How do you calculate the hourly output, measured in kWh, of a 1 kw solar photovoltaic panel given hourly irradiation data in kj/m2?
There's nothing complicated here, it's no more than book-keeping, and multiplication sums.

Output = Input * efficiency, when everything is expressed consistently.

If your hourly irradiance is expressed in kJ/m2, then you need to multiply by m2 of panel. If your panels are rated in kw each at a reference irradiance, then you need to scale the panel output by the ratio of actual irraadiance to the reference.

Amongst the efficiency terms are light to electricity, oblique incidence  effects, panel fouling reducing light, they may be a temperature dependence, and DC to AC conversion losses. You will get out of your simulation an accuracy that relates to the care with which you find all the efficiency terms, and account for them.

If you calculate for a single panel, at reference irradiance, you will get a nominal figure. That's rather like taking your hourly pay rate times your daily hours and arriving at a nominal income for the day.

But you have to pay income taxes, travel to and from work, buy work tools, all of which reduces your disposable income. But these are just book-keeping, and simple sums. In the same way you lose power through non-standard irradiance, oblique illumination, dirty panels, DC-AC losses. The more detailed you are keeping track of them, the more realistic power output estimate you'll have.

If your sole problem is turning kJ into kWh, then 1kWh = 3600kJ = 3.6MJ, as one hour = 3600 seconds, and 1J = 1 watt.second.