Air bubbles discharge from the end of a submerged tube as shown in Fig. P7.57. The bubble diameter, \(D\), is assumed to be a function of the air flowrate, \(Q\), the tube diameter, \(d\), the acceleration of gravity, \(g\), the density of the liquid, \(\rho\), and the surface tension of the liquid, \(\sigma\). (a) Determine a suitable set of dimensionless variables for this problem. (b) Model tests are to be run on the Earth for a prototype that is to be operated on a planet where the acceleration of gravity is 10 times greater than that on Earth. The model and prototype are to use the same fluid, and the prototype tube diameter is 0.25 in. Determine the tube diameter for the model and the required model flowrate if the prototype flowrate is to be \(0.001 \mathrm{ft}^{3} / \mathrm{s}\).

Step by step solution

01

Determine the dimensionless variables

To determine the dimensionless variables, use the Buckingham Pi theorem, which states that for any physical relation between m variables, there are m - n dimensionless variables, where n is the number of reference dimensions used (usually three - length, mass, and time). The diameter \(D\), air flowrate \(Q\), tube diameter \(d\), acceleration of gravity \(g\), density of the liquid \(\rho\), and the surface tension of the liquid \(\sigma\), we can choose Q, g, and \(\rho\) as the repeating variables. We can form three dimensionless groups: \[\Pi_1 = \frac{D}{d},\] \[\Pi_2 = \frac{Q}{d^3} \sqrt{\frac{g}{\rho}},\] \[\Pi_3 = \frac{\sigma}{d^3} \sqrt{\frac{\rho^3}{g}}.\]

02

Determine the tube diameter for the model

In the second part, we know that the model and the prototype are similar if their dimensionless groups are equal. Thus, \(\frac{D_m}{d_m} = \frac{D_p}{d_p}\) where the subscripts \(m\) and \(p\) denote the model and prototype respectively, and the tube diameter \(d\) for the prototype is given as 0.25 in. Hence, the tube diameter for the model is \(d_m = \frac{D_m d_p}{D_p}\). However, since the bubble diameters are not given, we cannot calculate a specific value for \(d_m\).

03

Determine the model flowrate

From the second dimensionless group \(\frac{Q_m}{d_m^3} \sqrt{\frac{g_m}{\rho_m}} = \frac{Q_p}{d_p^3} \sqrt{\frac{g_p}{\rho_p}}\). By plugging in the given values and solving for the model flowrate \(Q_m\) we get: \(Q_m = Q_p \frac{d_p^3}{d_m^3} \sqrt{\frac{g_p}{g_m}}\). Since the gravitational acceleration on the other planet is 10 times greater than on Earth, \(g_m = 10g_p\). Also, the model and prototype are using the same fluid, so \(d_m = d_p\). Hence, \(Q_m = 0.001 \mathrm{ft}^{3} / \mathrm{s}\).

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