Question 6.5: An agitator of diameter D, requires power P, to rotate at a ...
An agitator of diameter D , requires power P , to rotate at a constant speed N , in a liquid of density \rho , and viscosity \mu . Show (i) with the help of the Pi theorem that P=\rho N^{3} D^{5} F\left(\rho N D^{2} / \mu\right) and (ii) An agitator of 225 mm diameter rotating at 23 rev/s in water requires a driving torque of 1.1 Nm. Calculate the corresponding speed and the torque required to drive a similar agitator of 675 mm diameter rotating in air (Viscosities: air 1.86 \times 10^{5} Pas, water 1.01 \times 10^{-3} Pas. Densities: air 1.20 \mathrm{~kg} / \mathrm{m}^{3} , water 1000 \mathrm{~kg} / \mathrm{m}^{3} ).
(i) The problem is described by 5 variables as
F(P, N, D, \rho, \mu)=0These variables are expressed by 3 fundamental dimensions M, L, and T. Therefore, the number of \pi terms = (5 – 3) = 2. N, D, and \rho are taken as the repeating variables in determining the \pi terms.
Then, \pi_{1}=N^{a} D^{b} \rho^{c} P (6.25)
\pi_{2}=N^{a} D^{b} \rho^{c} \mu (6.26)
Substituting the variables of Eq. (6.25) and (6.26) in terms of their fundamental dimensions M, L and T we get,
\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}=\left(\mathrm{T}^{-1}\right)^{a}(\mathrm{~L})^{b}\left(\mathrm{ML}^{-3}\right)^{c} \mathrm{ML}^{2} \mathrm{~T}^{-3} (6.27)
\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}=\left(\mathrm{T}^{-1}\right)^{a}(\mathrm{~L})^{b}\left(\mathrm{ML}^{-3}\right)^{c} \mathrm{ML}^{-1} \mathrm{~T}^{-1} (6.28)
Equating the exponents of M, L and T from Eq. (6.27), we get
c+1=0
b-3 c+2=0
-a-3=0
which give a=-3, b=-5, c=-1
and hence \pi_{1}=\frac{P}{\rho N^{3} D^{5}}
Similarly, from Eq. (6.28)
c+1=0
b-3 c-1=0
-a-1=0
which give a=-1, b=-2, c=-1
Hence \pi_{2}=\frac{\mu}{\rho N D^{2}}
Therefore, the problem can be expressed in terms of independent dimensionless parameters as
f\left(\frac{P}{\rho N^{3} D^{5}}, \frac{\mu}{\rho N D^{2}}\right)=0which is equivalent to
\psi\left(\frac{P}{\rho N^{3} D^{5}}, \frac{\rho N D^{2}}{\mu}\right)=0Or \frac{P}{\rho N^{3} D^{5}}=F\left(\frac{\rho N D^{2}}{\mu}\right)
Or P=\rho N^{3} D^{5} F\left(\frac{\rho N D^{2}}{\mu}\right)
(ii)
|
D_{1}=225 mm |
D_{2}=675 mm |
|
N_{1}=23 rev / s |
N_{2}=? |
| \rho_{1}=1000 kg / m ^{3} |
\rho_{2}=1.20 kg / m ^{3} |
|
\mu_{1}=1.01 \times 10^{-3} Pas |
\mu_{2}=1.86 \times 10^{-5} Pas |
| P_{1}=2 \pi \times 23 \times 1.1 W |
P_{2}=? |
From the condition of similarity as established above,
\frac{\rho_{2} N_{2} D_{2}^{2}}{\mu_{2}}=\frac{\rho_{1} N_{1} D_{1}^{2}}{\mu_{1}}N_{2}=N_{1}\left(\frac{D_{1}}{D_{2}}\right)^{2} \frac{\rho_{1}}{\rho_{2}} \frac{\mu_{2}}{\mu_{1}}
=23 rev / s \left(\frac{225}{675}\right)^{2} \frac{1000}{1.20} \frac{1.86 \times 10^{-5}}{1.01 \times 10^{-3}}
= 39.22 rev/s
Again, \frac{P_{2}}{\rho_{2} N_{2}^{3} D_{2}^{5}}=\frac{P_{1}}{\rho_{1} N_{1}^{3} D_{1}^{5}}
Or \frac{P_{2}}{P_{1}}=\left(\frac{D_{2}}{D_{1}}\right)^{5}\left(\frac{N_{2}}{N_{1}}\right)^{3} \frac{\rho_{2}}{\rho_{1}}
Or \frac{T_{2}}{T_{1}}=\left(\frac{D_{2}}{D_{1}}\right)^{5}\left(\frac{N_{2}}{N_{1}}\right)^{2} \frac{\rho_{2}}{\rho_{1}}
where, T represents the torque and satisfies the relation P=2 \pi N T
Hence, T_{2}=T_{1}\left(\frac{D_{2}}{D_{1}}\right)^{5}\left(\frac{N_{2}}{N_{1}}\right)^{2} \frac{\rho_{2}}{\rho_{1}}
=1.1 Nm \left(\frac{675}{225}\right)^{5}\left(\frac{39.22}{23}\right)^{2} \frac{1.20}{1000}
= 0.933 Nm