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BILANS ENERGETYCZNY:Zerowymiarowy model energetyczny. Zima nuklearna |
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Celem ćwiczenia było wyznaczenie temperatury atmosfery dla modelu cieplarnianego „zima nuklearna”. |
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2. |
Zima nuklearna |
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1.Dane: |
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Szukane: |
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Dane: |
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Szukane: |
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ta |
0,53 |
Ta |
191,38452713195 |
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ta |
0,43 |
Ta |
248,5 |
Wartośc Ta odczytana z wykresu. |
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as |
0,11 |
Ts |
235,1 |
Wartość Ts odczytana z wykresu. |
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as |
0,11 |
Ts |
283,6 |
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S |
1370 |
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S |
1370 |
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c |
2,7 |
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c |
2,7 |
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σ |
5,672E-08 |
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σ |
5,672E-08 |
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aa' |
0,31 |
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aa' |
0,31 |
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ta' |
0,06 |
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ta' |
0,06 |
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aa |
0,11 |
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aa |
0,11 |
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W celu wykonania ćwiczenia należało skorzystać z metody bisekcji opartej na metodzie Bolzano- Cauchy’ego. |
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Wyznaczenie pierwiastków funkcji: |
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xmin |
245 |
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Wyznaczona wartość Ta z porównanych wzorów 3.1 i 3.2 |
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xmax |
250 |
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Ta= |
((S/(4σ))*(1-aa)-((Ts^4)*((-2)*a'a-t'a+2)))^0,25 |
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Lp |
Ts |
f(Ts) |
f(x) |
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Wyznaczenie pierwiastków funkcji w przedziale (xmin, xmax) |
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1 |
245 |
21,9518754591766 |
f(x)>0 |
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xmin |
233 |
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2 |
245,263157894737 |
20,3619004057335 |
f(x)>0 |
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xmax |
236 |
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Ts=xmin+(xmax-xmin)*(Lp-1)/19 |
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3 |
245,526315789474 |
18,7690899480158 |
f(x)>0 |
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4 |
245,789473684211 |
17,1734379982014 |
f(x)>0 |
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Lp |
Ts |
f(Ts) |
f(x) |
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5 |
246,052631578947 |
15,5749384619405 |
f(x)>0 |
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1 |
233 |
-31,1839450898232 |
f(x)<0 |
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6 |
246,315789473684 |
13,9735852383545 |
f(x)>0 |
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2 |
233,157894736842 |
-28,9008534461805 |
f(x)<0 |
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7 |
246,578947368421 |
12,3693722200364 |
f(x)>0 |
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3 |
233,315789473684 |
-26,6088776951855 |
f(x)<0 |
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8 |
246,842105263158 |
10,7622932930503 |
f(x)>0 |
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4 |
233,473684210526 |
-24,3079164648704 |
f(x)<0 |
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9 |
247,105263157895 |
9,15234233693226 |
f(x)>0 |
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5 |
233,631578947368 |
-21,997866148869 |
f(x)<0 |
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10 |
247,368421052632 |
7,53951322468976 |
f(x)>0 |
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6 |
233,78947368421 |
-19,678620836747 |
f(x)<0 |
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11 |
247,631578947368 |
5,92379982280187 |
f(x)>0 |
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7 |
233,947368421053 |
-17,3500722415566 |
f(x)<0 |
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12 |
247,894736842105 |
4,30519599121865 |
f(x)>0 |
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8 |
234,105263157895 |
-15,0121096244686 |
f(x)<0 |
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13 |
248,157894736842 |
2,68369558336227 |
f(x)>0 |
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9 |
234,263157894737 |
-12,6646197163486 |
f(x)<0 |
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14 |
248,421052631579 |
1,05929244612605 |
f(x)>0 |
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10 |
234,421052631579 |
-10,3074866361203 |
f(x)<0 |
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15 |
248,684210526316 |
-0,568019580124769 |
f(x)<0 |
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11 |
234,578947368421 |
-7,94059180575719 |
f(x)<0 |
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16 |
248,947368421053 |
-2,19824666155415 |
f(x)<0 |
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12 |
234,736842105263 |
-5,56381386172914 |
f(x)<0 |
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17 |
249,210526315789 |
-3,83139497085389 |
f(x)<0 |
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13 |
234,894736842105 |
-3,1770285627304 |
f(x)<0 |
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18 |
249,473684210526 |
-5,46747068724463 |
f(x)<0 |
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14 |
235,052631578947 |
-0,780108693489254 |
f(x)<0 |
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19 |
249,736842105263 |
-7,10647999647514 |
f(x)<0 |
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15 |
235,210526315789 |
1,62707603553793 |
f(x)>0 |
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20 |
250 |
-8,74842909082338 |
f(x)<0 |
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16 |
235,368421052632 |
4,04465909280501 |
f(x)>0 |
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17 |
235,526315789474 |
6,47277723487639 |
f(x)>0 |
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18 |
235,684210526316 |
8,91157062043359 |
f(x)>0 |
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Szukana wartość temperatury atmosfery wynosi około 248,5 K, wartość odczytano z wykresu,do którego sprządzenia wykorzystano metodę bisekcji. |
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19 |
235,842105263158 |
11,3611829293401 |
f(x)>0 |
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20 |
236 |
13,8217614870501 |
f(x)>0 |
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WNIOSKI |
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W zadaniu zoatała wyznaczona prawodopodobna temperatura atmosfery Ziemi . |
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Wyznaczenie temperatury dla modelu cieplarnianego ZIMA NUKLEARNA zakładało ograniczenie wspólczynnika |
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atmosfery do ta=0,43 oraz oziębienie temperatury powierzchni Ziemi o 4,4 st. Celsjusza(Ts=283,6 st.C). |
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Spowodowane miało to być skutkami wojny atomowej, podczas której wybuch glowic jadrowych doprowadzić miały |
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do olbrzymich pożarów, które wynosząc w górę atmosfery duże ilości dymu i pyłów zachamują promienie słoneczne |
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docierające do Ziemi. |
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W tak rozpatrywanym scenariuszu obliczono,że temperatura atmosfery wnosić będzie 248,5 st. C. |
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