The intake in the figure has cross-sectional area of 0.75 m2 and water flow at 0.41 m/s. At the outlet, distance D = 180 m below the intake, the cross-sectional area is smaller than at the intake and the water flows out at 9.6 m/s. What is the pressure difference between inlet and outlet?
p + rho g h + (1/2)rho v^2 = constant
Vin = .41 m/s
Vout = 9.6 m/s
hinlet = 180
houtlet = 0
rho = 1000 kg/m^2
g = 9.8 m/s^2
so
Pinlet + 1000(9.8)(180)+(1/2)1000(.41)^2
= Poutlet +1000(9.8)(0)+(1/2)1000(9.8)^2
solve for Poutlet - Pinlet
To find the pressure difference between the inlet and outlet, we can use Bernoulli's equation, which states that the total energy per unit mass of a flowing fluid is constant along a streamline. In this equation, the terms include the pressure, potential energy, and kinetic energy.
Bernoulli's equation can be expressed as follows:
P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2
Where:
P1 and P2 are the pressures at the inlet and outlet respectively,
ρ is the density of the fluid,
v1 and v2 are the velocities at the inlet and outlet respectively,
g is the acceleration due to gravity, and
h1 and h2 are the heights of the fluid column above a reference point at the inlet and outlet respectively.
In this case, since the heights of the fluid column are not given, we can assume that the reference point is at the same height for both the inlet and outlet. Therefore, the potential energy terms cancel out, and we are left with the following equation:
P1 + (1/2)ρv1^2 = P2 + (1/2)ρv2^2
We are given the cross-sectional area at the inlet (A1 = 0.75 m^2) and the velocity at the inlet (v1 = 0.41 m/s). We are also given the velocity at the outlet (v2 = 9.6 m/s).
To find the pressure difference (ΔP = P2 - P1), we need to rearrange the equation:
ΔP = (1/2)ρ(v2^2 - v1^2)
To calculate the pressure difference, we need to know the density of the fluid. Assuming it is water, the density (ρ) is approximately 1000 kg/m^3.
Substituting the given values into the equation, we have:
ΔP = (1/2)(1000 kg/m^3)((9.6 m/s)^2 - (0.41 m/s)^2)
Calculating this expression will give us the pressure difference (ΔP) between the inlet and outlet.
To find the pressure difference between the inlet and outlet, we can use Bernoulli's equation, which states that the sum of the pressure, kinetic energy, and potential energy per unit volume is constant along a streamline.
Mathematically, Bernoulli's equation can be expressed as:
P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2
Where:
P1 and P2 are the pressures at the inlet and outlet respectively,
ρ is the density of the fluid (water in this case),
v1 and v2 are the velocities at the inlet and outlet respectively,
g is the acceleration due to gravity, and
h1 and h2 are the heights at the inlet and outlet respectively.
In this case, we need to find the pressure difference between the inlet (P1) and outlet (P2). Let's break down the equation step-by-step to solve for it.
Step 1: Find the velocity at the inlet (v1).
Given: Cross-sectional area at the intake (A1) = 0.75 m^2 and water flow velocity at the intake (v1) = 0.41 m/s.
Since A1 = A2 (cross-sectional areas at the inlet and outlet respectively), the equation can be rewritten as:
A1v1 = A2v2
Therefore, v1 = (A2v2) / A1
Substituting the given values:
v1 = (A2 * 9.6 m/s) / 0.75 m^2
Step 2: Find the height difference between the inlet and outlet (h1 - h2).
Given: Distance D = 180 m.
Since the inlet is above the outlet, the height difference is negative and can be calculated as:
h1 - h2 = -D
Step 3: Simplify Bernoulli's equation based on the above calculations:
P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2
Since the pressure at the outlet (P2) is atmospheric pressure, the equation becomes:
P1 + (1/2)ρv1^2 + ρgh1 = atmospheric pressure + (1/2)ρv2^2 + ρgh2
Let's assume atmospheric pressure as Patm.
Step 4: Simplify the equation further:
P1 - Patm = (1/2)ρ(v2^2 - v1^2) - ρg(D)
Step 5: Calculate the pressure difference (ΔP):
ΔP = P1 - Patm = (1/2)ρ(v2^2 - v1^2) - ρg(D)
Substituting the values calculated in steps 1 and 2:
ΔP = (1/2)ρ[9.6^2 - ((A2 * 9.6 / A1)^2)] - ρg(-D)
Given:
Density of water (ρ) = 1000 kg/m^3
Acceleration due to gravity (g) = 9.8 m/s^2
Substituting these values into the equation, you can calculate the pressure difference (ΔP).