In the study of vortex shedding phenomenon due to the presence of a bluff body in a flow through a closed duct, the following parameters are found to be important: velocity of flow V, density of liquid ρ, coefficient of dynamic viscosity of liquid μ, hydraulic diameter of the duct Dh, the width

Question

In the study of vortex shedding phenomenon due to the presence of a bluff body in a flow through a closed duct, the following parameters are found to be important: velocity of flow [latex] V[/latex] , density of liquid [latex] \rho [/latex], coefficient of dynamic viscosity of liquid [latex] \mu [/latex], hydraulic diameter of the duct [latex] D_{h} [/latex], the width of the body [latex] B[/latex], and the frequency of vortex shedding [latex] n[/latex]. Obtain the dimensionless parameters governing the phenomenon.

Accepted Answer

The problem is described by 6 variables [latex] V, \rho, \mu, D_{h}, B, [/latex] and [latex] n [/latex]. The number of fundamental dimensions in which the variables can be expressed = 3. Therefore, the number of independentp terms is (6–3) = 3. We use the Buckingham’s [latex] \pi [/latex] theorem to find the [latex] \pi [/latex] terms and choose [latex] \rho, V, D_{h} [/latex] as the repeating variables.

Hence, [latex] \pi_{1}=\rho^{a} V^{b} D_{h}^{c} \mu[/latex]

[latex] \pi_{2}=\rho^{a} V^{b} D_{h}^{c} B[/latex]

[latex] \pi_{3}=\rho^{a} V^{b} D_{h}^{c} n[/latex]

Expressing the equations in terms of the fundamental dimensions of the variables we have

[latex] \mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}=\left(\mathrm{ML}^{-3}\right)^{a}\left(\mathrm{LT}^{-1}\right)^{b}(\mathrm{~L})^{c}\left(\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right) [/latex] (6.42)

[latex] \mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}=\left(\mathrm{ML}^{-3}\right)^{a}\left(\mathrm{LT}^{-1}\right)^{b}(\mathrm{~L})^{c} \mathrm{~L} [/latex] (6.43)

[latex] \mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}=\left(\mathrm{ML}^{-3}\right)^{a}\left(\mathrm{LT}^{-1}\right)^{b}(\mathrm{~L})^{c} \mathrm{~T}^{-1} [/latex] (6.44)

Equating the exponents of M, L and T in the above equations we get, From Eq. (6.42),

a+1 =0

-3 a+b+c-1 =0,-b-1=0

which give a = –1, b = –1 and c = –1

Hence, [latex] p_{1}=m / r V D_{h} [/latex]

From Eq. (6.43),

a=0

-3 a+b+c+1 &=0

-b =0

which give a = b = 0, c = –1

Hence, [latex] p_{2}=B / D_{h} [/latex]

From Eq. (6.44),

a=0

-3 a+b+c=0

-b-1=0

Which give a = 0, b = –1, c = 1

Hence, [latex] p_{3}=\left(n D_{h}\right) / V [/latex]

Therefore, the governing dimensionless parameters are

[latex] \left(\frac{\rho V D_{h}}{\mu}\right)\left(=1 / p_{1}\right) [/latex] the Reynolds number

[latex] \frac{B}{D_{h}} \quad\left(=p_{2}\right) [/latex] ratio of the width of the body to hydraulic diameter of the duct

[latex] \frac{n D_{h}}{V} \quad\left(=p_{3}\right) [/latex] the Strouhal number

Question 6.9: In the study of vortex shedding phenomenon due to the presen...

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