A Natural Nuclear Reactor — Question 6

Section 43-4 · Physics pages 7222–7223

Question 6 Water feedback can produce a repeating heating and cooling cycle A schematic water-feedback model heats rock from temperature excess 10 K to 40 K at net rate 20 kW with heat capacity 1.00 MJ/K. Heating then switches off. Cooling follows θ = 40e−t/(1000 s) K until 10 K, when heating resumes. These are illustrative parameters. t θ heat cool Schematic; labels, not drawing lengths, specify the data. (a) Find heating and cooling durations. (b) Find the fraction of each cycle spent heating, and identify the negative-feedback sequence.
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Question 6 - Solution Water feedback can produce a repeating heating and cooling cycle t θ heat cool (a) Method and calculation th = C(40 −10)/P ⇒th = 106(30)/(2.00 × 104) = 1500 s . 10 = 40e−tc/1000 ⇒tc = 1000 ln 4 = 1386.29 s . (b) Interpretation and further calculation d = th th + tc = .51970 . heating →water loss →reduced moderation →less heating . Physical check The model illustrates possible water-mediated cycling, not measured Oklo timing. The feedback sign depends on the system and is supplied here.

Question 6

Water feedback can produce a repeating heating and cooling cycle

A schematic water-feedback model heats rock from temperature excess 10 K to 40 K at net rate 20 kW with heat capacity 1.00 MJ/K. Heating then switches off. Cooling follows θ=40e−t/(1000s)\theta=40e^{-t/(1000\,\mathrm s)} K until 10 K, when heating resumes. These are illustrative parameters.

Diagram shown in the original SVG worksheet.

  1. Find heating and cooling durations.

  2. Find the fraction of each cycle spent heating, and identify the negative-feedback sequence.

Question 6 - Solution

Water feedback can produce a repeating heating and cooling cycle

Diagram shown in the original SVG worksheet.

(a) Method and calculation

th=C(40−10)/P⇒th=106(30)/(2.00×104)=1500s.t_h=C(40-10)/P\Rightarrow\boxed{t_h=10^6(30)/(2.00\times10^4)=1500\,\mathrm s}.10=40e−tc/1000⇒tc=1000ln⁡4=1386.29s.10=40e^{-t_c/1000}\Rightarrow\boxed{t_c=1000\ln4=1386.29\,\mathrm s}.

(b) Interpretation and further calculation

d=thth+tc=.51970.\boxed{d=\frac{t_h}{t_h+t_c}=.51970}.heating→water loss→reduced moderation→less heating.\boxed{\text{heating}\to\text{water loss}\to\text{reduced moderation}\to\text{less heating}}.

Physical check

The model illustrates possible water-mediated cycling, not measured Oklo timing. The feedback sign depends on the system and is supplied here.