Water flows through a garden hose that goes up a step 20.0 cm high. The cross sectional area of the hose on top of the step is half that at the bottom of the step. If the water pressure is 143 kPa at the bottom of the step, and the speed of the water at the bottom of the step is 1.20 m/s, what is the watter pressure at the top of the step?
To find the water pressure at the top of the step, we can use the principle of conservation of energy. The total energy at the bottom of the step is equal to the total energy at the top of the step.
1. First, let's calculate the cross-sectional area at the bottom of the step. Let's denote it as A1.
2. Given that the cross-sectional area at the top of the step is half that at the bottom, we can call it A2 = 0.5A1.
3. The velocity at the bottom of the step is given as v1 = 1.20 m/s.
4. The height of the step is given as h = 20.0 cm = 0.20 m.
5. The water pressure at the bottom of the step is given as P1 = 143 kPa.
Now, let's calculate the water pressure at the top of the step:
6. The potential energy at the bottom of the step is Ep1 = m * g * h, where m is the mass of the water and g is the acceleration due to gravity.
7. The kinetic energy at the bottom of the step is Ek1 = (1/2) * m * v1^2.
8. The total energy at the bottom of the step is E1 = Ep1 + Ek1.
9. The potential energy at the top of the step is Ep2 = 0 (since the height is 0 at the top).
10. The kinetic energy at the top of the step is Ek2 = (1/2) * m * v2^2, where v2 is the velocity at the top.
11. The total energy at the top of the step is E2 = Ep2 + Ek2.
Since the total energy is conserved, we have E1 = E2.
12. Setting Ep1 + Ek1 = Ep2 + Ek2, we get m * g * h + (1/2) * m * v1^2 = 0 + (1/2) * m * v2^2.
13. We can cancel out the mass (m) from both sides of the equation.
14. Rearranging the equation, we get v2^2 = v1^2 + 2 * g * h.
15. Substituting the given values, v1 = 1.20 m/s and h = 0.20 m, we can solve for v2^2.
v2^2 = (1.20 m/s)^2 + 2 * 9.8 m/s^2 * 0.20 m.
v2^2 = 1.44 m^2/s^2 + 0.392 m^2/s^2.
v2^2 = 1.832 m^2/s^2.
v2 ≈ 1.352 m/s.
16. Now, let's calculate the water pressure at the top of the step, P2.
P1 + (1/2) * ρ * v1^2 = P2 + (1/2) * ρ * v2^2, where ρ is the density of water.
17. Cancelling out the common terms, and substituting the given values, we can solve for P2.
143 kPa + (1/2) * ρ * (1.20 m/s)^2 = P2 + (1/2) * ρ * (1.352 m/s)^2.
18. Simplifying the equation further, we get:
143 kPa + 0.72 * ρ = P2 + 0.917 * ρ.
19. Rearranging the equation, we can isolate P2:
P2 = (143 kPa - 0.72 * ρ) + 0.917 * ρ.
20. Since we do not have the value of the density of water, we cannot solve for P2 without that information.
To find the water pressure at the top of the step, we need the density of water (ρ).
To find the water pressure at the top of the step, we can apply the principle of conservation of energy. The total mechanical energy of the water is conserved as it moves through the hose, neglecting any energy losses due to friction or other factors.
First, let's compute the water speed at the top of the step using the principle of conservation of mass. According to the principle, the mass flow rate of the water should be constant throughout the hose. The cross-sectional area of the hose at the bottom is twice that at the top. Therefore, the water speed at the top is given by:
A₁v₁ = A₂v₂
Where:
A₁ is the cross-sectional area at the bottom of the step
v₁ is the water speed at the bottom of the step
A₂ is the cross-sectional area at the top of the step
v₂ is the water speed at the top of the step
We are given that v₁ = 1.20 m/s, A₁/A₂ = 2, and we need to find v₂.
Considering the given values, we can rewrite the equation as:
(2A₂)v₁ = A₂v₂
Simplifying the equation, we get:
2v₁ = v₂
Now we can substitute the given values and solve for v₂:
2(1.20 m/s) = v₂
2.40 m/s = v₂
Now that we know the water speed at the top of the step is 2.40 m/s, we can proceed to find the water pressure at that point. We can use Bernoulli's equation, which relates the pressure, density, gravity, and velocity of a fluid along a streamline:
P₁ + ρgh₁ + 1/2ρv₁² = P₂ + ρgh₂ + 1/2ρv₂²
Where:
P₁ is the water pressure at the bottom of the step
ρ is the density of water
g is the acceleration due to gravity
h₁ is the height at the bottom of the step
h₂ is the height at the top of the step
v₁ is the water speed at the bottom of the step
v₂ is the water speed at the top of the step
We are given P₁ = 143 kPa, ρ = the density of water (approximately 1000 kg/m³), g = 9.8 m/s², h₁ = 0 m (since it's at the bottom), h₂ = 20.0 cm (converting to meters, h₂ = 0.20 m), and v₁ = 1.20 m/s. We need to find P₂.
Substituting the given values and solving for P₂, we get:
(143,000 Pa) + (1000 kg/m³)(9.8 m/s²)(0 m) + (1/2)(1000 kg/m³)(1.20 m/s)² = P₂ + (1000 kg/m³)(9.8 m/s²)(0.20 m) + (1/2)(1000 kg/m³)(2.40 m/s)²
Simplifying the equation, we have:
(143,000 Pa) + (1/2)(1000 kg/m³)(1.44 m²/s²) = P₂ + (1000 kg/m³)(1.96 m²/s²)
143,720 Pa = P₂ + 1,960 Pa
P₂ + 1,960 Pa = 143,720 Pa
P₂ = 143,720 Pa - 1,960 Pa
P₂ = 141,760 Pa
Therefore, the water pressure at the top of the step is approximately 141,760 Pa (or 141.76 kPa).
google Bernoulli
constant=
v^2/2 + g z + p/rho
in your units
v in m/s
p in Pascals(N/m^2)
rho = 1000 kg/m^3 for water
g about 9.8
v at bottom = 1.2 m/s
v at top = 2 v at bottom =2.4 m/s
z at top = z at bottom + .2 meters
p/rho bottom =143000/1000
so
2.4^2/2 + 9.8(.2)+ 143 = 1.2^2+0+p/1000