Question

A wind tunnel is used to study the flow around a car. The air is drawn at 60 mph into the tunnel. (a) Determine the pressure in the test section as determined by the manometer. (b) Determine the pre

Answer 1

The answer is below

Step-by-step explanation:

The complete question is attached.

a) Bernoulli equation is given as:

`P+(1)/(2)\rho V^2+ \rho gz=constant\\\\(P)/(\rho g) +(V^2)/(2g) +z=constant\\`

Where P = pressure, V = velocity, z = height, g = acceleration due to gravity and ρ = density.

`(P)/(\rho g) +(V^2)/(2g) +z=constant\\\\(P)/(\gamma) +(V^2)/(2g) +z=constant\\\\(P_1)/(\gamma) +(V_1^2)/(2g) +z_1=(P_2)/(\gamma) +(V_2^2)/(2g) +z_2\\\\but \ z_1=z_2,P_1=0,V_1=0,V_2=60\ mph=88\ ft/s. Hence:\\\\(P_2)/(\gamma) =-(V_2^2)/(2g) \\\\P_2=\gamma*-(V_2^2)/(2g) =\rho g*-(V_2^2)/(2g) \\\\P_2=-(V_2^2)/(2)*\rho=-((88.8\ ft/s)^2)/(2) * 0.00238\ slug/ft^3=-9.22\ lb/ft^2\\\\P_2+\gamma_(H_2O)h-\gamma_(oil)(1/12 \ ft)=0\\\\`

`\gamma_(oil)=0.9\gamma_(H_2O)=0.9*62.4\ lb/ft^3=56.2\ lb/ft^3\\\\Therefore:\\\\-9.22\ lb/ft^2+62.4\ lb/ft^3(h)-56.2\ lb/ft^3(1/12\ ft)=0\\\\h=0.223\ ft`

b)

`(P)/(\gamma) +(V^2)/(2g) +z=constant\\\\(P_2)/(\gamma) +(V_2^2)/(2g) +z_2=(P_3)/(\gamma) +(V_3^2)/(2g) +z_3\\\\but \ z_3=z_2,V_3=0,V_2=60\ mph=88\ ft/s. \\\\(P_2)/(\gamma)+(V_2^2)/(2g) = (P_3)/(\gamma)\\\\(P_3-P_2)/(\gamma)=(V_2^2)/(2g) \\\\P_3-P_2=(V_2)/(2g)*\gamma=(V_2^2)/(2g)*\rho g\\\\P_3-P_2=(V_2)/(2)*\rho=((88\ ft/s^2)^2)/(2)*0.00238\ slg\ft^3\\\\P_3-P_2=9.22\ lb/ft^2`