Water is pumped through a pipe of
diameter 15.0 cm from theColorado River
up to Grand Canyon Village, on the rim of
thecanyon. The river is at 564 m
elevation and the village isat 2096 m.
(a) At what minimum pressure must the water be pumped to arrive atthe village? (b) If 4500 m3 are pumped per day, what
is the speed of the waterin the pipe? (c) What additional pressure is necessary
to deliver this flow?Note: You may
assume that the free-fall acceleration
and thedensity of air are constant over
the given range of elevations.
a) Pressure will be (rho)gh with rho being the density of water (which I don't know offhand)
b)Mass flow rate (4500/seconds in a day) = (rho)Av. Where A is Pi r^2.
c) I'm guessing this is a Bernoulli with (rho)gy1 = 1/2(rho)v^2 + (rho)gy2
Sorry this is a little sketchy, I'm doing this off the top of my head.
To solve this problem, we can make use of Bernoulli's principle, which states that the total energy of a fluid flowing through a pipe remains constant along a streamline. This principle can help us determine the minimum pressure, speed, and additional pressure needed to deliver the water.
Let's solve each part of the problem separately:
(a) To find the minimum pressure, we need to consider the change in elevation of the water from the river to the village. The difference in elevation is 2096 m - 564 m = 1532 m.
Using Bernoulli's principle, we can equate the initial and final energies of the water:
P₁ + (1/2)ρv₁² + ρgh₁ = P₂ + (1/2)ρv₂² + ρgh₂
Here,
P₁ = pressure at the river's elevation (unknown)
P₂ = pressure at the village's elevation (minimum pressure required)
ρ = density of water
v₁ = velocity of water at the river
v₂ = velocity of water at the village
g = acceleration due to gravity
h₁ = height of the river's elevation
h₂ = height of the village's elevation
Since the water is pumped, we can assume that the velocity at the river is negligible (v₁ ≈ 0). Also, at the village, the water will come to rest (v₂ = 0).
This simplifies the equation to:
P₁ + ρgh₁ = P₂
Plugging in the given values:
P₂ = P₁ + ρgh₁
= 0 + (1000 kg/m³)(9.8 m/s²)(1532 m)
≈ 1.49 x 10⁷ Pa
Therefore, the minimum pressure at which the water must be pumped to arrive at the village is approximately 1.49 x 10⁷ Pa.
(b) To find the speed of the water in the pipe, we can use the volumetric flow rate equation:
Q = Av
where Q is the volumetric flow rate, A is the cross-sectional area of the pipe, and v is the velocity of the water.
The cross-sectional area can be found using the formula:
A = πr²
= π(d/2)²
= π(0.15 m/2)²
≈ 0.0177 m²
Given that 4500 m³ are pumped per day, we can convert it to the volumetric flow rate per second:
Q = (4500 m³/day)(1 day/86400 s)
≈ 0.0521 m³/s
Now, we can rearrange the equation Q = Av to solve for v:
v = Q/A
= (0.0521 m³/s)/(0.0177 m²)
≈ 2.94 m/s
Therefore, the speed of the water in the pipe is approximately 2.94 m/s.
(c) To find the additional pressure necessary to deliver this flow, we can once again use Bernoulli's principle. However, this time we will consider the initial velocity at the river (v₁ ≠ 0) and the final velocity at the village (v₂ ≠ 0).
The equation becomes:
P₁ + (1/2)ρv₁² + ρgh₁ = P₂ + (1/2)ρv₂² + ρgh₂
Rearranging the equation and substituting in the given values:
P₂ = P₁ + (1/2)ρv₁² + ρgh₁ - (1/2)ρv₂²
≈ 1.49 x 10⁷ Pa + (1/2)(1000 kg/m³)(0 m/s)² + (1000 kg/m³)(9.8 m/s²)(1532 m) - (1/2)(1000 kg/m³)(2.94 m/s)²
≈ 1.49 x 10⁷ Pa + 0 Pa + 1.49 x 10⁷ Pa - 0.00906 x 10⁷ Pa
≈ 2.97 x 10⁷ Pa
Therefore, the additional pressure necessary to deliver this flow is approximately 2.97 x 10⁷ Pa.