Solve Examples on Continuity Equation
Example 1: Imagine a pipe with water flowing through it. The pipe has a cross-sectional area of 0.1 square meters. Find the mass flow rate into pipe.
Solution
According to the continuity equation, mass flow rate into the pipe must be equal to the mass flow rate out of the pipe.
Use the equation:
Mass flow rate = Density * Velocity * Area
So, the mass flow rate into the pipe would be:
Mass flow rate = 1000 kg/m^3 * 2 m/s * 0.1 m^2 = 200 kg/s
Hence, the mast flow rate is 200 kg/s.
Example 2: A tube has a cross-sectional area of 5 m². Water flows through the tube with the velocity of 50 m/s. Calculate the volume flow rate of water.
Solution
The continuity equation tells that the volume flow rate (Q) is equal to the product of the cross-sectional area (A) and the velocity (v). So, we can use the formula Q = A × V.
Given:
A = 5 m2
V = 50 m/s
Now, the values in the formula,
we get:
Q = A × V
Q = 5 m2 × 50 m/s
Q = 250 m3/s
Therefore, the volume flow rate of water is 250 m3/s.
Example 3: A river has a width of 20 m and a depth of 15 m. The water is flowing with a velocity of 4 m/s. If the river is 200 m long, what is the volume flow rate of water?
Solution
Here, we need to use the continuity equation, which tells that the volume flow rate (Q) is equal to the product of the cross-sectional area (A) and the velocity (v).
Given:
Width (w) = 20 m
Depth (d) = 15 m
Velocity (v) = 4 m/s
Length (L) = 200 m
Now, first find area of cross-sectional (A) of river
A = Width × Depth
A = 20 m × 15 m
A = 300 m2
Now, we can use the formula of flow rate Q = A × V:
Q = 300 m2 × 4 m/s
Q = 1200 m3/s
Therefore, the volume flow rate of water is 1200 m3/s.
Example 4: A cylindrical tube with a diameter of 0.5 m carries a fluid with a velocity of 10 m/s. The pipe gradually narrows down to a diameter of 0.2 m. If the volume flow rate at the wider end is 20 m/s, what is the volume flow rate at the narrower end?
Solution
Here, we need to use the continuity equation, which tells that the volume flow rate (Q) is remains same throughout a pipe of varying diameter. So, the volume flow rate at the wider end is equal to the volume flow rate at the narrower end.
Given:
Diameter at wider end (D1) = 0.5 m
Diameter at narrower end (D2) = 0.2 m
Velocity (v) = 10 m
Volume flow rate at wider end (Q1) = 20 m/s
Now, first find area of cross-sectional (A) at each end using the formula A = πr2:
Area at wider end (A1) = π × (D1/2)²
Area at narrower end (A2) = π × (D2/2)²
Now, we can use the continuity equation to find the volume flow rate at the narrower end:
Q1 = A1 × V
Q2 = A1 × V
Since, Q1 = Q2 then,
A1 × V = A1 × V
Now, substitute the formulas for A1 and A2:
π × (D1/2)² × V = π × (D2/2)² × V
Simplify the equation, we get:
(D1/2)² = (D2/2)²
Taking the square root of both sides, we have:
D1/2 = D2/2
Since D1 = 0.5 m and D2 = 0.2 m, we can solve for the volume flow rate at the narrower end:
Q2 = Q1 × (D2/D1)
Q2 = 10 m3/s × (0.2 m / 0.5 m)
Q2 = 4 m3/s
Therefore, the volume flow rate at the narrower end is 4 m3/s.
Continuity Equation
This equation is a fundamental principle in physics and liquid elements or fluid dynamics that describes the transport of some quantity. It tells the preservation of mass within that system. It is expressed as a mass that is neither created nor destroyed but is conserved for a liquid flowing through a given area.
Table of Content
- Continuity Equation
- Derivation
- Continuity Equation in Different Fields
- Flow Rate Formula
- Assumptions
- Applications
- Solve Examples