Answer to Question #125513 in Physics for Afira Arifff

Question

c) The flow rate of a chocolate syrup is 4 m3s-1 and the cylindrical pipe in which the fluid flows has a diameter of 6 cm. Determine the Reynold’s number for the syrup which has a density of 1268 kgm-3 and a viscosity of 17 Pas. Indicate if the flow is laminar or turbulent

Answer 1

The Reynold’s number for the cylindrical pipe is given by the following expression:

$$Re = \dfrac{QD\rho}{A\mu}$$

where $Q = 4m^3/s$ is the volumetric flow rate, $D = 0.06m$ is the pipe diameter, $A = \pi D^2/4$ is the pipe's cross-sectional area, $\rho = 1268kg/m^3$ is a density and $\mu = 17Pa\cdot s$ is is the dynamic viscosity.

Substituting this values to the equation, get:

$$Re = \dfrac{4QD\rho}{\pi D^2\mu} = \dfrac{4Q\rho}{\pi D\mu} = \dfrac{4\cdot4\cdot 1268}{\pi \cdot 0.06\cdot 17} \approx 6331$$

A flow can be considered as turbulent if $Re>2900$. Thus, this flow is turbulent.

Answer. Re = 6331, turbulent.