Affinity Law and Efficiency of a Pump Curve

[Adam Bourque]

Homework Statement

(Pump Curve given as attachment) [/B]
Find the point where the pump is delivering 1,200 gpm at 45 ft of head.
I. What is the pump efficiency (read from the pump curve data)?
ii. What is the hp delivered to the water?
iii. What is the shaft hp required to drive the pump?

Homework Equations

ii.[/B] HP = CFM * Inches of Water Column/6356 * Efficiency

The Attempt at a Solution

I.[/B] I believe the efficiency is at 72.5% looking at the graph.
ii. 1200 * 30/6356 * .72 = 36,000/457,632 7.866 hp (I used 30 inches of water from the graph is that correct?)
iii. plot pump hp is 19hp, shaft hp=19/0.725=26.2 hp

 

[KataKoniK]

Hello there,
Thank you for your post. To find the point where the pump is delivering 1,200 gpm at 45 ft of head, we can use the pump curve provided. The point where the pump delivers 1,200 gpm at 45 ft of head can be found by drawing a horizontal line at 45 ft of head and a vertical line at 1,200 gpm. The point where these lines intersect on the pump curve is the desired point. In this case, it appears to be at around 72.5% efficiency.

I. To find the pump efficiency, we can read the efficiency value from the graph at the desired point. As mentioned, it appears to be around 72.5%.

II. To find the hp delivered to the water, we can use the formula provided: HP = CFM * Inches of Water Column/6356 * Efficiency. In this case, we can use the values of 1,200 gpm and 45 ft of head. Converting 45 ft of head to inches gives us 540 inches. Plugging these values into the formula gives us: 1200 * 540/6356 * 0.725 = 65.26 hp delivered to the water.

III. To find the shaft hp required to drive the pump, we can use the formula: Shaft HP = Pump HP/Efficiency. From the pump curve, we can see that the pump hp at the desired point is around 19 hp. Plugging this value into the formula gives us: 19/0.725 = 26.2 hp required to drive the pump.

I hope this helps! Let me know if you have any further questions.