Test: Continuity Equation - Civil Engineering (CE) MCQ

According to the Continuity Equation,

where a represents flow area, v represents flow velocity, i is for inlet conditions and o is for outlet conditions.
If, 
v_o = 4v

According to the condition for continuity,

2ax + cx = 0
2a + c = 0.

The condition for the flow field to be continuous is:

2ax + by + 2ay + bx = 0
x + y = 0
Hence, the condition for the flow field to be continuous is independent of the constants (a; b) and dependent only on the variables (x; y).

According to the Continuity Equation,

where a represents flow area, v represents flow velocity, i is for inlet conditions and o is for outlet conditions.
A₁v₁ + A₂v₂ = A₃v₃ + A₄v₄
Since d₁ = d₂ = d₃ = d₄, v₁ + v₂ = v₃ + v₄.

According to the Continuity Equation, if no fluid is added or removed from the pipe in any length then the mass passing across different sections shall be the same. This is in accordance with the principle of conservation of mass which states that matter can neither be created nor be destroyed.

According to the Continuity Equation,

where a represents flow area, v represents flow velocity, i is for inlet conditions and o is for outlet conditions. Thus,

St. Venant Equations:
1. Continuity equation

2. Momentum Equation

Assumptions for St. Venant Equations

• Flow is one-dimensional
• Hydrostatic pressure prevails and vertical accelerations are negligible
• Streamline curvature is small.
• Bottom slope of the channel is small.
• Manning’s equation is used to describe resistance effects
• The fluid is incompressible

Hence, St. venant equations represent continuity and momentum equations.

Concept:
According to continuity, the liquid flow rate will be conserved.
Q₁ = Q₂ + Q₃
In a pipe, flow rate is given by
Q = AV
Calculation:
Given V₁ = 3 m/s, D₁ = 450 mm, V₂ = 2.5 m/s, D₂ = 300 mm, D₃ = 200 mm;
From continuity,
D₁² V₁ = D₂² V₂ + D₃² V₃
⇒ 450² × 3 = 300² × 2.5 + 200² × V₃
⇒ V₃ = 9.56 m/s 

Concept:
The principle used here is Conservation of mass
∴ Q₁ = Q₂ + Q₃
Calculation:
Q₁ = Q₂ + Q₃
A₁V₁ = A₂V₂ + A₃V₃
⇒ 0.9 = 300 × 10^-3 × V₂ + 200 × 10^-3 × 3
⇒ 0.9 = 0.3 V₂ + 0.6
⇒ V₂ = 1 m/s