Question

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

Answer 1

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

Answer 2

The pressure drop is equal to 80.99 Pa

Given information:

d1 = 2 cm = 0.02 m

d2 = 1 cm = 0.01 m

v = 3 m/s

p = 1.25 kg/m^3

Here we use Bernoulli's principle for the Venturi Tube:

Calculation of pressure drop:

$P1 - P2 = ((v^2\times p)\div 2)\times ((A1^2\div A2^2)-1)\\\\P1 - P2 = \Delta P = ((v1^2\times p)\div 2)\times ((A1^2\div A2^2)-1)$

Now the following formula for area calculation should be used:

$A1 = (\pi\times d1^2)\div 4 = (\pi\times (0.02 m)^2)\div 4 = 0.00031 m^2\\\\A2 = (\pi\times (0.01 m)^2)\div 4 = 0.000079 m^2\\\\\Delta P = ((3 m/s)^2 \times1.25 kg/m^3)\div 2) \times ((0.00031 m^2)^2/(0.000079 m^2)^2)-1)$

= 80.99

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