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Numerical · Q18

Q.Water (density 1000 kg/m31000\ \text{kg/m}^3, coefficient of viscosity 1.0×10−3 Pa⋅s1.0\times10^{-3}\ \text{Pa·s}) is made to flow through a pipe of diameter 2.0 cm2.0\ \text{cm}. Taking the critical value of the Reynolds number for this pipe to be 20002000, calculate the critical velocity of flow above which the motion of the water would become turbulent.

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✓ Free question

Given: ρ=1000 kg/m3\rho = 1000\ \text{kg/m}^3, η=1.0×10−3 Pa⋅s\eta = 1.0\times10^{-3}\ \text{Pa·s}, d=2.0 cm=0.02 md = 2.0\ \text{cm} = 0.02\ \text{m}, Rec=2000Re_c = 2000.

Rearranging Rec=ρvcdηRe_c = \dfrac{\rho v_c d}{\eta} for the critical velocity vcv_c:

vc=Rec ηρd=2000×1.0×10−31000×0.02=2.020=0.10 m/sv_c = \frac{Re_c\,\eta}{\rho d} = \frac{2000\times1.0\times10^{-3}}{1000\times0.02} = \frac{2.0}{20} = 0.10\ \text{m/s}

✓Final answer

Critical velocity vc=0.10 m/sv_c = 0.10\ \text{m/s}; flow through this pipe would remain streamline below this speed and become turbulent above it.

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