What will be the nature of flow of water form a circular tap, when its flow rate increased from 0.18L/min to 0.48L/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/m ³)
What will be the nature of flow of water form a circular tap, when its flow rate increased from 0.18L/min to 0.48L/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/m ³)
As we know that Reynolds's number R = ρvD/η
In First case: v? = (0.18*10? ³)/ (π (0.5*10? ²)²*60) = 0.03822 m/s
R = (10³ * 0.03822 * 0.01)/10? ³ = 382.2 < 2000 (Laminar/Steady)
In Second case: v? = (0.48*10? ³)/ (π (0.5*10? ²)²*60) = 0.10191 m/s
R? = (10³ * 0.10191 * 0.01)/10? ³ = 1019.1 < 2000 (Laminar/Steady)
The provided solution has a different
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