1.

Refer to Q6, and calculate the power dissipated per unit mass of fluid.(a) 0.050 m^2 s^-3(b) 0.042 m^2 s^-3(c) 0.040 m^2 s^-3(d) 0.052 m^2 s^-3

Answer» Right choice is (d) 0.052 m^2 s^-3

Explanation: (lambda^4 = frac{v^3}{varepsilon})

(varepsilon = frac{v^3}{λ^4})

Therefore,

(varepsilon = frac{(1.29×10^{-6})^3 ,m^6 ,s^{-3}}{(8×10^{-5})^4 m^4}) = 0.052 m^2 s^-3.


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