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)
Option 1 -
Remains steady flow
Option 2 -
Unsteady to steady flow
Option 3 -
Remains turbulent flow
Option 4 -
Steady flow to unsteady flow
-
1 Answer
-
Correct Option - 4
Detailed Solution: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 calculation for R, leading to a different conclusion.
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 65k Colleges
- 1.2k Exams
- 688k Reviews
- 1800k Answers