# Chapter 1 Introduction¶

## Exa 1.1¶

In :
#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"

Rate of heat transfer per unit area is 3.7 MW/sq meter


## Exa 1.2¶

In :
#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"

Rate of heat transfer is 2.156 kW


## Exa 1.3¶

In :
#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

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"

Inside plate temperature is 253.05 degree C


## Exa 1.4¶

In :
#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"

Heat transfer is: 22.0 W
This is equal to the electric power which must be applied


## Exa 1.5¶

In :
#Example Number 1.5
# 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"

Heat transfer per unit area is: 69.03  kW/sq meter


## Exa 1.6¶

In :
#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

Total heat loss is: 55.67 W/m