How Is the Throat's Cross-Sectional Area Determined Using Bernoulli's Principle?

[lnbanzon]

Homework Statement

air is flowing horizontally at a speed of 100 mph through a duct 4 square feet in cross section. the duct gradually narrows down to a throat section. if a U-tube shows a difference in pressure between the throat and main sections of 7 inch of water, what is the cross sectional area of the throat? (assume that the air is incompressible and has a density of 0.002378 slug per cubic feet)

Homework Equations

The Attempt at a Solution

i already have the final answer and it is

2.57 square/feet

but i need the solution... T_T.. help

Homework Statement

Homework Equations

The Attempt at a Solution

 

[rock.freak667]

For a horizontal pipe what is Bernoulli's equation reduced to?

When you get that formula you can find the velocity leaving the throat section.

Do you know the equation of continuity?

 

[tomcue]

Bernoulli's Principle states that in a fluid flow, an increase in the speed of the fluid results in a decrease in pressure. This can be represented by the equation:

P1 + 1/2ρv1^2 = P2 + 1/2ρv2^2

Where:
P1 = pressure at point 1
P2 = pressure at point 2
ρ = density of the fluid
v1 = velocity at point 1
v2 = velocity at point 2

In this problem, we can use Bernoulli's Principle to find the cross sectional area of the throat. We know that the air is flowing at a speed of 100 mph, which is equivalent to 147.33 ft/s. We also know that the density of air is 0.002378 slug per cubic feet.

At the main section of the duct, the pressure is equal to the atmospheric pressure, which is 14.7 psi. Using the equation above, we can solve for the velocity at the throat:

P1 + 1/2ρv1^2 = P2 + 1/2ρv2^2
14.7 + 1/2(0.002378)(147.33)^2 = P2 + 1/2(0.002378)(v2)^2
P2 = 4.33 psi

Now, we can use the same equation to find the velocity at the throat:

P1 + 1/2ρv1^2 = P2 + 1/2ρv2^2
14.7 + 1/2(0.002378)(147.33)^2 = 4.33 + 1/2(0.002378)(v2)^2
v2 = 192.33 ft/s

Next, we can use the continuity equation, which states that the mass flow rate at any point in a fluid flow is constant. This can be represented by the equation:

A1v1 = A2v2

Where:
A1 = cross sectional area at point 1
A2 = cross sectional area at point 2
v1 = velocity at point 1
v2 = velocity at point 2

We know that the cross sectional area at the main section is 4 square feet, and we can solve for the cross sectional area at the throat:

A1v1 =