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Hint: Reynolds number is a dimensionless value which is applied in fluid mechanics to represent whether the fluid flow in a duct or part of a body is steady or turbulent. Reynolds number is given by ratio of inertial force to viscous force. Use the following formula, $ \left[ 1L={{10}^{-3}}m \right] $
$ R=\dfrac{\rho \nu d}{\eta } $ Here, $ \rho \to $ Fluid density
$ v\to $ Fluid velocity
$ \eta \to $ Fluid viscosity
$ d\to $ Diameter or Length of fluid.
Complete step by step solution
We have given,
( $ \rho $ ) density of water = $ {{10}^{3}}{kg}/{{{m}^{3}}}\; $
( $ \eta $ ) viscosity of water= 10-3 pa.s
We have to find, Reynolds’s number which is given by,
$ R=\dfrac{\rho \nu d}{\eta } $
Here, $ \nu $ is the fluid velocity which can be obtained by following,
$ \nu =\dfrac{4Q}{\pi {{d}^{2}}} $
Q is the volume of water flowing out per second which is given by ,
$ Q=\dfrac{15L}{5\min } $
$ Q=\dfrac{15\times {{10}^{-3}}{{m}^{3}}}{5\times 60s} $ $ \left[ 1L={{10}^{-3}}m \right] $ 1 min =60 sec
$ Q=\dfrac{3\times {{10}^{-3}}{{m}^{3}}}{60s} $
$ Q=5\times {{10}^{-5}}{{m}^{3}} $
Now, d is diameter = $ \dfrac{2}{\sqrt{\pi }} $ cm = $ \dfrac{2}{\sqrt{\pi }} $ ×10-2m
Reynolds’s number is given by,
$ R=\dfrac{\rho \nu d}{\eta } $ = $ \dfrac{\rho }{\eta }\dfrac{4Qd}{\pi {{d}^{2}}} $
$ R=\dfrac{\rho 4Q}{\eta \pi d} $
Put all the values in above equation
$ R=\dfrac{{{10}^{3}}\times 4\times 5\times {{10}^{-5}}}{{{10}^{-3}}\times 3.14\times \dfrac{2}{\sqrt{3.14}}\times {{10}^{-2}}} $ ….. Use $ \left[ \pi =3.14 \right] $
= $ \dfrac{4\times 5}{\sqrt{3.14}\text{ }\times \text{2}}\times \dfrac{{{10}^{-2}}}{{{10}^{-5}}} $
= $ \dfrac{20}{\text{ 2}\times \text{1}\text{.773}}\times {{10}^{3}} $
$ R\approx 5500 $ , This is the approximate value of Reynolds’s number.
Note
Reynolds number formula is used to determine the diameter, velocity and viscosity of the fluid
If Re $ \text{ 2}000 $ , the flow is called Laminar
If Re $ >4000 $ , the flow is called turbulent
If 2000 $ < $ Re $ <4000 $ , the flow is called transition.
Here, Re is Reynolds number,
$ R=\dfrac{\rho \nu d}{\eta } $ Here, $ \rho \to $ Fluid density
$ v\to $ Fluid velocity
$ \eta \to $ Fluid viscosity
$ d\to $ Diameter or Length of fluid.
Complete step by step solution
We have given,
( $ \rho $ ) density of water = $ {{10}^{3}}{kg}/{{{m}^{3}}}\; $
( $ \eta $ ) viscosity of water= 10-3 pa.s
We have to find, Reynolds’s number which is given by,
$ R=\dfrac{\rho \nu d}{\eta } $
Here, $ \nu $ is the fluid velocity which can be obtained by following,
$ \nu =\dfrac{4Q}{\pi {{d}^{2}}} $
Q is the volume of water flowing out per second which is given by ,
$ Q=\dfrac{15L}{5\min } $
$ Q=\dfrac{15\times {{10}^{-3}}{{m}^{3}}}{5\times 60s} $ $ \left[ 1L={{10}^{-3}}m \right] $ 1 min =60 sec
$ Q=\dfrac{3\times {{10}^{-3}}{{m}^{3}}}{60s} $
$ Q=5\times {{10}^{-5}}{{m}^{3}} $
Now, d is diameter = $ \dfrac{2}{\sqrt{\pi }} $ cm = $ \dfrac{2}{\sqrt{\pi }} $ ×10-2m
Reynolds’s number is given by,
$ R=\dfrac{\rho \nu d}{\eta } $ = $ \dfrac{\rho }{\eta }\dfrac{4Qd}{\pi {{d}^{2}}} $
$ R=\dfrac{\rho 4Q}{\eta \pi d} $
Put all the values in above equation
$ R=\dfrac{{{10}^{3}}\times 4\times 5\times {{10}^{-5}}}{{{10}^{-3}}\times 3.14\times \dfrac{2}{\sqrt{3.14}}\times {{10}^{-2}}} $ ….. Use $ \left[ \pi =3.14 \right] $
= $ \dfrac{4\times 5}{\sqrt{3.14}\text{ }\times \text{2}}\times \dfrac{{{10}^{-2}}}{{{10}^{-5}}} $
= $ \dfrac{20}{\text{ 2}\times \text{1}\text{.773}}\times {{10}^{3}} $
$ R\approx 5500 $ , This is the approximate value of Reynolds’s number.
Note
Reynolds number formula is used to determine the diameter, velocity and viscosity of the fluid
If Re $ \text{ 2}000 $ , the flow is called Laminar
If Re $ >4000 $ , the flow is called turbulent
If 2000 $ < $ Re $ <4000 $ , the flow is called transition.
Here, Re is Reynolds number,
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