Question
Download Solution PDFA sphere of diameter 30 cm is moving with a uniform velocity of 4 m/s. The dynamic viscosity and specific gravity of the liquid are given as 0.8 poises and 0.9 respectively. What is the value of the Reynolds number?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Reynolds Number Re is defined as the ratio of the Inertia force to viscous force.
Inertia force (Fi) = mass × acceleration
\({{\rm{F}}_{\rm{i}}} = {\rm{ρ }} \times {\rm{volume}} \times \frac{{{\rm{velocity}}}}{{{\rm{time}}}} = {\rm{ρ }} \times {\rm{velocity}} \times \frac{{{\rm{volume}}}}{{{\rm{time}}}} = {\rm{ρ }} \times {\rm{V}} \times {\rm{AV}}\)
Viscous force (Fv) = shear stress × area \(= {\rm{\;}}\tau \times {\rm{A}} = \mu \times \frac{{\rm{V}}}{{\rm{L}}} \times {\rm{A}}\)
Reynolds Number (Re) = \(\frac{{ρ \times {\bf{V}} \times {\rm{L}}}}{\mu }\)
Calculation:
\({{\rm{R}}_{\rm{e}}} = \frac{{{\rm{ρ }} \times {\rm{V}} \times {\rm{d}}}}{{\rm{\mu }}}\)
ρ = 0.9 g/cm3 = 900 kg/m3, V = 4 m/s, d = 30 cm = 0.3 m, μ = 0.8 poise = 0.08 Ns/m2
\({{\rm{R}}_{\rm{e}}} = \frac{{900 \times 4 \times 0.3}}{{0.08}} = 13500\)
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