What will be the nature of flow of water from a circular tap, when its flow rate increased from 0.18 L/min to 0.48 L/min? The radius of the tap and viscosity of water are 0.5 cm and 10$-$3 Pa s, respectively. (Density of water : 103 kg/m3)
Solution
The nature of flow is determined by reynolds no
<br><br>$R = {{\rho VD} \over \eta }$
<br><br>If R < 1000 $\to$ flow is steady
<br><br>1000 < R < 2000 $\to$ flow becomes unsteady
<br><br>R > 2000 $\to$ flow is turbulent
<br><br>$${R_1} = {{4 \times {{10}^3} \times 0.18 \times {{10}^{ - 3}}} \over {60 \times \pi \times {{10}^{ - 2}} \times {{10}^{ - 3}}}} = {{4 \times {{10}^5} \times 0.18} \over {60\pi }}$$<br><br>$= 0.0038 \times {10^5} = 380$<br><br>${R_2} = {{0.48} \over {0.18}} \times 380 = 1018$
About this question
Subject: Physics · Chapter: Properties of Solids and Liquids · Topic: Elasticity
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