A pipe carrying water is laid out over level ground. One segment of the pipe has a diameter of 2.22 cm and the next segment has a diameter of 1.82 cm. In the wider segment, the water pressure is 83,700 Pa and the water is moving at 2.77m/sec.
1-what is the speed of the water in the narrow segment?
2-what is the pressure of the water in the narrow segment?
I was going to use Bernoulli's equation to find P2 but I wasn't given the density so I'm lost for what to put in place of it. Would this be a place for the law of continuity? anything helps, thank you!!
It is water. The density is 1 gram/cc or 1000 kg/m^3
Q = flow rate , volume/second.
Q is constant, the same amount of water goes by each point each seconds (this is continuity)
therefore
V * area = constant
2.77 * pi (1.11^2) = v * pi (0.91^2)
or
v = 2.77(1.11/0.91)^2
============================
p/rho + (1/2)v^2 = constant
rho is constant here (incompressible fluid)
83700/1000 + (1/2)(2.77)^2 = p/1000 + (1/2)v^2
To solve this problem, you are correct that you can utilize the principles of fluid dynamics, including Bernoulli's equation and the law of continuity.
First, let's address your concern about not having the density (ρ) of water provided in the problem. While it is true that Bernoulli's equation involves density, we can actually eliminate the need for it in this particular scenario by using the law of continuity.
The law of continuity states that the volume flow rate of an incompressible fluid is constant within a closed system. In this case, the fluid is water, which is considered incompressible for most practical purposes.
Now let's go step by step to find the answers to the given questions:
1. Speed of water in the narrow segment:
According to the law of continuity, the volume flow rate should be the same in both segments of the pipe. Since the pipe is carrying water over level ground, we can assume that there is no change in height, so gravitational potential energy terms can be ignored.
Using the equation for volume flow rate:
A1⋅v1 = A2⋅v2
Where:
A1 = cross-sectional area of the wider segment
v1 = speed of water in the wider segment
A2 = cross-sectional area of the narrow segment
v2 = speed of water in the narrow segment (what we need to find)
We are given:
d1 = diameter of the wider segment = 2.22 cm = 0.0222 m
d2 = diameter of the narrow segment = 1.82 cm = 0.0182 m
v1 = 2.77 m/sec
Since the cross-sectional area (A) of a pipe is proportional to the square of its diameter (d), we can write:
A1 = (π/4)⋅d1²
A2 = (π/4)⋅d2²
Plug in the values and solve for v2:
(π/4)⋅d1²⋅v1 = (π/4)⋅d2²⋅v2
(π/4)⋅(0.0222 m)²⋅(2.77 m/sec) = (π/4)⋅(0.0182 m)²⋅v2
v2 = ((0.0222 m)²⋅(2.77 m/sec)) / (0.0182 m)²
v2 ≈ 3.38 m/sec
So, the speed of the water in the narrow segment is approximately 3.38 m/sec.
2. Pressure of water in the narrow segment:
As you mentioned, Bernoulli's equation can be used to calculate the pressure in the narrow segment.
Bernoulli's equation, neglecting energy losses and elevation differences, is given by:
P1 + 0.5⋅ρ⋅v1² = P2 + 0.5⋅ρ⋅v2²
Since the pipe is horizontal and the water is moving at the same level, we can safely assume the elevation terms cancel out.
As we discussed earlier, we don't have the density (ρ) of water, but in this case, we can eliminate the need for it by simplifying the equation further.
Dividing the equation by ρ⋅g, where g is the acceleration due to gravity, we get:
(P1/ρ⋅g) + (0.5⋅v1²/g) = (P2/ρ⋅g) + (0.5⋅v2²/g)
Since the segment heights are the same, the pressure difference due to height can be neglected, resulting in:
P1 + 0.5⋅ρ⋅v1² = P2 + 0.5⋅ρ⋅v2²
But now, since we have shown that we can eliminate the need for the density (ρ) term, we can rewrite the equation as:
P1 + 0.5⋅v1² = P2 + 0.5⋅v2²
We know:
P1 = 83,700 Pa
v1 = 2.77 m/sec
v2 = 3.38 m/sec (calculated earlier)
Plug in the values and solve for P2:
83,700 Pa + 0.5⋅(2.77 m/sec)² = P2 + 0.5⋅(3.38 m/sec)²
P2 = (83,700 Pa + 0.5⋅(2.77 m/sec)²) - 0.5⋅(3.38 m/sec)²
Calculating this expression, we find:
P2 ≈ 67,233 Pa
Therefore, the pressure of the water in the narrow segment is approximately 67,233 Pa.
By using the law of continuity and Bernoulli's equation, we were able to solve both questions without explicitly determining the density of water.