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Question: If it takes 5 minutes to fill a 15 L bucket from a water tap of diameter \( \dfrac{2}{\sqrt{\pi }} \...

If it takes 5 minutes to fill a 15 L bucket from a water tap of diameter 2π\dfrac{2}{\sqrt{\pi }} cm then the Reynolds number for the flow is ( density of water = 103kg/m3  {{10}^{3}}{kg}/{{{m}^{3}}}\; and viscosity of water =103=\text{1}{{0}^{-\text{3}}} pa.s) close to
(A) 11,000
(B) 550
(C) 1100
(D) 5500

Explanation

Solution

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, [1L=103m]\left[ 1L={{10}^{-3}}m \right]
R=ρνdηR=\dfrac{\rho \nu d}{\eta } Here, ρ\rho \to Fluid density
vv\to Fluid velocity
η\eta \to Fluid viscosity
dd\to Diameter or Length of fluid.

Complete step by step solution
We have given,
( ρ\rho ) density of water = 103kg/m3  {{10}^{3}}{kg}/{{{m}^{3}}}\;
( η\eta ) viscosity of water= 10-3 pa.s
We have to find, Reynolds’s number which is given by,
R=ρνdηR=\dfrac{\rho \nu d}{\eta }
Here, ν\nu is the fluid velocity which can be obtained by following,
ν=4Qπd2\nu =\dfrac{4Q}{\pi {{d}^{2}}}
Q is the volume of water flowing out per second which is given by ,
Q=15L5minQ=\dfrac{15L}{5\min }
Q=15×103m35×60sQ=\dfrac{15\times {{10}^{-3}}{{m}^{3}}}{5\times 60s} [1L=103m]\left[ 1L={{10}^{-3}}m \right] 1 min =60 sec
Q=3×103m360sQ=\dfrac{3\times {{10}^{-3}}{{m}^{3}}}{60s}
Q=5×105m3Q=5\times {{10}^{-5}}{{m}^{3}}
Now, d is diameter = 2π\dfrac{2}{\sqrt{\pi }} cm = 2π\dfrac{2}{\sqrt{\pi }} ×10-2m
Reynolds’s number is given by,
R=ρνdηR=\dfrac{\rho \nu d}{\eta } = ρη4Qdπd2\dfrac{\rho }{\eta }\dfrac{4Qd}{\pi {{d}^{2}}}
R=ρ4QηπdR=\dfrac{\rho 4Q}{\eta \pi d}
Put all the values in above equation
R=103×4×5×105103×3.14×23.14×102R=\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 [π=3.14]\left[ \pi =3.14 \right]
= 4×53.14 ×2×102105\dfrac{4\times 5}{\sqrt{3.14}\text{ }\times \text{2}}\times \dfrac{{{10}^{-2}}}{{{10}^{-5}}}
= 20 2×1.773×103\dfrac{20}{\text{ 2}\times \text{1}\text{.773}}\times {{10}^{3}}
R5500R\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  2000\text{ 2}000 , the flow is called Laminar
If Re >4000>4000 , the flow is called turbulent
If 2000 << Re <4000<4000 , the flow is called transition.
Here, Re is Reynolds number,