MCP3909 Energy metering IC , How to use the HFOUT in it?
From the Reference Design User's Guide, page 42:


  "The meter constant is typically given in units of impulses per kilo-watt hour. As an
  example, the calibration output frequency of CF, METER_CONSTANT =
  3200 imp/kWh or 6400 imp/kWh"


Section 5.1.1 on page 41 describes the calibration procedure.

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Also had a look at the datasheet, and on the same page 20  Mike refers to there's this equation:  



Channel 1 is the voltage sense channel, channel 0 is the current sense. Multiplied, including the current sense amplification you get instantaneous power. Input voltages are converted relative to a reference voltage, so you have to divide both channels by that (the denominator). HFc is a constant you can program by setting F0, F1 and F2, as per the table in Mike's answer (page 20 of the datasheet).  

Note that this gives you a frequency proportional to power. Integrating this over time means counting periods of that frequency gives you a measure of consumed energy, and that's where the 3200 imp/kWh from the Reference Design User's Guide comes in.

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A concrete example. Let's suppose you want to measure energy of a 230 V/ 10 A device. Maximum input for the voltage channel is 660 mV, so we choose a 1:1000 divider for that. At a gain of 1 for the current channel maximum input voltage is 470 mV, so we can choose a 10 m&Omega; shunt for that, which gives us 10 mV/A. We use the internal Vref of 2.4 V, and set F0, F1 and F2 to 0. Then for a 230 V/ 10 A input we get

\$ HF_{OUT} = \dfrac{8.06 \times 230 mV \times 100 mV \times 1 \times 109.25 Hz}{(2.4 V)^2} = 3.516 Hz \$

That's for 2300 W, or 1.529 Hz per kW. For 1 kWh we'll have 

\$ \text{pulse count} = 3600 s \times 1.529 Hz = 5503 \$

That's 5503 pulses/kWh.