Energy Loss in Pipe Flow – Exit Loss to Atmosphere What is the loss of head at the exit of a pipe discharging to a large reservoir or atmosphere?

Difficulty: Easy

Correct Answer: v^2 / (2g)

Explanation:


Introduction:
Minor losses in pipe flow include losses at entrances, exits, bends, valves, and fittings. The exit loss is a standard case when a pipe discharges into a large tank or the atmosphere.


Given Data / Assumptions:

  • Pipe exits into a large reservoir or free atmosphere.
  • Velocity in the reservoir is effectively zero.
  • v denotes the average velocity inside the pipe at the exit section.


Concept / Approach:

The kinetic energy per unit weight carried by the jet inside the pipe is v^2 / (2g). On discharge to a large reservoir, this kinetic energy is dissipated as turbulence, so the corresponding head is lost. Hence the exit loss coefficient K_exit = 1.


Step-by-Step Solution:

Step 1: Write exit loss formula: h_exit = K_exit * v^2 / (2g).Step 2: For discharge to a large tank or atmosphere, K_exit = 1.Step 3: Therefore, h_exit = v^2 / (2g).


Verification / Alternative check:

Energy equation between a section just inside the pipe and a point in the reservoir at the same elevation shows the velocity head is not recovered in pressure but is dissipated, confirming the result.


Why Other Options Are Wrong:

0.5 * v^2 / (2g): Entrance, not exit, typically uses approximately 0.5 for sharp edges.1.5 * v^2 / (2g) and 2 * v^2 / (2g): No standard basis for such high coefficients at a plain exit.Negligible: The velocity head is not recovered; the loss is significant.


Common Pitfalls:

Confusing entrance and exit coefficients or assuming kinetic energy is recuperated in pressure at the outlet to a large tank.


Final Answer:

v^2 / (2g)

More Questions from Hydraulics and Fluid Mechanics

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion