Water flows at speed of 4.2 m/s through a horizontal pipe of diameter 3.4 cm. The gauge pressure P1 of the water in the pipe is 1.5 atm. A short segment of the pipe is constricted to a smaller diameter of 2.4 cm. What is the gauge pressure of the water flowing through the constricted segment? Atmospheric pressure is 1.013 × 105 Pa. The density of water is 1000 kg/m3. The viscosity of water is negligible. Answer in units of atm.
To find the gauge pressure of the water flowing through the constricted segment of the pipe, we can use the principle of continuity and Bernoulli's equation.
Let's start by finding the velocity of the water flowing through the constricted segment. According to the principle of continuity, the flow rate of an incompressible fluid remains constant along a pipe. Mathematically, this is expressed as:
A1 * V1 = A2 * V2
Where A1 and A2 are the cross-sectional areas of the pipe before and after the constriction, and V1 and V2 are the corresponding velocities of the water.
The diameter of the first section of the pipe is 3.4 cm, which means the radius is half of that. So, the cross-sectional area A1 can be calculated as:
A1 = π * (r1)^2
= π * (3.4 cm / 2)^2
Similarly, the cross-sectional area A2 for the constricted segment can be calculated using the diameter of 2.4 cm:
A2 = π * (r2)^2
= π * (2.4 cm / 2)^2
Now, we can calculate the velocity V2 of the water flowing through the constricted segment by rearranging the continuity equation:
V2 = (A1 * V1) / A2
Next, we can use Bernoulli's equation to find the gauge pressure of the water in the constricted segment. Bernoulli's equation relates the pressure, velocity, and height (relative to a reference point) of a fluid. It can be expressed as:
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, ρ is the density of the fluid, V1 and V2 are the velocities at the respective points, g is the acceleration due to gravity, and h1 and h2 are the heights above the reference point.
Since the water is flowing horizontally, the height terms (h1 and h2) can be considered equal. Also, we can assume that the potential energy term (ρ * g * h) is negligible compared to the other terms since the change in height is small.
Therefore, the Bernoulli's equation simplifies to:
P1 + (1/2) * ρ * V1^2 = P2 + (1/2) * ρ * V2^2
Now, rearrange the equation to solve for P2, the gauge pressure in the constricted segment:
P2 = P1 + (1/2) * ρ * (V1^2 - V2^2)
Plug in the given values:
P1 = 1.5 atm
ρ = 1000 kg/m³
V1 = 4.2 m/s
V2 = (A1 * V1) / A2 (calculated earlier)
Finally, convert the pressure to units of atmospheres (atm) by dividing by the atmospheric pressure (1.013 × 10^5 Pa).
With these calculations, you will be able to find the gauge pressure of the water flowing through the constricted segment of the pipe.