Practice Problem 13.9

Let's begin by looking at a flow problem in which the diameter of a pipe changes along the flow path. We will need to use both the continuity equation and Bernoulli's equation for this problem. Water enters a house through a pipe with an inside diameter of 2.0 cm at a gauge pressure of 4.0×105Pa (about 4 atm, or 60 lb/in.2). The cold-water pipe leading to the second-floor bathroom 5.0 m above is 1.0 cm in diameter (Figure 1) . Find the flow speed and gauge pressure in the bathroom when the flow speed at the inlet pipe is 2.0 m/s. How much time would be required to fill a 100 L bathtub with cold water?

Part A - Practice Problem:

Determine the gauge pressure for a 2-cm-diameter faucet located next to the washing machine in the basement, 2.0 m below where the pipe enters the house.
Express your answer to two significant figures and include the appropriate units.

Ok I'll answer it for all you physics students that want to cheat.

The answer is 425000 Pa

its been 5 years.....almost 400 people have seen this. No responses. How you doing?

To determine the gauge pressure at the 2-cm-diameter faucet located in the basement, we can use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid at two different points in a 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 points 1 and 2, respectively.
ρ is the density of the fluid.
v1 and v2 are the velocities at points 1 and 2, respectively.
g is the acceleration due to gravity.
h1 and h2 are the heights of points 1 and 2, respectively.

In this case, we have the following information:
- Diameter of the pipe at the inlet (point 1): 2.0 cm
- Gauge pressure at the inlet (point 1): 4.0×10^5 Pa (4 atm or 60 lb/in.2)
- Flow speed at the inlet (point 1): 2.0 m/s
- Diameter of the pipe at the faucet (point 2): 2.0 cm
- Height difference between points 1 and 2: 2.0 m

To find the gauge pressure at the faucet (point 2), we need to find the velocity at point 2 (v2). We can assume that the height at point 2 (h2) is the same as at point 1 (h1).

Using the continuity equation, which states that the flow rate is constant in an incompressible fluid, we can relate the velocities at points 1 and 2:

A1 * v1 = A2 * v2

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

Given that the diameters of the pipes are 2.0 cm at both points 1 and 2, we can calculate the areas:

A1 = π * (d1/2)^2 = π * (2.0 cm/2)^2 = π * (1.0 cm)^2
A2 = π * (d2/2)^2 = π * (2.0 cm/2)^2 = π * (1.0 cm)^2

Since the areas are the same, we can simplify the continuity equation to:

v1 = v2

Now, let's use Bernoulli's equation:

P1 + 1/2 * ρ * v1^2 + ρ * g * h1 = P2 + 1/2 * ρ * v2^2 + ρ * g * h2

Since v1 = v2, we can simplify the equation to:

P1 + ρ * g * h1 = P2 + ρ * g * h2

Rearranging the equation:

P2 = P1 + ρ * g * (h1 - h2)

Substituting the given values:

P2 = 4.0×10^5 Pa + (1000 kg/m^3) * (9.8 m/s^2) * (2.0 m)

Calculating the gauge pressure at the faucet:

P2 = 4.0×10^5 Pa + 19600 Pa

P2 ≈ 419600 Pa

Expressing the gauge pressure at the faucet to two significant figures:

P2 ≈ 4.2×10^5 Pa

Therefore, the gauge pressure for a 2-cm-diameter faucet located next to the washing machine in the basement, 2.0 m below where the pipe enters the house, is approximately 4.2×10^5 Pa.

To solve this problem, we can use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid in a flowing system.

Bernoulli's equation is given by:

P1 + ½ρv1^2 + ρgh1 = P2 + ½ρv2^2 + ρgh2

where P1 and P2 are the gauge pressures, v1 and v2 are the velocities, ρ is the density of the fluid, g is the acceleration due to gravity, and h1 and h2 are the heights.

In this case, we are given the values for P1 (4.0x10^5 Pa), v1 (2.0 m/s), and h1 (0 m).

We can rearrange Bernoulli's equation to solve for P2:

P2 = P1 + ½ρv1^2 - ½ρv2^2 - ρgh2

To find the gauge pressure for the faucet located in the basement, we need to determine the value of h2. We are told that the bathroom is 5.0 m above the inlet pipe. Since the faucet is located 2.0 m below where the pipe enters the house, the height difference between the faucet and the bathroom is 5.0 m + 2.0 m = 7.0 m.

Let's plug in the given values into the equation:

P2 = 4.0x10^5 Pa + ½ρ(2.0 m/s)^2 - ½ρv2^2 - ρg(7.0 m)

To express the gauge pressure to two significant figures, we need to know the density of water (ρ). The density of water is approximately 1000 kg/m^3.

Now we have all the necessary information to calculate the gauge pressure for the faucet located in the basement.