A test tube is spun in a centrifuge. The tube supportis mounted on a pivot so that the tube swings outward asrotation speed increases. At high speeds, the tube is nearlyhorizontal. Find (a) an expression for the radial componentof acceleration of a liquid element located at radius r, (b) theradial pressure gradient dp/dr, and (c) the required angularvelocity to generate a pressure of 250 MPa in the bottom of atest tube containing water. (The free surface and bottomradii are 50 and 130 mm, respectively.)
This text is generated by AI trained by Professor Kirby on his lectures, videos, notes, message board responses, and textbook
PK Bot’s answer:
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ENGINEERING CONTEXT: We are analyzing the pressure buildup in a water‐filled test tube spinning nearly horizontally in a centrifuge to design for high-pressure generation via centripetal acceleration.
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PHYSICS: The dominant effect is the imposed centripetal (radial) acceleration producing a pressure gradient in the rotating fluid. It is a specific application of the Euler equation in normal coordinates.
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ASSUMPTIONS: We assume that the tube is nearly horizontal so that the gravitational influence is negligible compared to the centrifugal forces and that water rotates as a solid body.
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SOLUTION TECHNIQUE:
(a) The Euler equation for the normal coordinate is
\frac{\partial k}{\partial n} = -\rho \frac{U^2}{R}
(b) Rearranging this for the radial direction r=-n, setting U=\omega r, setting R=r for solid-body rotation, and ignoring gravity gives
\frac{dp}{dr}=\rho\,\omega^2 r\,.
Integrating from the free surface (at r=r_f=0.050 m, where pressure is taken as ambient) to an arbitrary radius r yields
p(r)=p(r_f)+\frac{1}{2}\rho\,\omega^2\left(r^2-r_f^2\right)\,.
Thus, the pressure gradient is
\frac{dp}{dr}=\rho\,\omega^2 r\,.
(c) To generate a pressure difference \Delta p=250 MPa at the bottom (r=r_b=0.130 m) relative to the free surface, set
\Delta p=\frac{1}{2}\rho\,\omega^2 (r_b^2-r_f^2)\,.
Substitute \rho=1000~kg/m^3, r_b=0.130~m, and r_f=0.050~m:
250\times10^6=\frac{1}{2}(1000)\,\omega^2\left(0.130^2-0.050^2\right)\\[1mm]
250\times10^6=\frac{1}{2}\times1000\,\omega^2(0.0169-0.0025)\\[1mm]
250\times10^6=500\,\omega^2(0.0144)\\[1mm]
\omega^2=\frac{250\times10^6}{7.2} \approx 34.722\times10^6\\[1mm]
\omega\approx\sqrt{34.722\times10^6}\approx5888~\text{rad/s}\,.
- REFLECTION: The analysis shows that the centripetal acceleration produces a quadratic pressure rise with radius; the extremely high angular velocity (nearly 5900 rad/s) required to reach 250 MPa indicates that such pressures in a laboratory centrifuge setting would demand extraordinary design and materials as well as an extreme rotation rate. This is routinely achieved with ultracentrifuges, but they are expensive and operating them safely requires care.
To formulate this response, I reviewed the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2). Below, I have organized the sections based on their relevance. If they are marked as relevant, I encourage you to explore them:
Textbook sections:
- Chapter 9.3: Why We Use Irrotational Flow Analysis Part 5 (HIGHLY RELEVANT)
- Chapter 9.6: Centrifugal Pump Design Part 5 (HIGHLY RELEVANT)
- Chapter 9.6: Centrifugal Pump Design Part 1 (HIGHLY RELEVANT)
- Appendix F.3: Volumetric Viscous Force Part 1 (RELEVANT)
- Chapter 3 Engineering Models: Hydrostatic Equation (RELEVANT)
Disclaimer:
See What is Professor Kirby Bot for information about Professor Kirby Bot and its answers.
why is the Euler equation in normal coordinates the one to use
This text is generated by AI trained by Professor Kirby on his lectures, videos, notes, message board responses, and textbook
PK Bot’s answer:
The Euler equation in normal coordinates directly relates the curvature of a streamline (and thus the centripetal acceleration) to the pressure gradient acting normal to that streamline. In the centrifuge (or rotating test tube) case, the fluid moves in circular paths, so its acceleration is dominated by the centripetal term U^2/R. Writing the momentum balance in the direction normal to the streamline yields
\frac{\partial k}{\partial n} = -\frac{\rho U^2}{R}\,,
with k=p+\rho g z. When gravity is negligible (as in this nearly horizontal tube), this reduces to an equation for how pressure must vary radially to provide the necessary centripetal force.
The Euler equation for the normal coordinate comes directly from Newton’s second law applied in the direction perpendicular to the streamline in the direction of curvature. It is especially useful in situations where the fluid is forced to turn—like in a centrifuge—because the radial (or normal) acceleration is exactly what creates the pressure variation that sustains the curved motion.
The normal Euler equation gives you the direct relationship between curvature (through U^2/R) and the pressure gradient. This makes it the natural choice for analyzing the pressure differences driven by centripetal acceleration in the rotating test tube.
To formulate this response, I reviewed the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2). Below, I have organized the sections based on their relevance. If they are marked as relevant, I encourage you to explore them:
Textbook sections:
- Chapter 5.5: Euler Equation: Normal Direction (HIGHLY RELEVANT)
- Chapter 5 Engineering Models: Steady Euler Equations (HIGHLY RELEVANT)
- Chapter 5.7: Using the Euler Equations Correctly Part 2 (HIGHLY RELEVANT)
- Chapter 5.6: Euler Equations: Summary of Streamwise, Normal, and Binormal Equations (HIGHLY RELEVANT)
- Chapter 5.1: Inviscid Incompressible Flow (HIGHLY RELEVANT)
Disclaimer:
See What is Professor Kirby Bot for information about Professor Kirby Bot and its answers.