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Considere um fluido incompressível, de massa específica \(\rho\), que escoa em regime permanente com velocidade uniforme \(v\) em um trecho de duto consistindo de uma curva a 90º, no plano horizontal, tendo uma extremidade rigidamente flangeada e outra acoplada a uma junta de expansão (veja Figura 2). Deseja-se determinar os esforços presentes no flange, admitindo desconsideráveis os esforços na junta de expansão e o peso do fluido em face das outras forças envolvidas.

Escoamento em um duto com curva de 90°

Agora vamos fazer uma análise de um fluido incompressível escoando em regime permanente através de duto com uma curva de 90°, considerando efeitos hidrodinâmicos, térmicos e estruturais. O sistema consiste em um duto horizontal com flange rígido na entrada e junta de expansão na saída.

1. Definição do Problema

• Fluido incompressível: Massa específica \(\rho\)=constante em regime isotérmico.

• Regime permanente: As propriedades do escoamento não variam no tempo.

• Velocidade uniforme \(v\): A velocidade é a mesma na entrada e na saída (assumindo área transversal constante).

• Plano horizontal: Forças gravitacionais são desprezíveis.

• Junta de expansão: Não transmite esforços (livre para se mover).

• Flange rígido: Transmite todos os esforços para a estrutura.

2. Equações

2.1. Conservação da Massa:

\begin{equation}
\int \rho \vec{v} d \vec{A} = \rho A (\vec{v_e} – \vec{v_s}) = 0
\end{equation}

2.2. Conservação da Quantidade de Movimento Linear:

A equação integral da conservação da quantidade de movimento para um volume de controle (VC) fixo é:

\begin{equation}
\sum \vec{F} = \int_{VC} \rho \vec{v} (\vec{v} \cdot \vec{n}) d \vec{A} = \dot{m}(\vec{v_e} – \vec{v_s})
\end{equation}

Onde:

\(\sum \vec{F}\): Forças externas atuando no \(VC\) (pressão e reação no flange).
\(\vec{F}\): Vetor velocidade.
\(\vec{n}\): Vetor normal à superfície de controle (apontando para fora).
\(\dot{m}=\rho v A\): Vazão mássica
\(\vec{v_e} ​= v_e \hat{i}\) e \(\vec{v_s} ​= v_s \hat{j}\): Velocidades na entrada e saída.

A força exercida pelo fluido no duto (ação) é: 

\(\vec{R}_{flange} = \dot{m}(\vec{v_e} – \vec{v_s}) + (p_e ​A \hat{i} + p_s ​A \hat{j})\)

Se \(p_e=p_s=p\) e \(v_e = v_s = v\)

\(\vec{R}_{flange} = \rho v A(v\hat{i} – v\hat{j}) + (pA \hat{i} + p​A \hat{j})\)

\(\vec{R}_{flange} = \rho v^2 A(\hat{i} – \hat{j}) + pA(\hat{i} + \hat{j})\)

A força resultante no flange é:

\[
\boxed{\vec{R}_{flange} = – (\rho v^2 + p ) A (\hat{i} + \hat{j})}
\]

2.3. Caso com Variação de Área ou Velocidade

Se a área da seção transversal mudar (\(A_e \ne A_s\)​), a velocidade também muda (\(v_e A_e=v_s A_s\)), por continuidade). A equação de Bernoulli pode ser usada para relacionar pressões:

\begin{equation}
p_e + \frac{1}{2+}\rho {v_e}^2 =p_s + \frac{1}{2+}\rho {v_s}^2 +\Delta_{perdas​}
\end{equation}

Onde \(\Delta p_{perdas​}$ são perdas por atrito e turbulência.

Forças resultantes atualizadas:

\(R_x = \rho {v_e}^2 A_e −p_e A_e\)

\(R_y=\rho {v_s}^2 A_s −p_s A_s\)

Aplicação Prática: Dimensionamento do Flange
Para garantir a resistência do flange, um engenheiro deve considerar:

Força estática máxima:

\(R_{max} = \rho v^2 A + p A\)    (considerando pressão e velocidade máximas operacionais).

1. Fator de segurança: Multiplicar \(R_{max}\) por um coeficiente.
2. Material e parafusos: O flange deve ser fabricado com material resistente (aço carbono, aço inox) e os parafusos devem ser dimensionados para suportar a carga.
3. Vibrações e fadiga: Se o escoamento for pulsante, deve-se avaliar efeitos dinâmicos.

 

3. Conservação do Momento Angular

3.1. Momento Torsor (Torque) no Flange

Além da força resultante, a curva pode induzir um momento torsor no flange devido à distribuição assimétrica de pressões e velocidades. 

A equação integral para o momento angular é:

\begin{equation}
\sum \vec{M}_{ext} = \int \rho (\vec{r} \times\vec{v})(\vec{v}\cdot d \vec{A})
\end{equation}

• Para uma curva simétrica e flange rígido:

Como o escoamento é simétrico e o flange está rigidamente fixado, o torque líquido no flange é zero em condições ideais. Porém, em situações reais (escoamento turbulento, assimetrias), pode haver um pequeno torque residual.

Entrada: Momento angular \(-\rho r v^2A \hat{k}\).
Saída: Momento angular \(+\rho r v^2 A \hat{k}\).

Assim:

\begin{equation}
\sum \vec{M}_ext = \int \rho (\vec{r} \times\vec{v})(\vec{v}\cdot d \vec{A} = 0
\end{equation}

\[
\boxed{\vec{T} = 0}
\]

• Com assimetria (ex.: junta excêntrica em z):

O torque \(\vec{T}\) em relação ao centro da curva é dado por:

\begin{equation}
\vec{T} = \vec{r_e}\vec{F_e} \times \vec{r_s}\vec{F_s}
\end{equation}

​Onde:

 \(\vec{r_e}\) e \(\vec{r_e}\) são vetores posição dos pontos de aplicação das forças.
\(\vec{F_s}​\) e \(\vec{F_s}\) são as forças de pressão e quantidade de movimento nas seções de entrada e saída.

Torque no flange:

\[
\boxed{\vec{T}= – (\rho v^2 + p) r A \hat{k}}
\]

onde \(r\) é a distância do centro de curvatura ao ponto de aplicação da força.

4. Ajustes para Cenários Reais

Perdas por Atrito e Queda de Pressão

Em sistemas reais, o atrito entre o fluido e as paredes do duto causa perdas de energia, levando a uma queda de pressão ( \(\Delta p\) ). Isso pode ser modelado pela equação de Darcy-Weisbach ou por correlações empíricas.

Equação de Darcy-Weisbach:

\begin{equation}
\Delta p = f \frac{L}{D} \frac{\rho v^2}{2}
\end{equation}

Onde:

\(f\): Fator de atrito (depende do regime de escoamento e rugosidade da parede).
\(L\): Comprimento da curva ou trecho analisado.
\(D\): Diâmetro hidráulico do duto.

Impacto no modelo:

A pressão na saída (\(p_s\)​) será menor que na entrada (\(p_e\)​):

\(p_s = p_e – \Delta p\)

A força resultante no flange deve incluir essa diferença:

\[
\boxed{\vec{F}_{flange} = -[(\rho v^2+p_e)A \hat{i} +(\rho v^2+p_s)A \hat{j}]}
\]

Distribuição Não Uniforme de Pressão

Em curvas reais, a pressão não é constante ao longo da seção transversal devido a efeitos centrífugos:

Parede externa: Pressão mais alta.
Parede interna: Pressão mais baixa (efeito de sucção).

Modelo para Pressão Assimétrica:
A pressão pode ser aproximada por uma distribuição parabólica:

\begin{equation} p (\theta) = p_{avg} + \Delta p_{max} cos (2 \theta) \end{equation}

Onde:

\(\theta\): Ângulo ao longo da curva (0° a 90°).

\(\Delta p_{max​}\): Diferença máxima de pressão (empiricamente determinada).

A força de pressão agora requer uma integração angular:

\begin{equation}
\vec{F}_p=\int_{0}^{\pi/2} p (\theta) \cdot \vec{n} (\theta) r d\theta
\end{equation}

Isso pode introduzir torques adicionais se a distribuição não for simétrica.

Turbulência e Vórtices

– Escoamentos reais em curvas geram vórtices secundários e turbulência, que:

– Aumentam as perdas de energia.Podem causar vibrações no duto e no flange.

Modelagem via Número de Reynolds (\(Re$):
\begin{equation}
Re=\frac{\rho v D}{\mu}
\end{equation}

Se \(Re> 4000\) escoamento é turbulento, e o fator de atrito \(f\) deve ser ajustado (usando, por exemplo, o diagrama de Moody).

Variação de Área ou Velocidade

Se o duto tem áreas diferentes na entrada ( \(A_e\) ) e saída (\(A_s\)​), a velocidade muda (\(v_e A_e = v_s A_s\)), afetando o balanço de momento.

1. Conservação da massa com áreas diferentes: 

   \( v_s = v_e \frac{A_e}{A_s}\) ​​

2. Força no flange:

   Conservação da massa com áreas diferentes: \( v_2 = v_1 \frac{A1}{A2} \)

   Força resultante atualizada:

   \begin{equation} \vec{F} = -[(\rho {v_1}^2 + p_1)A_1 \hat{i} + (\rho {v_2}^2 + p_2)A_2 \hat{j}] \end{equation}

Efeitos Dinâmicos Transiente

Em cenários como fechamento de válvulas ou partida do sistema, forças adicionais surgem devido a acelerações do fluido.

Equação de Momentum Transiente:

\begin{equation} \vec{F}_{inercia} = \rho L A \frac{\vec{F}}{dt} \end{equation}

$\backslash (L\backslash )$: Comprimento característico do duto.

\(\frac{dv}{dt}\)​: Aceleração do fluido.

Variação das Propriedades do Fluido com a Temperatura

Para líquidos (ex.: água, óleo) e gases (ex.: ar, vapor), \(\rho\) e \(\mu\) dependem da temperatura (\(T\)):

• Líquidos:

\(\rho (T)= \rho_0[1 – \beta(T – T_0)]\),   \(\mu(T) \propto e^{-k(T – T_0)}\)

(\(\beta\) = coeficiente de expansão térmica, \(k\) = constante empírica).

• Gases (lei dos gases ideais):

\begin{equation} \rho(T)=\frac{p}{RT} \end{equation}

\(\mu (T) \propto T^{0.7}\)

Dilatação Térmica do Duto

O material do duto (aço, PVC, etc.) expande-se com a temperatura:

\begin{equation} \Delta L=\alpha L \Delta T \end{equation}

$\backslash (\backslash alpha\$$ = coeficiente de dilatação térmica.

Critério de falha:

Tensão térmica:

\(\sigma_{termico} = E \alpha \Delta T\)

Se \(\sigma_{termico} > \sigma_{adm}\)​, há risco de trincas ou vazamentos.

Convecção Natural

Número de Grashof:

\begin{equation} Gr = \frac{g\beta \Delta T L^3}{\nu} \end{equation}

Força e Torque com Efeitos Térmicos

Força no flange:

\begin{equation} \vec{F} = -[(\rho(T) v^2 + p(T))A](\hat{i}+ \hat{j)} \end{equation}

Torque térmico:

\begin{equation} \vec{T}_{termico} = \int r \times (\alpha E \Delta T)d \end{equation}

Recomendações para Projeto

Para escoamento:

Use CFD para simular distribuição de pressão e turbulência.

Adote um fator de segurança de 20-50 % nas forças calculadas.

Para efeitos térmicos:

Materiais com baixo \(\alpha\) (ex.: aço inox).

Juntas de expansão termicamente isoladas.

Para transientes:

Amortecedores de pressão para golpes de aríete.