Respuesta :
Answer:
Velocity in the smaller pipe should not be included as an additional variable.
Explanation:
[tex]$\Delta P = f(D_1, D_2, V, \rho, \mu)$[/tex]
The dimensional formula of the variables are
[tex]$\Delta P = FL^{-2} , D_1 = L, D_2=L, V=LT_{-1}, \rho =FL^{-4}T^2, \mu = FL^{-2}T$[/tex]
Now using Buchingham's Pi Theorem, 6 - 3 = 3 dimensional parameters are required.
Use, D_1, V, \mu as the repeating variables.
Therefore, [tex]$\pi = \Delta PD_1^aV^b \mu^c$[/tex]
[tex]$(FL^{-2})(L)^a(LT^{-1})^b(FL^{-2}T)^c = F^0L^0T^0$[/tex]
From this
1+c=0
-2+a+b-2c=0
-b+c=0
c=-1, b = -1, a = 1
Now, [tex]$\pi_1=\frac{\Delta PD_1}{V\mu}=\frac{(ML^{-1}T^{-2})L}{(LT^{-1})(ML^{-1}T^{-1})}=M^0L^0T^0$[/tex]
For [tex]$\pi_2 = D_2D_1^aV^b\mu^c$[/tex]
[tex]$F^0L^0T^0=L(L)^a(LT^{-1})^b(FL^{-2}T)^c$[/tex]
c = 0
1 + a + b - 2c = 0
-b + c = 0
Therefore, a = -1, b = 0, c = 0
[tex]$\pi_2 = \frac{D_2}{D_1}$[/tex]
For [tex]$\pi_3$[/tex]
[tex]$\pi_3 = \rho, D_1^a V^b\mu^c$[/tex]
[tex]$F^0L^0T^0 = (FL^{-4}T^2)(L)^a(LT^{-1})^b(FL^{-2}T)^c$[/tex]
1 + c = 0
-4 + a + b - 2c = 0
2-b+c=0
c=-1, b=1, a = 1
Therefore, [tex]$\pi_3 = \frac{\rho D_1V}{\mu}$[/tex]
Now checking,
[tex]$\pi_3 = \frac{(ML^{-3})(L)(LT^{-1})}{ML^{-1}T^{-1}} = M^0L^0T^0$[/tex]
Therefore, [tex]$\frac{\Delta P D_1}{V \mu} = \phi (\frac{D_2}{D_1}, \frac{\rho D_1 V}{\mu})$[/tex]
From continuity equation
[tex]$V\frac{\pi}{4}D_1^2 = V_s \frac{\pi}{4}D_2^2$[/tex]
[tex]$V_s = V (\frac{D_1}{D_2})^2$[/tex]
[tex]$V_s$[/tex] is not independent of [tex]$D_1,D_2, V$[/tex]
Therefore it should not be included as an additional variable.
