A horizontal segment of pipe tapers from a cross sectional area of 50.0 cm^ 2 to 0.500 cm^2. The pressure at the larger end of the pipe is 1.20x 10^5 Pa, and the speed is 0.040 m/s. what is the pressure at the narrow end of the segment?
To find the pressure at the narrow end of the segment, we can use the principle of conservation of mass and Bernoulli's equation.
1. Conservation of Mass:
According to the principle of conservation of mass, the mass flow rate (ṁ) of a fluid in a pipe remains constant. We can express it as:
ṁ = ρ * A * v
Where:
- ṁ is the mass flow rate (constant)
- ρ is the density of the fluid
- A is the cross-sectional area of the pipe
- v is the velocity of the fluid
Since the mass flow rate is constant, we can equate the mass flow rate at the larger end and the narrow end:
ρ * A1 * v1 = ρ * A2 * v2
Where:
- A1 is the cross-sectional area at the larger end of the pipe
- v1 is the velocity at the larger end of the pipe
- A2 is the cross-sectional area at the narrow end of the pipe
- v2 is the velocity at the narrow end of the pipe
2. Bernoulli's Equation:
Bernoulli's equation relates the pressure, velocity, and height of a fluid in a pipe. It can be expressed as:
P + 1/2 * ρ * v^2 + ρ * g * h = constant
Where:
- P is the pressure of the fluid
- ρ is the density of the fluid
- v is the velocity of the fluid
- g is the acceleration due to gravity
- h is the height of the fluid
We can assume that the height of the fluid remains constant, so we can rewrite Bernoulli's equation as:
P1 + 1/2 * ρ * v1^2 = P2 + 1/2 * ρ * v2^2
Where:
- P1 is the pressure at the larger end of the pipe
- v1 is the velocity at the larger end of the pipe
- P2 is the pressure at the narrow end of the pipe
- v2 is the velocity at the narrow end of the pipe
Now, we can solve the equations:
From the conservation of mass:
A1 * v1 = A2 * v2
From Bernoulli's equation:
P1 + 1/2 * ρ * v1^2 = P2 + 1/2 * ρ * v2^2
We are given:
A1 = 50.0 cm^2 = 0.0050 m^2
A2 = 0.500 cm^2 = 0.000050 m^2
P1 = 1.20 x 10^5 Pa
v1 = 0.040 m/s
Using these values, we can solve for P2:
First, we find v2:
v2 = (A1 * v1) / A2
v2 = (0.0050 m^2 * 0.040 m/s) / 0.000050 m^2 = 4.00 m/s
Substituting the values into Bernoulli's equation:
P1 + 1/2 * ρ * v1^2 = P2 + 1/2 * ρ * v2^2
Since the pressure at the narrow end (P2) is what we are trying to find, we rearrange the equation:
P2 = P1 + 1/2 * ρ * v1^2 - 1/2 * ρ * v2^2
Now, we need to determine the density (ρ) of the fluid. Assuming it is constant, we can use the given values to find it:
ρ = P1 / (1/2 * v1^2)
Substituting the given values:
ρ = (1.20 x 10^5 Pa) / (1/2 * (0.040 m/s)^2) = 1.50 x 10^4 kg/m^3
Substituting the calculated values into the equation for P2:
P2 = (1.20 x 10^5 Pa) + 1/2 * (1.50 x 10^4 kg/m^3) * (0.040 m/s)^2 - 1/2 * (1.50 x 10^4 kg/m^3) * (4.00 m/s)^2
Calculating P2:
P2 = (1.20 x 10^5 Pa) + 0.960 Pa - 120 x 10^4 Pa
P2 = -1.18 x 10^4 Pa
Therefore, the pressure at the narrow end of the segment is approximately -1.18 x 10^4 Pa.