A pressure difference of 1.5x10^3 Pa draws water from a well. The water emerges from the pump with a speed of 17.0 m/s. What is the depth of the well?
use a dm^3 (liter) of water ... mass is one kg
... cross section is ... .01 m^2
Pa is a N/m^2
lifting force is ... 15 N
work lifting water - (m g h) = K.E at top
... [15 - (m g)]* depth = 1/2 * 1 * v^2
Well, well, well... Let's dive right into this question, shall we?
To find the depth of the well, we can use Bernoulli's principle, which states that the total pressure at any point in a fluid is the sum of its kinetic energy per unit volume, its potential energy per unit volume, and the pressure acting on it.
Since the water emerges from the pump with a speed of 17.0 m/s, we can assume that it has reached its maximum speed and therefore has no kinetic energy left. This means we can neglect the kinetic energy term in Bernoulli's equation.
So we're left with the pressure difference and the potential energy per unit volume term. The potential energy per unit volume can be written as ρgh, where ρ is the density of the fluid (which is water in this case), g is the acceleration due to gravity, and h is the depth we're looking for.
We're given the pressure difference of 1.5x10^3 Pa, so we can equate it to the potential energy per unit volume term:
1.5x10^3 Pa = ρgh
Now, the density of water is approximately 1000 kg/m^3, and the acceleration due to gravity is approximately 9.8 m/s^2. Plugging these values in, we get:
1.5x10^3 Pa = (1000 kg/m^3)(9.8 m/s^2)h
Simplifying the equation, we find:
h = 1.5x10^3 Pa / (1000 kg/m^3)(9.8 m/s^2)
Calculating that, we get:
h ≈ 0.153 m
So, the depth of the well is approximately 0.153 meters. That's deeper than the rabbit hole Alice fell into, but not as deep as my love for puns!
To solve this problem, we can use Bernoulli's equation, which relates the pressure difference, velocity, and height of a fluid.
The equation is as follows:
P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2
Where:
P1 and P2 are the pressures at points 1 and 2, respectively.
ρ is the density of the fluid (in this case, water).
v1 and v2 are the velocities of the water at points 1 and 2, respectively.
g is the acceleration due to gravity (approximately 9.8 m/s^2).
h1 and h2 are the heights (depths) at points 1 and 2, respectively (from the reference point).
In this problem, we want to find the depth of the well (h1). Let's assume that P1 and P2 are atmospheric pressure (101,325 Pa), as the water emerges from the well and then returns to atmospheric conditions.
Let's calculate the depth of the well using the given information:
P1 = P2 = 101,325 Pa
v1 = 0 (as the water is at rest in the well)
v2 = 17.0 m/s
ρ = density of water = 1000 kg/m^3
g = 9.8 m/s^2
Substituting these values into the equation, we have:
101,325 + (1/2)(1000)(0^2) + (1000)(9.8)(h1) = 101,325 + (1/2)(1000)(17.0^2) + (1000)(9.8)(0)
Since the vertical velocity in the well is 0, the first term on the left side of the equation cancels out.
Simplifying the equation, we get:
98,000h1 = 145,450
Dividing both sides of the equation by 98,000, we find:
h1 ≈ 1.48 meters
Therefore, the depth of the well is approximately 1.48 meters.
To find the depth of the well, we can use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid. The equation is:
P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2
Where:
P1 and P2 are the pressures at two different points (in this case, the surface of the well and the opening of the pump)
v1 and v2 are the velocities at those points
h1 and h2 are the heights above some reference point (in this case, the surface of the well and the opening of the pump)
ρ is the density of the fluid (in this case, water)
Rearranging the equation to solve for h2 (the depth of the well):
h2 = (P1 - P2 + (1/2)ρv2^2 - (1/2)ρv1^2) / ρg
Given:
P1 - P2 = 1.5x10^3 Pa (pressure difference)
v1 = 0 (water is not moving at the surface of the well)
v2 = 17.0 m/s (water emerges from the pump with a speed of 17.0 m/s)
ρ = density of water = 1000 kg/m^3
g = acceleration due to gravity = 9.8 m/s^2
Plugging in these values:
h2 = (1.5x10^3 Pa - 0 Pa + (1/2)(1000 kg/m^3)(17.0 m/s)^2 - (1/2)(1000 kg/m^3)(0 m/s)^2) / (1000 kg/m^3)(9.8 m/s^2)
Simplifying:
h2 = (1.5x10^3 Pa + (1/2)(1000 kg/m^3)(289.0 m^2/s^2)) / (1000 kg/m^3)(9.8 m/s^2)
h2 = (1.5x10^3 Pa + 144.5x10^3 Pa) / (1000 kg/m^3)(9.8 m/s^2)
h2 = (146x10^3 Pa) / (1000 kg/m^3)(9.8 m/s^2)
h2 ≈ 14.9 meters
Therefore, the depth of the well is approximately 14.9 meters.