3: ACOUSTICS OF BUILDINGS AND SUPERCONDUCTIVITY

Example number 3.1, Page number 3.32

In [22]:
#importing modules
import math
from __future__ import division

#Variable declaration
V=2265
A=92.9
x=2               #The absorption become 2*A of open window

#Calculation
T=(0.16*V)/A      #Sabine's formula  
T2=(0.16*V)/(x*A)

#Result
print"Reverbration time =",round(T,1),"s"
print"Final Reverbration time =",round(T2,2),"s"
print"Thus the reverbration time is reduced to one-half of its initial value"

Reverbration time = 3.9 s
Final Reverbration time = 1.95 s
Thus the reverbration time is reduced to one-half of its initial value

Example number 3.2, Page number 3.32

In [19]:
#importing modules
import math
from __future__ import division

#Variable declaration
a1=450    #Area of plastered wall
a2=360    #Area of wooden floor and wooden doors
a3=24     #Area of Glass
a4=600    #Area of seats
a5=500    #Area of audience when they are in seats
c1=0.03   #Coefficient of absorption of plastered wall
c2=0.06   #Coefficient of absorption of wooden floor and wooden doors
c3=0.025   #Coefficient of absorption of Glass
c4=0.3    #Coefficient of absorption of seats
c5=0.43   #Coefficient of absorption of audience when they are in seats
l=12
b=30
h=6

#Calculation
V=l*b*h        #volume of the hall
A=(a1*c1)+(a2*c2)+(a3*c3)+(a4*c4)+(a5*c5)  #Total absorption
T=(0.16*V)/A   #Reverbration time

#Result
print"Volume of the hall =",V,"m**3"
print"Total absorption =",A,"m**2"
print"Reverbration time =",round(T,1),"second"
print"Answer given for the Reverbration time in the text book is wrong"

Volume of the hall = 2160 m**3
Total absorption = 430.7 m**2
Reverbration time = 0.8 second
Answer given for the Reverbration time in the text book is wrong

Example number 3.3, Page number 3.33

In [21]:
#importing modules
import math
from __future__ import division

#Variable declaration
T=1.2
V=7500

#Calculation
A=(0.16*V)/T

#Result
print"Total absorpttion =",A," m**2 of O.W.U."

Total absorpttion = 1000.0  m**2 of O.W.U.

Example number 3.4, Page number 3.34

In [26]:
#importing modules
import math
from __future__ import division

#Variable declaration
V=12*10**4
A=13200
x=2               #The absorption become 2*A of open window

#Calculation
T1=(0.16*V)/A      #Sabine's formula  
T2=(0.16*V)/(x*A)
Td=T1-T2

#Result
print"T1 =",round(T1,2),"second"
print"T2 =",round(T2,2),"second"
print"Change in Reverbration time =",round(Td,3),"second"

T1 = 1.45 second
T2 = 0.73 second
Change in Reverbration time = 0.727 second

Example number 3.6, Page number 3.34

In [1]:
#importing modules
import math
from __future__ import division

#Variable declaration
H0=64*10**3;    #initial field(ampere/m)
T=5;    #temperature(K)
Tc=7.26;   #transition temperature(K)

#Calculation
H=H0*(1-(T/Tc)**2);     #critical field(ampere/m)

#Result
print "critical field is",round(H/10**3,2),"*10**3 ampere/m"

critical field is 33.64 *10**3 ampere/m

Example number 3.7, Page number 3.34

In [4]:
#importing modules
import math
from __future__ import division

#Variable declaration
e=1.6*10**-19
V=1*10
h=6.625*10**-34

#Calculations
v=(2*e*V**-3)/h 

#Result
print"Frequency of generated microwaves=",round(v/10**9),"*10**9 Hz"

Frequency of generated microwaves= 483.0 *10**9 Hz

Example number 3.8, Page number 3.34

In [2]:
#importing modules
import math
from __future__ import division

#Variable declaration
d=7300                  #density in (kg/m**3)
N=6.02*10**26           #Avagadro Number
A=118.7                 #Atomic Weight
E=1.9                 #Effective mass
e=1.6*10**-19

#Calculations
n=(d*N)/A
m=E*9.1*10**-31
x=4*math.pi*10**-7*n*e**2
lamda_L=math.sqrt(m/x)
      
#Result
print "Number of electrons per unit volume =",round(n/10**28,1),"*10**28/m**3"
print"Effective mass of electron 'm*' =",round(m*10**31,1),"*10*-31 kg"
print"Penetration depth =",lamda_L*10**8,"Angstroms"
print"#The answer given in the text book is wrong"

Number of electrons per unit volume = 3.7 *10**28/m**3
Effective mass of electron 'm*' = 17.3 *10*-31 kg
Penetration depth = 3.81011659367 Angstroms
#The answer given in the text book is wrong

Example number 3.9, Page number 3.35

In [18]:
#importing modules
import math
from __future__ import division

#Variable declaration
lamda_L1=39.6*10**-9
lamda_L2=173*10**-9
T1=7.1
T2=3

#Calculations
x=(lamda_L1/lamda_L2)**2
Tc4=(T1**4)-((T2**4)*x)/(1-x)
Tc=(Tc4)**(1/4)
print"Tc =",round(Tc,4),"K"
print"lamda0=",round((math.sqrt(1-(T2/Tc)**4)*lamda_L1)*10**9),"nm"

Tc = 7.0969 K
lamda0= 39.0 nm

Example number 3.10, Page number 3.35

In [24]:
#importing modules
import math
from __future__ import division

#Variable declaration
H0=6.5*10**4           #(ampere/metre)
T=4.2                  #K
Tc=7.18                #K
r=0.5*10**-3

#Calculations
Hc=H0*(1-(T/Tc)**2)
Ic=(2*math.pi*r)*Hc
A=math.pi*r**2
Jc=Ic/A                #Critical current density

#Result
print"Hc =",round(Hc/10**4,4),"*10**4"
print "Critical current density,Jc =",round(Jc/10**8,2),"*10**8 ampere/metre**2"

Hc = 4.2759 *10**4
Critical current density,Jc = 1.71 *10**8 ampere/metre**2

Example number 3.11, Page number 6.36

In [26]:
#importing modules
import math
from __future__ import division

#Variable declaration
Tc1=4.185
M1=199.5
M2=203.4

#Calculations
Tc2=Tc1*(M1/M2)**(1/2)

#Result
print"New critical temperature for mercury =",round(Tc2,3),"K"

New critical temperature for mercury = 4.145 K