ME 533 Introduction to ICF

Assignment 4 Solutions

Problem 1

Starting from hot-spot mass equation

𝒅

𝒅𝒕(

𝑷𝑹𝟑

𝑻𝟎) = 𝟎. 𝟖𝟔 ∗ 𝜿𝟎𝑻𝟎

𝟓/𝟐𝑹

We define normalized variables 𝑷′, 𝑹′, 𝒕′ normalized to no-α stagnation

quantities 𝑷𝒔, 𝑹𝒔 and 𝒕𝒔(𝑹𝒔

𝑽𝒊) as found in assignment 3. Also define

𝑻𝟎′ =

𝑻𝟎

𝑻𝟎𝒔 ( for now 𝑻𝟎𝒔 unknown)

We get

(𝟏

𝒕𝒔𝑻𝟎𝒔)

𝒅

𝒅𝒕′(

𝑷′𝑹′𝟑

𝑻𝟎′ ) 𝑷𝒔𝑹𝒔

𝟑 = 𝟎. 𝟖𝟔 ∗ 𝜿𝟎𝑻𝟎𝒔𝟓/𝟐𝑻𝟎

′𝟓/𝟐𝑹𝒔𝑹′

To simplify this as

𝒅

𝒅𝒕′ (𝑷′𝑹′𝟑) = 𝑹′𝑻𝟎

′𝟓/𝟐 ( ……. Eqn. 1) , we need 𝑻𝟎𝒔 = (

𝑷𝒔𝑹𝒔𝑽𝒊

𝟎.𝟖𝟔∗𝜿𝟎)𝟐/𝟕

Let 𝑿 ≡𝟏

𝑻𝟎′𝑹′𝟐 we can write (1) as 𝑿𝟓/𝟐 𝒅𝑿

𝒅𝒕′ =𝟏

𝑹′𝟒

⟹ 𝒅 (𝑿

𝟕𝟐)

𝒅𝒕′= (

𝟕

𝟐)

𝟏

𝑹′𝟒

⟹ 𝒅 (𝑿

𝟕𝟐)

𝒅𝒕′= (

𝟕

𝟐)

𝟏

(𝟏 + 𝒕′𝟐)𝟐

Integrating in mathematica and using B.C 𝑿 ⟶ ∞ 𝒂𝒔 𝑹′ → 𝟎

⟹ 𝑿 = (𝟕

𝟒(

𝒕′

𝟏 + 𝒕′𝟐+ 𝑻𝒂𝒏−𝟏(𝒕′) +

𝝅

𝟐))𝟐/𝟕

⟹ 𝑻𝟎′ =

𝟏

(𝟏 + 𝒕′𝟐)𝟐(𝟕

𝟒(

𝒕′

𝟏 + 𝒕′𝟐+ 𝑻𝒂𝒏−𝟏(𝒕′) +

𝝅

𝟐))−𝟐/𝟕

Problem 2

Normalization of equation for Newton`s second law for shell

motion is straight forward, and we derived normalized hot-spot

mass/temperature equation in problem 1.

Starting from hot-spot energy equation with alpha heating

𝒅

𝒅𝒕(𝑷𝑹𝟓) = 𝝁𝑷𝟐𝑹𝟓𝑻𝟎

𝟏.𝟏

Expanding the derivative and defining normalized variables as

in problem 1, we get

𝐝𝐏′

𝐝𝐭′+

𝟓𝐏′

𝐑′

𝐝𝐑′

𝐝𝐭′= 𝛍𝐭𝐬𝐏𝐬(𝐓𝟎𝐬𝐓𝟎

′)𝟏.𝟎𝟏𝐏′𝟐

Also, at stagnation 𝑰. 𝑬𝒉𝒐𝒕−𝒔𝒑𝒐𝒕 = 𝑲. 𝑬𝒔𝒉𝒆𝒍𝒍−𝒎𝒂𝒙

⟹ 𝑷𝒔𝑹𝒔𝟑 = (

𝟏

𝟒𝝅)𝑴𝒔𝒉𝑽𝒊𝒎𝒑

𝟐

Also using from problem 1, 𝑻𝟎𝒔 = (𝑷𝒔𝑹𝒔𝑽𝒊

𝟎.𝟖𝟔∗𝜿𝟎)𝟐/𝟕 we get,

𝒅

𝒅𝒕(𝐏′𝐑′𝟓

) = 𝝃𝒏𝒐−𝜶𝐏′𝟐𝐑′𝟓

𝐓𝟎′𝟏.𝟎𝟏

With 𝝃𝒏𝒐−𝜶 = 𝛍𝐭𝐬𝐏𝐬𝐓𝟎𝐬𝟏.𝟎𝟏 =

𝛍𝐑𝐬𝐏𝐬𝐓𝟎𝐬𝟏.𝟎𝟏

𝑽𝒊

𝝃𝒏𝒐−𝜶 = 𝝁(𝟎. 𝟖𝟔𝜿𝟎)−𝟐.𝟎𝟐/𝟕𝑽𝒊

𝟏𝟑.𝟎𝟔/𝟕(𝝆𝚫)𝟗.𝟎𝟐/𝟕

Problem 3

𝜉𝑐𝑟𝑖𝑡 = 1.07 (𝑠𝑒𝑒 𝑎𝑡𝑡𝑎𝑐ℎ𝑒𝑑 𝑚𝑎𝑡ℎ𝑒𝑚𝑎𝑡𝑖𝑐𝑎 𝑛𝑜𝑡𝑒𝑏𝑜𝑜𝑘)

Singular solution at 𝜉𝑛𝑜−𝛼 = 1.07

Problem 4

Plots of hot-spot pressure for various values of 𝜉𝑛𝑜−𝛼

Problem 5

𝝌𝒏𝒐−𝜶 ≡𝜉𝑛𝑜−𝛼

𝜉𝑐𝑟𝑖𝑡

> 1

Substituting for 𝜉𝑛𝑜−𝛼

⟹ 𝝌𝒏𝒐−𝜶 = (𝟏

𝟏. 𝟎𝟕) ∗ (

𝜺𝜶𝑺𝒇𝐓𝟎𝐬𝟏.𝟎𝟏

𝟐𝟒)

Substituting for 𝐓𝟎𝐬, we get

⟹ 𝝌𝒏𝒐−𝜶 = 𝟎. 𝟖𝟖 ∗ 𝜿𝟎𝟏/𝟑 ∗ (

𝜺𝜶𝑺𝒇(𝝆𝚫)𝟐/𝟑𝐓𝟎𝐬𝟐.𝟏𝟖

𝟐𝟒)

Use neutron averaged quantities

< 𝑇 >= 0.53𝐓𝟎𝐬 & < 𝝆𝐑 >= 𝟎. 𝟖𝟖 𝝆𝚫

and 𝜺𝜶 = 𝟑. 𝟓 𝑴𝒆𝒗, 𝑺𝒇 = 𝟕. 𝟓 ∗ 𝟏𝟎−𝟐𝟑 𝒎𝟑

𝒔 𝑲𝒆𝑽𝟑.𝟎𝟏 , 𝜿𝟎 = 𝟑. 𝟕𝟒 ∗ 𝟏𝟎𝟔𝟗 𝑱𝟐/𝟓

𝒎∗𝒔

We finally get

𝝌𝒏𝒐−𝜶 = 𝟎. 𝟎𝟑 < 𝝆𝐑𝒏𝒐−𝜶 >𝟐/𝟑< 𝑻𝒏𝒐−𝜶 >𝟐.𝟏𝟖

Problem 3D[P[t] R[t]^3 / T[t], t]

R[t]3 P′[t]

T[t]+3 P[t] R[t]2 R′[t]

T[t]-P[t] R[t]3 T′[t]

T[t]2

xi = 1.07eps = 10^-2NDSolve[{R''[t] ⩵ R[t]^2 P[t], P'[t] + 5 P[t] / R[t] R'[t] ⩵ xi P[t]^2 T[t]^1.01,

D[P[t] R[t]^3 / T[t], t] ⩵ T[t]^5 / 2 R[t], R[0] ⩵ 1 / eps, P[0] ⩵ eps^5,R'[0] ⩵ -1, T[0] ⩵ eps}, {R[t], P[t] , T[t]}, {t, 0, 500}]

1.07

1

100

NDSolve::ndsz : At t == 100.53605397419585` , step size is effectively zero; singularity or stiff system suspected.

R[t] → InterpolatingFunctionDomain: {{0., 101.}}Output: scalar

[t],

P[t] → InterpolatingFunctionDomain: {{0., 101.}}Output: scalar

[t],

T[t] → InterpolatingFunctionDomain: {{0., 101.}}Output: scalar

[t]

xi = 1.04eps = 10^-1NDSolve[{R''[t] ⩵ R[t]^2 P[t], P'[t] + 5 P[t] / R[t] R'[t] ⩵ xi P[t]^2 T[t]^1.01,

D[P[t] R[t]^3 / T[t], t] ⩵ T[t]^5 / 2 R[t], R[0] ⩵ 1 / eps, P[0] ⩵ eps^5,R'[0] ⩵ -1, T[0] ⩵ eps}, {R[t], P[t] , T[t]}, {t, 0, 50}]

1.04

1

10

R[t] → InterpolatingFunctionDomain: {{0., 50.}}Output: scalar

[t],

P[t] → InterpolatingFunctionDomain: {{0., 50.}}Output: scalar

[t],

T[t] → InterpolatingFunctionDomain: {{0., 50.}}Output: scalar

[t]

Problem 4xiCrit = 1.05getRandP[xi_, eps_] :=

NDSolve[{R''[t] ⩵ R[t]^2 P[t], P'[t] + 5 P[t] / R[t] R'[t] ⩵ xi P[t]^2 T[t]^1.01,D[P[t] R[t]^3 / T[t], t] ⩵ T[t]^5 / 2 R[t], R[0] ⩵ 1 / eps, P[0] ⩵ eps^5,R'[0] ⩵ -1, T[0] ⩵ eps}, {R[t], P[t] , T[t]}, {t, 0, 50}]

1.05

2 assn_4 (1).nb

S = getRandP[0 / xiCrit, 10^-1];Plot[Evaluate[{R[t]} /. S], {t, 0, 20}, PlotStyle → Automatic]a = Plot[Evaluate[{P[t]} /. S], {t, 0, 20}, PlotRange → {{5, 15}, {0, 2.5}}]Plot[Evaluate[{T[t]} /. S], {t, 0, 20}, PlotRange → All]

5 10 15 20

2

4

6

8

10

6 8 10 12 14

0.5

1.0

1.5

2.0

2.5

5 10 15 20

0.2

0.4

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0.8

1.0

assn_4 (1).nb 3

S1 = getRandP[0.1 * xiCrit, 10^-1];Plot[Evaluate[{R[t]} /. S], {t, 0, 20}, PlotStyle → Automatic]a1 = Plot[Evaluate[{P[t]} /. S1], {t, 0, 20}, PlotRange → {{5, 15}, {0, 2.5}}]Plot[Evaluate[{T[t]} /. S], {t, 0, 20}, PlotRange → All]

5 10 15 20

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4

6

8

10

6 8 10 12 14

0.5

1.0

1.5

2.0

2.5

5 10 15 20

0.2

0.4

0.6

0.8

1.0

4 assn_4 (1).nb

S2 = getRandP[0.5 * xiCrit, 10^-1];Plot[Evaluate[{R[t]} /. S], {t, 0, 20}, PlotStyle → Automatic]a2 = Plot[Evaluate[{P[t]} /. S2], {t, 0, 20}, PlotRange → {{5, 15}, {0, 2.5}}]Plot[Evaluate[{T[t]} /. S], {t, 0, 20}, PlotRange → All]

5 10 15 20

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4

6

8

10

6 8 10 12 14

0.5

1.0

1.5

2.0

2.5

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1.0

assn_4 (1).nb 5

S3 = getRandP[0.9 * xiCrit, 10^-1];Plot[Evaluate[{R[t]} /. S], {t, 0, 20}, PlotStyle → Automatic]a3 = Plot[Evaluate[{P[t]} /. S3], {t, 0, 20}, PlotRange → {{5, 15}, {0, 2.5}}]Plot[Evaluate[{T[t]} /. S], {t, 0, 20}, PlotRange → All]

5 10 15 20

2

4

6

8

10

6 8 10 12 14

0.5

1.0

1.5

2.0

2.5

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0.4

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1.0

6 assn_4 (1).nb

S = getRandP[xiCrit, 10^-1];Plot[Evaluate[{R[t]} /. S], {t, 0, 20}, PlotStyle → Automatic]a4 = Plot[Evaluate[{P[t]} /. S], {t, 0, 20}, PlotRange → All,

AxesLabel → {Style["P", Bold, 14], Style["t", Bold, 14]}, PlotStyle → FontSize]Plot[Evaluate[{T[t]} /. S], {t, 0, 20}, PlotRange → All]

NDSolve::ndsz : At t == 10.402023057183255` , step size is effectively zero; singularity or stiff system suspected.

5 10 15 20

-1.5×1054

-1.0×1054

-5.0×1053

5 10 15 20P

5.0×1044

1.0×1045

1.5×1045

2.0×1045

2.5×1045

3.0×1045

t

5 10 15 20

5.0×1039

1.0×1040

1.5×1040

2.0×1040

2.5×1040

assn_4 (1).nb 7

Show[{a1, a2, a3}]

6 8 10 12 14

0.2

0.4

0.6

0.8

1.0

1.2

1.4

Plot[{Evaluate[{P[t]} /. S], Evaluate[{P[t]} /. S1], Evaluate[{P[t]} /. S2],Evaluate[{P[t]} /. S3]}, {t, 0, 20}, PlotRange → {{5, 15}, {0, 2.5}},

AxesLabel → {Style["P'", Bold, 14], Style["t'", Bold, 14]},PlotLegends → Placed[{"0", "0.1", "0.525", "0.945"}, Above]]

0 0.1 0.525 0.945

6 8 10 12 14P'

0.5

1.0

1.5

2.0

2.5t'

1.05 * 0.9

0.945

ealpha = 3.5 * 1.6 * 10^-13; (to Joules)sf = 7.5 * 10^-23;k0 = 3.74 * 10^69;

coeff = (0.88 * k0^(1 / 3) * sf * 0.53^2.18 * 0.88^2 / 3) / 24

0.0276093

8 assn_4 (1).nb