Answer:
the pressure drop is equal to 80.99 Pa
Explanation:
we have the following data:
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