{"id":9290,"date":"2024-07-03T07:00:34","date_gmt":"2024-07-03T12:00:34","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/analisisnumerico\/?p=9290"},"modified":"2026-04-05T19:58:35","modified_gmt":"2026-04-06T00:58:35","slug":"s1eva2024paoi_t1-vaciado-de-reservorio-semiesferico","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s1eva30\/s1eva2024paoi_t1-vaciado-de-reservorio-semiesferico\/","title":{"rendered":"s1Eva2024PAOI_T1 Vaciado de reservorio semiesf\u00e9rico"},"content":{"rendered":"\n

Ejercicio<\/strong>: 1Eva2024PAOI_T1 Vaciado de reservorio semiesf\u00e9rico<\/a><\/p>\n\n\n\n

literal a. Planteamiento<\/h3>\n\n\n\n

A partir de la soluci\u00f3n expresi\u00f3n propuesta y simplificada del ejercicio, se reemplaza los valores de R, K, a y g que son constantes. Como se da el valor de t<\/strong>=(15 min)(60 s\/min), se unifica la unidad de medida de tiempo a segundos pues g=9.6 m\/s2<\/sup>.<\/p>\n\n\n\n

\n\n\n -\\frac{4}{3}R h^{3\/2} + \\frac{2}{5} h^{5\/2} = -\\frac{Ka}{\\pi} t \\sqrt{2g}<\/span>\n\n\n\n
\n\n\n -\\frac{4}{3}(3) h^{3\/2} + \\frac{2}{5} h^{5\/2} = -\\frac{(0.85)(0.01)}{\\pi} (15)(60) \\sqrt{2(9.8)}<\/span>\n\n\n\n

la funci\u00f3n para encontrar la ra\u00edz se reordena como f(h)=0<\/p>\n\n\n f(h) = -\\frac{4}{3}(3) h^{3\/2} + \\frac{2}{5} h^{5\/2} +\\frac{(0.85)(0.01)}{\\pi} (15)(60) \\sqrt{2(9.8)}<\/span>\n\n\n f(h) = -4h^{3\/2} + 0.4 h^{5\/2} +10.7805 <\/span>\n\n\n\n

\"tanque<\/figure>\n\n\n\n
\"tanque<\/figure>\n\n\n\n

Usando la gr\u00e1fica proporcionada del reservorio semiesf\u00e9rico, el valor de h<\/strong> no puede superar la altura R. Al vaciar el reservorio h<\/strong>=0.<\/p>\n\n\n\n

Por lo que el intervalo a usar es [0,3]. Se verifica la existencia de una ra\u00edz al con la gr\u00e1fica y Python.<\/p>\n\n\n\n

b. M\u00e9todo de Newton Raphson<\/h3>\n\n\n\n

El m\u00e9todo requiere la derivada f'(h)<\/p>\n\n\n f(h) = -4h^{3\/2} + 0.4 h^{5\/2} +10.7805 <\/span>\n\n\n f'(h) = -6\\frac{ h^{3\/2}}{h} + \\frac{h^{5\/2}}{h} <\/span>\n\n\n f'(h) = -6 h^{1\/2}+ h^{3\/2} <\/span>\n\n\n\n

El valor inicial puede ser por ejemplo, el centro del intervalo:<\/p>\n\n\n x_0= \\frac{a+b}{2} = \\frac{0+3}{2} = 1.5 <\/span>\n\n\n\n

itera=0<\/p>\n\n\n f(1.5) = -4(1.5)^{3\/2} + 0.4 (1.5)^{5\/2} +10.7805= 4.5343<\/span>\n\n\n f'(1.5) = -6\\frac{ 1.5^{3\/2}}{1.5} + \\frac{1.5^{5\/2}}{1.5} = -5.5113<\/span>\n\n\n x_1= x_0 - \\frac{f(1.5)}{f'(1.5)}= 1.5-\\frac{4.5343}{-5.5113} =2.3227<\/span>\n\n\n error_1 =|x1-x0| = |2.3227-1.5| =0.8227<\/span>\n\n\n\n

itera=1<\/p>\n\n\n f(2.3227) = -4(2.3227)^{3\/2} + 0.4 (2.3227)^{5\/2} +10.7805= -0.09019 <\/span>\n\n\n f'(2.3227) = -6\\frac{ 2.3227^{3\/2}}{2.3227} + \\frac{2.3227^{5\/2}}{2.3227} = -5.6043 <\/span>\n\n\n x_2= 2.3227-\\frac{-0.09019}{-5.6043} = 2.3066 <\/span>\n\n\n error_2 = |2.3066-2.3227| =0.01611 <\/span>\n\n\n\n

itera=2<\/p>\n\n\n f(2.3227) = -4(2.3066)^{3\/2} + 0.4 (2.3066)^{5\/2} +10.7805= -0.0007104 <\/span>\n\n\n f'(2.3227) = -6\\frac{ 2.3227^{3\/2}}{2.3227} + \\frac{2.3227^{5\/2}}{2.3227} = -5.6093<\/span>\n\n\n x_3= 2.3227-\\frac{-0.0007104}{-5.6093} = 2.3066 <\/span>\n\n\n error_3 = |2.3066-2.3066| =0.000007241 <\/span>\n\n\n\n

literal c, convergencia<\/h3>\n\n\n\n

considerando tolerancia de 10-3 (milim\u00e9trica), solo ser\u00e1n necesarias 3 iteraciones.<\/p>\n\n\n\n

El m\u00e9todo converge al mostrarse que los errores disminuyen en cada iteraci\u00f3n.<\/p>\n\n\n\n

El resultado se encuentra dentro del intervalo y es x = 2.3066<\/p>\n\n\n\n

Algoritmo con Python<\/h3>\n\n\n\n

resultados:<\/p>\n\n\n\n

fh:\n         ____          ____                   \n        \/  3          \/  5                    \n- 4.0*\\\/  h   + 0.4*\\\/  h   + 10.7805172327811\ndfh:\n         ____          ____\n        \/  3          \/  5 \n  6.0*\\\/  h     1.0*\\\/  h  \n- ----------- + -----------\n       h             h     \n['xi', 'xnuevo', 'tramo']\n[[1.5000e+00 2.3227e+00 8.2272e-01]\n [2.3227e+00 2.3066e+00 1.6118e-02]\n [2.3066e+00 2.3066e+00 7.2412e-06]]\nraiz en:  2.306612665792577\ncon error de:  7.241218404896443e-06\n\n>>> f(1.5)\n4.534318388683996\n>>> df(1.5)\n-5.5113519212621505\n>>> f(2.3227)\n-0.09019959114247378\n>>> df(2.3227)\n-5.604354799446854\n>>> f(2.3066)\n7.104683901282272e-05\n>>> df(2.3066)\n-5.609349350102558<\/code><\/pre>\n\n\n\n
Algoritmo en Python:<\/p>\n\n\n
\n# 1Eva_2024PAOI_T1 Vaciado de reservorio semiesf\u00e9rico\nimport numpy as np\nimport matplotlib.pyplot as plt\nimport sympy as sym\n\n# INGRESO\nh = sym.Symbol('h')\nt = 15*60\nR = 3 \ng = 9.8 \narea = 0.01\nK = 0.85 \nfh = -(4\/3)*R*sym.sqrt(h**3)+(2\/5)*sym.sqrt(h**5)+(K*area\/np.pi)*np.sqrt(2*g)*(t)\n\n# Grafica de f(h)\na = 0\nb = 3\nmuestras = 15\n\n# PROCEDIMIENTO\nhi = np.linspace(a,b,muestras)\n# para evaluaci\u00f3n num\u00e9rica con numpy\nf = sym.lambdify(h,fh)\nfi = f(hi)\n\n# derivada con sympy\ndfh = sym.diff(fh,h,1)\ndf = sym.lambdify(h,dfh)\n\n# SALIDA\nprint('fh:')\nsym.pprint(fh)\nprint('dfh:')\nsym.pprint(dfh)\n\nplt.plot(hi,fi)\nplt.axhline(0, color='black')\nplt.xlabel('h')\nplt.ylabel('f')\nplt.title(fh)\nplt.show()\n\n## M\u00e9todo de Newton-Raphson\n\n# INGRESO\nfx  = f\ndfx = df\nx0 = (a+b)\/2 # mitad del intervalo (ejemplo)\ntolera = 0.001\n\n# PROCEDIMIENTO\ntabla = []\ntramo = abs(2*tolera)\nxi = x0\nwhile (tramo>=tolera):\n    xnuevo = xi - fx(xi)\/dfx(xi)\n    tramo  = abs(xnuevo-xi)\n    tabla.append([xi,xnuevo,tramo])\n    xi = xnuevo\n\n# convierte la lista a un arreglo.\ntabla = np.array(tabla)\nn = len(tabla)\n\n# SALIDA\nprint(['xi', 'xnuevo', 'tramo'])\nnp.set_printoptions(precision = 4)\nprint(tabla)\nprint('raiz en: ', xi)\nprint('con error de: ',tramo)\n<\/pre><\/div>\n\n\n
 <\/p>\n","protected":false},"excerpt":{"rendered":"
Ejercicio: 1Eva2024PAOI_T1 Vaciado de reservorio semiesf\u00e9rico literal a. Planteamiento A partir de la soluci\u00f3n expresi\u00f3n propuesta y simplificada del ejercicio, se reemplaza los valores de R, K, a y g que son constantes. Como se da el valor de t=(15 min)(60 s\/min), se unifica la unidad de medida de tiempo a segundos pues g=9.6 m\/s2. […]<\/p>\n","protected":false},"author":8043,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"wp-custom-template-entrada-mn","format":"standard","meta":{"footnotes":""},"categories":[46],"tags":[58,54],"class_list":["post-9290","post","type-post","status-publish","format-standard","hentry","category-mn-s1eva30","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/9290","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/users\/8043"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/comments?post=9290"}],"version-history":[{"count":3,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/9290\/revisions"}],"predecessor-version":[{"id":23827,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/9290\/revisions\/23827"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=9290"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=9290"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=9290"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}