Power Transmission Through a Pipeline Condition for Maximum Power: The power delivered at the outlet is maximized when the head lost due to friction equals one-third of the total supply head.
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Aone-fourth of the total supply head
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Bone-third of the total supply head
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Cone-half of the total supply head
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Dtwo-third of the total supply head
Answer
Correct Answer: one-third of the total supply head
Explanation
Introduction:This item examines the condition for maximizing hydraulic power transmitted through a pipeline when head losses occur due to friction. It is a classic optimization result in fluid power systems.
Given Data / Assumptions:
- Total supply head H at the pipeline inlet.
- Head loss due to friction h_f; remaining head at outlet H_d = H - h_f.
- Discharge Q depends on head loss via Darcy–Weisbach or equivalent; proportional reasoning suffices.
Concept / Approach:
Power transmitted P is proportional to Q * H_d. For a given pipeline, Q increases with head loss, but H_d decreases as h_f increases. Maximizing P with respect to h_f yields the optimal split.
Step-by-Step Solution:
1) Let P = rho * g * Q * (H - h_f).2) For a given pipe, Q is proportional to sqrt(h_f) (for turbulent regimes with head loss proportional to V^2), so Q = k * h_f^(1/2).3) Then P = rho * g * k * h_f^(1/2) * (H - h_f).4) Differentiate with respect to h_f and set derivative to zero to maximize: dP/dh_f = 0 gives h_f = H / 3.5) Therefore maximum power occurs when friction loss is one-third of supply head.Verification / Alternative check:
Substituting h_f = H/3 gives delivery head H_d = 2H/3 and confirms that small deviations reduce P on either side of this optimum.
Why Other Options Are Wrong:
One-fourth, one-half, two-third: These do not satisfy the first-order optimality condition and produce lower transmitted power.
Common Pitfalls:
Assuming maximum power at minimum losses; forgetting that Q depends on h_f; confusing efficiency optimization with power maximization.
Final Answer:
one-third of the total supply head