Question

Air at 25 C and 1 atm is flowing over a long flat plate with a velocity of 8 m/s. Determine the distance from the leading edge of the plate where the flow becomes turbulent, and the thickness of the boundary layer at that location. Repeat your solution for water flowing at 25 C and 1 atm.

Answer 1

Answer:

Explanation:

Given data

Temprature of air=$25^{\circ}$

pressure =1 atm

velocity(V)=8m/s

From the table for air at $25^{\circ}$ and 1 atm

kinematic viscosity$\left ( \nu\right )=1.562\times 10^{-5} m^{2}/s$

Reynolds number for turbulent flow$=5\times 10^{5}$

and Re.no.=$\frac{V \times X}{\nu }$

therefore length where turbulent flows start is

X=$\frac{\left ( Re.no.\right )\times \nu }{V}$

X=0.976 m

Thickness of boundary layer is given by

$\delta _x=\frac{5X}{\sqrt{Re_x}}$

$\delta _x=\frac{5\times 0.976}{\sqrt{5\times 10^5}}$

$\delta _x=6.901mm$

for water

kinematic viscosity$\left ( \nu\right )=8.91\times 10^{-7} m^{2}/s$

Reynolds number for turbulent flow$=5\times 10^{5}$

and Re.no.=$\frac{V \times X}{\nu }$

therefore length where turbulent flows start is

X=$\frac{\left ( Re.no.\right )\times \nu }{V}$

X=0.055m

Thickness of boundary layer is given by

$\delta _x=\frac{5X}{\sqrt{Re_x}}$

$\delta _x=\frac{5\times 0.055}{\sqrt{5\times 10^5}}$

$\delta _x=0.3889mm$

Answer 2

Answer:

The thickness of air flow will be 0.923 m

The thickness of water flow will be 0.056 m.

Explanation:

For flow over flat plate, the flow becomes turbulent at the reynold number value of 5 x 10^5

Reynold's number is given as:

Re = ρvx/μ

where,

ρ = density of fluid

v = velocity of flow

x = thickness of boundary layer

Re = reynold's number

For Air:

ρ = 1.225 kg/m^3

μ = 1.81 x 10^(-5) Pa.s

v = 8 m/s

therefore,

Re = 5 x 10^5 = (1.225 kg/m^3)(8 m/s)(x)/1.81 x 10^(-5) Pa.s

x = 0.923 m

For Water:

ρ = 997 kg/m^3

μ = 8.9 x 10^(-4) Pa.s

v = 8 m/s

therefore,

Re = 5 x 10^5 = (997 kg/m^3)(8 m/s)(x)/8.9 x 10^(-4) Pa.s

x = 0.056 m