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Chapter 8: Problem 22

Consider fully developed laminar flow between infinite parallel plates separated by gap width \(d=0.2\) in. The upper plate moves to the right with speed \(U_{2}=5 \mathrm{ft} / \mathrm{s}\); the lower plate moves to the left with speed \(U_{1}=2 \mathrm{ft}\) /s. The pressure gradient in the direction of flow is zero. Develop an expression for the velocity distribution in the gap. Find the volume flow rate per unit depth (gpm/ft) passing a given cross section.

Short Answer

Expert verified
The velocity distribution in the gap is \(u(y) = 2 + y [(5 - 2) / 0.01667]\) and the total volume flow rate per unit depth passing a given cross section is 26.186 gal/min/ft.

Step by step solution

01

Identify Relevant Equations and Variables

This is a laminar flow problem in fluid dynamics. As the flow is fully developed and the pressure gradient is zero (so it's a Couette flow), we can make use of the equation for velocity distribution in such flow conditions which is given by \(u(y) = U_1 + y [(U_2-U_1)/d]\) where \(u(y)\) is the velocity at a distance \(y\) from the bottom plate, \(U_1\) is the velocity of the bottom plate, \(U_2\) is the velocity of the top plate and \(d\) is the distance between the plates.
02

Develop the Velocity Distribution

Replacing given variables into the formula for velocity distribution, we have \(u(y) = 2 + y [(5 - 2) / 0.2 ]\). However, we need to convert the gap width from inches to feet. Given that 1 inch equals 0.08333 feet, \(d = 0.2 * 0.08333 = 0.01667\) feet. So, the velocity distribution becomes \(u(y) = 2 + y [(5 - 2) / 0.01667]\).
03

Calculate the Volume Flow Rate

To calculate the volume flow rate (Q) we use the formula for flow rate Q = \(\int_0^d u dy\). Substituting the developed expression for \(u(y)\) into the flow rate equation we obtain \(Q = \int_0^{0.01667} (2 + y [(5 - 2) / 0.01667]) dy\). Doing the integration, the solution to the above integral is \(Q = (2d + [(3 * d^2) / 2])\). Substituting the value of \(d = 0.01667\) feet, we find \(Q = 0.05834 ft^3/s\). To convert to gallons per minute, use the fact that 1 cubic foot = 7.48 gallons, and 1 minute = 60 seconds. Therefore, \(Q = 0.05834 ft^3/s * 7.48 gal/ft^3 * 60 s/min = 26.186 gal/min\). Therefore, the total volume flow rate per unit depth passing a given cross section is 26.186 gal/min/ft.

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Most popular questions from this chapter

Two immiscible fluids are contained between infinite parallel plates. The plates are separated by distance \(2 h\), and the two fluid layers are of equal thickness \(h=5 \mathrm{mm}\). The dynamic viscosity of the upper fluid is four times that of the lower fluid, which is \(\mu_{10 \text { wer }}=0.1 \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}\). If the plates are stationary and the applied pressure gradient is \(-50 \mathrm{kPa} / \mathrm{m}\) find the velocity at the interface. What is the maximum velocity of the flow? Plot the velocity distribution.

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Consider fully developed laminar flow in a circular tube. Evaluate the kinetic energy coefficient for this flow.

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