To answer this question properly requires the flowrate (Q) -- head (H) chart for the pump.
I suppose that this is for an irrigation pump.
From the Q - H chart, you can get the RPM that the pump has to operate at to deliver the Q and H. From this you will obtain the HP that is needed to power the pump. You can also ensure that the pump is operating where it is most efficient.
From the engine HP - Speed graph you can then determine which RPM will supply this HP. Be sure to use the continuous ratings.
However; since the RPM of the engine will probably not match the pump RPM, you will then need to determine the driver and driven pulley sizes and belt requirements. (5 Hp is well within the capability of a single belt, but I would use two for reliability. Nothing worse than a belt breaking just after starting and you don't find out until afternoon.)
You will then need to figure about a 10 to 15% power loss by the belt drive. Refigure the engine speed and recalculate the pulley sizes.
Also, check your pump application. If the pump is generating too much head for the application, then you are wasting power and as a result fuel.
If less head is needed, you can put a smaller diameter impeller in (or machine down the existing one). Most pumps have different Q -- H curves for different diameter impellers.
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Today's Featured Article - The Nuts and Bolts of Fasteners - Part 2 - by Curtis Von Fange. In our previous article we discussed capscrews, bolts, and nuts along with their relative hardness and thread sizes. In this segment we will finish up on our fasteners and then work with ways to keep them from loosening up in the field. Capscrews, bolts and nuts are not the only means of holding two parts together. When dealing with thinner metals like sheet tin, a long bolt and
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