First, the annual solar irradiance is calculated by taking the power incident on a square meter when perpendicular to the sun, and multiplying by 24 hours and then by 365 days. Since the irradiance is typically about 1.366 kW/sq m, when you multiply it out you get 11.9 kW-hr/sq m, which is your number.
But you might have noticed a problem. This assumes the sun is shining 24 hours per day, which is hardly proper.
Second, your second calculation also assumes 24 hours per day sunlight, which is not right. Since the average length of day is about 12 hours, your 25.6 Wh should be divided by 2, giving 12.8 Wh per day.
Finally, this number tells you how much power is falling on the solar cell, not how much the cell puts out. It also assumes the cell will be perpendicular to the sun at all times, and that the atmosphere does not attenuate sunlight. The first may be true, the second is not. Consider that the sun is much dimmer at sunset than at noon.
Let's take the case of noon sunlight. Assuming the air is very clear and you are getting the nominal irradiance, the power falling on the cell will be 1366 w/sq m x .0078, or about 10 watts. Since you have a solar cell rated for 1 watt, this says that you can assume the cell has an efficiency of about 10%, which is about right.
For a 1-watt solar cell which is fixed in position, and is perpendicular to the sun at noon, you can figure on a total output of 4-6 Wh on a clear day. On the one hand, as the sun moves away from the noon position, the effective area decreases (reaching zero at sunrise and sunset when the cell is edge-on to the sun). Also, sunlight gets attenuated by the atmosphere at lower angles, and finally the cells themselves will typically become less efficient at lower intensities.
The reading of energy in the top of the final column is accumulated energy taken from power measurements that occured prior to this table. The 2nd energy reading is 1 watt hour larger than the previous reading.
From that and given that the time interval is 30 seconds, I calculate that the average power over that time period (30 seconds) is 120 watts i.e. 1 watt x 3600 / 30.
Does that help?
The fact that the power reading at 54:52 is 123.45 watts doesn't mean that it remained stable at value throughout the previous 30 seconds but it may have and the energy reading may not show enough decimal places to register this slight discrepancy.
Best Answer
If each register is numerically in kW then add all three and integrate over one hour to get kilo watt hours.
You might find that the register is updated on a cycle of AC. If so, then average the power for each second and that figure can be used to integrate over 3600 seconds.
If the registers are in watt hours as the name implies then just add all three.
Somewhere in the data sheet it will give you a value for each bit in watts or watt hours. You need to make conversions using this figure into proper units.