In a machine, an ideal fluid with a density of 0.9*10^3 kg/m^3 and a pressure of 1.3*10^5 Pa flows at a velocity of 6.0 m/s through a level tube with a readius of 0.5 cm. This tube connects to a level tube that has a radius of 1.5 cm. How fast does the water flow in the larger tube?
a. 54 m/s
b. 18 m/s
c. 0.67 m/s
d. 0.33 m/s
To solve this problem, we can use the principle of continuity, which states that the mass flow rate of an incompressible fluid remains constant as it flows through a tube of varying cross-sectional area.
The mass flow rate (m_dot) is calculated using the formula:
m_dot = density * velocity * cross-sectional area
In the first tube, we have a density of 0.9 * 10^3 kg/m^3, a velocity of 6.0 m/s, and a radius of 0.5 cm (or 0.005 m). Using these values, we can calculate the mass flow rate in the first tube:
m_dot1 = (0.9 * 10^3 kg/m^3) * (6.0 m/s) * π * (0.005 m)^2
Now, according to the principle of continuity, the mass flow rate should remain constant as the fluid flows into the larger tube. Therefore, the mass flow rate in the second tube (m_dot2) should be equal to m_dot1.
In the second tube, we have a radius of 1.5 cm (or 0.015 m). We need to find the velocity (v2) in the second tube.
Using the formula for mass flow rate, we can setup the equation:
m_dot2 = (0.9 * 10^3 kg/m^3) * v2 * π * (0.015 m)^2
Since we know m_dot1 is equal to m_dot2, we can set up the equation:
m_dot1 = m_dot2
(0.9 * 10^3 kg/m^3) * (6.0 m/s) * π * (0.005 m)^2 = (0.9 * 10^3 kg/m^3) * v2 * π * (0.015 m)^2
Now, solving for v2:
v2 = (6.0 m/s) * (0.005 m / 0.015 m)^2
v2 = (6.0 m/s) * (1/9)
v2 = 0.67 m/s
Therefore, the water flows at a velocity of 0.67 m/s in the larger tube. Therefore, the correct answer is option c. 0.67 m/s.