Problem Statement
The water is supplied at the rate of 30 litres per second from a height of 4 m to a hydraulic ram, which raises 3 litres per second to a height of 18 m from the ram. Determine D' Aubuisson's and Rankine's efficiencies of the hydraulic ram.
Given Data & Constants
- Supply flow rate, \(Q = 30 \, \text{L/s}\)
- Supply height, \(H = 4 \, \text{m}\)
- Delivery flow rate, \(q = 3 \, \text{L/s}\)
- Delivery height, \(h = 18 \, \text{m}\)
Solution
1. D'Aubuisson's Efficiency (\(\eta_D\))
This efficiency is the simple ratio of the energy output (lifting water) to the energy input (supply water).
2. Rankine's Efficiency (\(\eta_R\))
This efficiency considers the net work done. The input energy comes from the "waste" water (\(Q-q\)), and the output work is lifting the delivered water through the net height (\(h-H\)).
D'Aubuisson's Efficiency: \( \eta_D = 45\% \)
Rankine's Efficiency: \( \eta_R \approx 38.9\% \)
Explanation of a Hydraulic Ram and its Efficiencies
A hydraulic ram is a clever pump that uses the energy of a large amount of water falling a small height to lift a small amount of that water to a much greater height. It works by using the "water hammer" effect. The flowing supply water is suddenly stopped by a valve, creating a high-pressure spike that forces a small portion of the water up the delivery pipe.
- D'Aubuisson's Efficiency: This is a straightforward energy balance. It compares the total potential energy of the water that was lifted to the total potential energy of the water that was supplied.
- Rankine's Efficiency: This is often considered a more practical measure. It recognizes that the water being lifted was already at the supply height, so the useful work is only in lifting it the *additional* height (\(h-H\)). It also correctly identifies that the energy to do this work comes from the water that was *not* lifted, i.e., the "waste" water (\(Q-q\)). This generally results in a lower, but more realistic, efficiency value.
