Water is moving with a speed of 5.0 m/s through a pipe with a cross-sectional area of 4.2 cm2. The water gradually descends 8.2 m as the pipe increases to 9.0 cm2. (a) What is the speed at the lower level? (b) If the pressure at the upper level is 1.8 × 105 Pa, what is the pressure at the lower level?
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To answer these questions, we can make use of the principle of continuity and Bernoulli's equation.
(a) The principle of continuity states that the flow rate of an incompressible fluid remains constant as it passes through a pipe of varying cross-sectional area. Mathematically, this can be expressed as:
A1v1 = A2v2
where A1 and v1 are the cross-sectional area and velocity at the upper level, and A2 and v2 are the cross-sectional area and velocity at the lower level.
In this case, A1 = 4.2 cm^2 = 4.2 × 10^-4 m^2, v1 = 5.0 m/s, A2 = 9.0 cm^2 = 9.0 × 10^-4 m^2 (as given in the question), and we need to find v2.
Using the continuity equation, we can solve for v2:
A1v1 = A2v2
(4.2 × 10^-4 m^2)(5.0 m/s) = (9.0 × 10^-4 m^2)(v2)
Simplifying the equation, we get:
v2 = (4.2 × 10^-4 m^2)(5.0 m/s) / (9.0 × 10^-4 m^2)
v2 ≈ 2.333 m/s
Therefore, the speed at the lower level is approximately 2.333 m/s.
(b) To find the pressure at the lower level, we can use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid at different points in a flow.
Bernoulli's equation is given by:
P1 + ρgh1 + (1/2)ρv1^2 = P2 + ρgh2 + (1/2)ρv2^2
where P1 and P2 are the pressures at the upper and lower levels, ρ is the density of the fluid (assumed to be water in this case), g is the acceleration due to gravity, h1 and h2 are the heights at the upper and lower levels, and v1 and v2 are the velocities at the upper and lower levels.
In this case, P1 = 1.8 × 10^5 Pa (as given in the question), v1 = 5.0 m/s, h1 = 0 m (since the height is not changing), ρ is the density of water (approximately 1000 kg/m^3), v2 ≈ 2.333 m/s (as calculated in part (a)), and h2 = 8.2 m (as given in the question).
Using Bernoulli's equation, we can solve for P2:
P1 + (1/2)ρv1^2 = P2 + ρgh2 + (1/2)ρv2^2
Substituting the known values and calculating:
1.8 × 10^5 Pa + (1/2)(1000 kg/m^3)(5.0 m/s)^2 = P2 + (1000 kg/m^3)(9.8 m/s^2)(8.2 m) + (1/2)(1000 kg/m^3)(2.333 m/s)^2
1.8 × 10^5 Pa + 12500 Pa = P2 + 803360 Pa + 2716.5 Pa
P2 ≈ -8200 Pa
So, the pressure at the lower level is approximately -8200 Pa relative to the pressure at the upper level.