Through a pipe of diameter 16cm, water is pumped from the Colorado River up to Grand Canyon Village, on the rim of the canyon. The river is at 564m elevation and the village is at 2096m elevation.
a) At what minimum pressure must the water be pumped to arrive at the village? The acceleration of gravity is 9.8m/s^3
b)If 4800m^3 are pumped per day, what is the speed of water in the pipe?
c) What additional pressure is necessary to deliver this flow? You may assume that the free-fall acceleration and the density for air are constant over this range of elevations.
a) (density)*g*(elevation change)
= 1.00*10^3 kg/m^3*9.8 m/s^2*1532 m = 15.0*10^6 Pascals
That is about 150 atmospheres.
b) Velocity = (Volume flow rate)/Area
c) Add (1/2)*(density)*V^2 dynamic pressure to the answer in (a)
The density here is the density of water . Air has a neglible effect on the difference between inlet and outlet pressure
THANK YOU!!!
To solve these problems, we can apply Bernoulli's principle, which states that the sum of the pressure energy, kinetic energy, and potential energy per unit volume of a fluid flowing along a streamline remains constant. By using Bernoulli's equation, we can calculate the minimum pressure, speed of water, and additional pressure required.
a) To find the minimum pressure required, we need to consider the potential energy change between the river and the village. The potential energy per unit volume is given by the formula P = ρgh, where P is the pressure, ρ is the density of water, g is the acceleration due to gravity, and h is the height difference.
Given:
River elevation (h1) = 564m
Village elevation (h2) = 2096m
Acceleration due to gravity (g) = 9.8m/s^2
Let's calculate the pressure difference:
ΔP = ρg(h2 - h1)
= ρg(2096m - 564m)
The density of water (ρ) is approximately 1000 kg/m^3.
Substituting the values into the equation:
ΔP = 1000 kg/m^3 * 9.8 m/s^2 * (2096m - 564m)
= 1000 kg/m^3 * 9.8 m/s^2 * 1532m
Therefore, the minimum pressure required to pump water to the village is 15,027,200 Pascals (Pa), which is equivalent to 15.03 megapascals (MPa).
b) To find the speed of water in the pipe, we can use the principle of conservation of mass, which states that the mass flow rate (m_dot) entering a section of a pipe is equal to the mass flow rate leaving that section.
Given:
Volume pumped per day = 4800 m^3
To calculate the speed of water (v), we can use the formula:
m_dot = ρAv
where A is the cross-sectional area of the pipe.
The cross-sectional area of the pipe (A) can be found using the formula:
A = πr^2
where r is the radius of the pipe, which is half the diameter.
Given:
Diameter of the pipe = 16cm
Converting the diameter to meters:
Diameter = 16cm = 0.16m
Radius (r) = 0.08m
Substituting the values into the equation:
A = π(0.08m)^2
= 0.0201m^2
Now, let's calculate the mass flow rate (m_dot):
m_dot = ρAv
= 1000 kg/m^3 * 0.0201m^2 * v
Given:
Volume pumped per day = 4800 m^3
Converting the volume pumped per day to volume per second:
Volume per second = (4800 m^3) / (24 hours * 3600 seconds)
= 0.0556 m^3/s
Now, we can equate the mass flow rate to the volume per second and solve for v:
0.0556 m^3/s = 1000 kg/m^3 * 0.0201m^2 * v
Simplifying the equation, we find:
v = (0.0556 m^3/s) / (1000 kg/m^3 * 0.0201m^2)
≈ 2.76 m/s
Therefore, the speed of water in the pipe is approximately 2.76 m/s.
c) To find the additional pressure required to maintain this flow, we need to consider the energy loss due to friction in the pipe. This is known as the head loss.
Given that the density of air can be assumed constant, we can use the Darcy-Weisbach equation to calculate the head loss:
ΔP = λ(ρv^2/2)(L/D)(v^2/2g)
= λρv^2(L/D)(1/2g)
Where λ is the Darcy friction factor, L is the length of the pipe, and D is the diameter of the pipe.
Let's assume a typical value for λ:
λ = 0.02 (for smooth pipes)
Given:
Length of pipe (L) = Unknown
Diameter of pipe (D) = 16cm = 0.16m
Speed of water (v) = 2.76 m/s
Acceleration due to gravity (g) = 9.8m/s^2
Substituting the values into the equation:
ΔP = (0.02)(1000 kg/m^3)(2.76 m/s)^2 (L/0.16m)(1/19.6m/s^2)
To maintain the given flow rate, ΔP must be equal to the pressure difference calculated in part a (15.03 MPa):
(0.02)(1000 kg/m^3)(2.76 m/s)^2 (L/0.16m)(1/19.6 m/s^2) = 15.03 MPa
Solving for L, we find:
L = (15.03 MPa)(0.16m)(19.6 m/s^2) / [(0.02)(1000 kg/m^3)(2.76 m/s)^2]
Therefore, the additional pressure necessary to deliver the given flow is approximately 15.03 megapascals (MPa).