Answer:
a). 8.67 x
m
b).0.3011 m
c).0.0719 m
d).0.2137 N
e).1.792 N
Explanation:
Given :
Temperature of air, T = 293 K
Air Velocity, U = 5 m/s
Length of the plate is L = 6 m
Width of the plate is b = 5 m
Therefore Dynamic viscosity of air at temperature 293 K is, μ = 1.822 X
Pa-s
We know density of air is ρ = 1.21 kg /![m^{3}](https://tex.z-dn.net/?f=m%5E%7B3%7D)
Now we can find the Reyonld no at x = 1 m from the leading edge
Re = ![\frac{\rho .U.x}{\mu }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Crho%20.U.x%7D%7B%5Cmu%20%7D)
Re = ![\frac{1.21 \times 5\times 1}{1.822\times 10^{-5} }](https://tex.z-dn.net/?f=%5Cfrac%7B1.21%20%5Ctimes%205%5Ctimes%201%7D%7B1.822%5Ctimes%2010%5E%7B-5%7D%20%7D)
Re = 332052.6
Therefore the flow is laminar.
Hence boundary layer thickness is
δ = ![\frac{5.x}{\sqrt{Re}}](https://tex.z-dn.net/?f=%5Cfrac%7B5.x%7D%7B%5Csqrt%7BRe%7D%7D)
= ![\frac{5\times 1}{\sqrt{332052.6}}](https://tex.z-dn.net/?f=%5Cfrac%7B5%5Ctimes%201%7D%7B%5Csqrt%7B332052.6%7D%7D)
= 8.67 x
m
a). Boundary layer thickness at x = 1 is δ = 8.67 X
m
b). Given Re = 100000
Therefore the critical distance from the leading edge can be found by,
Re = ![\frac{\rho .U.x}{\mu }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Crho%20.U.x%7D%7B%5Cmu%20%7D)
100000 = ![\frac{1.21\times5\times x}{1.822 \times10^{-5}}](https://tex.z-dn.net/?f=%5Cfrac%7B1.21%5Ctimes5%5Ctimes%20x%7D%7B1.822%20%5Ctimes10%5E%7B-5%7D%7D)
x = 0.3011 m
c). Given x = 3 m from the leading edge
The Reyonld no at x = 3 m from the leading edge
Re = ![\frac{\rho .U.x}{\mu }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Crho%20.U.x%7D%7B%5Cmu%20%7D)
Re = ![\frac{1.21 \times 5\times 3}{1.822\times 10^{-5} }](https://tex.z-dn.net/?f=%5Cfrac%7B1.21%20%5Ctimes%205%5Ctimes%203%7D%7B1.822%5Ctimes%2010%5E%7B-5%7D%20%7D)
Re = 996158.06
Therefore the flow is turbulent.
Therefore for a turbulent flow, boundary layer thickness is
δ = ![\frac{0.38\times x}{Re^{\frac{1}{5}}}](https://tex.z-dn.net/?f=%5Cfrac%7B0.38%5Ctimes%20x%7D%7BRe%5E%7B%5Cfrac%7B1%7D%7B5%7D%7D%7D)
= ![\frac{0.38\times 3}{996158.06^{\frac{1}{5}}}](https://tex.z-dn.net/?f=%5Cfrac%7B0.38%5Ctimes%203%7D%7B996158.06%5E%7B%5Cfrac%7B1%7D%7B5%7D%7D%7D)
= 0.0719 m
d). Distance from the leading edge upto which the flow will be laminar,
Re = ![\frac{\rho \times U\times x}{\mu }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Crho%20%5Ctimes%20U%5Ctimes%20x%7D%7B%5Cmu%20%7D)
5 X
= ![\frac{1.21 \times 5\times x}{1.822\times 10^{-5}}}](https://tex.z-dn.net/?f=%5Cfrac%7B1.21%20%5Ctimes%205%5Ctimes%20x%7D%7B1.822%5Ctimes%2010%5E%7B-5%7D%7D%7D)
x = 1.505 m
We know that the force acting on the plate is
= ![\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20C_%7BD%7D%5Ctimes%20%5Crho%20%5Ctimes%20A%5Ctimes%20U%5E%7B2%7D)
and
at x= 1.505 for a laminar flow is = ![\frac{1.328}{\sqrt{Re}}](https://tex.z-dn.net/?f=%5Cfrac%7B1.328%7D%7B%5Csqrt%7BRe%7D%7D)
= ![\frac{1.328}{\sqrt{5\times10 ^{5}}}](https://tex.z-dn.net/?f=%5Cfrac%7B1.328%7D%7B%5Csqrt%7B5%5Ctimes10%20%5E%7B5%7D%7D%7D)
= 1.878 x ![10^{-3}](https://tex.z-dn.net/?f=10%5E%7B-3%7D)
Therefore,
= ![\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20C_%7BD%7D%5Ctimes%20%5Crho%20%5Ctimes%20A%5Ctimes%20U%5E%7B2%7D)
= ![\frac{1}{2}\times 1.878\times 10^{-3}\times 1.21\times (5\times 1.505)\times 5^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Ctimes%201.878%5Ctimes%2010%5E%7B-3%7D%5Ctimes%201.21%5Ctimes%20%285%5Ctimes%201.505%29%5Ctimes%205%5E%7B2%7D)
= 0.2137 N
e). The flow is turbulent at the end of the plate.
Re = ![\frac{\rho \times U\times x}{\mu }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Crho%20%5Ctimes%20U%5Ctimes%20x%7D%7B%5Cmu%20%7D)
= ![\frac{1.21 \times 5\times 6}{1.822\times 10^{-5} }](https://tex.z-dn.net/?f=%5Cfrac%7B1.21%20%5Ctimes%205%5Ctimes%206%7D%7B1.822%5Ctimes%2010%5E%7B-5%7D%20%7D)
= 1992316
Therefore
= ![\frac{0.072}{Re^{\frac{1}{5}}}](https://tex.z-dn.net/?f=%5Cfrac%7B0.072%7D%7BRe%5E%7B%5Cfrac%7B1%7D%7B5%7D%7D%7D)
= ![\frac{0.072}{1992316^{\frac{1}{5}}}](https://tex.z-dn.net/?f=%5Cfrac%7B0.072%7D%7B1992316%5E%7B%5Cfrac%7B1%7D%7B5%7D%7D%7D)
= 3.95 x ![10^{-3}](https://tex.z-dn.net/?f=10%5E%7B-3%7D)
Therefore
= ![\frac{1}{2}\times C_{D}\times \rho\times A\times U^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20C_%7BD%7D%5Ctimes%20%5Crho%5Ctimes%20A%5Ctimes%20U%5E%7B2%7D)
= ![\frac{1}{2}\times 3.95\times 10^{-3}\times 1.21\times (5\times 6)\times 5^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Ctimes%203.95%5Ctimes%2010%5E%7B-3%7D%5Ctimes%201.21%5Ctimes%20%285%5Ctimes%206%29%5Ctimes%205%5E%7B2%7D)
= 1.792 N