![]() ![]() It has been determined that, above 600Hz, the conversion efficiency of the alternator would drop on account of skin effect and result in flattening of the speed/power curve.Ī 12 pole alternator would generate 600 Hz at 6000 RPM. Best fuel efficiency would be obtained at around that RPM making it the optimum coasting RPM.Īutomobile alternators are designed for higher frequencies to achieve size reduction and for ease of filtering out output ripple (with the battery itself functioning as a capacitor at the higher frequencies). The following additional information may be of use.Īn automobile engine generates maximum torque at around 2000 RPM. Otherwise, the engine would require to be throttled in order to generate a lower current. It would be necessary to ascertain that the magnitude of the charging current would not be detrimental to the battery. It is assumed that the batteries would be sufficiently depleted after feeding external loads. The charging current would also not be very high during constant-voltage charging by the alternator (no current control).įor the current project, the extent of discharge of the batteries would be a key factor in estimating the maximum charging current that would be required.Ĭharging current = (Alternator voltage - Battery voltage) / Battery internal resistance Since the battery would get discharged and charged in quick succession, its depletion would be quite low. Likewise, after discharge with the engine stationary. The highest current would be drawn from the battery during cranking, with the alternator making up for that after starting the engine. The purpose of a vehicle alternator is to feed various electrical loads as well as to charge the vehicle battery. The goal being to charge the batteries, as the mower engine is running, for that hour duration. The expected engine run-time is around 1 hour. If it is peak RPM, can you just linearly extrapolate and say 3,000 RPM the alternator will provide half the current (20A)?įrom an application standpoint, the 24v alternator will be used to charge 2 12v 55Ah (wired in series 24v). If you know the engine runs at 3,000 RPM, with a max of 3,200 RPM Do you gear your alternator pulley in such a way that maps the running engine RPM (3,000) to the alternators rated RPM (6,000) to provide the 40A regularly? Or, will pushing 40A continuously burn the alternator out? In other words, if the measurements provided for the alternator are based on 6K RPM, is that peak (highest RPM the alternator should see)? or average (average running RPM)? However, when looking at alternators online, these metrics are often left out, which leads me to believe I may be incorrectly interpreting their meaning.įor example, and as reference to the questions below, take this Delco 10SI (1102916), which is rated for 40A 6,000 RPM. Where IL is the idle amperage output, IR is the rated amperage output, and VT is the test voltage. I did a bit of research and see that alternators, which conform to ISO 8854, are rated as IL/IRA VTV. ![]() The main reason for this question is that I am trying to understand the power output relationship between the alternator and engine. This engine in particular runs around 3,000 RPM, and does not exceed 3,200 RPM (based on a few quick tests with a cheap tach). I am currently working on a project that requires a 24v alternator, which is powered by a mower engine. ![]()
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