# Chapter 3: Tem¶

## Exa 3.1¶

In :
#Make the following conversions : (a) -3 C to F (b) 650 R to C (c) 650 R to K
#initialisation of variables
T1= -3 					#degrees
T2= 650. 				#Rankine
T3= 650. 				#Rankine
#CALCULATIONS
t1= (9./5.)*T1+32 		#In F
t2= T2-459.67 			#In F
t21= (5./9.)*(t2-32) 	#In C
t3= t21+273.15 			#In K
#RESULTS
print '%s %.2f' % ('T (F) = ',t1)
print '%s %.2f' % (' \n T (C) = ',t21)
print '%s %.2f' % (' \n T (K) = ',t3)
raw_input('press enter key to exit')

T (F) =  26.60

T (C) =  87.96

T (K) =  361.11
press enter key to exit

Out:
''

## Exa 3.2¶

In :
#The temperatuer of an object drops from 40 to 30C. Determine the temperature
#drop in (a) F and (b) R.
#initialisation of variables
T1= 40. 				#degrees
T2= 30. 				#degrees
#CALCULATIONS
d1= (T1-T2)*(9./5.) 	#drop in F
d2= d1 					#drop in R
#RESULTS
print '%s %.2f' % ('T (F) = ',d1)
print '%s %.2f' % (' \n T (R) = ',d2)
raw_input('press enter key to exit')

T (F) =  18.00

T (R) =  18.00
press enter key to exit

Out:
''

## Exa 3.3¶

In :
#A brass wire has a length of 400mm at 20C. What is the length of the wire at 90C?
#initialisation of variables
l= 400 					#mm
t1= 20 					#degrees
t2= 90 					#degrees
alpha= 19.3/1000000. 	#degrees^-1
#CALCULATIONS
L= alpha*(t2-t1)*l 		#Change in Length
L1= L+l 				#Final Length
#RESULTS
print '%s %.2f' % ('L (mm) = ',L1)
raw_input('press enter key to exit')

L (mm) =  400.54
press enter key to exit

Out:
''

## Exa 3.4¶

In :
#An Al 2024-T3 plate has a 2.980-in diameter hole at 69F. What is the
#diameter if the temperature is lowered to -15F?
import math
#initialisation of variables
d= 2.98 					#in
T1= 69 						#F
T2= -15 					#F
alpha= 22.7/1000000. 		#C^-1
#CALCULATIONS
A0= math.pi*d*d/4. 			#Initial Area
alpha1= alpha/1.8
A= 2*alpha1*A0*(T1-T2)  	#Change in Area
A1= A0-A
d1= math.sqrt(4*A1/math.pi) #diameter at -15F
#RESULTS
print '%s %.2f' % ('diameter at -15 (in) = ',d1)
raw_input('press enter key to exit')

diameter at -15 (in) =  2.98
press enter key to exit

Out:
''