/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 Oil flows steadily in a thin lay... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 33

Oil flows steadily in a thin layer down an inclined plane. The velocity profile is \\[ u=\frac{\rho g \sin \theta}{\mu}\left[h y-\frac{y^{2}}{2}\right] \\] Express the mass flow rate per unit width in terms of \(\rho, \mu, g\) \(\theta,\) and \(h\)

Short Answer

Expert verified
The mass flow rate per unit width of the oil is given by \(\frac{\rho^2 g h^3 \sin \theta}{3\mu}\).

Step by step solution

01

Identify Relevant Variables

Firstly, all the relevant variables are identified from the velocity profile. These are: fluid density (\(\rho\)), acceleration due to gravity (\(g\)), angle of inclination (\(\theta\)), dynamic viscosity (\(\mu\)), flow thickness (\(h\)), and vertical height (\(y\)).
02

Define Mass Flow Rate

The mass flow rate (\(m\)) per unit width is given by multiplying the velocity profile by the density of the fluid, and then integrating over the thickness (\(h\)) of the fluid layer. So, \(m = \int_0^h \rho u\, dy\).
03

Substitute the Velocity Profile

Substitute the given velocity profile into the formula: \(m = \int_0^h \rho \left[\frac{\rho g \sin \theta}{\mu}\left(h y - \frac{y^{2}}{2}\right)\right] dy\). Simplifying, \(m = \frac{\rho^2 g \sin \theta}{\mu} \int_0^h \left(h y - \frac{y^{2}}{2}\right) dy\).
04

Compute the Integral

Using the power rule for integration, evalute the integral (\(\int_0^h (h y - \frac{1}{2} y^2) dy\)). This results in: \(\Big[\frac{h}{2}y^2 - \frac{1}{6}y^3\Big]_0^h\). The result of the integral from 0 to \(h\) is: \(\frac{h^3}{2} - \frac{h^3}{6}\). And simplifying the fraction we got: \(\frac{h^3}{3}\).
05

Substitute Back

Substitute this value back into the flow rate formula to obtain the final solution: \(m = \frac{\rho^2 g \sin \theta}{\mu} . \frac{h^3}{3} = \frac{\rho^2 g h^3 \sin \theta}{3\mu}\).

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Most popular questions from this chapter

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