Maximum efficiency of power transmission through a pipeline If the total head available at the pipe inlet is H and the head lost due to friction is h_f, what is the maximum possible efficiency of power transmission through the pipe?
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A1/2 (50%)
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B2/3 (66.7%)
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C3/4 (75%)
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D4/5 (80%)
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E1/3 (33.3%)
Answer
Correct Answer: 2/3 (66.7%)
Explanation
Introduction / Context:When transmitting power by water through a long pipe to a turbine/nozzle, frictional losses reduce the head at the outlet. The efficiency depends on how much head is lost en route versus delivered for useful work.
Given Data / Assumptions:
- Total head at inlet = H.
- Head loss due to friction along pipe = h_f.
- No other losses or shaft work until the outlet.
Concept / Approach:
Hydraulic power delivered is proportional to the product of discharge and available head at outlet. Optimization (with velocity related to head) gives a specific ratio of loss to available head at maximum efficiency: h_f = H/3, yielding η_max = 2/3.
Step-by-Step Solution:
Let head at outlet H_o = H − h_f.For a given pipe, h_f ∝ V^2; discharge Q ∝ V; delivered power P_d ∝ H_o * Q ∝ (H − k V^2) * V.Differentiate with respect to V, set dP_d/dV = 0 ⇒ optimal when h_f = H/3.Then η_max = H_o / H = (H − H/3) / H = 2/3.Verification / Alternative check:
Classical derivation in fluid mechanics texts confirms η_max = 66.7% for power transmission by a pipeline without other losses.
Why Other Options Are Wrong:
Other percentages do not correspond to the optimal friction–head trade-off; they arise if h_f differs from H/3 and are suboptimal.
Common Pitfalls:
Maximizing outlet head alone (min V) or flow alone (max V) does not maximize power; the product must be optimized.
Final Answer:
2/3 (66.7%)