Hydraulic power transmission in a pipeline: if inlet pressure is 90 kg/cm² and the pressure drop along the line is 10 kg/cm², what is the efficiency of transmission (based on head ratio)?

Introduction / Context:
In long pipeline power transmission, useful head at the outlet is reduced by frictional losses. A common performance metric is the efficiency of transmission, defined as the ratio of the available head (or pressure) at delivery to the head supplied at the inlet, ignoring mechanical conversion losses at the receiver.

Given Data / Assumptions:

• Inlet pressure = 90 kg/cm² (proportional to head supplied).
• Pressure drop along the line = 10 kg/cm².
• Steady incompressible flow with negligible elevation change and no additional minor losses considered beyond the stated drop.

Concept / Approach:

Efficiency of transmission η_t is typically taken as η_t = Head at outlet / Head at inlet = (H_in − h_loss) / H_in. With pressures used as head proxies for the same fluid, η_t = (P_in − ΔP) / P_in.

Step-by-Step Solution:

Verification / Alternative check:

This matches the general optimum-power result where the maximum transmitted power occurs when h_loss = H/3 (η_t = 2/3 ≈ 66.7%). Since here h_loss/H = 10/90 ≈ 0.111 < 1/3, the efficiency is higher than 66.7%, consistent with 88.8%.

Why Other Options Are Wrong:

(a) 66.6% corresponds to the special case of maximum power, not this scenario. (b) and (c) arise from incorrect ratios. (e) is unrelated to the given numbers.

Common Pitfalls:

Mixing gauge and absolute pressures (here only differences matter); confusing efficiency of transmission with turbine or pump efficiency.

Final Answer:

88.8%