The Equation of Continuity states that the mass flow rate has the same value at every position along a tube that has a single entry and a single exit point for fluid flow.
Basically it boils down to the idea that the fluid doesn’t magically disappear or appear. If 2.0 kg of fluid flows past a point in a tube in a time of 1.0 s, then 2.0 kg of fluid flows past another point in that tube in 1.0 s as well. If the tube is getting larger or smaller, the velocity adjusts to keep the mass flow rate the same.
For a definition of mass flow rate, check the textbook.
At Location One, fluid with a density of 1.27×103 kg/m3 is flowing at speed of 3.38 m/s through a circular pipe which has a radius of 7.400×10-2 m. As the fluid flows along the pipe, the pipe gets larger. At Location Two the pipe has a radius of 2.464×10-1 m (it is still circular in nature).
What is the cross-sectional area of the pipe at Location One? What is the speed of the fluid at Location Two?
A = pi r^2
A1v1 = A2v2
To find the cross-sectional area of the pipe at Location One, we can use the formula for the area of a circle: A = πr^2, where A is the area and r is the radius.
Given that the radius at Location One is 7.400×10^-2 m, we can substitute this value into the formula:
A1 = π(7.400×10^-2)^2
A1 = π(5.4656×10^-3)
A1 ≈ 1.7175×10^-2 m^2
So, the cross-sectional area of the pipe at Location One is approximately 1.7175×10^-2 m^2.
Now, to find the speed of the fluid at Location Two, we can use the equation of continuity, which states that the mass flow rate is constant throughout the pipe. The formula for mass flow rate is:
m_dot = ρ * A * v
Where m_dot is the mass flow rate, ρ is the density of the fluid, A is the cross-sectional area, and v is the velocity of the fluid.
Since the mass flow rate is constant, we can set the mass flow rate at Location One equal to the mass flow rate at Location Two:
m_dot1 = m_dot2
Using the given information:
m_dot1 = ρ * A1 * v1
m_dot2 = ρ * A2 * v2
Where ρ = 1.27×10^3 kg/m^3, A1 ≈ 1.7175×10^-2 m^2, and v1 = 3.38 m/s.
We also know the radius at Location Two is 2.464×10^-1 m, so we can calculate A2:
A2 = π(2.464×10^-1)^2
A2 = π(6.0475×10^-2)
A2 ≈ 1.9074×10^-1 m^2
Now we can solve for v2:
m_dot1 = m_dot2
ρ * A1 * v1 = ρ * A2 * v2
Substituting the known values:
(1.27×10^3)(1.7175×10^-2)(3.38) = (1.27×10^3)(1.9074×10^-1)(v2)
v2 = (1.27×10^3)(1.7175×10^-2)(3.38) / (1.27×10^3)(1.9074×10^-1)
v2 ≈ 0.379 m/s
So, the speed of the fluid at Location Two is approximately 0.379 m/s.