Mechanical Maintenance Flashcards

A centrifugal pump is experiencing cavitation. Which of the following conditions is MOST likely causing this problem?

Answer: Net Positive Suction Head Available (NPSHa) falling below Net Positive Suction Head Required (NPSHr)

Cavitation occurs when the fluid's local pressure drops below its vapor pressure, forming vapor bubbles that collapse violently. This happens when NPSHa < NPSHr — the available suction head is insufficient to keep the fluid in a liquid state at the pump inlet. Excessive discharge pressure would reduce flow but not cause cavitation. Impeller diameter affects head and flow, not directly cavitation onset. Misalignment causes vibration and bearing wear, not cavitation.

When inspecting a V-belt drive system, you find that the belt deflects 3/4 inch under a 10-pound force at the midpoint of a 24-inch span. The manufacturer specifies 1/2 inch deflection under the same force. What is the CORRECT corrective action?

Answer: Increase center distance to tension the belt until deflection meets spec

Belt deflection greater than specified (3/4" vs. 1/2") indicates the belt is too loose — insufficient tension. The correct fix is to increase center distance (moving the motor or tensioning the idler) until proper deflection is achieved. Belt replacement is not warranted unless the belt shows wear or damage. Reducing sheave diameter changes speed ratio and does not fix tension. Belt dressing masks the problem and can swell belts, causing further issues.

A maintenance technician is measuring runout on a lathe spindle using a dial indicator. The total indicator reading (TIR) shows 0.004 inches. What does this value represent?

Answer: The total variation in spindle position is 0.004 inches, meaning eccentricity is 0.002 inches

TIR (Total Indicator Reading) represents the full swing of the indicator needle — the difference between maximum and minimum readings. Since eccentricity (the offset of center) produces displacement on both sides of the true center, the actual eccentricity or out-of-roundness is TIR ÷ 2 = 0.002 inches. A TIR of 0.004" does not mean a 0.004" diameter error or a 0.004" centerline deviation; those would each imply double the actual runout.

A hydraulic system uses a fixed-displacement gear pump rated at 10 GPM at 1800 RPM. The system relief valve is set to 2000 PSI. If the prime mover runs at 1800 RPM and the actuator is stalled, approximately how much horsepower is being dissipated as heat?

Answer: 11.67 HP

When the actuator stalls, all pump flow bypasses through the relief valve at the relief pressure setting. Hydraulic horsepower = (GPM × PSI) ÷ 1714. HP = (10 × 2000) ÷ 1714 ≈ 11.67 HP. This entire amount is dissipated as heat in the fluid and relief valve — a fixed-displacement pump cannot unload; it continuously produces flow that must go somewhere. This is why fixed-displacement systems are thermally inefficient at stall.

A technician is replacing a failed deep-groove ball bearing on a shaft. After installation, the bearing runs hot (180°F+) within 30 minutes under normal load. The most probable cause is:

Answer: The bearing was press-fitted onto the shaft with excessive interference, preloading the races

Excessive interference fit during press installation compresses the bearing races, reducing internal clearance below the designed operating clearance. The resulting preload generates significant friction and heat. Deep-groove ball bearings should only be filled 1/3 to 1/2 with grease — overpacking (not underpacking) causes heat, but this is listed as 'insufficient.' Shaft-race hardness incompatibility is not a standard failure mode. Load capacity is not indicated as a problem by the scenario description.

During a precision shaft alignment using reverse-dial indicator method, the technician records the following readings on the stationary machine's bracket (at 12, 3, 6, and 9 o'clock positions): +0.000, +0.008, +0.016, +0.008. What condition does this indicate?

Answer: Parallel (offset) misalignment only — the shaft centerlines are parallel but not collinear

In reverse-dial alignment, a pure parallel (offset) misalignment produces a consistent bottom reading that equals the top reading but with opposite sign (or double magnitude when using the convention shown). Here, 12 o'clock = 0.000 and 6 o'clock = +0.016, giving a total variation of 0.016" — this is a classic offset-only pattern. The 3 o'clock and 9 o'clock readings are equal (+0.008), confirming no twist/angular component in the horizontal plane. Angular misalignment would show the 3 and 9 o'clock readings differing from each other.