for low-speed (laminar) flow through a circular pipe, eh vleocity distribution takes the form u=(B/mu)(r0^2-r^2), where mu is viscosity. What is the max velocity and mass in terms of B, mu, rho, and r0
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PK Bot’s answer:
Using the given velocity profile
u(r)=\frac{B}{\mu}\Bigl(r_0^2-r^2\Bigr),
the maximum velocity occurs at the centerline, where r=0. Hence,
u_{\max}=\frac{B}{\mu}\,r_0^2.
We assume that by “mass” the problem refers to the mass flow rate \dot{m}. For an incompressible fluid, the mass flow rate is the density times the volumetric flow rate, i.e.,
\dot{m}=\rho Q,
where
Q=\int_0^{r_0} u(r)\, 2\pi r\,dr.
Substitute u(r):
Q=\frac{B}{\mu}\,2\pi\int_0^{r_0}\bigl(r_0^2-r^2\bigr)r\,dr.
Evaluate the integral:
\int_0^{r_0}\bigl(r_0^2r-r^3\bigr)\,dr
=\; r_0^2\int_0^{r_0}r\,dr-\int_0^{r_0}r^3\,dr
=\; r_0^2\left[\frac{r^2}{2}\right]_0^{r_0}-\left[\frac{r^4}{4}\right]_0^{r_0}
=\; r_0^2\frac{r_0^2}{2}-\frac{r_0^4}{4}
=\; \frac{r_0^4}{2}-\frac{r_0^4}{4}
=\; \frac{r_0^4}{4}\,.
Then
Q=\frac{B}{\mu}\,2\pi\left(\frac{r_0^4}{4}\right)
=\; \frac{B\pi\,r_0^4}{2\mu}\,.
Thus the mass flow rate is
\dot{m}=\rho Q=\frac{B\pi\,\rho\,r_0^4}{2\mu}\,.
Answers:
• Maximum velocity: u_{\max}=\dfrac{B\,r_0^2}{\mu}
• Mass flow rate: \displaystyle \dot{m}=\dfrac{B\pi\,\rho\,r_0^4}{2\mu}
To formulate this response, I reviewed the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I have ranked their significance below. If they are marked as relevant, I encourage you to take a look at them:
Textbook sections:
- Chapter 11.3: Poiseuille Flow Part 3 (HIGHLY RELEVANT)
- Chapter 11.3: Poiseuille Flow Part 1 (HIGHLY RELEVANT)
- Chapter 9.3: Why We Use Irrotational Flow Analysis Part 5 (HIGHLY RELEVANT)
- Chapter 6 Fundamentals: Area-averaged speed (HIGHLY RELEVANT)
- Chapter 11 Fundamentals: Laminar and Turbulent Flow (HIGHLY RELEVANT)
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