Water is flowing with a speed of 2.0 m/s in a horizontal pipe with cross-sectionalarea decreasing from 2 × 10−2𝑚2 𝑡𝑜 1 × 10−2𝑚2at pressure 4 × 104 𝑃𝑎.What will be the pressure at smaller cross-section?
Question
Water is flowing with a speed of 2.0 m/s in a horizontal pipe with cross-sectionalarea decreasing from 2 × 10−2𝑚2 𝑡𝑜 1 × 10−2𝑚2at pressure 4 × 104 𝑃𝑎.What will be the pressure at smaller cross-section?
Solution
To solve this problem, we can use Bernoulli's equation, which states that the sum of the pressure energy, kinetic energy, and potential energy per unit volume is constant for an incompressible, non-viscous fluid in steady flow. In this case, since the pipe is horizontal, the potential energy term is constant and can be ignored.
Bernoulli's equation is:
P1 + 1/2ρv1² = P2 + 1/2ρv2²
where: P1 = initial pressure = 4 × 10^4 Pa v1 = initial velocity = 2.0 m/s P2 = final pressure (what we're trying to find) v2 = final velocity ρ = density of the fluid (for water, ρ = 1000 kg/m³)
We also know that the velocity of fluid flow changes with the cross-sectional area of the pipe according to the equation of continuity:
A1v1 = A2v2
where: A1 = initial cross-sectional area = 2 × 10^-2 m² A2 = final cross-sectional area = 1 × 10^-2 m²
We can solve this equation for v2:
v2 = A1v1/A2 = (2 × 10^-2 m² * 2.0 m/s) / (1 × 10^-2 m²) = 4.0 m/s
Now we can substitute P1, v1, ρ, and v2 into Bernoulli's equation and solve for P2:
P2 = P1 + 1/2ρv1² - 1/2ρv2² = 4 × 10^4 Pa + 1/2 * 1000 kg/m³ * (2.0 m/s)² - 1/2 * 1000 kg/m³ * (4.0 m/s)² = 4 × 10^4 Pa + 2000 Pa - 8000 Pa = 4 × 10^4 Pa - 6000 Pa = 3.4 × 10^4 Pa
So, the pressure at the smaller cross-section is 3.4 × 10^4 Pa.
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