ME 533 Introduction to ICF Assignment 4 SolutionsAssignment 4 Solutions Problem 1 Starting from...
Transcript of ME 533 Introduction to ICF Assignment 4 SolutionsAssignment 4 Solutions Problem 1 Starting from...
ME 533 Introduction to ICF
Assignment 4 Solutions
Problem 1
Starting from hot-spot mass equation
π
π π(
π·πΉπ
π»π) = π. ππ β πΏππ»π
π/ππΉ
We define normalized variables π·β², πΉβ², πβ² normalized to no-Ξ± stagnation
quantities π·π, πΉπ and ππ(πΉπ
π½π) as found in assignment 3. Also define
π»πβ² =
π»π
π»ππ ( for now π»ππ unknown)
We get
(π
πππ»ππ)
π
π πβ²(
π·β²πΉβ²π
π»πβ² ) π·ππΉπ
π = π. ππ β πΏππ»πππ/ππ»π
β²π/ππΉππΉβ²
To simplify this as
π
π πβ² (π·β²πΉβ²π) = πΉβ²π»π
β²π/π ( β¦β¦. Eqn. 1) , we need π»ππ = (
π·ππΉππ½π
π.ππβπΏπ)π/π
Let πΏ β‘π
π»πβ²πΉβ²π we can write (1) as πΏπ/π π πΏ
π πβ² =π
πΉβ²π
βΉ π (πΏ
ππ)
π πβ²= (
π
π)
π
πΉβ²π
βΉ π (πΏ
ππ)
π πβ²= (
π
π)
π
(π + πβ²π)π
Integrating in mathematica and using B.C πΏ βΆ β ππ πΉβ² β π
βΉ πΏ = (π
π(
πβ²
π + πβ²π+ π»ππβπ(πβ²) +
π
π))π/π
βΉ π»πβ² =
π
(π + πβ²π)π(π
π(
πβ²
π + πβ²π+ π»ππβπ(πβ²) +
π
π))βπ/π
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
π
π π(π·πΉπ) = ππ·ππΉππ»π
π.π
Expanding the derivative and defining normalized variables as
in problem 1, we get
ππβ²
ππβ²+
ππβ²
πβ²
ππβ²
ππβ²= πππ¬ππ¬(πππ¬ππ
β²)π.πππβ²π
Also, at stagnation π°. π¬πππβππππ = π². π¬πππππβπππ
βΉ π·ππΉππ = (
π
ππ )π΄πππ½πππ
π
Also using from problem 1, π»ππ = (π·ππΉππ½π
π.ππβπΏπ)π/π we get,
π
π π(πβ²πβ²π
) = πππβπΆπβ²ππβ²π
ππβ²π.ππ
With πππβπΆ = πππ¬ππ¬πππ¬π.ππ =
πππ¬ππ¬πππ¬π.ππ
π½π
πππβπΆ = π(π. πππΏπ)βπ.ππ/ππ½π
ππ.ππ/π(ππ«)π.ππ/π
Problem 3
πππππ‘ = 1.07 (π ππ ππ‘π‘ππβππ πππ‘βππππ‘πππ πππ‘πππππ)
Singular solution at πππβπΌ = 1.07
Problem 4
Plots of hot-spot pressure for various values of πππβπΌ
Problem 5
πππβπΆ β‘πππβπΌ
πππππ‘
> 1
Substituting for πππβπΌ
βΉ πππβπΆ = (π
π. ππ) β (
πΊπΆπΊππππ¬π.ππ
ππ)
Substituting for πππ¬, we get
βΉ πππβπΆ = π. ππ β πΏππ/π β (
πΊπΆπΊπ(ππ«)π/ππππ¬π.ππ
ππ)
Use neutron averaged quantities
< π >= 0.53πππ¬ & < ππ >= π. ππ ππ«
and πΊπΆ = π. π π΄ππ, πΊπ = π. π β ππβππ ππ
π π²ππ½π.ππ , πΏπ = π. ππ β ππππ π±π/π
πβπ
We finally get
πππβπΆ = π. ππ < ππππβπΆ >π/π< π»ππβπΆ >π.ππ
Problem 3D[P[t] R[t]^3 / T[t], t]
R[t]3 Pβ²[t]
T[t]+3 P[t] R[t]2 Rβ²[t]
T[t]-P[t] R[t]3 Tβ²[t]
T[t]2
xi = 1.07eps = 10^-2NDSolve[{R''[t] β©΅ R[t]^2 P[t], P'[t] + 5 P[t] / R[t] R'[t] β©΅ xi P[t]^2 T[t]^1.01,
D[P[t] R[t]^3 / T[t], t] β©΅ T[t]^5 / 2 R[t], R[0] β©΅ 1 / eps, P[0] β©΅ eps^5,R'[0] β©΅ -1, T[0] β©΅ 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] β InterpolatingFunctionDomain: {{0., 101.}}Output: scalar
[t],
P[t] β InterpolatingFunctionDomain: {{0., 101.}}Output: scalar
[t],
T[t] β InterpolatingFunctionDomain: {{0., 101.}}Output: scalar
[t]
xi = 1.04eps = 10^-1NDSolve[{R''[t] β©΅ R[t]^2 P[t], P'[t] + 5 P[t] / R[t] R'[t] β©΅ xi P[t]^2 T[t]^1.01,
D[P[t] R[t]^3 / T[t], t] β©΅ T[t]^5 / 2 R[t], R[0] β©΅ 1 / eps, P[0] β©΅ eps^5,R'[0] β©΅ -1, T[0] β©΅ eps}, {R[t], P[t] , T[t]}, {t, 0, 50}]
1.04
1
10
R[t] β InterpolatingFunctionDomain: {{0., 50.}}Output: scalar
[t],
P[t] β InterpolatingFunctionDomain: {{0., 50.}}Output: scalar
[t],
T[t] β InterpolatingFunctionDomain: {{0., 50.}}Output: scalar
[t]
Problem 4xiCrit = 1.05getRandP[xi_, eps_] :=
NDSolve[{R''[t] β©΅ R[t]^2 P[t], P'[t] + 5 P[t] / R[t] R'[t] β©΅ xi P[t]^2 T[t]^1.01,D[P[t] R[t]^3 / T[t], t] β©΅ T[t]^5 / 2 R[t], R[0] β©΅ 1 / eps, P[0] β©΅ eps^5,R'[0] β©΅ -1, T[0] β©΅ 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 β Automatic]a = Plot[Evaluate[{P[t]} /. S], {t, 0, 20}, PlotRange β {{5, 15}, {0, 2.5}}]Plot[Evaluate[{T[t]} /. S], {t, 0, 20}, PlotRange β All]
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assn_4 (1).nb 3
S1 = getRandP[0.1 * xiCrit, 10^-1];Plot[Evaluate[{R[t]} /. S], {t, 0, 20}, PlotStyle β Automatic]a1 = Plot[Evaluate[{P[t]} /. S1], {t, 0, 20}, PlotRange β {{5, 15}, {0, 2.5}}]Plot[Evaluate[{T[t]} /. S], {t, 0, 20}, PlotRange β All]
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4 assn_4 (1).nb
S2 = getRandP[0.5 * xiCrit, 10^-1];Plot[Evaluate[{R[t]} /. S], {t, 0, 20}, PlotStyle β Automatic]a2 = Plot[Evaluate[{P[t]} /. S2], {t, 0, 20}, PlotRange β {{5, 15}, {0, 2.5}}]Plot[Evaluate[{T[t]} /. S], {t, 0, 20}, PlotRange β All]
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assn_4 (1).nb 5
S3 = getRandP[0.9 * xiCrit, 10^-1];Plot[Evaluate[{R[t]} /. S], {t, 0, 20}, PlotStyle β Automatic]a3 = Plot[Evaluate[{P[t]} /. S3], {t, 0, 20}, PlotRange β {{5, 15}, {0, 2.5}}]Plot[Evaluate[{T[t]} /. S], {t, 0, 20}, PlotRange β All]
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6 assn_4 (1).nb
S = getRandP[xiCrit, 10^-1];Plot[Evaluate[{R[t]} /. S], {t, 0, 20}, PlotStyle β Automatic]a4 = Plot[Evaluate[{P[t]} /. S], {t, 0, 20}, PlotRange β All,
AxesLabel β {Style["P", Bold, 14], Style["t", Bold, 14]}, PlotStyle β FontSize]Plot[Evaluate[{T[t]} /. S], {t, 0, 20}, PlotRange β All]
NDSolve::ndsz : At t == 10.402023057183255` , step size is effectively zero; singularity or stiff system suspected.
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-1.5Γ1054
-1.0Γ1054
-5.0Γ1053
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5.0Γ1044
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t
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5.0Γ1039
1.0Γ1040
1.5Γ1040
2.0Γ1040
2.5Γ1040
assn_4 (1).nb 7
Show[{a1, a2, a3}]
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Plot[{Evaluate[{P[t]} /. S], Evaluate[{P[t]} /. S1], Evaluate[{P[t]} /. S2],Evaluate[{P[t]} /. S3]}, {t, 0, 20}, PlotRange β {{5, 15}, {0, 2.5}},
AxesLabel β {Style["P'", Bold, 14], Style["t'", Bold, 14]},PlotLegends β Placed[{"0", "0.1", "0.525", "0.945"}, Above]]
0 0.1 0.525 0.945
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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