A model (1/10 of scale) of a tractor-trailer rig is tested in a wind tunnel. The

Question

A model (1/10 of scale) of a tractor-trailer rig is tested in a wind tunnel. The cross section area of the model is 0.1 m^2, the flow velocity in the wind tunnel is 75 m/s and the measured drag force is 334 N. (Note: the model length scale is 1/10 of the prototype)

1. Assuming that the drag force is a function of the cross-sectional area of the vehicle (which is the square of a length), the flow velocity, the fluid density, the fluid viscosity and the speed of sound, identify and name all the non-dimensional parameters that govern the problem.

2. Calculate the coefficient of drag assuming that the density of air is 1.25 kg/m^3.

3. Assuming that the coefficient of drag is constant, calculate the drag of the prototype for 90 km/h.

4. In order to achieve incomplete similarity, at least two non-dimensional parameters (including the coefficient of drag) need to be equal for the prototype and the model. Assuming a speed of 90 km/h for the prototype and assuming that the Reynolds numbers (prototype and model) are equal, what is the flow speed required for the model? If a flow is considered incompressible for Mach number lower than 0.2, do you think that the proposed incomplete similarity work? Why?

Explanation / Answer

1)Fd = Fn ( A,V,p,u,Vs)

Fd - Drag Force - kg m/s^2

A - cross sectional area - m^2

V - Fluid velocity - m/s

p - Density - kg/m^3

u - Viscosity- kg/m.s

Vs - Speed of sound- m/s

Taking p,A,V as Fundamental Variables

Pi 1: Fd/p*V^2*A

Pi 2 : u/p*V*A^1/2

Pi 3 : Vs/V

2) Drag Coefficient,Cd = Fd/(1/2*p*Vm^2*A) = 334*2/(1.25*75^2*0.1) = 0.95

3) Vp = 90 km/hr = 25 m/s

A is proportional to L^2

Ap/Am = (Lp/Lm)^2

But Lp/Lm = 10

Ap = Am*100

Ap = 10 m^2

Fd = Cd *1/2*p*Vp^2*Ap = 0.95*1/2*1.25*25^2*10 = 3711.11 N

4) Equating Reynolds number for model and prototype

p*Vm*Am^1/2/u = p*Vp*Ap^1/2/u

Vm*Am^1/2 = Vp*Ap^1/2

Given A = L^2

Vm*Lm = Vp*Lp

Given Lm = Lp/10

Vm = 10*Vp = 900 km/hr = 250 m/s

Mach no = V/Vs

Mach no for the Model = Vm/Vs = 250/343 = 0.73 > 0.2

Mach no for the prototype = Vp/Vs = 25/343 = 0.073 < 0.2

No the proposed incomplete similarity doesn't work becuase Reynolds number is not constant for model and prototype becuase the model has mach no > 0.2 the fluid is compressible and density changes

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