A horizontal segment of pipe tapers from a cross sectional area of 50 cm2 to 0.5 cm2. The pressure at the larger end of the pipe is 1.26 105 Pa and the speed is 0.029 m/s. What is the pressure at the narrow end of the segment? Pa

Please help!

physics - drwls, Monday, April 9, 2012 at 8:09pm

Due to the flow contraction, the velocity at the exit (Vout) will be 100 times the velocity at the entrance (Vin). That results from the incompresible-flow continuity equation.

The pressure at the entrance, combined with the Bernoulli Equation, Vin and Vout, will tell you the pressure at the exit.

You should use a ^ before exponents, when typing equations and scientific-notation numbers. I am sure you mean 10^5 and not 105.

Can someone or drwls please give a more detailed explaination please. I don't understand it still i keep getting it wrong.

To find the pressure at the narrow end of the pipe segment, we can use the Bernoulli equation and the principle of conservation of mass.

1. The first step is to convert the areas given in square centimeters to square meters. We have:

A1 = 50 cm^2 = 50 * 10^-4 m^2
A2 = 0.5 cm^2 = 0.5 * 10^-4 m^2

2. The law of conservation of mass states that the mass flow rate at any point in a pipe is constant. This can be expressed as:

A1 * V1 = A2 * V2

Where V1 and V2 are the velocities at the larger and smaller ends of the pipe segment, respectively.

3. From the given information, we know that V1 = 0.029 m/s. Rearranging the equation from step 2, we can solve for V2:

V2 = (A1 * V1) / A2

Substituting the values:

V2 = (50 * 10^-4 m^2 * 0.029 m/s) / (0.5 * 10^-4 m^2)
= 2.9 m/s

4. Now we can use the Bernoulli equation to find the pressure at the narrow end of the pipe segment. The equation is:

P1 + (1/2) * ρ * V1^2 + ρ * g * h1 = P2 + (1/2) * ρ * V2^2 + ρ * g * h2

Where P1 and P2 are the pressures at the larger and smaller ends of the pipe segment, respectively, ρ is the density of the fluid, g is the acceleration due to gravity, h1 and h2 are the heights of the fluid columns above some reference point.

Assuming the pipe is horizontal, we can drop the vertical height term (ρ * g * h) from the equation. Also, as the fluid is incompressible, the density (ρ) remains constant. Simplifying the equation, we get:

P1 + (1/2) * ρ * V1^2 = P2 + (1/2) * ρ * V2^2

5. Plugging in the given values:

P1 = 1.26 * 10^5 Pa
V1 = 0.029 m/s
V2 = 2.9 m/s

The final equation becomes:

1.26 * 10^5 + (1/2) * ρ * (0.029)^2 = P2 + (1/2) * ρ * (2.9)^2

Since the fluid density cancels out, we can solve for P2:

P2 = (1.26 * 10^5) + (1/2) * ρ * (0.029)^2 - (1/2) * ρ * (2.9)^2

Note: We don't have the value for the fluid density (ρ) given, so we cannot find the exact pressure at the narrow end of the segment without that information.

Sure! I can help explain the steps to solve this problem in more detail.

First, let's understand the concept behind the problem. The question mentions a pipe that tapers horizontally, meaning the cross-sectional area changes from a larger value to a smaller value. Due to the principle of continuity, the velocity of the fluid increases as the cross-sectional area decreases.

Now, let's break down the steps to find the pressure at the narrow end of the segment:

Step 1: Determine the velocity at the entrance (Vin) of the pipe.
The problem states that the speed at the larger end of the pipe is 0.029 m/s. This is the velocity at the entrance (Vin).

Step 2: Calculate the velocity at the exit (Vout) of the pipe.
According to the principle of continuity, the velocity at the exit is 100 times the velocity at the entrance. Therefore, Vout = 100 * Vin.

Step 3: Calculate the pressure at the entrance (P1).
The problem provides the pressure at the larger end of the pipe, which is 1.26 * 10^5 Pa. This is the pressure at the entrance (P1).

Step 4: Use the Bernoulli Equation to find the pressure at the exit (P2).
The Bernoulli Equation states that the total pressure at any point in a fluid flow consists of the static pressure and the dynamic pressure. Since the fluid is incompressible, the equation can be simplified as:

P + (1/2) * ρ * V^2 = constant

Where:
P is the pressure
ρ is the density of the fluid
V is the velocity of the fluid

Applying this equation at the entrance (P1) and the exit (P2) of the pipe, we have:

P1 + (1/2) * ρ * V1^2 = P2 + (1/2) * ρ * V2^2

Since the problem doesn't provide the density of the fluid, we can assume it to be constant. Therefore, this equation can be rewritten as:

P1 + (1/2) * V1^2 = P2 + (1/2) * V2^2

Step 5: Substitute the values into the equation and solve for P2.
Plug in the values we know:
P1 = 1.26 * 10^5 Pa
V1 = 0.029 m/s
V2 = 100 * V1 (since we know Vout = 100 * Vin)

Now, substitute these values into the equation and solve for P2:

P1 + (1/2) * V1^2 = P2 + (1/2) * (100 * V1)^2

Simplifying the equation gives:

P1 + (1/2) * V1^2 = P2 + (1/2) * 100^2 * V1^2

Rearranging terms:

P1 - P2 = (1/2) * 100^2 * V1^2 - (1/2) * V1^2

Now, substitute the given values and solve:

1.26 * 10^5 Pa - P2 = (1/2) * 100^2 * (0.029 m/s)^2 - (1/2) * (0.029 m/s)^2

Calculating the right-hand side of the equation gives:

(1/2) * 100^2 * (0.029 m/s)^2 - (1/2) * (0.029 m/s)^2 = 42.118 Pa

Simplifying the equation:

1.26 * 10^5 Pa - P2 = 42.118 Pa

Finally, solve for P2:

P2 = 1.26 * 10^5 Pa - 42.118 Pa

P2 ≈ 1.26 * 10^5 Pa - 42 Pa

P2 ≈ 1.26 * 10^5 Pa - 42 Pa ≈ 1.2596 * 10^5 Pa

So, the pressure at the narrow end of the segment is approximately 1.2596 * 10^5 Pa.

I hope this detailed explanation helps clarify the steps to solve the problem!