Question

A hydraulic ram operates at a drive head of 3 m and a delivery head of 20 m. The flow through the drive pipe is 0.6 $\text{m}^3\text{/min}$ and the discharge at the outlet of the delivery pipe is 0.07 $\text{m}^3\text{/min}$. Compute the efficiency of the ram adopting Rankine's formula.

• A. 74.8 %
• B. 77.3 %
• C. 80.0 %
• D. 72.8 %

Hint:

Be careful to distinguish between the two formulas:
- Rankine's uses net lift $(h - H)$ and waste flow $(Q - q)$ in the denominator.
- Aubuisson's uses total lift $h$ and total flow $Q$ in the denominator: $\eta = \frac{q \cdot h}{Q \cdot H}$.
Select the formula specifically requested in the question.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A hydraulic ram uses the kinetic energy of a large volume of water falling through a small supply head to lift a small volume of water to a much higher delivery head.
The efficiency of a hydraulic ram can be calculated using two classical formulas: Rankine's formula and Aubuisson's formula.
Rankine's formula assumes the useful work done is lifting the delivered water from the supply level (not the ram level) to the delivery level.
Key Formula or Approach:
Rankine's efficiency ($\eta_R$) is given by: \[ \eta_R = \frac{q \times (h - H)}{(Q - q) \times H} \times 100 \] where: - $Q$ is the total supply flow entering the drive pipe.
- $q$ is the useful delivery flow.
- $H$ is the supply or drive head.
- $h$ is the delivery head.
- $(h - H)$ is the net lift above the supply reservoir water level.
- $(Q - q)$ is the waste water discharged through the waste valve.

Step 2: Detailed Explanation:
Let us plug in the given values:
- Drive head ($H$) $= 3\text{ m}$
- Delivery head ($h$) $= 20\text{ m}$
- Net lift ($h - H$) $= 20\text{ m} - 3\text{ m} = 17\text{ m}$
- Supply flow ($Q$) $= 0.6\text{ m}^3\text{/min}$
- Delivery flow ($q$) $= 0.07\text{ m}^3\text{/min}$
- Waste water flow ($Q - q$) $= 0.6 - 0.07 = 0.53\text{ m}^3\text{/min}$
Substitute these values into Rankine's efficiency equation: \[ \eta_R = \frac{0.07 \times 17}{0.53 \times 3} \times 100 \] \[ \eta_R = \frac{1.19}{1.59} \times 100 \] \[ \eta_R \approx 0.7484 \times 100 = 74.84\% \]

Step 3: Final Answer:
Adopting Rankine's formula, the efficiency of the hydraulic ram is approximately $74.8\%$.
Hence, the correct option is (A).