For Air Glycerin and Liquid Water at 50 degree F each flowin

For Air, Glycerin, and Liquid Water at 50 degree F each flowing separately over a flat plate at a free stream velocity of 6 ft./s, determine the velocity and thermal boundary layer thicknesses at a distance of 0.5 ft. from the leading edge. Calculate the ratio of the velocity boundary layer to the thermal boundary layer thickness for each of the three flows.

Solution

a)

For air at 50 deg F, kinematic viscosity neu = 1.535*10^-4 ft² /s , Pr = 0.7336

Re = V*L / neu

= 6 * 0.5 / 1.535*10^-4

= 19544

Since Re < 500,000, flow is laminar.

For laminar flow over flat plate, Velocity Boundary layer thickness = 5*x / sqrt (Re)

= 5*0.5 / sqrt (19544)

= 0.0178 ft = 0.2146 inches

Thermal Boundary layer thickness = Velocity BL thickness / Pr^1/3

= 0.2146 / 0.7336^1/3

= 0.238 inches

Ratio of Velocity to thermal BL thickness = Pr^1/3 = 0.7336^1/3 = 0.902

b)

For glycerin at 50 deg F, kinematic viscosity neu = 0.036 ft2 /s, Pr = 34561

Re = V*L / neu

= 6 * 0.5 / 0.036

= 0.0927

Since Re < 500,000, flow is laminar.

For laminar flow over flat plate, Velocity Boundary layer thickness = 5*x / sqrt (Re)

= 5*0.5 / sqrt (0.0927)

= 8.21 ft

Thermal Boundary layer thickness = Velocity BL thickness / Pr^1/3

= 8.21 / 34561^1/3

= 0.252 ft

Ratio of Velocity to thermal BL thickness = Pr^1/3 = 34561^1/3 = 32.57

c)

For water at 50 deg F, we have kinematic viscosity neu = 14.07*10^-6 ft² /s, Prandtl number Pr = 9.45

Re = V*L / neu

= 6 * 0.5 / 14.07*10^-6

= 213,220

Since Re < 500,000, flow is laminar.

For laminar flow over flat plate, Velocity Boundary layer thickness = 5*x / sqrt (Re)

= 5*0.5 / sqrt (213220)

= 0.00541 ft = 0.065 inches

Thermal Boundary layer thickness = Velocity BL thickness / Pr^1/3

= 0.065 / 9.45^1/3

= 0.031 inches

Ratio of Velocity to thermal BL thickness = Pr^1/3 = 9.45^1/3 = 2.11