Acoustic Impedance
Pipe Acoustic Impedance
Let us suppose that we want to send an acoustic wave from left towards right through a series of connected pipes having different areas. What is the acoustic impedance of each pipe?
a) For two pipes connected in series first having an area of 4A and the second an area of A, what is the percentage of the incoming wave into 4A pipe that comes out from the small pipe of area A?
b) If between the pipes of area 4A and A there is another “intermediate” pipe of area 2A, find the transmission coefficient into the intermediate pipe and from the intermediate pipe into the smaller pipe.
Answer
The coefficient of reflection between two mediums of acoustic impedance $Z1$ and $Z2$ is
$R = [(Z2-Z1)/(Z2+Z1)]^2$
The coefficient of transmission is just $T = 1-R$. The acoustic impedance of an open tube is given by $Z =\rho*v/A$ where $\rho$ is the density of air, $v$ is the speed of wave and $A$ is the area of the pipe
(http://www.animations.physics.unsw.edu.au/jw/compliance-inertance-impedance.htm)
Thus the acoustic impedances are $Z1= C/A1$ and $Z2 = C/A2$
$R = [(1/A -1/4A)/(1/A +1/4A)]^2 = [(3/4)/(5/4)]^2 =9/25$
and $T =1-9/25 =16/25$ for the first case.
For the second case
$R1 = [(1/A-1/2A)/(1/A+1/2A)]^2 =(1/2)/(3/2)]^2 =1/9$
$T1 = 1-1/9 =8/9$
$R2 = R1 =1/9$ since the raport of the areas is the same
$T2 =8/9$
$T(total) =T1*T2 =8/9*8/9 =64/81$
Hydrogen Fusion
Please find the necessary energy o remove one neutrons from the tritium atom to obtain a deuterium and a neutron.
$3H +energy = 2H +1n$
atomic masses are (all are taken from wiki)
M(3H) =3.01604 uam =2817.143 MeV
M(2H) = 2.0141 uam = 1881.27 MeV
M(1n) = 1.00866 uam = 942.142 MeV
because
$1uam =1.6605*10^-27 kg =1.6605*10^-27*(3*10^8)^2 / 1.6*10^-19 =934.053 MeV$
($E =m*c^2$ )
$Energy = 1881.27 +942.142 -2817.143 = 6.269 MeV$
