/*! 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 120 A test tube is spun in a centrif... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 120

A test tube is spun in a centrifuge. The tube support is mounted on a pivot so that the tube swings outward as rotation speed increases. At high speeds, the tube is nearly horizontal. Find (a) an expression for the radial component of acceleration of a liquid clement located at radius \(r,\) (b) the radial pressure gradient \(d p / d r,\) and (c) the required angular velocity to generate a pressure of \(250 \mathrm{MPa}\) in the bottom of a test tube containing water. (The free surface and bottom radii are 50 and \(130 \mathrm{mm},\) respectively.

Short Answer

Expert verified
The radial component of acceleration of a liquid element located at radius \(r\) is \(a_{r}=\omega^{2}r\), the radial pressure gradient is \( \frac{dp}{dr} = \rho \omega^2 r\) where \(\rho\) is the density of the liquid, and the required angular velocity to generate a pressure of 250 MPa is \(\omega= \sqrt[3]{\frac{dp}{2 \rho (r_2^2-r_1^2)}}\) where \(r_1\) is 50 mm, \(r_2\) is 130 mm, \(\rho\) is the density of water (1000 kg/m^3), and \(dp\) is the difference in pressure (250 MPa or 250e6 Pa).

Step by step solution

01

Calculate the radial acceleration

The radial (or centripetal) acceleration in circular motion is given by \(a_{r}=\omega^{2}r\), where \(\omega\) is the angular velocity and \(r\) is the radius.
02

Determine the radial pressure gradient

Under centrifugation, the pressure in the liquid changes due to the acting force. This leads to the gradient of pressure which can be found from the equilibrium condition for the liquid mass element in radial direction. Hence, the radial pressure gradient is given by: \( \frac{dp}{dr} = \rho \omega^2 r\) where \(\rho\) is the density of the liquid.
03

Calculate required angular velocity for a given pressure at the bottom

In order to find the angular velocity to generate the pressure of 250 MPa, we can solve the pressure gradient equation for \(\omega\) setting the pressure difference from the free surface (where pressure is atmospheric) to the bottom equal to the given value. We also consider that our radius is varying from 50 mm (free surface) to 130 mm (bottom). thus, we integrate pressure gradient from \(r=0.05\) m to \(r=0.13\) m, and equate this to 250e6 Pa. We get: \(\omega= \sqrt[3]{\frac{dp}{2 \rho (r_2^2-r_1^2)}}\) where \(r_1\) is 50 mm, \(r_2\) is 130 mm, \(\rho\) is the density of water (1000 kg/m^3), and \(dp\) is the difference in pressure (250 MPa or 250e6 Pa).

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