The last lesson scored a retort in $F_0$: minutes at 121.1 °C, with a z of 10 C°. That scale was built for bacterial spores in low-acid canned food. Beer is a different product with a different enemy. It is acidic, alcoholic, hopped and carbonated, and the organisms that spoil it are feeble against heat. So the brewing trade keeps its own scale. It is the same equation with two different constants packed in.

One pasteurisation unit is one minute at 60 °C. That sentence is the whole definition. Every other temperature is converted into that currency by a factor of ten for each 7 C°: $\mathrm{PU} = t \times 10^{\,(T - 60)/7}$, read aloud P-U equals t times ten to the T minus sixty, over seven. $t$ is the real time held, in minutes. $T$ is the temperature of the beer itself in °C, at the slowest-heating point of the pack. The 60 is the reference temperature in °C, and the 7 is the z-value in C°. Both are fixed by the convention and nobody at the line chooses them. $\mathrm{PU}$ is the lethality delivered, in minutes at 60 °C, and it is what you are solving for.

Set it beside the F-value and the family likeness is complete. $F = t \times 10^{\,(T - T_{\mathrm{ref}})/z}$ becomes the PU the moment $T_{\mathrm{ref}} = 60$ and $z = 7$ go in, exactly as it becomes $F_0$ with 121.1 and 10. The constants are not interchangeable. Score a beer hold on the $F_0$ scale and a sound process comes out as a vanishing fraction of a minute. Borrow only the z of 10 and the answer is wrong by a factor that grows with every degree.

Read the rate before the total. At 67 °C, one z above the reference, every real minute is worth ten PU. At 74 °C it is worth a hundred. At 53 °C it is worth a tenth, so twenty PU down there costs over three hours. That steepness is why flash pasteurisation works: a few seconds in a plate heat exchanger does what a tunnel needs many minutes of warm spray to do. Turned around for the schedule, $t = \mathrm{PU} \times 10^{-(T - 60)/7}$.

What the number should be: ordinary packaged beer is treated to somewhere between 15 and 30 PU, and low-alcohol products need considerably more, because they have lost most of the hurdles that protect a normal beer. To turn PU into a log reduction, divide by the spoilage organism's D-value at 60 °C: $\mathrm{LR} = \dfrac{\mathrm{PU}}{D}$. It works because a PU already is a minute at 60 °C, so both figures share the one reference.

Two measurement cautions, and they are the ones that matter. $T$ is the temperature of the beer, not the spray water and not the zone setpoint. The two differ by several degrees and many minutes, and that lag is where under-processing comes from. And a real pasteuriser accumulates PU through every zone, come-up and cool-down included, so the honest figure is the sum a PU logger reports after travelling with the pack. The target is a floor to clear, not a score to maximise: heat above 60 °C also speeds up staling.