Question

A small wind tunnel in a university’s undergraduate fluid flow laboratory has a test section that is 20 in. by 20 in. in cross section and is 4.0 ft long. Its maximum speed is 145 ft/s. Some students wish to build a model 18-wheeler to study how aerodynamic drag is affected by rounding off the back of the trailer. A full-size (prototype) tractor-trailer rig is 52 ft long, 8.33 ft wide, and 12 ft high. Both the air in the wind tunnel and the air flowing over the prototype are at 80 °F and atmospheric pressure.What is the largest scale model they can build to stay within the rule-of-thumb guidelines for blockage? (Convert the scaling ratio to percentage. Round the final answer to two decimal places.) ______What are the dimensions of the model truck in inches?The width of the model truck is _____ .The height of the model truck is _____ .The length of the model truck is _____ .What is the maximum model truck Reynolds number (based on truck width) achievable by the students?The maximum Reynolds number is ______ × 10–5.

Answer 1

a. Wm = 4.56 in, Hm = 6.57 in, and Lm = 28.5 in, x = 182.7%, b . Reynolds number is 3.25 x 10⁵

Step-by-step explanation:

a. The rule of thumb guideline given is that blockage should be kept below 7.5 %. Therefore, the frontal area of the model truck must be no more than 0.075 then the area of the wind tunnel.

Notice the ratio of the height H and width W of the full-sized truck

H(p)/W(p) = 12/8.33 = 1.44

Realize that the area of the model truck is equal to 1.44W²p

So, from rule of thumb

0.075A(wind tunnel) = 1.44W²(m)

W²(m) = (0.075/1.44) (20 x 20)

W(m) = √20.832 = 4.56in

Calculate the ratio of the prototype to model

X = W(p)/W(m)

X = 8.33/4.56 = 1.827 or 182.7%

Calculate the length of truck of model

L(m) = L(p)/x

= 52/1.827 = 28.46 in

Calculate the height of the truck of model

H(m) = H(p)/x

= 12/1.827 = 6.57 in

Thus, the largest scale by rule of thumb is W(p)/W(m) =1.827

Thus, the dimension of model truck is Wm = 4.56 in, Hm = 6.57 in, and Lm = 28.5 in

b. The Reynolds Number Re, can be calculated using the density of air, the width of the model W(m) the maximum velocity of the wind tunnel V(max) and the viscosity of air μ

Re = ρ x W(m) x V(max)/μ

Where ρ = 0.0735lbm/ft³, μ = 1.248 x 10⁻⁵, Wm = 4.56 in = 0.38ft, V(max) = 145ft/s

Re = (0.0735) x (0.38) x (145)/(1.248 x 10⁻⁵)

= 3.25 x 105

Thus,. The maximum Reynolds number is 3.25 x 10⁵