Question

Laminar flow occurs between extensive stationary plates. The kinetic energy correction factor is nearly

• A. 1.0
• B. 1.5
• C. 2.0
• D. 2.3

Answer 1

The correct option is B 1.5


Assume width = 1

$U=-\frac{1}{2\mu }\left(\frac{\partial P}{\partial x}\right)(Hy-y^2)$

$\text{Kinetic energy correction factor}\left(\alpha \right)$

$\alpha =\frac{{\int }_0^H\frac{1}{2}d\dot{m}{u}^2}{\frac{1}{2}\dot{m}{\overline{u}}^2}=\frac{{\int }_0^H\rho \left(dy1\right)u.u^2}{\rho \left(H.1\right)\overline{u}.{\overline{u}}^2}$

$=\frac{{\int }_0^H{u}^{3}dy}{H{\overline{u}}^3}=1.543$

$\text{where}\overline{U}=-\frac{1}{12\mu }\left(\frac{\partial P}{\partial x}\right)H^2$