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Standard XII

Physics

Speed of Efflux

A flocculatio...

Question

A flocculation tank contains $1800 m^{3}$ of water, which is mixed using paddles at an average velocity gradient G of $100 / s$ . The water temperature and the corresponding dynamic viscosity are $30^{\circ} C$ and $0.798 \times 10^{- 3} N s / m^{2}$ , respectively. The theoretical power required to achieve the stated value of G (in kW, up to two decimal places) is
14.36

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Solution

The correct option is A 14.36
Velocity Gradient,

$G = \sqrt{\frac{P}{μ V}}$

Given,
$G = 100 S^{- 1}$
$P = ?, μ = 0.798 \times 10^{- 3} N s / m^{2}$
$V = 1800 m^{3}$

$\Rightarrow P = 100^{2} \times 0.798 \times 10^{- 3} \times 1800$

$= 14.364 k W$

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The correct option is A 14.36Velocity Gradient, G=√PμV Given, G=100 S−1 P=?,μ=0.798×10−3Ns/m2 V=1800 m3 ⇒P=1002×0.798×10−3×1800 =14.364 kW

The correct option is A 14.36
Velocity Gradient,

$G = \sqrt{\frac{P}{μ V}}$

Given,
$G = 100 S^{- 1}$
$P = ?, μ = 0.798 \times 10^{- 3} N s / m^{2}$
$V = 1800 m^{3}$

$\Rightarrow P = 100^{2} \times 0.798 \times 10^{- 3} \times 1800$

$= 14.364 k W$