Pressure difference across an inclined pipe – contributing heads In a steady flow through an inclined pipe with entrance and exit, the pressure difference between the two ends arises due to which components?

Introduction / Context:Applying the energy equation between two pipe sections reveals that several terms contribute to head differences. Recognizing all contributors (major losses, minor losses, and elevation) is essential for accurate sizing and pump selection.

Given Data / Assumptions:

• Steady, incompressible flow.
• Uniform pipe diameters at sections, with defined inlet and outlet.
• Significant friction and minor (local) losses at fittings and ends.

Concept / Approach:

Bernoulli with head-loss terms: z1 + p1/γ + V1^2/(2g) = z2 + p2/γ + V2^2/(2g) + h_f + Σh_m. For equal diameters, velocities may cancel, leaving elevation head (z1 − z2), frictional loss h_f, and local losses (entrance/exit). Thus the pressure difference is due to all three contributions listed.

Step-by-Step Solution:

Verification / Alternative check:

Compute p1 − p2 = γ[(z2 − z1) + h_f + Σh_m] when velocities are equal; each term contributes to the total pressure difference.

Why Other Options Are Wrong:

Choosing only one source underestimates the actual pressure difference; real systems include all listed effects.

Common Pitfalls:

Ignoring exit loss to atmosphere or assuming friction is the only contributor on long lines.

Final Answer:

all the above