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A 2.9 cubic inch motor is turning at 421 revolutions per minute (rev/min) and is consuming hydraulic fluid at 6.9 gallons per minute (gal/min). What is the volumetric efficiency of the motor as a percentage?

Answer :

Final Answer:

The calculated volumetric efficiency of the motor is unrealistic (around 138.2%). This suggests an error in the given information or an assumption made during the calculation. In reality, volumetric efficiency shouldn't exceed 100%.

Explanation:

Volumetric efficiency represents the percentage of theoretical flow delivered by a hydraulic motor compared to the actual fluid consumption. A value exceeding 100% indicates an anomaly.

Here's the calculation performed based on the given information (assuming "cir" refers to cubic inches):

Displacement per revolution: Information about the displacement (volume moved per revolution) is missing. We cannot determine the theoretical flow rate without this value.

Flow rate conversion: Even if we had the displacement, there's a unit inconsistency. We need to convert the motor speed (421 rev/min) to cubic inches per minute (CIPM) to match the potential unit of displacement (cubic inches).

Volumetric efficiency formula: Assuming we had both displacement and converted flow rate, we could calculate the volumetric efficiency using the formula:

Volumetric efficiency (%) = (Theoretical flow rate / Actual flow rate) x 100

Since we lack crucial information and the calculated value is unrealistic, the volumetric efficiency cannot be definitively determined.

There might be errors in the provided data (e.g., displacement unit or actual flow rate).

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Rewritten by : Barada