Answer to Question #125513 in Physics for Afira Arifff

Question #125513
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
1
Expert's answer
2020-07-07T10:02:40-0400

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


Re=QDρAμRe = \dfrac{QD\rho}{A\mu}

where Q=4m3/sQ = 4m^3/s is the volumetric flow rate, D=0.06mD = 0.06m is the pipe diameter, A=πD2/4A = \pi D^2/4 is the pipe's cross-sectional area, ρ=1268kg/m3\rho = 1268kg/m^3 is a density and μ=17Pas\mu = 17Pa\cdot s is is the dynamic viscosity.

Substituting this values to the equation, get:


Re=4QDρπD2μ=4QρπDμ=441268π0.06176331Re = \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>2900Re>2900. Thus, this flow is turbulent.


Answer. Re = 6331, turbulent.


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