Water flows into one end of a horizontal, cylindrical pipe at 2.0 m/s. The pipe then narrows until

its diameter at the opposite end is only 1/3 of that at the beginning.
(a) What is the flow speed at the narrow end of the pipe?
(b) What is the pressure difference between the wide and narrow ends of the pipe?

a: area1*velocity1=area2*velocity2

b) you have the difference in velocity1 and 2, solve for pressure difference.

P1*speed1=p2*speed2
p1*speed1=P1*area1/area2*speed1
P1=P2*9
P1-P2=pressure difference
P1-1/9P1=pressure differnce

To find the flow speed at the narrow end of the pipe, we can use the principle of conservation of mass, which states that the mass of fluid entering a section of the pipe must equal the mass of fluid leaving that section. Since the pipe is horizontal and there are no external forces affecting the fluid, this principle can be expressed as:

ρ₁A₁V₁ = ρ₂A₂V₂

Where:
- ρ₁ and ρ₂ are the densities of the fluid at the wide and narrow ends of the pipe, respectively.
- A₁ and A₂ are the cross-sectional areas of the wide and narrow ends of the pipe, respectively.
- V₁ and V₂ are the flow speeds at the wide and narrow ends of the pipe, respectively.

To find the flow speed at the narrow end of the pipe (V₂), we need to know the density ratio (ρ₁/ρ₂) and the area ratio (A₁/A₂).

The problem states that the diameter at the opposite end is only 1/3 of that at the beginning. Since the area of a circle is proportional to the square of its diameter, we can calculate the area ratio (A₁/A₂) as follows:

(A₂/A₁) = (D₂/D₁)²

Given that D₂ = 1/3 D₁, we have:

(A₂/A₁) = [(1/3 D₁)/D₁]² = (1/3)² = 1/9

Now, let's consider the density ratio. Since water is an incompressible fluid, its density remains constant throughout the pipe. Therefore, the density ratio (ρ₁/ρ₂) is equal to 1.

We can now substitute these values into the conservation of mass equation and solve for V₂:

ρ₁A₁V₁ = ρ₂A₂V₂
(1)(A₁)(2.0 m/s) = (1/9)(A₁)(V₂)

Simplifying the equation gives:

2.0 m/s = (1/9)V₂

To obtain V₂, we multiply both sides of the equation by 9/1:

V₂ = (9/1)(2.0 m/s) = 18 m/s

So, the flow speed at the narrow end of the pipe is 18 m/s.

To find the pressure difference between the wide and narrow ends of the pipe, we can use Bernoulli's equation, which relates the pressure, speed, and height of a fluid in steady flow conditions. For an incompressible, non-viscous fluid (like water), Bernoulli's equation can be expressed as:

P₁ + (1/2)ρV₁² + ρgh₁ = P₂ + (1/2)ρV₂² + ρgh₂

Where:
- P₁ and P₂ are the pressures at the wide and narrow ends of the pipe, respectively.
- ρ is the density of the fluid (in this case, water).
- V₁ and V₂ are the flow speeds at the wide and narrow ends of the pipe, respectively.
- g is the acceleration due to gravity.
- h₁ and h₂ are the heights (or elevations) at the wide and narrow ends of the pipe, respectively.

Since both ends of the pipe are at the same height (horizontal pipe), the height terms cancel out. Additionally, the problem does not provide information about the pressures or elevations, so we can assume they are equal and cancel them out as well. This leaves us with:

(1/2)ρV₁² = (1/2)ρV₂²

Simplifying the equation gives:

V₁² = V₂²

Taking the square root of both sides, we find:

V₁ = V₂

Therefore, the flow speeds at the wide and narrow ends of the pipe are the same, and the pressure difference between the two ends is zero.

To solve this problem, we can apply the principle of conservation of mass and Bernoulli's equation. Let's tackle each part separately:

(a) To find the flow speed at the narrow end of the pipe, we can use the principle of conservation of mass, which states that the mass flow rate remains constant at any point in an incompressible fluid:

Mass flow rate = ρ₁A₁V₁ = ρ₂A₂V₂

Where:
- ρ₁ and ρ₂ are the densities of the fluid at the wide and narrow ends, respectively.
- A₁ and A₂ are the cross-sectional areas of the wide and narrow ends, respectively.
- V₁ and V₂ are the flow speeds at the wide and narrow ends, respectively.

The cross-sectional area of a pipe is related to its diameter by the formula: A = πr², where r is the radius of the pipe.

Given:
- V₁ = 2.0 m/s
- The diameter at the narrow end is 1/3 of the diameter at the wide end.

Let's denote the diameter at the wide end as D, and the diameter at the narrow end as d.
Thus, the radius at the wide end is r₁ = D/2, and the radius at the narrow end is r₂ = d/2.

We know that D = 3d (since the narrow end has a diameter that is 1/3 of the wide end).

So, we have:
A₁ = π(D/2)² = π(D²/4)
A₂ = π(d/2)² = π(d²/4)

Now, we can rewrite the equation for mass flow rate as:

ρ₁π(D²/4)(2.0) = ρ₂π(d²/4)(V₂)

Simplifying the equation:

(D²/4) = (ρ₂/ρ₁)(d²/4)(V₂/V₁)

Since we are interested in the flow speed at the narrow end (V₂), we can rearrange the equation:

V₂ = (V₁)(D²/d²)(ρ₁/ρ₂)

Substituting the known values:

V₂ = (2.0)(D²/d²)(ρ₁/ρ₂)

Since we do not know the values of ρ₁ and ρ₂, we cannot determine the exact value of V₂. However, we now have a relation between V₂ and the other variables in the problem.

(b) To find the pressure difference between the wide and narrow ends of the pipe, we can apply Bernoulli's equation, which relates the pressure, density, and velocity of a fluid in a streamline:

P₁ + (1/2)ρ₁V₁² = P₂ + (1/2)ρ₂V₂²

Since the problem does not provide information about the pressures at either end, we cannot determine the exact pressure difference between the wide and narrow ends.