What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t

Question

pentagon

3 months ago

8

What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t

he pressure difference caused by the radius change?

Physics

1 answer:

Yuliya22 [3.3K] · Answer 1 · 2026-02-21T03:04:50+03:00

Details that are not provided: the problem figure is included.

We can address the exercise by applying Poiseuille's law. This law indicates that for a fluid flowing in a laminar manner within a confined pipe,

$\Delta P = \frac{8 \mu L Q}{\pi r^4}$

where:

$\Delta P$ represents the pressure difference across the two ends

$\mu$ denotes the viscosity of the fluid

L signifies the length of the pipe

$Q=Av$ indicates the volumetric flow rate, where $A=\pi r^2$ is the cross-sectional area of the tube and $v$ refers to the fluid's velocity

r stands for the pipe's radius.

This law can be utilized for the needle, allowing us to compute the pressure difference between point P and the needle's end. In this scenario, we have:

$\mu=0.001 Pa/s$ is the dynamic viscosity of water at $20^{\circ}$

L=4.0 cm=0.04 m

$Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s$

and r=1 mm=0.001 m

Substituting these values into the formula yields:

$\Delta P = 3200 Pa$

This pressure difference specifies the value between point P and the needle's termination. As the end of the needle is under atmospheric pressure, the gauge pressure at point P, relative to atmospheric pressure, is exactly 3200 Pa.