A Venturi tube may be used as the inlet to an automobile carburetor. If an inlet pipe with a diameter of 2.0 cm diameter narrows to diameter of 1.0 cm, determine the pressure drop in the constricted section for an initial airflow of 3.0 m/s in the 2-cm section? (Assume air density is 1.25 kg/m

I think it's 81

the pressure drop is equal to 80.99 Pa

Explanation:

we have the following dа ta:

d1 = 2 cm = 0.02 m

d2 = 1 cm = 0.01 m

v = 3 m/s

p = 1.25 kg/m^3

ΔP = ?

For the calculation of the pressure drop we will use Bernoulli's principle for the Venturi Tube:

P1 - P2 = ((v^2*p) / 2) * ((A1^2/A2^2) - 1)

where A = area

P1 - P2 = ΔP = ((v1^2*p) / 2) * ((A1^2/A2^2) - 1)

for the calculation of the areas we will use the following formula:

A1 = (pi*d1^2) / 4 = (pi * (0.02 m) ^2) / 4 = 0.00031 m^2

A2 = (pi * (0.01 m) ^2) / 4 = 0.000079 m^2

ΔP = ((3 m/s) ^2*1.25 kg/m^3) / 2) * ((0.00031 m^2) ^2 / (0.000079 m^2) ^2) - 1) = 80.99 N/m^2 = Pa