Water flows past a flat plate that is orientedparallelto the flow with an upstream velocity of 0.5 m/s. (a) Determine the approximate location downstream from the leadingedge where the boundary layer becomes turbulent. (b) What is the boundary layer thickness at this location?

A) 1.12m

B) 7.92 x 10^(-3) m

Explanation:

A) Formula for reynolds number to reflect this question is given as;

R_e = Vx/v

Where;

V is velocity

x is polar coordinate where the boundary layer will become turbulent.

R_e is reynolds number.

Reynolds number for turbulent flow of flat plates has a value of 5 x 10^(5)

v is kinematic velocity

From water tables at standard conditions, kinematic viscosity is 1.12 x 10^(-6) m²/s

Now, R_e = Vx/v

We are looking for x, so let's make x the subject.

x = v•R_e/V

Thus, plugging in the relevant values to obtain;

x = [1.12 x 10^(-6) x 5 x 10^(5)]/0.5

x = 1.12m

B) The boundary layer thickness at the point where the flow becomes turbulent is given by;

δ = 5√(vx/V)

Where;

V is velocity

x is polar coordinate where the boundary layer will become turbulent.

R_e is reynolds number.

Reynolds number for turbulent flow of flat plates has a value of 5 x 10^(5)

v is kinematic velocity

Plugging in the relevant values to get ;

δ = 5√(1.12 x 10^(-6)•1.12/0.5)

δ = 5√(2.5088 x 10^(-6)

δ = 7.92 x 10^(-3) m

answer: chicken nuggets.

explanation: