Consider a U-shaped tube filled with water as shown below. The left arm of the tube has a diameter of 2 cm, while the right arm has a diameter of 1 cm. The length of each arm is 15 cm. The tube is open to the atmosphere at both ends.
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(a) Calculate the height difference, h, between the water levels in each arm of the tube.
(b) Calculate the pressure difference, ΔP, between the water levels in each arm of the tube.
(c) If the water velocity in the left arm is 2 m/s, what is the velocity of the water in the right arm?
(d) Is this system in equilibrium? Justify your answer.
(a) To calculate the height difference between the water levels in each arm of the tube, we'll use the principle of Pascal's law. According to Pascal's law, the pressure at any given point in an incompressible fluid is the same in all directions.
Since the tubes are open to the atmosphere, the pressure at each water level is equal to the atmospheric pressure, which we'll denote as P_atm.
Next, we'll use the equation for pressure in a fluid:
Pressure = density * gravitational acceleration * height
Let's denote the height difference between the water levels as h. In the left arm of the tube, the pressure is given by:
P_left = P_atm + (density * g * h)
In the right arm of the tube, the pressure is also given by:
P_right = P_atm + (density * g * h)
Since the pressures are equal, we can set these two equations equal to each other:
P_atm + (density * g * h) = P_atm + (density * g * h)
Simplifying, we find:
density * g * h = density * g * h
The density and gravitational acceleration are constant, and thus cancel out:
h = h
Therefore, the height difference, h, between the water levels in each arm of the tube is the same. Hence,
h = 0 cm
(b) Since the height difference is zero, the pressure difference, ΔP, between the water levels in each arm of the tube is also zero. Hence,
ΔP = 0 Pa
(c) To find the velocity of water in the right arm, we'll apply the principle of continuity. According to the principle of continuity, the product of the area of a pipe and the velocity of fluid flowing through it is constant.
Let's denote the velocity of water in the left arm as v_left and the velocity of water in the right arm as v_right. The cross-sectional areas of the left arm and right arm are A_left and A_right, respectively.
The continuity equation can be written as:
A_left * v_left = A_right * v_right
We are given the diameter of the left arm as 2 cm, so the radius is 1 cm (0.01 m). Therefore, the area of the left arm is:
A_left = π * (0.01 m)^2
= 0.000314 m^2
The diameter of the right arm is given as 1 cm, so the radius is 0.5 cm (0.005 m). Therefore, the area of the right arm is:
A_right = π * (0.005 m)^2
= 0.0000785 m^2
Substituting these values into the continuity equation:
0.000314 m^2 * (2 m/s) = 0.0000785 m^2 * v_right
Simplifying, we find:
v_right = (0.000314 m^2 * (2 m/s)) / 0.0000785 m^2
= 4 m/s
Therefore, the velocity of water in the right arm is 4 m/s.
(d) Yes, this system is in equilibrium. The height difference between the water levels in each arm is zero, indicating that there is no net pressure difference. Additionally, the velocities of the water in each arm are equal, indicating a continuous and steady flow. Therefore, the system is in equilibrium.