Water moves through a constricted pipe in steady, ideal flow. At the lower point shown in the figure below, the pressure is 1.70 105 Pa and the pipe radius is 2.50 cm. At the higher point located at y = 2.50 m, the pressure is 1.30 105 Pa and the pipe radius is 1.60 cm. (a) Find the speed of flow in the lower section.

(b) Find the speed of flow in the upper section.
(c) Find the volume flow rate through the pipe.

To answer this question, 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 an ideal, steady flow of an incompressible fluid.

Mathematically, Bernoulli's equation can be written as:

P1 + (1/2)ρv1^2 + ρgy1 = P2 + (1/2)ρv2^2 + ρgy2

Where:
P1 and P2 are the pressures at points 1 and 2, respectively.
ρ is the density of the fluid.
v1 and v2 are the velocities of the fluid at points 1 and 2, respectively.
g is the acceleration due to gravity.
y1 and y2 are the heights of points 1 and 2, respectively.

In this case, we are given the pressures and radii at two different points along the pipe. We need to find the velocities at these points and calculate the volume flow rate.

Let's solve this step-by-step:

(a) Find the speed of flow in the lower section (point 1):

Given:
Pressure at point 1 (P1) = 1.70 x 10^5 Pa
Radius at point 1 (r1) = 2.50 cm = 0.025 m

To find the speed of flow at point 1 (v1), we need to rearrange Bernoulli's equation and solve for v1:

P1 + (1/2)ρv1^2 + ρgy1 = P2 + (1/2)ρv2^2 + ρgy2

(The height difference between points 1 and 2 is not given, so we can ignore the potential energy term.)

P2 is not given, but we can assume that the pressure at point 2 is the same as at point 1 because the fluid is in a steady, ideal flow.

Therefore, the equation becomes:

P1 + (1/2)ρv1^2 = P2 + (1/2)ρv2^2

Since P1 = P2:

(1/2)ρv1^2 = (1/2)ρv2^2

We can cancel out ρ (density) from both sides:

v1^2 = v2^2

Taking the square root of both sides:

v1 = v2

So, the speed of flow in the lower section (point 1) is equal to the speed of flow in the upper section (point 2).

Therefore, v1 = v2

(b) Find the speed of flow in the upper section (point 2):

Since v1 = v2, the speed of flow in the upper section (point 2) is the same as in the lower section (point 1).

Therefore, v2 = v1

(c) Find the volume flow rate through the pipe:

To find the volume flow rate (Q), we can use the equation:

Q = A1v1 = A2v2

Where:
A1 and A2 are the cross-sectional areas of the pipe at points 1 and 2, respectively.

The cross-sectional area (A) of a pipe can be calculated using the formula:

A = πr^2

Given:
Radius at point 1 (r1) = 2.50 cm = 0.025 m
Radius at point 2 (r2) = 1.60 cm = 0.016 m

Using the formula for the cross-sectional area, we can calculate the areas at each point:

A1 = πr1^2
A2 = πr2^2

Now, we can substitute the values into the equation for the volume flow rate:

Q = A1v1 = A2v2

= (πr1^2)v1 = (πr2^2)v2

Substituting the values of r1, r2, v1, and v2:

Q = (π(0.025)^2)v1 = (π(0.016)^2)v1

Calculating the volume flow rate (Q) will require additional information, such as the pipe length or the cross-sectional area at another point along the pipe.

To solve this problem, we can use Bernoulli's equation, which states that the total mechanical energy per unit volume of liquid flowing through a pipe is constant. We can use this equation to relate the pressure and speed of flow at different points in the pipe.

(a) To find the speed of flow in the lower section, we can use the following equation derived from Bernoulli's equation:

P + 1/2 ρv^2 + ρgh = constant

Where P is the pressure, ρ is the density of the liquid (in this case, water), v is the speed of flow, g is the acceleration due to gravity, and h is the height of the point above a reference point.

At the lower point, we have:
P1 = 1.70 * 10^5 Pa (pressure)
r1 = 2.50 cm = 0.025 m (radius)

At this point, the height above the reference point is y1 = 0 m.
Substituting these values into the equation, we have:

1.70 * 10^5 + 1/2 * ρ * v1^2 + ρ * g * y1 = constant

Since the two points are at the same height, the term ρ * g * y1 cancels out, and we are left with:

1.70 * 10^5 + 1/2 * ρ * v1^2 = constant

Now let's find the speed of flow in the upper section.

(b) At the upper point, we have:

P2 = 1.30 * 10^5 Pa (pressure)
r2 = 1.6 cm = 0.016 m (radius)
y2 = 2.5 m (height above the reference point)

Using the same equation as before, we get:

1.30 * 10^5 + 1/2 * ρ * v2^2 + ρ * g * y2 = constant

Again, since the height of the point is different from the reference point, the term ρ * g * y2 is not canceled.

(c) To find the volume flow rate through the pipe, we can use the equation:

Q = A * v

Where Q is the volume flow rate, A is the cross-sectional area of the pipe, and v is the speed of flow.

For the lower section, the cross-sectional area is given by:

A1 = π * r1^2

For the upper section, the cross-sectional area is given by:

A2 = π * r2^2

Substituting the respective values, we can find the volume flow rate through the pipe.

(a) P + (1/2)*rho*V^2 + (rgo)*g*y = constant. That's Bernoulli's law, neglecting fluid frictional pressure drops, for incompressible flow. Elevation is y = 2.50 m on the high side, and is zero on the low side.

You know the pressure on both sides, and that
V(high) = V(low)*[(2.5)/(1.6)]^2
= 2.44 V(low)
Now you can solve for V(low).
(b) I already told you the relation between V(high) and V(low). It comes from the continuity equation
(c)(Volume flow rate) = pi*R^2*V
You can used data from either the high or low side.