Question

Water at 20 °C is flowing with velocity of 0.5 m/s between two parallel flat plates placed 1 cm apart. Determine the distances from the entrance at which the velocity and thermal boundary layers meet.

Answer 1

Therefore, the distance from the entrance at which the velocity and thermal boundary layer meets is x = 0.516 m, $x_{r}$ = 1.889 m

Explanation:

The value for the property of water was obtained at 20° C from the table.

ρ = 998 Kg/$m^{3}$

μ = 1.002 × $10^{-3}$ Kg/m.s

Pr = 7.01

The waters kinematic viscosity is calculated at 20° C using the relation.

$v=\frac{\mu}{\rho}$

$v=\frac{1.002 \times 10^{-3}}{998}$

v = 1.004 × $10^{-6}$ $m^{2}$/s

The thickness of thermal boundary layer is calculated using the relations.

$\delta_{r}=\frac{y}{2}$

$\delta_{r}$ = $\frac{0.01}{2}$

$\delta_{r}$ = 0.005 m

$\delta_{v}$ = $\delta_{r}$ = 0.005 m

Predict the flow is laminar,

The formula for the thickness of the velocity boundary layer is

$\delta_{v}=\frac{4.91}{\sqrt{\frac{V}{(v x)}}}$

The entrance by the distance was calculated in which the velocity boundary layer meets as shown.

The equation is rearranged.

$\sqrt{\frac{(v x)}{V}}=\frac{\delta_{v}}{4.91}$

Square Both the equations:

$(\sqrt{\frac{(v x)}{V}})^{2}=\left(\frac{\delta_{v}}{4.91}\right)^{2}$

$\frac{(v x)}{V}=\left(\frac{\delta_{v}}{4.91}\right)^{2}$

$x=\frac{V}{v}\left(\frac{\delta_{v}}{4.91}\right)^{2}$

$x=\frac{0.5}{1.004 \times 10^{-6}}\left(\frac{0.005}{4.91}\right)^{2}$

= $0.498 \times 10^{6} \times 1.036 \times 10^{-6}$

x = 0.516 m

The formula for the thermal boundary layers thickness is

$\delta_{r}$ = $\frac{4.91}{Pr^{1/3}\sqrt{\frac{v}{vx_{r} } } }$

The entrance by the distance was calculated in which the velocity boundary layer meets as shown.

$\sqrt{\frac{\left(v x_{r}\right)}{V}}=\frac{\delta_{r} \times \mathrm{Pr}^{1 / 3}}{4.91}$

Square Both the equations:

$(\sqrt{\frac{\left(v x_{r}\right)}{V}})^{2}=\left(\frac{\delta_{r} \times \mathrm{Pr}^{1/3}}{4.91}\right)^{2}$

$\frac{\left(v x_{r}\right)}{V}=\left(\frac{\delta_{r}}{4.91}\right)^{2}\left(\mathrm{Pr}^{1/3}\right)^{2}$

$x_{r}=\frac{V}{v}\left(\frac{\delta_{r}}{4.91}\right)^{2}\left(\mathrm{Pr}^{2 / 3}\right)$

$x_{r}=\frac{0.5}{1.004 \times 10^{-6}}\left(\frac{0.005}{4.91}\right)^{2}(7.01)^{2 / 3}$

$=0.498 \times 10^{6} \times 1.036 \times 10^{-6} \times 3.662$

$x_{r}$ = 1.889 m

The Reynolds number value is calculated, x = 0.516 m.

$\mathrm{Re}=\frac{V x}{v}$

$\mathrm{Re}=\frac{0.5 \times 0.516}{1.004 \times 10^{-6}}$

Re = 2.57 ×$10^{5}$ < 5 × $10^{5}$