Question #4: Coastdown Recovery with fill then trickle (hard).

Consider the problem of Question #2 above. In that question there was a coastdown and the issue of whether to recover with trickle or with filling was addressed. However in question #2 is was presumed that the filling approach would be used for the entire recovery. This question asks what if filling and then trickle are used.

Consider the picture to the right: At time t=0 ability to inject is lost. At time t=1 luminosity has decayed to a level L=Lmin but injection is again available. Injection proceeds at time t=1 with two courses shown. In the course shown in green, trickle injection is used, and data logging continues. The trickle luminosity increase rate is Rtrickle. At time t=5 the peak luminosity is again achieved. The green dashed line shows the running sum of integrated luminosity while under deliverable conditions for the green course. In the course shown in yellow, data collection is initially abandoned and beams are injected under (faster) filling conditions. A luminosity increase rate Rfill occurs. At time t=3 the the luminosity reaches the level Lpeak(1-Rtrickle/Rfill) and trickle conditions with data logging continue (as question 3 above would indicate is appropriate). At t=4 LPeak is again achieved and the recovery from the interruption is complete. The yellow dashed line shows the running sum of integrated luminosity while under deliverable conditions for the yellow course.

Can you show that if ever L_min decays below the level L_peak(1-R_trickle/R_fill) that less luminosity is lost if you recover with the yellow course rather than the green course?

Solution to Question #4

Let us use the same approach as was used for question #2. We can write down an expression for the luminosity lost in each of the two courses. We can then ask for what value L_min would the luminosity lost be equal for the two courses.

In question #3 it was shown that the interval before t=1 does not effect the result, so we need only look at the luminosity lost between when injection is again available (t=1) and when recovery is fully completed (t=4 for the yellow course, t=5 for the green course).

In question #3 we found that in a circumstance such as this the optimal value for L_deliver is L_deliver= L_peak(1-R_trickle/R_fill).

For convenience let's define a parameter α ≡(1-R_trickle/R_fill) such that we can replace L_deliver with αL_peak. Doing this substitution yields:

Next we can clean up the denominators by multiplying both sides of this equation by 2R_trickleR_fill which yields:

This equation is getting messy. One can see from the second term on the left hand side that it will be quadratic in L_min. I'll let the determined reader follow through with all of the algebra. This equation can be manipulated into a standard quadratic form:

L²_min - 2αL_peakL_min + α²L²_peak = 0

The left hand side is immediately seen as a perfect square:

(L_min-αL_peak)² = 0

With solution

L_min = αL_peak.

And with our definition of α from above:

L_min=L_peak(1-R_trickle/R_fill)