22*d*l/7 #[square meter] surface area of wire
h = 5000 #[W/square meter] heat transfer coefficient
Twall = 114 # [degree celsius]
Twater = 100 # [degree celsius]
#total convection loss is given by equation(1-8)
#Calculation
Q = h*Sarea*(Twall-Twater) # [W]
#Results
print"Heat transfer is:",Q,"W"
print" This is equal to the electric power which must be applied"
#Example Number 1.5
# radiation heat transfer
# Calculate the heat transfer per unit area
#Variable declaration
sigma = 5.669*10**(-8) #[W/square meter*k^(4)] universal constant
T1 = 273+800 # [k] first plate temperature
T2 = 273+300 # [k] second plate temperature
#equation(1-10) may be employed for this problem
#Calculation
Q = sigma*(T1**4-T2**4) # [W/square meter]
#Results
print"Heat transfer per unit area is:",round(Q/1000,2)," kW/sq meter"
#Example Number 1.6
# total heat loss by convection and radiation
#Variable declaration
d = 0.05 #[m] diameter of pipe
Twall = 50 #[degree celsius]
Tair = 20 #[degree celsius]
emi = 0.8 #emissivity
h = 6.5 #[W/square meter] heat transfer coefficient for free convection
import math
Q1 = h*math.pi*d*(Twall-Tair) #[W/m] convection loss per unit length
sigma = 5.669*10**(-8) # [W/square meter*k^(4)] universal constant
T1 = 273+Twall # [k]
T2 = 273+Tair # [k]
Q2 = emi*math.pi*d*sigma*((T1**(4))-(T2**(4))) # [W/m] heat loss due to radiation per unit length
Qtotal = Q1+Q2 # [W/m] total heat loss per unit length
print"Total heat loss is:",round(Qtotal,2),"W/m"