Water at a pressure of 4.20 atm at street level flows into an office building at a speed of 0.65 m/s through a pipe 6.60 cm in diameter. The pipes taper down to 2.20 cm in diameter by the top floor, 28.0 m above. Calculate the water pressure in such a pipe on the top floor in Pa.
I attempted the problem and got 271465 Pa but it is incorrect.
first, use the law of continuity to determine outlet flow
(area1*velocity1)=area2*velocity2
then straight Bernoluli's formula. You should get the answer ok. If not post your work and I will check later.
To calculate the water pressure on the top floor of the building, we can use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid.
Bernoulli's equation:
P₁ + ρgh₁ + (1/2)ρv₁² = P₂ + ρgh₂ + (1/2)ρv₂²
where:
P₁ and P₂ are the pressures at two different points in the fluid
ρ is the density of the fluid
g is the acceleration due to gravity
h₁ and h₂ are the heights of the fluid at the two points
v₁ and v₂ are the velocities of the fluid at the two points
In this problem, we have:
P₁ = 4.20 atm
P₂ = ? (what we need to find)
h₁ = 0 m (street level)
h₂ = 28.0 m (top floor)
v₁ = 0.65 m/s
v₂ = ? (what we need to find)
First, let's convert the pressure at street level to Pascals:
P₁ = 4.20 atm * 101325 Pa/atm = 426030 Pa
Next, we need to find the velocity v₂ on the top floor. To do this, we'll use the principle of continuity, which states that the flow rate of an incompressible fluid remains constant as it passes through a pipe with varying diameters.
The formula is given as:
A₁v₁ = A₂v₂
where A₁ and A₂ are the cross-sectional areas of the pipe at two different points, and v₁ and v₂ are the velocities of the fluid at those points.
Given that the diameters of the pipes are 6.60 cm and 2.20 cm respectively, we can calculate the areas:
A₁ = pi * (0.0660/2)^2 = 0.003415 m²
A₂ = pi * (0.0220/2)^2 = 0.000379 m²
We now have two equations:
P₁ + ρgh₁ + (1/2)ρv₁² = P₂ + ρgh₂ + (1/2)ρv₂² (Bernoulli's equation)
A₁v₁ = A₂v₂ (continuity equation)
Rearranging the continuity equation, we get:
v₂ = A₁v₁ / A₂ = (0.003415 m² * 0.65 m/s) / 0.000379 m² = 5.882 m/s
Now, substitute the values into Bernoulli's equation:
426030 Pa + ρgh₁ + (1/2)ρv₁² = P₂ + ρgh₂ + (1/2)ρv₂²
Since the pressure on the top floor (P₂) is what we're trying to find, we isolate it:
P₂ = 426030 Pa + ρgh₁ + (1/2)ρv₁² - ρgh₂ - (1/2)ρv₂²
The density of water (ρ) is approximately 1000 kg/m³.
Now, substitute the known values:
P₂ = 426030 Pa + (1000 kg/m³ * 9.8 m/s² * 0 m) + (1/2)(1000 kg/m³ * 0.65 m/s)² - (1000 kg/m³ * 9.8 m/s² * 28.0 m) - (1/2)(1000 kg/m³ * 5.882 m/s)²
Calculating the expression gives:
P₂ = 426030 Pa + 0 Pa + 0.2113 Pa - 274400 Pa - 16876 Pa = 134.741 Pa
Therefore, the water pressure on the top floor of the building is approximately 134.741 Pa.