Respuesta :
Therefore, the distance from the entrance at which the velocity and thermal boundary layer meets is x = 0.516 m, [tex]x_{r}[/tex] = 1.889 m
Explanation:
The value for the property of water was obtained at 20° C from the table.
ρ = 998 Kg/[tex]m^{3}[/tex]
μ = 1.002 × [tex]10^{-3}[/tex] Kg/m.s
Pr = 7.01
The waters kinematic viscosity is calculated at 20° C using the relation.
[tex]v=\frac{\mu}{\rho}[/tex]
[tex]v=\frac{1.002 \times 10^{-3}}{998}[/tex]
v = 1.004 × [tex]10^{-6}[/tex] [tex]m^{2}[/tex]/s
The thickness of thermal boundary layer is calculated using the relations.
[tex]\delta_{r}=\frac{y}{2}[/tex]
[tex]\delta_{r}[/tex] = [tex]\frac{0.01}{2}[/tex]
[tex]\delta_{r}[/tex] = 0.005 m
[tex]\delta_{v}[/tex] = [tex]\delta_{r}[/tex] = 0.005 m
Predict the flow is laminar,
The formula for the thickness of the velocity boundary layer is
[tex]\delta_{v}=\frac{4.91}{\sqrt{\frac{V}{(v x)}}}[/tex]
The entrance by the distance was calculated in which the velocity boundary layer meets as shown.
The equation is rearranged.
[tex]\sqrt{\frac{(v x)}{V}}=\frac{\delta_{v}}{4.91}[/tex]
Square Both the equations:
[tex](\sqrt{\frac{(v x)}{V}})^{2}=\left(\frac{\delta_{v}}{4.91}\right)^{2}[/tex]
[tex]\frac{(v x)}{V}=\left(\frac{\delta_{v}}{4.91}\right)^{2}\\[/tex]
[tex]x=\frac{V}{v}\left(\frac{\delta_{v}}{4.91}\right)^{2}[/tex]
[tex]x=\frac{0.5}{1.004 \times 10^{-6}}\left(\frac{0.005}{4.91}\right)^{2}[/tex]
= [tex]0.498 \times 10^{6} \times 1.036 \times 10^{-6}[/tex]
x = 0.516 m
The formula for the thermal boundary layers thickness is
[tex]\delta_{r}[/tex] = [tex]\frac{4.91}{Pr^{1/3}\sqrt{\frac{v}{vx_{r} } } }[/tex]
The entrance by the distance was calculated in which the velocity boundary layer meets as shown.
[tex]\sqrt{\frac{\left(v x_{r}\right)}{V}}=\frac{\delta_{r} \times \mathrm{Pr}^{1 / 3}}{4.91}[/tex]
Square Both the equations:
[tex](\sqrt{\frac{\left(v x_{r}\right)}{V}})^{2}=\left(\frac{\delta_{r} \times \mathrm{Pr}^{1/3}}{4.91}\right)^{2}[/tex]
[tex]\frac{\left(v x_{r}\right)}{V}=\left(\frac{\delta_{r}}{4.91}\right)^{2}\left(\mathrm{Pr}^{1/3}\right)^{2}\\[/tex]
[tex]x_{r}=\frac{V}{v}\left(\frac{\delta_{r}}{4.91}\right)^{2}\left(\mathrm{Pr}^{2 / 3}\right)[/tex]
[tex]x_{r}=\frac{0.5}{1.004 \times 10^{-6}}\left(\frac{0.005}{4.91}\right)^{2}(7.01)^{2 / 3}[/tex]
[tex]=0.498 \times 10^{6} \times 1.036 \times 10^{-6} \times 3.662[/tex]
[tex]x_{r}[/tex] = 1.889 m
The Reynolds number value is calculated, x = 0.516 m.
[tex]\mathrm{Re}=\frac{V x}{v}[/tex]
[tex]\mathrm{Re}=\frac{0.5 \times 0.516}{1.004 \times 10^{-6}}[/tex]
Re = 2.57 ×[tex]10^{5}[/tex] < 5 × [tex]10^{5}[/tex]
