University Physics V · topic UPV-16.05
Fission energy, decay heat, and the waste inventory
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
Fission energy, decay heat and waste inventory: predict a finite change
Decay heat is the thermal power a reactor keeps producing after the chain reaction has stopped, supplied by the beta and gamma decay of the fission-product and actinide inventory that its operating history built up. Use the supplied topic model Y_out = 10^u, where Y_out is TIME AFTER SCRAM (s). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. u is Cursor time after scram, measured in log₁₀ s. Keep w (Operating time before scram) = 4 log₁₀ s fixed. For this exercise use only 0.74375 log₁₀ s ≤ u ≤ 6.25625 log₁₀ s on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at u = 1.9565 log₁₀ s and 2.72825 log₁₀ s. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δu across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
TIME AFTER SCRAM against u
Predictions from the supplied Fission energy, decay heat and waste inventory model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): TIME AFTER SCRAM / s
Horizontal axis (x): u: Cursor time after scram / log₁₀ s
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| u: Cursor time after scram / log₁₀ s | TIME AFTER SCRAM / s |
|---|---|
| 0.74375 | 5.543065366083506 |
| 2.1218749999999997 | 132.39604139104352 |
| 3.4999999999999996 | 3162.2776601683763 |
| 4.878125 | 75530.95919585766 |
| 6.25625 | 1804055.9400917848 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of u separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At u = 1.9565 log₁₀ s, substitute into Y_out = 10^u: Y₀ = 90.469 s.
- At u = 2.72825 log₁₀ s, the same fixed-parameter model gives Y₁ = 534.872 s.
- Signed change: ΔY_out = Y₁ − Y₀ = 444.403 s; the input change is Δu = 0.77175 log₁₀ s.
- Average sensitivity is ΔY_out/Δu = 575.838 s per log₁₀ s. Its positive sign means that this response increases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 90.469 s, Y₁ = 534.872 s; ΔY_out = 444.403 s; average sensitivity = 575.838 s per log₁₀ s.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Fission energy, decay heat and waste inventory: infer an input with uncertainty
Decay heat is the thermal power a reactor keeps producing after the chain reaction has stopped, supplied by the beta and gamma decay of the fission-product and actinide inventory that its operating history built up. Use the supplied topic model Y_out = 10^u, where Y_out is TIME AFTER SCRAM (s). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. u is Cursor time after scram, measured in log₁₀ s. Keep w (Operating time before scram) = 4 log₁₀ s fixed. For this exercise use only 0.74375 log₁₀ s ≤ u ≤ 6.25625 log₁₀ s on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 12775.415 s, with a bounded reading uncertainty ±1915.5 s. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown u by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00055125 log₁₀ s, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
TIME AFTER SCRAM against u
Predictions from the supplied Fission energy, decay heat and waste inventory model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): TIME AFTER SCRAM / s
Horizontal axis (x): u: Cursor time after scram / log₁₀ s
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| u: Cursor time after scram / log₁₀ s | TIME AFTER SCRAM / s |
|---|---|
| 0.74375 | 5.543065366083506 |
| 2.1218749999999997 | 132.39604139104352 |
| 3.4999999999999996 | 3162.2776601683763 |
| 4.878125 | 75530.95919585766 |
| 6.25625 | 1804055.9400917848 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(u) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(0.74375) = 5.54307 s and Y_out(6.25625) = 1804060 s. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives u_line = 0.74375 + (12775.4 − 5.54307)(6.25625 − 0.74375)/(1804060 − 5.54307) = 0.78277 log₁₀ s. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives u = 4.10638 log₁₀ s; substituting back gives Y_out = 12775.415 s.
- The interpolation error in the input is |u_line − u| = 3.32361 log₁₀ s. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 10859.915 s and 14690.915 s. Sort the roots even if the response decreases: 4.03583 log₁₀ s ≤ u ≤ 4.16705 log₁₀ s.
- The lower and upper input deviations are 0.0705486 log₁₀ s and 0.0606738 log₁₀ s. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer u ≈ 4.10638 log₁₀ s; compatible interval [4.03583, 4.16705] log₁₀ s. Endpoint interpolation gives 0.78277 log₁₀ s.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.