What is the Correct Equation for Flow Rate?

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SUMMARY

The correct equation for flow rate is given by the formula \(\frac{\pi r_o^4}{8\mu} \left[-\frac{d}{ds}(\rho + yz)\right]\). The user initially miscalculated the flow rate but later confirmed the correct answer after simplifying the integral \(\int_0^{r_0} (r_0^2 - r^2) r \, dr\). This simplification helped avoid complications arising from higher powers of \(r_0\), leading to the correct solution.

PREREQUISITES
  • Understanding of fluid dynamics principles
  • Familiarity with calculus, particularly integration
  • Knowledge of variables such as density (\(\rho\)), viscosity (\(\mu\)), and radius (\(r_o\))
  • Ability to manipulate differential equations
NEXT STEPS
  • Study the derivation of the Hagen-Poiseuille equation for flow rate
  • Learn about the implications of viscosity on fluid flow
  • Explore advanced integration techniques in calculus
  • Investigate the effects of varying radius on flow rate in cylindrical pipes
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Students in engineering or physics, particularly those studying fluid mechanics, as well as professionals involved in fluid dynamics analysis.

foo9008
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Homework Statement


the equation of flow aret given is [(pi)((r_o)^4 ) / 8μ ] [-d/ds (ρ +yz ) ] , but the ans that i got is - [(pi)((r_o)^4 ) / 8μ ] [-d/ds (ρ +yz ) ] , what's wrong with my working ?
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Homework Equations

The Attempt at a Solution

 
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Try the simpler $$\int_0^{r_0} (r_0^2 -r^2)\, r dr $$to avoid things like the ##r_0^3## (?) in your first step
 
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BvU said:
Try the simpler $$\int_0^{r_0} (r_0^2 -r^2)\, r dr $$to avoid things like the ##r_0^3## (?) in your first step
thanks , i got the ans already
 

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