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Chapter 7: Problem 9

At a sudden contraction in a pipe the diameter changes from \(D_{1}\) to \(D_{2} .\) The pressure drop, \(\Delta p,\) which develops across the contraction is a function of \(D_{1}\) and \(D_{2},\) as well as the velocity, \(V\), in the larger pipe, and the fluid density, \(\rho,\) and viscosity, \(\mu .\) Use \(D_{1}, V,\) and \(\mu\) as repeating variables to determine a suitable set of dimensionless parameters. Why would it be incorrect to include the velocity in the smaller pipe as an additional variable?

Short Answer

Expert verified
The dimensionless parameters are: \(\pi_1 = \frac{\Delta p \cdot D_1}{\mu V}\), \(\pi_2 = \frac{D_2}{D_1}\), and \(\pi_3 = \frac{\rho V D_1}{\mu}\). The velocity in the smaller pipe is not included as it would introduce an additional dimension, violating the Buckingham Pi theorem.

Step by step solution

01

Identify the repeating variables

Choosing repeating variables to be the diameter of the larger pipe \(D_1\), the velocity in the larger pipe \(V\), and the fluid viscosity \(\mu\).
02

Forming the dimensionless terms

We will form our dimensionless groups by attaching the remaining variables one by one to our repeating group. Our goal is to eliminate the dimensions of the new variable in each group.\n1. Use \(\Delta p, D_1, V, \mu\) to get group 1: \(\pi_1 = \frac{\Delta p \cdot D_1}{\mu V}\)\n2. Use \(D_2, D_1, V, \mu\) to get group 2: \(\pi_2 = \frac{D_2}{D_1}\)\n3. Use \(\rho, D_1, V, \mu\) to get Group 3: \(\pi_3 = \frac{\rho V D_1}{\mu}\)
03

Examine the question of velocity in the smaller pipe

If we were to include the velocity in the smaller pipe as an additional variable, this would introduce an additional dimension (velocity) into the problem. However, according to the Buckingham Pi theorem we already have a complete set of dimensionless parameters. Adding an additional dimension would violate the theorem.

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Most popular questions from this chapter

The dimensionless parameters for a ball released and falling from rest in a fluid are $$C_{D}, \quad \frac{g t^{2}}{D}, \quad \frac{\rho}{\rho_{b}}, \quad \text { and } \quad \frac{V_{t}}{D}$$ where \(\left.C_{\mathbf{D}} \text { is a drag coefficient (assumed to be constant at } 0,4\right)\) \(g\) is the acceleration of gravity, \(D\) is the ball diameter, \(t\) is the time after it is released, \(\rho\) is the density of the fluid in which it is dropped, and \(\rho_{b}\) is the density of the ball. Ball 1 , an aluminum ball \(\left(\rho_{b_{1}}=2710 \mathrm{kg} / \mathrm{m}^{3}\right)\) having a diameter \(D_{1}=1.0 \mathrm{cm},\) is dropped in water \(\left(\rho=1000 \mathrm{kg} / \mathrm{m}^{3}\right) .\) The ball velocity \(V_{\mathrm{t}}\) is \(0.733 \mathrm{m} / \mathrm{s}\) at \(t_{1}=\) 0.10 s. Find the corresponding velocity \(V_{2}\) and time \(t_{2}\) for ball 2 having \(D_{2}=2.0 \mathrm{cm}\) and \(\rho_{b_{2}}=\rho_{b_{1}} .\) Next, use the computed value of \(V_{2}\) and the equation of motion,$$\rho_{b} V_{g}-\frac{\mathrm{C}_{\mathrm{D}}}{2} \rho A V^{2}=\rho_{b} V \frac{d V}{d t}$$ where \(Y\) is the volume of the ball and \(A\) is its cross-sectional area. to verify the value of \(t_{2}\). Should the two values of \(t_{2}\) agree?

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