Exemplos

Paulo Jabardo

Exemplos de Análise dimensional usando o pacote Python BuckinghamPi¶

In [2]:

from buckinghampy import BuckinghamPi

Transferência de calor de um corpo aquecido¶

Um corpo com dimensão característica $a$ submetido a um fluido com capacidade térmica $c$ e condutividade térmica $k$ é mantido a uma temperatura $\theta$ acima da temperatura do fluido. Qual a taxa de transferência de calor deste corpo?

Nome da grandeza	Símbolo	Dimensões
Taxa de transferência de calor	$h$	$HT^{-1}$
Dimensão linear do corpo	$a$	$L$
Velocidade do escoamento	$v$	$LT^{-1}$
Diferença de temperatura	$\theta$	$\Theta$
Capacidade de calor do fluido	$c$	$HL^{-3}\Theta^{-1}$
Condutividade térmica do fluido	$k$	$HL^{-1}T^{-1}\Theta^{-1}$

In [6]:

calor = BuckinghamPi()
calor.add_variable(name="h", dimensions='H*T^(-1)')
calor.add_variable(name="a", dimensions='L')
calor.add_variable(name="v", dimensions='L*T^(-1)')
calor.add_variable(name="θ", dimensions='Θ')
calor.add_variable(name="c", dimensions='H*L^(-3)*Θ^(-1)')
calor.add_variable(name="k", dimensions='H*L^(-1)*T^(-1)*Θ^(-1)')
calor.generate_pi_terms()
calor.print_all()

$\displaystyle \text{Set }1: \quad\pi_1 = \frac{a \sqrt{c} \sqrt{v} \sqrt{θ}}{\sqrt{h}}\quad\pi_2 = \frac{k \sqrt{θ}}{\sqrt{c} \sqrt{h} \sqrt{v}}\quad$

$\displaystyle \text{Set }2: \quad\pi_1 = \frac{a k θ}{h}\quad\pi_2 = \frac{c h v}{k^{2} θ}\quad$

$\displaystyle \text{Set }3: \quad\pi_1 = \frac{a c v}{k}\quad\pi_2 = \frac{k^{2} θ}{c h v}\quad$

$\displaystyle \text{Set }4: \quad\pi_1 = \frac{a^{2} c v θ}{h}\quad\pi_2 = \frac{a k θ}{h}\quad$

$\displaystyle \text{Set }5: \quad\pi_1 = \frac{a c v}{k}\quad\pi_2 = \frac{a k θ}{h}\quad$

$\displaystyle \text{Set }6: \quad\pi_1 = \frac{a^{2} c v θ}{h}\quad\pi_2 = \frac{k}{a c v}\quad$

Escolhemos um conjunto¶

Conjunto 5:

$$\Pi_1 = \frac{a c v}{k} \qquad \Pi_2 = \frac{a k \theta}{h}$$

$$\Pi_2 = \Pi_2\left(\Pi_1\right)$$

Ou seja: $$ h = k a \theta F\left(\frac{a c v}{k} \right) $$

Problemas: aqui consideramos 4 grandezas independentes¶

Mas e se considerarmos apenas 3 independentes?

Como? Temperatura como energia cinética média, então temos um resultado diferente!

In [11]:

calor1 = BuckinghamPi()
calor1.add_variable(name="h", dimensions='H*T^(-1)')
calor1.add_variable(name="a", dimensions='L')
calor1.add_variable(name="v", dimensions='L*T^(-1)')
calor1.add_variable(name="θ", dimensions='H')  # SÓ MODIFIQUEI AQUI!!!!
calor1.add_variable(name="c", dimensions='H^(-3)')
calor1.add_variable(name="k", dimensions='L^(-1)*T^(-1)')
calor1.generate_pi_terms()
calor1.print_all()

$\displaystyle \text{Set }1: \quad\pi_1 = \frac{a h}{v θ}\quad\pi_2 = c θ^{3}\quad\pi_3 = \frac{a^{2} k}{v}\quad$

$\displaystyle \text{Set }2: \quad\pi_1 = \frac{a \sqrt[3]{c} h}{v}\quad\pi_2 = \sqrt[3]{c} θ\quad\pi_3 = \frac{a^{2} k}{v}\quad$

$\displaystyle \text{Set }3: \quad\pi_1 = \frac{h}{a k θ}\quad\pi_2 = \frac{v}{a^{2} k}\quad\pi_3 = c θ^{3}\quad$

$\displaystyle \text{Set }4: \quad\pi_1 = \frac{\sqrt[3]{c} h}{a k}\quad\pi_2 = \frac{v}{a^{2} k}\quad\pi_3 = \sqrt[3]{c} θ\quad$

$\displaystyle \text{Set }5: \quad\pi_1 = \frac{h}{\sqrt{k} \sqrt{v} θ}\quad\pi_2 = \frac{a \sqrt{k}}{\sqrt{v}}\quad\pi_3 = c θ^{3}\quad$

$\displaystyle \text{Set }6: \quad\pi_1 = \frac{\sqrt[3]{c} h}{\sqrt{k} \sqrt{v}}\quad\pi_2 = \frac{a \sqrt{k}}{\sqrt{v}}\quad\pi_3 = \sqrt[3]{c} θ\quad$

$\displaystyle \text{Set }7: \quad\pi_1 = \frac{a h}{v θ}\quad\pi_2 = c θ^{3}\quad\pi_3 = \frac{k v θ^{2}}{h^{2}}\quad$

$\displaystyle \text{Set }8: \quad\pi_1 = \frac{a \sqrt[3]{c} h}{v}\quad\pi_2 = \sqrt[3]{c} θ\quad\pi_3 = \frac{k v}{c^{\frac{2}{3}} h^{2}}\quad$

$\displaystyle \text{Set }9: \quad\pi_1 = \frac{a \sqrt{k}}{\sqrt{v}}\quad\pi_2 = \frac{\sqrt{k} \sqrt{v} θ}{h}\quad\pi_3 = \frac{c h^{3}}{k^{\frac{3}{2}} v^{\frac{3}{2}}}\quad$

$\displaystyle \text{Set }10: \quad\pi_1 = \frac{a k θ}{h}\quad\pi_2 = \frac{k v θ^{2}}{h^{2}}\quad\pi_3 = c θ^{3}\quad$

$\displaystyle \text{Set }11: \quad\pi_1 = \frac{a k}{\sqrt[3]{c} h}\quad\pi_2 = \frac{k v}{c^{\frac{2}{3}} h^{2}}\quad\pi_3 = \sqrt[3]{c} θ\quad$

$\displaystyle \text{Set }12: \quad\pi_1 = \frac{v θ}{a h}\quad\pi_2 = c θ^{3}\quad\pi_3 = \frac{a k θ}{h}\quad$

$\displaystyle \text{Set }13: \quad\pi_1 = \frac{v}{a \sqrt[3]{c} h}\quad\pi_2 = \sqrt[3]{c} θ\quad\pi_3 = \frac{a k}{\sqrt[3]{c} h}\quad$

$\displaystyle \text{Set }14: \quad\pi_1 = \frac{v}{a^{2} k}\quad\pi_2 = \frac{a k θ}{h}\quad\pi_3 = \frac{c h^{3}}{a^{3} k^{3}}\quad$

$\displaystyle \text{Set }15: \quad\pi_1 = \frac{v θ}{a h}\quad\pi_2 = \frac{a^{3} c h^{3}}{v^{3}}\quad\pi_3 = \frac{a^{2} k}{v}\quad$

Selecionando o conjunto 3:¶

$$\Pi_1 = \frac{h}{k a \theta} \qquad \Pi_2 = \frac{v}{k a^2} \qquad \Pi_3 = c a^3 $$

ou seja:

$$ h = k a \theta F\left(\frac{v}{k a^2}, c a^3\right) $$

Aplicações em Remo¶

In [12]:

remo = BuckinghamPi()
remo.add_variable(name='v', dimensions='V')
remo.add_variable(name='G', dimensions='L^3*N^(-1)')
remo.add_variable(name='A', dimensions='R*V^3*L^2*N^(-1)')
remo.add_variable(name='rho', dimensions='R')
remo.add_variable(name='N', dimensions='N')
remo.generate_pi_terms()
remo.print_all()

$\displaystyle \text{Set }1: \quad\pi_1 = \frac{G^{\frac{2}{9}} \sqrt[3]{\rho} v}{\sqrt[3]{A} \sqrt[9]{N}}\quad$

$\displaystyle \text{Set }2: \quad\pi_1 = \frac{G \rho^{\frac{3}{2}} v^{\frac{9}{2}}}{A^{\frac{3}{2}} \sqrt{N}}\quad$

$\displaystyle \text{Set }3: \quad\pi_1 = \frac{G^{\frac{2}{3}} \rho v^{3}}{A \sqrt[3]{N}}\quad$

$\displaystyle \text{Set }4: \quad\pi_1 = \frac{A^{3} N}{G^{2} \rho^{3} v^{9}}\quad$

Escolhendo o conjunto 1¶

$$\Pi = \frac{G^{2/9}\rho^{1/3} v }{A^{1/3} N^{1/9}} \quad\longrightarrow\quad v = const \cdot \frac{A^{1/3}}{\rho^{1/3}G^{2/9}}\cdot N^{1/9}$$

Para uma distância fixa, 2000 m por exemplo, o tempo deve ser proporcional a $N^{-1/9}$

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