Momentum Equation

Why it matters

The momentum equation is crucial in fluid mechanics as it helps engineers analyze and predict the forces exerted by fluid flow on structures, such as pipes, nozzles, and turbines. Understanding these forces is essential for designing safe and efficient fluid systems in various engineering applications.

Key ideas

• Momentum Principle: The momentum equation is derived from Newton's second law, which states that the rate of change of momentum of a fluid is equal to the sum of external forces acting on it.
• Control Volume: A control volume is a defined region in space through which fluid flows. The momentum equation is applied to this control volume to analyze the forces.
• Forces in Fluid Flow: These include pressure forces, viscous forces, and body forces such as gravity.
• Applications: Used in designing piping systems, analyzing flow over structures, and in turbo machinery.

Formulas

• F_average = Δ(momentum_system)/Δt; instantaneously, ΣF = d(momentum_system)/dt
  • F: Net force acting on the fluid (N)
  • m: Mass of the fluid (kg)
  • v: Velocity of the fluid (m/s)
  • t: Time (s)
• ΣF = ρQ(v_out - v_in)
  • ΣF: Sum of external forces (N)
  • ρ: Density of the fluid (kg/m³)
  • Q: Volumetric flow rate (m³/s)
  • v_out: Velocity of fluid leaving the control volume (m/s)
  • v_in: Velocity of fluid entering the control volume (m/s)

The compact control-volume expression ΣF = ρQ(v_out-v_in) assumes steady flow, one inlet and outlet, constant density and uniform velocity profiles; it is a vector equation. Nonuniform profiles need momentum-flux integrals or correction factors. Include pressure forces at cuts and distinguish forces on the fluid from the opposite forces it exerts on hardware.

Worked example

Given: A water jet with a density of 1000 kg/m³ strikes a stationary flat plate perpendicularly at a velocity of 10 m/s. The cross-sectional area of the jet is 0.01 m². Calculate the force exerted by the jet on the plate.

1. Calculate the mass flow rate (ṁ):

   • Formula: ṁ = ρ·A·v
   • Calculation: ṁ = 1000 kg/m³ · 0.01 m² · 10 m/s = 100 kg/s

2. Apply the momentum equation:

   • Formula: F = ṁ·(v_out - v_in)
   • The jet spreads along the plate; its outlet velocity component normal to the plate is zero, not its total speed. Take the incident-jet direction as positive. Assume atmospheric pressure on the free jet and neglect other forces in that direction.
   • Calculation: F = 100 kg/s · (0 - 10 m/s) = -1000 N

3. Result:

   • The plate exerts -1000 N on the fluid. By action and reaction the jet exerts +1000 N on the plate, in the incoming-jet direction.

Common mistakes

• Confusing mass flow rate with volumetric flow rate.
• Not accounting for all forces acting on the control volume.
• Incorrectly assuming steady flow conditions when they do not apply.

For GATE ME

• Questions often involve calculating forces on submerged surfaces or within piping systems.
• Practice problems involving both steady and unsteady flow conditions.
• Be familiar with applying the momentum equation to different control volumes.

Quick check

1. What is the primary principle behind the momentum equation?
2. How does the momentum equation relate to Newton's second law?
3. What are the common forces considered in fluid flow analysis?

Answers: 1. Newton's second law; 2. It equates the rate of change of momentum to external forces; 3. Pressure, viscous, and body forces.