Question 6.6: A torpedo-shaped object 900 mm diameter is to move in air at...
A torpedo-shaped object 900 mm diameter is to move in air at 60 m/s and its drag is to be estimated from tests in water on a half scale model. Determine the necessary speed of the model and the drag of the full scale object if that of the model is 1140 N. (Fluid properties are same as in Example 6.5 (ii)).
The dimensionless parameters representing the criteria of similarity, have to be determined first. The drag force F on the object depends upon its velocity V , diameter D ,the density \rho , and viscosity \mu of air. Now we use Buckhingham’s \pi theorem to find the dimensionless parameters. The five variables F, V, D, \rho and \mu are expressed by three fundamental dimensions M, L and T. Therefore the number of \pi terms is (5-3) =2 .
We choose V, D \text { and } \rho as the repeating variables
Then, \pi_{1}=V^{a} D^{b} \rho^{c} F
\pi_{2}=V^{a} D^{b} \rho^{c} \mu
Expressing the variables in the equations above in terms of their fundamental dimensions, we have
\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}=\left(\mathrm{LT}^{-1}\right)^{a} \mathrm{~L}^{b}\left(\mathrm{ML}^{-3}\right)^{c} \mathrm{MLT}^{-2} (6.29)
\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}=\left(\mathrm{LT}^{-1}\right)^{a} \mathrm{~L}^{b}\left(\mathrm{ML}^{-3}\right)^{c} \mathrm{ML}^{-1} \mathrm{~T}^{-1} (6.30)
Equating the exponents of M, L and T on both the sides of the above equations, we get a=-2, b=-2, c=-1 from Eq. (6.29), and a=-1, b=-1, c=-1 from Eq. (6.30)
Hence, \pi_{1}=\frac{F}{\rho V^{2} D^{2}} and \pi_{2}=\frac{\mu}{\rho V D}
The problem can now be expressed mathematically as
f\left(\frac{F}{\rho V^{2} D^{2}}, \frac{\rho V D}{\mu}\right)=0Or \frac{F}{\rho V^{2} D^{2}}=\phi\left(\frac{\rho V D}{\mu}\right) (6.31)
For dynamic similarity, the Reynolds numbers (\rho V D / \mu) of both the model and prototype have to be same so that drag force F , of the model and prototype can be compared from the \pi_{1} term which is known as the drag coefficient. Therefore, we can write
\frac{\rho_{m} V_{m} D_{m}}{\mu_{m}}=\frac{\rho_{p} V_{p} D_{p}}{\mu_{p}}(Subscripts m and p refer to the model and prototype, respectively)
Or V_{m}=V_{p}\left(\frac{D_{p}}{D_{m}}\right)\left(\frac{\rho_{p}}{\rho_{m}}\right)\left(\frac{\mu_{m}}{\mu_{p}}\right)
Here, \frac{D_{m}}{D_{p}}=\frac{1}{2}
Hence, V_{m}=60 \times(2) \times\left(\frac{1.20}{1000}\right)\left(\frac{1.01 \times 10^{-3}}{1.86 \times 10^{-5}}\right)
=7.82 m / s
At the same value of Re, we can write from Eq. (6.31)
\frac{F_{p}}{\rho_{p} V_{p}^{2} D_{p}^{2}}=\frac{F_{m}}{\rho_{m} V_{m}^{2} D_{m}^{2}}Or F_{p}=F_{m}\left(\frac{D_{p}}{D_{m}}\right)^{2}\left(\frac{\rho_{p}}{\rho_{m}}\right)\left(\frac{V_{p}}{V_{m}}\right)^{2}
=1140(4)\left(\frac{60}{7.82}\right)^{2}\left(\frac{1.20}{1000}\right) N
= 322 N