A vapor bubble rises in a liquid. The relevant dimensional parameters are the liquid specific weight, \(\gamma_{\ell},\) the vapor specific weight, \(\gamma_{\nu},\) bubble velocity, \(V,\) bubble diameter, \(d,\) surface tension, \(\sigma,\) and liquid viscosity, \(\mu .\) Find appropriate dimensionless parameters.

Short Answer

Expert verified
The dimensionless parameters for the problem are: \(\Pi_1 = \frac{\gamma_{v}}{\gamma_{\ell}}\), \(\Pi_2 = \frac{V^2}{g d} = \frac{V^2 \gamma_{\ell}}{d \gamma_{\ell}}\), \( \Pi_3 = \frac{\sigma}{d \gamma_{\ell}}\), \( \Pi_4 = \frac{\mu V}{d \gamma_{\ell}}\).

Step by step solution

01

Identify the problem variables

Initially, six dimensional physical parameters are given, namely: the liquid specific weight (\(\gamma_{\ell}\)), the vapor specific weight (\(\gamma_{v}\)), bubble velocity (V), bubble diameter (d), surface tension (\(\sigma\)) and liquid viscosity (\(\mu\)).
02

Determine the base quantities

Identify the base quantities that these physical parameters relate to. They are: mass [M], length [L], time [T]. The following relations can be established: bubble velocity, V (LT^-1); bubble diameter, d [L]; liquid specific weight (\(\gamma_{\ell}\)) and vapor specific weight (\(\gamma_{v}\)) have similar dimensions (ML^-2T^-2); surface tension (\(\sigma\)) (MT^-2) and liquid viscosity (\(\mu\)) (ML^-1T^-1). This implies 3 fundamental dimensions are involved: M, L, T.
03

Apply the Buckingham Pi Theorem

The Buckingham Pi theorem states that the number of dimensionless parameters is the difference between the number of variables and the number of base quantities. Here, 6-3 = 3 dimensionless parameters need to be created.
04

Formulate the dimensionless parameters

Consider Liquid specific weight, Velocity and Bubble diameter as repeating variables. Now derive three dimensionless terms: 1. \(\Pi_1 = \frac{\gamma_{v}}{\gamma_{\ell}}\) (since both \( \gamma_{v} \) and \( \gamma_{\ell} \) have the same dimensions, this equals to 1), 2. \( \Pi_2 = \frac{V^2}{g d} = \frac{V^2 \gamma_{\ell}}{d \gamma_{\ell}}\), (this is dimensionless since the denominator and numerator have the same dimensions of square length per square time) 3. \( \Pi_3 = \frac{\sigma}{d \gamma_{\ell}}\), (this is dimensionless as both the numerator and denominator have dimensions of force per unit length) 4. \( \Pi_4 = \frac{\mu V}{d \gamma_{\ell}}\). (this is dimensionless since both numerator and denominator share dimensions of mass per unit time)

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