sludge drainage pump and filter

[FREE] A centrifugal pump has the performance characteristics of a pump with a 6-inch diameter impeller, as - Zhuo Mei

Centrifugal Pump Performance

A centrifugal pump has the performance characteristics of a pump with a 6-inch diameter impeller. The pump is operating at 3500 rpm. The expected head gained if the speed of this pump is reduced to 2760 rpm while operating at peak efficiency can be determined by comparing the performance characteristics of the pump at the two different speeds.

Head Gained Calculation

The head gained by a centrifugal pump is directly proportional to the square of the impeller speed. The original speed is 3500 rpm, and the new speed is 2760 rpm.

(Change in speed ratio)² = (New speed / Original speed)²

(Change in speed ratio)² = (2760 rpm / 3500 rpm)²

(Change in speed ratio)² = 0.7878

Change in speed ratio ≈ 0.887

Expected head gained = Change in speed ratio × Head gained at original speed

Without the specific values of the head gained or the performance characteristics of the pump, it is not possible to provide an exact numerical answer. However, you can use the given formula and the specific values of the pump's performance characteristics to calculate the expected head gained.

Affinity Laws

The expected head gained when reducing a centrifugal pump's speed from 3500 rpm to 2760 rpm can be calculated using the affinity law for head, which shows that head is proportional to the square of the speed ratio. Specifically, the new head will be about 62.1% of the original head at 3500 rpm. For an exact calculation, the head at 3500 rpm must be known.

The most relevant affinity law for head (H) is that head is proportional to the square of the speed (N) of the pump:

H1 / H2 = (N1 / N2)²

Where:

H1 = original head at original speed (3500 rpm)

H2 = new head at new speed (2760 rpm)

Speed Ratio

First, calculate the ratio of the new speed to the original speed:

N2 / N1 = 2760 rpm / 3500 rpm ≈ 0.7886

Calculating the New Head

If we let H1 (the head at 3500 rpm) be known or measured, we can then find H2 (the head at 2760 rpm):

H2 = H1 * (N2 / N1)²

H2 ≈ H1 * (0.7886)²

H2 ≈ H1 * 0.6219

Thus, the head gained at 2760 rpm would be approximately 62.1% of the head gained at 3500 rpm.

Conclusion

Without the specific head value at 3500 rpm, we cannot provide a numerical answer, but this formula can be used with known values to determine the head gained at reduced speed. The head will typically decrease due to the reduction in speed, reflecting a decrease in pump performance at lower speeds.

For instance, if the head at 3500 rpm is 100 feet, then the expected head at 2760 rpm would be approximately 62.1 feet using the aforementioned method.