Question 6.3: An agitator of diameter D requires power P to rotate at a co...
An agitator of diameter D requires power P to rotate at a constant speed N in a liquid of density ρ and viscosity μ (i) show with the help of Pi theorem that
P=\rho N^{3} D^{5} F\left(\rho N D^{2} / \mu\right)(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} Pa s, water 1.01 \times 10^{-3} Pa s. Densities: air 1.20 kg / m ^{3}, water 1000 kg / 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 π terms = (5 3) = 2. N, D, and ρ are taken as the repeating variables in determining the π terms.
Then, \pi_{1}=N^{a} D^{b} \rho^{c} P (6.30)
\pi_{2}=N^{a} D^{b} \rho^{c} \mu (6.31)
Substituting the variables of Eq. (6.30) and (6.31) in terms of their fundamental dimensions M, L and T we get,
M ^{0} L ^{0} T ^{0}=\left( T ^{-1}\right)^{a}( L )^{b}\left( ML ^{-3}\right)^{c} ML ^{2} T ^{-3} (6.32)
M ^{0} L ^{0} T ^{0}=\left( T ^{-1}\right)^{a}( L )^{b}\left( ML ^{-3}\right)^{c} ML ^{-1} T ^{-1} (6.33)
Equating the exponents of M, L and T from Eq. (6.32), we get
c + 1 = 0
b 3c + 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 equation (6.33)
c + 1 = 0
b 3c 1 = 0
a 1 = 0
which give a = 1 , b = 2, c = 1
and 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)
\begin{aligned}D_{1} &=225 \mathrm{~mm} & D_{2} &=675 \mathrm{~mm} \\N_{1} &=23 \mathrm{rev} / \mathrm{s} & N_{2} &=? \\\rho_{1} &=1000 \mathrm{~kg} / \mathrm{m}^{3} & \rho_{2} &=1.20 \mathrm{~kg} / \mathrm{m}^{3} \\\mu_{1} &=1.01 \times 10^{-3} \mathrm{Pas} & \mu_{2} &=1.86 \times 10^{-5} \mathrm{Pas} \\P_{1} &=2 \pi \times 23 \times 1.1 \mathrm{~W} & P_{2} &=?\end{aligned}
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