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Chapter 7: Problem 30

By inspection, arrange the following dimensional parameters into dimensionless parameters: (a) kinematic viscosity, \(v,\) length, \(\ell,\) and time, \(t ;\) and (b) volume flow rate, \(Q,\) pump diameter, \(D,\) and pump impeller rotation speed, \(N\)

Short Answer

Expert verified
The dimensionless parameters for problem (a) is \(\frac{v}{\ell t}\), and for problem (b) is \(\frac{Q}{D^3 N}\)

Step by step solution

01

Understand the variables for problem (a)

For problem (a), the given parameters are the kinematic viscosity (\(v\)), which has units of m²/sec, length (\(l\)), which is expressed in metres (m), and time (\(t\)), which is measured in seconds (sec).
02

Creating dimensionless parameters for problem (a)

To create a dimensionless parameter, the goal is to balance the units from the different parameters given. The variables could be multiplied or divided among each other to cancel the units out. So, dividing kinematic viscosity (\(v\)) by the product of length (\(l\)) and time (\(t\)) will result in a dimensionless parameter, because the units cancel each other out: \(\frac{v}{\ell t}\). This ratio doesn't have any units and is therefore dimensionless.
03

Understanding variables for problem (b)

For problem (b), the given parameters are the volume flow rate (\(Q\)), which has units of m³/sec, the pump diameter (\(D\)), which is expressed in metres (m), and pump impeller rotation speed (\(N\)), which is measured in revolution per second (rps).
04

Creating dimensionless parameters for problem (b)

For problem (b), dividing volume flow rate (\(Q\)) by the product of pump's diameter cubed (\(D^3\)) and impeller rotation speed (\(N\)) gives a dimensionless parameter because all of dimensions get cancelled out. So, the dimensionless parameter for problem (b) is \(\frac{Q}{D^3 N}\).

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Most popular questions from this chapter

The pressure rise, \(\Delta p=p_{2}-p_{1},\) across the abrupt expansion of Fig. \(\mathrm{P} 7.38\) through which a liquid is flowing can be expressed as $$\Delta p=f\left(A_{1}, A_{2}, \rho, V_{1}\right)$$ where \(A_{1}\) and \(A_{2}\) are the upstream and downstream cross-sectional areas, respectively, \(\rho\) is the fluid density, and \(V_{1}\) is the upstream velocity. Some experimental data obtained with \(A_{2}=1.25 \mathrm{ft}^{2}\) \(V_{1}=5.00 \mathrm{ft} / \mathrm{s},\) and using water with \(\rho=1.94\) slugs/ft \(^{3}\) are given in the following table: $$\begin{array}{l|l|l|l|l|r} A_{1}\left(\mathrm{ft}^{2}\right) & 0.10 & 0.25 & 0.37 & 0.52 & 0.61 \\ \hline \Delta p\left(\mathrm{lb} / \mathrm{ft}^{2}\right) & 3.25 & 7.85 & 10.3 & 11.6 & 12.3 \end{array}$$ Plot the results of these tests using suitable dimensionless parameters. With the aid of a standard curve fitting program determine a general equation for \(\Delta p\) and use this equation to predict \(\Delta p\) for water flowing through an abrupt expansion with an area ratio \(A_{1} / A_{2}=0.35\) at a velocity \(V_{1}=3.75 \mathrm{ft} / \mathrm{s}\).

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