In a vane (rotary) blower, some work is wasted due to backflow or slip. If W1 is the useful work due to compression and W2 is the extra work required due to backflow, what is the blower efficiency?

Introduction / Context:
Vane blowers experience internal leakage/backflow that adds to the work input. Defining efficiency properly helps quantify the fraction of input that is usefully expended on compression versus losses.

Given Data / Assumptions:

• W1 is the useful work for compressing the fluid to the delivery condition.
• W2 is the extra work consumed due to backflow/slip (loss).
• Total input work to the blower is W1 + W2.

Concept / Approach:
Device efficiency in this context is defined as useful output divided by input. For a blower, the useful effect is the theoretical/ideal compression work W1, while the actual input includes the loss W2. Therefore, efficiency must be W1 divided by the total W1 + W2.

Step-by-Step Solution:

Verification / Alternative check:
If W2 = 0 (no losses), η = W1 / W1 = 1, which makes physical sense. As W2 increases, η decreases, also consistent with intuition.

Why Other Options Are Wrong:

• (W1 + W2)/W1: >1 for any loss; efficiencies cannot exceed 1 in this definition.
• W2/(W1 + W2): This is a loss fraction, not efficiency.
• W1 - W2: Not nondimensional and can be negative; not an efficiency.
• W1/W2: A ratio of useful work to loss, not efficiency.

Common Pitfalls:
Confusing loss fraction with efficiency and forgetting to include all loss mechanisms in the energy balance when estimating actual power draw.

Final Answer:
η = W1 / (W1 + W2)