water flows along a horizontal pipe of cross-sectional area 48cm^2 which has a constriction of cross-sectional area 12cm^2 at one place. If the speed of the water at the constriction is 4 m/s, calculate the speed in the wider section. The pressure in the wider section is 1.0 x 10^5 Pa. Calculate pressure at the constriction. (density of water= 1000kg/m^3)

Please show your work. Think "Bernoulli equation"

To solve this problem, we can use the continuity equation and Bernoulli's equation.

Step 1: Use the continuity equation to relate the velocities in the wider section and the constriction.

The continuity equation states that the mass flow rate is conserved for an incompressible fluid. Since the pipe is horizontal, the mass flow rate is given by:

A1v1 = A2v2

where A1 and A2 are the cross-sectional areas of the wider section and the constriction, respectively, and v1 and v2 are the velocities in the wider section and the constriction, respectively.

Step 2: Calculate the velocity in the wider section.

We are given the cross-sectional areas as A1 = 48 cm^2 = 0.0048 m^2 and A2 = 12 cm^2 = 0.0012 m^2, and the velocity at the constriction as v2 = 4 m/s.

Substituting these values into the continuity equation, we can solve for v1:

A1v1 = A2v2
(0.0048 m^2)(v1) = (0.0012 m^2)(4 m/s)
v1 = (0.0012 m^2)(4 m/s) / (0.0048 m^2)
v1 = 1 m/s

Therefore, the speed in the wider section is 1 m/s.

Step 3: Calculate the pressure at the constriction.

We can use Bernoulli's equation, which states that the sum of the pressure, kinetic energy per unit volume, and potential energy per unit volume is constant along a streamline.

In the wider section, the speed is 1 m/s and the pressure is given as 1.0 × 10^5 Pa. At the constriction, the speed is given as 4 m/s.

Using Bernoulli's equation, we can set up the following equation:

P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2

where P1 and P2 are the pressures in the wider section and the constriction, respectively, ρ is the density of water, g is the acceleration due to gravity, and h1 and h2 are the heights of the points in the wider section and the constriction, respectively.

Since the pipe is horizontal, the heights h1 and h2 are equal and can be ignored. The kinetic energy per unit volume (1/2)ρv^2 can also be ignored as the heights are the same.

Therefore, the equation simplifies to:

P1 = P2 + (1/2)ρv2^2

Substituting the given values, we have:

P1 = (1.0 × 10^5 Pa) + (1/2)(1000 kg/m^3)(4 m/s)^2
P1 = 1.0 × 10^5 Pa + (1/2)(1000 kg/m^3)(16 m^2/s^2)
P1 = 1.0 × 10^5 Pa + 8000 Pa
P1 = 1.08 × 10^5 Pa

Therefore, the pressure at the constriction is 1.08 × 10^5 Pa.

To calculate the speed of water in the wider section, use the principle of conservation of mass. According to this principle, the mass of water flowing through the pipe per unit time is constant.

Step 1: Calculate the mass flow rate in the wider section.

Given:
Cross-sectional area of the wider section (A1) = 48 cm^2 = 48/10000 = 0.0048 m^2
Speed of water at the constriction (v2) = 4 m/s
Density of water (ρ) = 1000 kg/m^3

Mass flow rate (m dot) = ρ * A1 * v1

Rearranging the equation, we get:
v1 = (m dot) / (ρ * A1)

Step 2: Substitute the values and calculate v1.

Mass flow rate (m dot) is constant, so it is the same in both the wider section and the constriction.

Substituting the given values:
v1 = (4 m/s) * (12 cm^2 / 0.0048 m^2) / (1000 kg/m^3)

Converting cm^2 to m^2:
v1 = (4 m/s) * (0.0012 m^2) / (1000 kg/m^3)

v1 = 0.0048 m/s

The speed of water in the wider section is 0.0048 m/s.

To calculate the pressure at the constriction, use Bernoulli's equation, which relates the speed, pressure, and height of a fluid.

P1 + 0.5 * ρ * v1^2 + ρ * g * h1 = P2 + 0.5 * ρ * v2^2 + ρ * g * h2

P1 = Pressure in the wider section = 1.0 x 10^5 Pa
v1 = Speed in the wider section = 0.0048 m/s
v2 = Speed at the constriction = 4 m/s
ρ = Density of water = 1000 kg/m^3
g = Acceleration due to gravity = 9.8 m/s^2
h1 = Height in the wider section
h2 = Height at the constriction

Assuming the pipe is horizontal, the height in both sections is the same, so h1 = h2 = 0.

Substituting the given values and simplifying the equation, we get:

P2 = P1 + 0.5 * ρ * (v1^2 - v2^2)

Substituting the values:
P2 = 1.0 x 10^5 Pa + 0.5 * 1000 kg/m^3 * (0.0048 m/s)^2 - (4 m/s)^2

P2 = 1.0 x 10^5 Pa + 0.5 * 1000 kg/m^3 * 0.00002304 m^2/s^2 - 16 m^2/s^2

P2 = 1.0 x 10^5 Pa + 0.01152 Pa - 16 Pa

P2 = 1.0 x 10^5 Pa - 15.98848 Pa

P2 ≈ 1.0 x 10^5 Pa - 16 Pa

P2 ≈ 9.9984 x 10^4 Pa

Therefore, the pressure at the constriction is approximately 9.9984 x 10^4 Pa.