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Question: A liquid of coefficient of the viscosity \(\eta = 1\) poise in a pipe of the radius \(3cm\) such tha...

A liquid of coefficient of the viscosity η=1\eta = 1 poise in a pipe of the radius 3cm3cm such that the rate of volume flow is 1000l/min1000l/\min . Determine the Reynolds numbers.
(A)3563(A)3563
(B)3500(B)3500
(C)3400(C)3400
(D)3600(D)3600

Explanation

Solution

The ratio of the inertial forces to the viscosity force that is subjected to the internal movement which is related to the different fluid velocities is known as the Reynolds number. It is a dimensionless number that is used to determine the types of flow patterns which are laminar or turbulent.
Formula used:
To find the Reynolds number,
Re=ρvLμ{R_e} = \dfrac{{\rho vL}}{\mu }
Where,
ρ\rho is the density,
vv is the speed of the flow,
LL is the linear dimension characteristic,
μ\mu is the dynamic viscosity

Complete step by step answer:
The values are given in the question. The liquid of coefficient of the viscosity η=1\eta = 1 poise, pipe of the radius 3cm3cm, and the rate of volume flow are 1000l/min1000l/\min .
The rate of the flow is 1000l/min1000l/\min
The value of πr2\pi {r^2}is 160m3/s\dfrac{1}{{60}}{m^3}/s
Substitute all the values in the Reynolds formula. We have,
Re=ρvLμ{R_e} = \dfrac{{\rho vL}}{\mu }
Where,
ρ\rho is the density,
vv is the speed of the flow,
LL is the linear dimension characteristic,
μ\mu is the dynamic viscosity
Re=10000.1×160πr2×2r\Rightarrow {R_e} = \dfrac{{1000}}{{0.1}} \times \dfrac{1}{{60\pi {r^2}}} \times 2r
Re=10000.1×2r60πr2\Rightarrow {R_e} = \dfrac{{1000}}{{0.1}} \times \dfrac{{2r}}{{60\pi {r^2}}}
Re=10000.1×260πr\Rightarrow {R_e} = \dfrac{{1000}}{{0.1}} \times \dfrac{2}{{60\pi r}}
Re=20001.0×60π×r\Rightarrow {R_e} = \dfrac{{2000}}{{1.0 \times 60\pi \times r}}
The value of the radius is 3cm3cm converting the centimeter into the meter we get,
Re=20001.0×60π×3×102\Rightarrow {R_e} = \dfrac{{2000}}{{1.0 \times 60\pi \times 3 \times {{10}^{ - 2}}}}
Substituting the value of π\pi,
Re=20001.0×60(3.16)×3×102\Rightarrow {R_e} = \dfrac{{2000}}{{1.0 \times 60\left( {3.16} \right) \times 3 \times {{10}^{ - 2}}}}
Re=2000568.8×102\Rightarrow {R_e} = \dfrac{{2000}}{{568.8 \times {{10}^{ - 2}}}}
Re=3563\Rightarrow {R_e} = 3563
Therefore the Reynolds number is 3563.

Hence option (A)\left( A \right) is the correct answer

Note: If the Reynolds number has a high value the pipe has a turbulent flow. If the Reynolds number has a low value the pipe has a laminar flow. Numerically these values are acceptable though in general the laminar and turbulent flow can be classified according to the range.