# A water pipe is inclined 31.0° below the horizontal. The radius of the pipe at the upper end is 2.00 cm. If the gauge pressure at a point at the upper end is 0.122 atm, what is the gauge pressure at a point 2.70 m downstream, where the pipe has narrowed to a 1.00 cm radius? The flow rate is 20.0π cm3/s.

## To find the gauge pressure at a point 2.70 m downstream, we need to use the principles of fluid dynamics and hydrostatics.

First, let's calculate the velocity of the fluid at the upper end of the pipe. The flow rate (Q) is given as 20.0π cm3/s, and we can use the formula Q = A1v1, where A1 is the cross-sectional area at the upper end and v1 is the velocity at the upper end.

Given:
Cross-sectional area (A1) = π(r1)² = π(2.00 cm)² = 4π cm²
Flow rate (Q) = 20.0π cm³/s

Using the formula, we can solve for v1:
20.0π cm³/s = 4π cm² * v1
v1 = 20.0 cm/s

Now, let's find the velocity at the downstream point using the principle of continuity. According to the principle of continuity, the volume flow rate must remain constant along a streamline. Therefore, we have Q = A1v1 = A2v2, where A2 is the cross-sectional area at the downstream point and v2 is the velocity at the downstream point.

Given:
A1 = 4π cm²
A2 = π(r2)² = π(1.00 cm)² = π cm²

Using the formula, we can solve for v2:
A1v1 = A2v2
4π cm² * 20.0 cm/s = π cm² * v2
v2 = 80.0 cm/s

Now, let's move on to calculating the pressure at the downstream point. We can use the Bernoulli's equation, which relates the pressure, velocity, and height of a fluid along a streamline.

Bernoulli's equation: P1 + 1/2(ρv1²) + ρgh1 = P2 + 1/2(ρv2²) + ρgh2

Since the pipe is horizontal at the upper end (h1 = h2), and the velocity is nearly the same as the atmospheric pressure is small, we can simplify the equation to:

P1 + 1/2(ρv1²) = P2 + 1/2(ρv2²)

Where:
P1 = gauge pressure at the upper end (0.122 atm)
v1 = velocity at the upper end (20.0 cm/s)
P2 = gauge pressure at the downstream point (to be determined)
v2 = velocity at the downstream point (80.0 cm/s)

We need to convert cm/s to m/s and atm to Pa for consistent units. 1 atm = 101325 Pa.

P1 + 1/2((ρ * v1²) / 10000) = P2 + 1/2((ρ * v2²) / 10000)
0.122 atm + 1/2((1.00 kg/m³) * (20.0 m/s)²) * (10000 cm²) = P2 + 1/2((1.00 kg/m³) * (80.0 m/s)²) * (10000 cm²)

Now, let's solve the equation for P2:

0.122 atm + 1/2((100 kg/m³) * (20.0 m/s)²) * (10000 cm²) = P2 + 1/2((100 kg/m³) * (80.0 m/s)²) * (10000 cm²)

0.122 atm + 2.000 × 10⁻⁴ Pa = P2 + 6.400 × 10⁻³ Pa

Rearranging the equation:

P2 = 0.122 atm + 2.000 × 10⁻⁴ Pa - 6.400 × 10⁻³ Pa

P2 = 0.122 atm - 6.000 × 10⁻³ Pa

Finally, let's convert P2 back to atm:

P2 = 0.122 atm - 6.000 × 10⁻³ Pa / 101325 Pa/atm

P2 ≈ 0.121 atm

Therefore, the gauge pressure at a point 2.70 m downstream, where the pipe has narrowed to a 1.00 cm radius, is approximately 0.121 atm.