water is moving with a speed of 5 m/s through a pip with a cross-sectional area of 4cm^2 . the water gradual descends 10 m as the pipe cross-sectional area increases 8cm^2 . if the pressure at the upper level is 1.5*10^5 , the pressure at the lower level is ?
To find the pressure at the lower level, we can apply Bernoulli's equation, which states that the total pressure at any point in a fluid is the sum of the static pressure, dynamic pressure, and potential energy per unit volume.
Given information:
Velocity at upper level (v1) = 5 m/s
Cross-sectional area at upper level (A1) = 4 cm^2 = 4 * 10^(-4) m^2
Velocity at lower level (v2) = ?
Cross-sectional area at lower level (A2) = 8 cm^2 = 8 * 10^(-4) m^2
Pressure at upper level (P1) = 1.5 * 10^5 Pa
Pressure at lower level (P2) = ?
Bernoulli's equation can be written as:
P1 + 1/2 * ρ * v1^2 + ρ * g * h1 = P2 + 1/2 * ρ * v2^2 + ρ * g * h2
Where:
ρ is the density of water (approximately 1000 kg/m^3)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
h1 is the height at the upper level (10 m)
h2 is the height at the lower level (0 m)
Substituting the given values into the equation, we get:
P1 + 1/2 * ρ * v1^2 + ρ * g * h1 = P2 + 1/2 * ρ * v2^2 + ρ * g * h2
1.5 * 10^5 + 0.5 * 1000 * (5)^2 + 1000 * 9.8 * 10 = P2 + 0.5 * 1000 * v2^2 + 1000 * 9.8 * 0
150000 + 0.5 * 1000 * 25 + 98000 = P2 + 0.5 * 1000 * v2^2
150000 + 12500 + 98000 = P2 + 0.5 * 1000 * v2^2
260000 = P2 + 500 * v2^2
Since the pipe is gradually descending, the height at the lower level is 0, which means the term ρ * g * h2 becomes zero.
Rearranging the equation, we get:
P2 = 260000 - 500 * v2^2
Finally, substitute the given velocity at the lower level into the equation to determine P2.
To find the pressure at the lower level, we can use Bernoulli's equation, which states that the sum of the pressure energy, kinetic energy, and potential energy per unit volume of a fluid is constant along a streamline.
Bernoulli's equation is given by:
P₁ + 1/2 ρv₁² + ρgh₁ = P₂ + 1/2 ρv₂² + ρgh₂
Where:
P₁ and P₂ are the pressures at the two levels,
v₁ and v₂ are the velocities at the two levels,
ρ is the density of the fluid (water in this case),
g is the acceleration due to gravity,
h₁ and h₂ are the heights of the two levels.
Given:
P₁ = 1.5 × 10^5 Pa,
v₁ = 5 m/s,
h₁ = 0 m (the initial level),
h₂ = -10 m (the final level),
A₁ = 4 cm² = 4 × 10^(-4) m² (the initial cross-sectional area),
A₂ = 8 cm² = 8 × 10^(-4) m² (the final cross-sectional area).
First, let's find v₂, the final velocity. Since the cross-sectional area of the pipe increases, the velocity of the water decreases as it flows downward (due to the principle of continuity). We can use the equation of continuity to find v₂:
A₁v₁ = A₂v₂
Substituting the known values:
(4 × 10^(-4)) × 5 = (8 × 10^(-4)) × v₂
v₂ = (4 × 10^(-4)) × 5 / (8 × 10^(-4))
v₂ = 20 / 8
v₂ = 2.5 m/s
Now we can substitute the known values into Bernoulli's equation and solve for P₂:
P₁ + 1/2 ρv₁² + ρgh₁ = P₂ + 1/2 ρv₂² + ρgh₂
1.5 × 10^5 + 1/2 × ρ × (5)^2 + ρ × 9.8 × 0 = P₂ + 1/2 × ρ × (2.5)^2 + ρ × 9.8 × (-10)
1.5 × 10^5 + 1/2 × 1000 × 25 + 0 = P₂ + 1/2 × 1000 × 6.25 - 9800 ρ
1.5 × 10^5 + 12,500 = P₂ + 3,125 - 9.8 ρ
P₂ = 1.5 × 10^5 + 12,500 - 3,125 + 9.8 ρ
Now, we need to determine the density of water (ρ). The density of water varies slightly with temperature, but we can approximate it as 1000 kg/m³.
P₂ = 1.5 × 10^5 + 12,500 - 3,125 + 9.8 × 1000
P₂ = 1.5 × 10^5 + 12,500 - 3,125 + 9,800
P₂ = 1.5 × 10^5 + 12,500 - 3,125 + 9,800
P₂ = 1.5 × 10^5 + 19,175
P₂ = 1.69 × 10^5 Pa
Therefore, the pressure at the lower level is approximately 1.69 × 10^5 Pa.