
What will be the nature of the flow of water from a circular tap, when its flow rate increased from $0.18L{\min ^{ - 1}}$ to $0.48L{\min ^{ - 1}}$ ? The radius of the tap and viscosity of water is $0.5cm$ and ${10^{ - 3}}Pa\sec $, respectively. (Density of water:${10^3}Kg{m^{ - 3}}$)
(A) Remains steady flow
(B) unsteady to steady flow
(C) steady to unsteady flow
(D) Remains turbulent flow
Answer
164.4k+ views
Hint: In order to solve this question, the nature of water flow is determined by the Reynolds number and here we will find the initial and final Reynolds number and then using Reynolds number conditions we will determine the nature of the flow of water.
Formula used:
Reynolds Number is given by ${R_e} = \dfrac{{\rho vD}}{\eta }$ where, $\rho ,v,D$ is the density of water, rate of flow of water, D is the diameter of the tap and $\eta $ is the viscosity.
Complete answer:
We have given that the radius of the tap is $0.5cm$ so diameter is $D = 2r = 1 \times {10^{ - 2}}m$
initial rate of flow of water is ${v_{initial}} = 0.18L{\min ^{ - 1}} = \dfrac{{0.18}}{{60}} \times {10^{ - 3}}{m^3}{s^{ - 1}}$
Density of water is $\rho = {10^3}kg{m^{ - 3}}$
viscosity of water is $\eta = {10^{ - 3}}pa\sec $
Area of the pipe is $A = \pi {(0.5 \times {10^{ - 2}})^2}$
Since rate of flow of water is given in Volume change so we have to divide Reynolds number by area of the tap hence,
Initial Reynolds number is given by
$
{R_{initial}} = \dfrac{{({{10}^3})(0.18 \times {{10}^{ - 3}})(1 \times {{10}^{ - 2}})}}{{\pi {{(0.5 \times {{10}^{ - 2}})}^2} \times 60 \times {{10}^{ - 3}}}} \\
{R_{initial}} = 382.16 \\
$
So, when Reynolds number is less than $1000$ flow is steady so, Initial flow was steady.
Similarly, for final Reynolds number we have changed value as
${v_{final}} = 0.48L{\min ^{ - 1}} = \dfrac{{0.48}}{{60}} \times {10^{ - 3}}{m^3}{s^{ - 1}}$
so we get,
$
{R_{final}} = \dfrac{{({{10}^3})(0.48 \times {{10}^{ - 3}})(1 \times {{10}^{ - 2}})}}{{\pi {{(0.5 \times {{10}^{ - 2}})}^2} \times 60 \times {{10}^{ - 3}}}} \\
{R_{final}} = 1019.09 \\
$
so, when the Reynolds number is greater than $1000$ and less than $2000$ the flow is unsteady. So, The flow changes from steady to unsteady.
Hence, the correct answer is option (C) steady to unsteady flow
Note: The rate of flow of water was given in volume rate of change so always convert the volume rate of change in simple meter per the second form and also convert all quantities units in the same units to avoid any minor calculation errors.
Formula used:
Reynolds Number is given by ${R_e} = \dfrac{{\rho vD}}{\eta }$ where, $\rho ,v,D$ is the density of water, rate of flow of water, D is the diameter of the tap and $\eta $ is the viscosity.
Complete answer:
We have given that the radius of the tap is $0.5cm$ so diameter is $D = 2r = 1 \times {10^{ - 2}}m$
initial rate of flow of water is ${v_{initial}} = 0.18L{\min ^{ - 1}} = \dfrac{{0.18}}{{60}} \times {10^{ - 3}}{m^3}{s^{ - 1}}$
Density of water is $\rho = {10^3}kg{m^{ - 3}}$
viscosity of water is $\eta = {10^{ - 3}}pa\sec $
Area of the pipe is $A = \pi {(0.5 \times {10^{ - 2}})^2}$
Since rate of flow of water is given in Volume change so we have to divide Reynolds number by area of the tap hence,
Initial Reynolds number is given by
$
{R_{initial}} = \dfrac{{({{10}^3})(0.18 \times {{10}^{ - 3}})(1 \times {{10}^{ - 2}})}}{{\pi {{(0.5 \times {{10}^{ - 2}})}^2} \times 60 \times {{10}^{ - 3}}}} \\
{R_{initial}} = 382.16 \\
$
So, when Reynolds number is less than $1000$ flow is steady so, Initial flow was steady.
Similarly, for final Reynolds number we have changed value as
${v_{final}} = 0.48L{\min ^{ - 1}} = \dfrac{{0.48}}{{60}} \times {10^{ - 3}}{m^3}{s^{ - 1}}$
so we get,
$
{R_{final}} = \dfrac{{({{10}^3})(0.48 \times {{10}^{ - 3}})(1 \times {{10}^{ - 2}})}}{{\pi {{(0.5 \times {{10}^{ - 2}})}^2} \times 60 \times {{10}^{ - 3}}}} \\
{R_{final}} = 1019.09 \\
$
so, when the Reynolds number is greater than $1000$ and less than $2000$ the flow is unsteady. So, The flow changes from steady to unsteady.
Hence, the correct answer is option (C) steady to unsteady flow
Note: The rate of flow of water was given in volume rate of change so always convert the volume rate of change in simple meter per the second form and also convert all quantities units in the same units to avoid any minor calculation errors.
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