To solve this problem, we will use Bernoulli's equation, which states that the sum of the pressure energy, kinetic energy, and potential energy per unit volume is constant for an incompressible, non-viscous fluid in steady flow.

Bernoulli's equation is given by:

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 wide and narrow sections respectively,
ρ is the fluid density,
v1 and v2 are the velocities at the wide and narrow sections respectively,
g is the acceleration due to gravity,
h1 and h2 are the heights of the wide and narrow sections respectively.

First, we need to find the velocity at the narrow section (v2). We can use the principle of conservation of mass which states that the mass flow rate must be constant in a pipe. This gives us:

A1 * v1 = A2 * v2

where A1 and A2 are the cross-sectional areas of the wide and narrow sections respectively.

We can calculate A1 and A2 using the formula for the area of a circle (π * d^2 / 4), where d is the diameter of the section.

A1 = π * (12 cm)^2 / 4 = 113.1 cm^2
A2 = π * (9 cm)^2 / 4 = 63.6 cm^2

Substituting these values into the conservation of mass equation gives:

113.1 cm^2 * 4 m/s = 63.6 cm^2 * v2

Solving for v2 gives v2 = 7.1 m/s.

Next, we need to find the height difference between the wide and narrow sections (h1 - h2). We can use the formula for the height of a right triangle (h = l * sin(θ)), where l is the length of the incline and θ is the angle of the incline.

h1 - h2 = 10 m * sin(30 degrees) = 5 m

Finally, we can substitute all these values into Bernoulli's equation and solve for P2:

250 kPa + 1/2 * 1000 kg/m^3 * (4 m/s)^2 + 1000 kg/m^3 * 9.81 m/s^2 * 0 m = P2 + 1/2 * 1000 kg/m^3 * (7.1 m/s)^2 + 1000 kg/m^3 * 9.81 m/s^2 * -5 m

Solving for P2 gives P2 = 226.5 kPa.

Therefore, the pressure in the narrow section of the pipe is 226.5 kPa.

Question

To solve this problem, we will use Bernoulli's equation, which states that the sum of the pressure energy, kinetic energy, and potential energy per unit volume is constant for an incompressible, non-viscous fluid in steady flow.

Bernoulli's equation is given by:

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 wide and narrow sections respectively,
ρ is the fluid density,
v1 and v2 are the velocities at the wide and narrow sections respectively,
g is the acceleration due to gravity,
h1 and h2 are the heights of the wide and narrow sections respectively.

First, we need to find the velocity at the narrow section (v2). We can use the principle of conservation of mass which states that the mass flow rate must be constant in a pipe. This gives us:

A1 * v1 = A2 * v2

where A1 and A2 are the cross-sectional areas of the wide and narrow sections respectively.

We can calculate A1 and A2 using the formula for the area of a circle (π * d^2 / 4), where d is the diameter of the section.

A1 = π * (12 cm)^2 / 4 = 113.1 cm^2
A2 = π * (9 cm)^2 / 4 = 63.6 cm^2

Substituting these values into the conservation of mass equation gives:

113.1 cm^2 * 4 m/s = 63.6 cm^2 * v2

Solving for v2 gives v2 = 7.1 m/s.

Next, we need to find the height difference between the wide and narrow sections (h1 - h2). We can use the formula for the height of a right triangle (h = l * sin(θ)), where l is the length of the incline and θ is the angle of the incline.

h1 - h2 = 10 m * sin(30 degrees) = 5 m

Finally, we can substitute all these values into Bernoulli's equation and solve for P2:

250 kPa + 1/2 * 1000 kg/m^3 * (4 m/s)^2 + 1000 kg/m^3 * 9.81 m/s^2 * 0 m = P2 + 1/2 * 1000 kg/m^3 * (7.1 m/s)^2 + 1000 kg/m^3 * 9.81 m/s^2 * -5 m

Solving for P2 gives P2 = 226.5 kPa.

Therefore, the pressure in the narrow section of the pipe is 226.5 kPa.

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