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Universidade Federal Fluminense
Disciplina:
Aula 4 β Equaçáes Diferenciais
FENΓMENOS DE TRANSPORTE
Prof.: Gabriel Nascimento (Depto. de Eng. AgrΓcola e Meio Ambiente)Elson Nascimento (Depto. de Eng. Civil)
Escola de Engenharia
Aula 4 β Equaçáes Diferenciais
Equação da continuidade
CinemΓ‘tica
Equação da quantidade de movimento linear
Equação de Euler
Equação de Navier-Stokes
Métodos de solução de problemas com fluidos:
F
u(r)
Solução analΓtica ou numΓ©rica
(CFD β ComputationalFluid Dynamics)
VCGrandezas integrais (volume de controle β VC):
⒠Vazão⒠Força ⒠Energia
EQUAΓΓES INTEGRAIS
EQUAΓΓES DIFERENCIAIS
MΓTODOS EXPERIMENTAIS
Grandezas infinitesimais (pontual):
β’ Velocidade: π(π₯, π¦, π§, π‘)β’ TensΓ£o: π π₯, π¦, π§, π‘ , π π₯, π¦, π§, π‘
⒠Modelos reduzidos em laboratório, protótipos ou mediçáes em campo
β’ AnΓ‘lise dimensional
DisponΓvel em <http://www.canadianautoreview.ca/news/gm-reduced-scale-wind-tunnel.html>. Acesso em 27 mar. 2018.
Eq. da Continuidade
Volume de controle infinitesimal:Volume de controle finito:
Equação Integral: Equação Diferencial:
x
y
z
dz
dx
dy
Conservação da massa:
Eq. Integral:
Eq. Diferencial:
π
ππ‘ ππΆ
π πV + β Β± ππ = 0
Volume de controle infinitesimal:
ππ
ππ‘πV +
+ saΓda entrada
π = πππππ΄
= 0
x
y
z
dz
dx
dy
π ππ₯π β π ππ₯
π+
π ππ¦π β π ππ¦
π+
π ππ§π β π ππ§
π
Conservação da massa:
ππ
ππ‘πV +
π ππ₯π β π ππ₯
π+
π ππ¦π β π ππ¦
π+
π ππ§π β π ππ§
π
= 0 Eq. Diferencial:
Volume de controle infinitesimal:
x
y
z
dz
dx
dy
Conservação da massa:
ππ
ππ‘πV +
π ππ₯π β π ππ₯
π+
π ππ¦π β π ππ¦
π+
π ππ§π β π ππ§
π
= 0 Eq. Diferencial:
π ππ₯π β π ππ₯
π= π π₯ + ππ₯ β π π₯
= limπΏπ₯β0
π π₯ + πΏπ₯ β π π₯
πΏπ₯ππ₯
=ππ
ππ₯ππ₯
=π π₯ + ππ₯ β π π₯
ππ₯ππ₯
=π π ππ₯
ππ₯ππ₯ =
π ππππππ΄ π₯
ππ₯ππ₯
=π π π’ ππ¦ππ§
ππ₯ππ₯ =
π π π’
ππ₯ππ₯ππ¦ππ§
Volume de controle infinitesimal:
πV
=π π π’
ππ₯πV
x
y
z
dz
dx
dy
π π₯
β’ Em x:
π’
πβ² π₯
Conservação da massa:
Volume de controle infinitesimal:
=π π π’
ππ₯πV
π ππ₯π β π ππ₯
πβ’ Em x:
ππ
ππ‘πV +
π ππ₯π β π ππ₯
π+
π ππ¦π β π ππ¦
π+
π ππ§π β π ππ§
π
= 0 Eq. Diferencial:
x
y
z
dz
dx
dy
Conservação da massa:
Volume de controle infinitesimal: β’ Em y: π ππ¦π β π ππ¦
π=
π π π£
ππ¦πV
=π π π’
ππ₯πV
β’ Em z: π ππ§π β π ππ§
π=
π π π€
ππ§πV
π π π’
ππ₯+
π π π£
ππ¦+
π π π€
ππ§πV
π ππ₯π β π ππ₯
πβ’ Em x:
ππ
ππ‘πV +
π ππ₯π β π ππ₯
π+
π ππ¦π β π ππ¦
π+
π ππ§π β π ππ§
π
= 0 Eq. Diferencial:
x
y
z
dz
dx
dy
Conservação da massa:
Volume de controle infinitesimal:
βππ
ππ‘πV +
π π π’
ππ₯+
π π π£
ππ¦+
π π π€
ππ§πV
ππ
ππ‘πV +
π ππ₯π β π ππ₯
π+
π ππ¦π β π ππ¦
π+
π ππ§π β π ππ§
π
= 0 Eq. Diferencial:
x
y
z
dz
dx
dy
Conservação da massa:
βππ
ππ‘πV +
π π π’
ππ₯+
π π π£
ππ¦+
π π π€
ππ§πV = 0
βππ
ππ‘+
π π π’
ππ₯+
π π π£
ππ¦+
π π π€
ππ§= 0
divergente de ππ
ππ
ππ‘πV +
π ππ₯π β π ππ₯
π+
π ππ¦π β π ππ¦
π+
π ππ§π β π ππ§
π
= 0 Eq. Diferencial:
ππ
ππ‘+ π» β ππ = 0
Conservação da massa:
Fluido compressΓvel:ππ
ππ‘+ π» β ππ = 0
Fluido incompressΓvel:ππ
ππ‘+ π» β ππ = 0
0
β ππ» β π = 0
β π» β π = 0
divergente nulo(campo solenoidal)
Conservação da massa:
Exemplo:[PETROBRAS β ENG. EQP. JΓNIOR - 2008]
Considerando um escoamento permanente e incompressΓvel, cuja
distribuição de velocidades seja dada pela função π = 3π₯ π + πΆπ¦ π + 2π₯ π,
calcule o valor da constante C para que seja atendido o princΓpio da
continuidade.
β β V = 0
ββπ’βx
+βπ£βy
+βπ€βz
= 0 V = u π + v π + w π
β u = 3xv = Cyw = 2x
V = 3x i + Cy j + 2x k
β 3 + C + 0 = 0 β C = β3
ββ 3xβx
+β Cyβy
+β 2xβz
= 0
CinemΓ‘tica
x
y x
y translação
x
y rotação
x
ydeformação
angular (distorção)
x
y deformação linear
PartΓcula Pem (xp, yp, zp)
ππ π‘
ππ π‘Campo de velocidade:
π π₯, π¦, π§, π‘ = π’ π₯, π¦, π§, π‘ π + π£ π₯, π¦, π§, π‘ π + π€ π₯, π¦, π§, π‘ π
= π π₯π, π¦π, π§π, π‘
ππ π‘ =πππ π‘
ππ‘=
ππ π₯π, π¦π, π§π, π‘
ππ‘
=πππ π‘
ππ‘
=ππ
ππ₯
ππ₯π
ππ‘+
ππ
ππ¦
ππ¦π
ππ‘+
ππ
ππ§
ππ§π
ππ‘+
ππ
ππ‘
ππ‘
ππ‘
π’ π₯π, π¦π , π§π, π‘
π£ π₯π, π¦π , π§π, π‘
π€ π₯π, π¦π, π§π, π‘
1
= π’ππ
ππ₯+ π£
ππ
ππ¦+ π€
ππ
ππ§+
ππ
ππ‘ ππ =
π·π
π·π‘
aceleração convectiva aceleração localaceleração total
π = πΎπ₯ π β πΎπ¦ πEx. de campo de velocidade:
Linhas de corrente:
PartΓcula P
y
x
+1
+1 -1
-1
+2
+2 -2
-2
+3
C = +3 -3
C = -3
ππ
ππ‘= 0 β permanente ππ β 0
ππ =π·π
π·π‘= π’
ππ
ππ₯+ π£
ππ
ππ¦+ π€
ππ
ππ§+
ππ
ππ‘
0
Escoamento permanente não significa aceleração nula !
y
πΏπΌ
πΏπ½πΏπ
πΏπ₯
πΏπ¦
π’π₯πΏπ‘π’π₯+πΏπ₯πΏπ‘
πΏπ₯ + π’π₯+πΏπ₯ β π’π₯
πΏπ’π₯
πΏπ‘ = πΏπ₯ +ππ’
ππ₯πΏπ₯πΏπ‘
πΏπ₯
ππ’
ππ₯πΏπ₯
x
πΏπ‘
Rotação: πΏππ₯π¦ =πΏπΌ β πΏπ½
2
y
πΏπΌ
πΏπ½πΏπ
πΏπ₯
πΏπ¦
x
πΏπ‘
π£π₯πΏπ‘
π£π₯+πΏπ₯πΏπ‘
π£π₯+πΏπ₯ β π£π₯
πΏπ£π₯
πΏπ‘ =
ππ£
ππ₯πΏπ₯
ππ£
ππ₯πΏπ₯πΏπ‘
πΏπ₯ +ππ’
ππ₯πΏπ₯πΏπ‘
Rotação: πΏππ₯π¦ =πΏπΌ β πΏπ½
2
y
πΏπΌ
πΏπ½πΏπ
πΏπ₯
πΏπ¦
x
πΏπ‘
πΏπ₯ +ππ’
ππ₯πΏπ₯πΏπ‘
ππ£
ππ₯πΏπ₯πΏπ‘
πΏπ¦ +ππ£
ππ¦πΏπ¦πΏπ‘
ππ’
ππ¦πΏπ¦πΏπ‘
πΏπΌ = atan
ππ£ππ₯
πΏπ₯πΏπ‘
πΏπ₯ +ππ’ππ₯
πΏπ₯πΏπ‘ 0
=ππ£
ππ₯πΏπ‘
ππΌ
ππ‘= lim
πΏπ‘β0
πΏπΌ
πΏπ‘=
ππ£
ππ₯
ππ½
ππ‘= lim
πΏπ‘β0
πΏπ½
πΏπ‘=
ππ’
ππ¦
ππ§ =πππ₯π¦
ππ‘=
1
2
ππΌ
ππ‘β
ππ½
ππ‘=
1
2
ππ£
ππ₯β
ππ’
ππ¦
ππ₯ =πππ¦π§
ππ‘=
1
2
ππ€
ππ¦β
ππ£
ππ§
ππ¦ =πππ§π₯
ππ‘=
1
2
ππ’
ππ§β
ππ€
ππ₯
π = ππ₯ π + ππ¦ π + ππ§ π =
1
2π» Γ π
vorticidade
se escoamento irrotacional:
π» Γ π = 0
Rotação: πΏππ₯π¦ =πΏπΌ β πΏπ½
2
y
πΏπΌ
πΏπ½πΏπ
πΏπ₯
πΏπ¦
x
πΏπ‘
πΏπ₯ +ππ’
ππ₯πΏπ₯πΏπ‘
ππ£
ππ₯πΏπ₯πΏπ‘
πΏπ¦ +ππ£
ππ¦πΏπ¦πΏπ‘
ππ’
ππ¦πΏπ¦πΏπ‘
πΏπΌ = atan
ππ£ππ₯
πΏπ₯πΏπ‘
πΏπ₯ +ππ’ππ₯
πΏπ₯πΏπ‘ 0
=ππ£
ππ₯πΏπ‘
ππΌ
ππ‘= lim
πΏπ‘β0
πΏπΌ
πΏπ‘=
ππ£
ππ₯
ππ½
ππ‘= lim
πΏπ‘β0
πΏπ½
πΏπ‘=
ππ’
ππ¦ππ‘
ππ‘+πΏπ‘
Distorção: (deformação angular)
πΏππ₯π¦ = πΏπΌ + πΏπ½
νπ₯π¦ =πππ₯π¦
ππ‘
ππΌ
ππ‘+
ππ½
ππ‘
=ππ£
ππ₯+
ππ’
ππ¦
νπ¦π§ =πππ¦π§
ππ‘=
ππ€
ππ¦+
ππ£
ππ§
νπ§π₯ =πππ§π₯
ππ‘=
ππ’
ππ§+
ππ€
ππ₯
taxa de cisalhamento
βππππ
ππ‘=
ππ’π
ππ₯π+
ππ’π
ππ₯π
βπππ₯π¦
ππ‘=
νπ₯π¦
y
πΏπ₯
πΏπ¦
x
πΏπ‘
πΏπ₯ +ππ’
ππ₯πΏπ₯πΏπ‘
πΏπ₯ πΏππ₯
πΏππ₯ =ππ’
ππ₯πΏπ₯πΏπ‘
β νπ₯π₯ =πππ₯
ππ₯
νπ₯π₯ =ππ’
ππ₯
=ππ’
ππ₯πΏπ‘
νπ¦π¦ =ππ£
ππ¦
β νπ₯π₯ =πνπ₯
ππ‘=
ππ’
ππ₯
νπ§π§ =ππ€
ππ§
Taxa de deformação linear:
Taxa de dilatação volumétrica :
πΏπ₯ + πππ₯ β πΏπ¦ + πππ¦ β πΏπ§ + πππ§
1
πΏπ
π πΏπ
ππ‘=
πΏπ₯ πΏπ¦ πΏπ§ + πΏπ₯πΏπ¦πππ§ + πΏπ₯πΏπ§πππ¦ + πΏπ¦πΏπ§πππ₯
πππ₯πππ¦
πππ₯πππ§
πππ¦πππ§
β 0
πΏπ₯ πΏπ¦ πΏπ§1
πΏπlimπΏπ‘β0
πΏππ‘+πΏπ‘ β πΏππ‘
πΏπ‘=
= limπΏπ‘β0
πΏπ₯ πΏπ¦ πΏππ§ + πΏπ₯ πΏπ§ πΏππ¦ + πΏπ¦ πΏπ§ πΏππ₯
πΏπ₯ πΏπ¦ πΏπ§ πΏπ‘
= limπΏπ‘β0
πΏππ₯
πΏπ₯πΏπ‘+
πΏππ¦
πΏπ¦πΏπ‘+
πΏππ§
πΏπ§πΏπ‘
β1
πΏπ
π πΏπ
ππ‘=
ππ’
ππ₯+
ππ£
ππ¦+
ππ€
ππ§= π» β π
se incompressΓvel
β π» β π = 0
= νπ₯π₯ + νπ¦π¦ + νπ§π§
Exemplo:Dados o campos de velocidade abaixo (S.I.): a) classifique o respectivo escoamento quanto ao regime temporal e dimensionalidade; b) calcule a aceleração no ponto (1,1), quando t=0; c) verifique se sΓ£o rotacionais; d) e se sΓ£o incompressΓveis (possivelmente).
π = π₯π‘ + 2π¦ π + 3π₯ β π¦π‘ π
π’ π£
π€ = 0
ππ
ππ‘β 0
ππ
ππ₯β 0 π
ππ
ππ¦β 0,
ππ
ππ§= 0
2D transiente
π =π·π
π·π‘= π’
ππ
ππ₯+ π£
ππ
ππ¦+ π€
ππ
ππ§+
ππ
ππ‘
a)
π₯π‘ + 2π¦ π‘ π + 3 π + 3π₯ β π¦π‘ 2 π β π‘ π
π₯ π β π¦ π
= 6 π + π m/s
b)
c) π = ππ₯ π + ππ¦ π + ππ§ π
ππ§ =1
2
ππ£
ππ₯β
ππ’
ππ¦
3 2
=1
2rad/s
d) π» β π =ππ’
ππ₯+
ππ£
ππ¦+
ππ€
ππ§
π‘ βπ‘
= 0
possivelmente incompressΓvel
Resumo
Translação:
Rotação:
Deformação angular (distorção):
Deformação linear:
βͺ Taxa de dilatação volumΓ©trica:
π =π·π
π·π‘= π’
ππ
ππ₯+ π£
ππ
ππ¦+ π€
ππ
ππ§+
ππ
ππ‘
π =1
2π» Γ π
ππππ
ππ‘=
ππ’π
ππ₯π+
ππ’π
ππ₯π
νπ =ππ’π
ππ₯π
1
πΏπ
π πΏπ
ππ‘= π» β π
Eq. do Momentum
= πππ·π
π·π‘
Quantidade de movimento linear:
πΉ =π
ππ‘π π
Em uma partΓcula fluida:
π πΉ =π
ππ‘ππ π
π =ππ
πVβ ππ = π πV
= π πVπ·π
π·π‘
β π πΉ
πV= π
π·π
π·π‘= π π’
ππ
ππ₯+ π£
ππ
ππ¦+ π€
ππ
ππ§+
ππ
ππ‘
aceleraçãototal
aceleraçãoconvectiva
aceleraçãolocal
π πΉ
πV= π
ππ
ππ‘+ π’
ππ
ππ₯+ π£
ππ
ππ¦+ π€
ππ
ππ§
2Βͺ Lei de Newton:
Quantidade de movimento linear: Em uma partΓcula
fluida:
π πΉ
πV= π
ππ
ππ‘+ π’
ππ
ππ₯+ π£
ππ
ππ¦+ π€
ππ
ππ§
Forças de campo
Forças de contato
βͺ ForΓ§as viscosas
βͺ ForΓ§as de pressΓ£o
π πΉπ = ππ π = π πV π βπ πΉπ
πV= π π
ππ π₯: πππ₯
ππ π¦: πππ¦
ππ π§: πππ§π =
ππ
πVβ ππ = π πV
Quantidade de movimento linear:
Forças de campo
Forças de contato
βͺ ForΓ§as viscosas
βͺ ForΓ§as de pressΓ£o
x
y
z
dz
dx
dy
ππ₯π₯
ππ₯π¦
ππ₯π§
ππ¦π₯
ππ¦π¦
ππ¦π§
ππ§π₯
ππ§π¦
ππ§π§
πππ
face direção
Em uma partΓcula fluida:
π πΉ
πV= π
ππ
ππ‘+ π’
ππ
ππ₯+ π£
ππ
ππ¦+ π€
ππ
ππ§
π πΉπ
πV= π π
β’ em x:
ππ₯π₯π₯
ππ¦π₯π¦
ππ§π₯π§
x
y
z
dz
dx
dy
Quantidade de movimento linear:
Forças de campo
Forças de contato
βͺ ForΓ§as viscosas
βͺ ForΓ§as de pressΓ£o
ππ₯π₯π₯+ππ₯
ππ¦π₯π¦+ππ¦
ππ§π₯π§+ππ§
ππ₯π₯π₯+ππ₯
β ππ₯π₯π₯
ππ¦ππ§
ππ§π₯π§+ππ§
β ππ§π₯π§
ππ₯ππ¦
ππ¦π₯π¦+ππ¦
β ππ¦π₯π¦
ππ₯ππ§
Em uma partΓcula fluida:
π πΉ
πV= π
ππ
ππ‘+ π’
ππ
ππ₯+ π£
ππ
ππ¦+ π€
ππ
ππ§
π πΉπ
πV= π π
Quantidade de movimento linear:
Forças de campo
Forças de contato
βͺ ForΓ§as viscosas
βͺ ForΓ§as de pressΓ£o
ππ₯π₯π₯+ππ₯
β ππ₯π₯π₯
ππ¦ππ§
ππ§π₯π§+ππ§
β ππ§π₯π§
ππ₯ππ¦
ππ¦π₯π¦+ππ¦
β ππ¦π₯π¦
ππ₯ππ§ = ππ¦π₯ π¦+ππ¦
β ππ¦π₯ π¦
ππ¦ππ¦ ππ₯ππ§
= ππ₯π₯ π₯+ππ₯ β ππ₯π₯ π₯
ππ₯ππ₯ ππ¦ππ§
= ππ§π₯ π§+ππ§ β ππ§π₯ π§
ππ§ππ§ ππ₯ππ¦
Em uma partΓcula fluida:
π πΉ
πV= π
ππ
ππ‘+ π’
ππ
ππ₯+ π£
ππ
ππ¦+ π€
ππ
ππ§
β’ em x:
π πΉπ
πV= π π
ππΉπ₯ =πππ₯π₯
ππ₯+
πππ¦π₯
ππ¦+
πππ§π₯
ππ§πV
ππΉπ₯
πV=
πππ₯π₯
ππ₯+
πππ¦π₯
ππ¦+
πππ§π₯
ππ§
Quantidade de movimento linear:
Forças de campo
Forças de contato
βͺ ForΓ§as viscosas
βͺ ForΓ§as de pressΓ£o
= ππ¦π₯ π¦+ππ¦
β ππ¦π₯ π¦
ππ¦ππ¦ ππ₯ππ§
= ππ₯π₯ π₯+ππ₯ β ππ₯π₯ π₯
ππ₯ππ₯ ππ¦ππ§
= ππ§π₯ π§+ππ§ β ί¬π§π₯ π§
ππ§ππ§ ππ₯ππ¦
= limπΏπ¦β0
ππ¦π₯ π¦+πΏπ¦β ππ¦π₯ π¦
πΏπ¦ππ¦ππ₯ππ§
= limπΏπ₯β0
ππ₯π₯ π₯+πΏπ₯ β ππ₯π₯ π₯
πΏπ₯ππ₯ππ¦ππ§
= limπΏπ§β0
ππ§π₯ π§+πΏπ§ β ί¬π§π₯ π§
πΏπ§ππ§ππ₯ππ¦
=πππ₯π₯
ππ₯πV
πππ₯π₯ππ₯
πV
=πππ¦π₯
ππ¦πV
=πί¬π§π₯
ππ§πV
Em uma partΓcula fluida:
π πΉ
πV= π
ππ
ππ‘+ π’
ππ
ππ₯+ π£
ππ
ππ¦+ π€
ππ
ππ§
β’ em x:
π πΉπ
πV= π π
Quantidade de movimento linear:
Forças de campo
Forças de contato
βͺ ForΓ§as viscosas
βͺ ForΓ§as de pressΓ£o
ππΉπ£π₯
πV=
πππ₯π₯
ππ₯+
πππ¦π₯
ππ¦+
πππ§π₯
ππ§ππΉπ£π¦
πV=
πππ₯π¦
ππ₯+
πππ¦π¦
ππ¦+
πππ§π¦
ππ§
ππΉπ£π§
πV=
πππ₯π§
ππ₯+
πππ¦π§
ππ¦+
πππ§π§
ππ§
β’ em x:
β’ em y:
β’ em z:
π πΉπ
πV= π π
Em uma partΓcula fluida:
π πΉ
πV= π
ππ
ππ‘+ π’
ππ
ππ₯+ π£
ππ
ππ¦+ π€
ππ
ππ§
Quantidade de movimento linear:
Forças de campo
Forças de contato
βͺ ForΓ§as viscosas
βͺ ForΓ§as de pressΓ£o
ππΉπ£π₯
πV=
πππ₯π₯
ππ₯+
πππ¦π₯
ππ¦+
πππ§π₯
ππ§ππΉπ£π¦
πV=
πππ₯π¦
ππ₯+
πππ¦π¦
ππ¦+
πππ§π¦
ππ§
ππΉπ£π§
πV=
πππ₯π§
ππ₯+
πππ¦π§
ππ¦+
πππ§π§
ππ§
β’ em x:
β’ em y:
β’ em z:
π πΉπ
πV= π π
Em uma partΓcula fluida:
π πΉ
πV= π
ππ
ππ‘+ π’
ππ
ππ₯+ π£
ππ
ππ¦+ π€
ππ
ππ§
x
y
z
dz
dx
dy
ππ₯
ππ₯+ππ₯
β’ em x:
ππ₯
β ππ₯+ππ₯
ππ¦ππ§ = β π π₯+ππ₯ β π π₯
ππ₯ππ₯ ππ¦ππ§ =
= β limπΏπ₯β0
π π₯+πΏπ₯ β π π₯
πΏπ₯ππ₯ππ¦ππ§ = β
ππ
ππ₯πV
ππΉππ₯=
ππππ₯
πV
Quantidade de movimento linear:
Forças de campo
Forças de contato
βͺ ForΓ§as viscosas
βͺ ForΓ§as de pressΓ£o
ππΉπ£π₯
πV=
πππ₯π₯
ππ₯+
πππ¦π₯
ππ¦+
πππ§π₯
ππ§ππΉπ£π¦
πV=
πππ₯π¦
ππ₯+
πππ¦π¦
ππ¦+
πππ§π¦
ππ§
ππΉπ£π§
πV=
πππ₯π§
ππ₯+
πππ¦π§
ππ¦+
πππ§π§
ππ§
β’ em x:
β’ em y:
β’ em z:
π πΉπ
πV= π π
Em uma partΓcula fluida:
π πΉ
πV= π
ππ
ππ‘+ π’
ππ
ππ₯+ π£
ππ
ππ¦+ π€
ππ
ππ§
β’ em x:ππΉππ₯
πV= β
ππ
ππ₯ππΉππ¦
πV= β
ππ
ππ¦β’ em y:
ππΉππ§
πV= β
ππ
ππ§β’ em z:
Quantidade de movimento linear:
Forças de campo
Forças de contato
βͺ ForΓ§as viscosas
βͺ ForΓ§as de pressΓ£o
ππΉπ£π₯
πV=
πππ₯π₯
ππ₯+
πππ¦π₯
ππ¦+
πππ§π₯
ππ§ππΉπ£π¦
πV=
πππ₯π¦
ππ₯+
πππ¦π¦
ππ¦+
πππ§π¦
ππ§
ππΉπ£π§
πV=
πππ₯π§
ππ₯+
πππ¦π§
ππ¦+
πππ§π§
ππ§
β’ em x:
β’ em y:
β’ em z:
π πΉπ
πV= π π
Em uma partΓcula fluida:
π πΉ
πV= π
ππ
ππ‘+ π’
ππ
ππ₯+ π£
ππ
ππ¦+ π€
ππ
ππ§
β’ em x:ππΉππ₯
πV= β
ππ
ππ₯ππΉππ¦
πV= β
ππ
ππ¦β’ em y:
ππΉππ§
πV= β
ππ
ππ§β’ em z:
Quantidade de movimento linear:
βππ
ππ₯
βππ
ππ¦
βππ
ππ§
πππ₯
πππ¦
πππ§
+πππ₯π₯
ππ₯+
πππ¦π₯
ππ¦+
πππ§π₯
ππ§
+πππ₯π¦
ππ₯+
πππ¦π¦
ππ¦+
πππ§π¦
ππ§
+πππ₯π§
ππ₯+
πππ¦π§
ππ¦+
πππ§π§
ππ§
= πππ’
ππ‘+ π’
ππ’
ππ₯+ π£
ππ’
ππ¦+ π€
ππ’
ππ§
= πππ£
ππ‘+ π’
ππ£
ππ₯+ π£
ππ£
ππ¦+ π€
ππ£
ππ§
= πππ€
ππ‘+ π’
ππ€
ππ₯+ π£
ππ€
ππ¦+ π€
ππ€
ππ§
Em uma partΓcula fluida:
π πΉ
πV= π
ππ
ππ‘+ π’
ππ
ππ₯+ π£
ππ
ππ¦+ π€
ππ
ππ§
Quantidade de movimento linear:
βππ
ππ₯
βππ
ππ¦
βππ
ππ§
πππ₯
πππ¦
πππ§
+πππ₯π₯
ππ₯+
πππ¦π₯
ππ¦+
πππ§π₯
ππ§
+πππ₯π¦
ππ₯+
πππ¦π¦
ππ¦+
πππ§π¦
ππ§
+πππ₯π§
ππ₯+
πππ¦π§
ππ¦+
πππ§π§
ππ§
= πππ’
ππ‘+ π’
ππ’
ππ₯+ π£
ππ’
ππ¦+ π€
ππ’
ππ§
= πππ£
ππ‘+ π’
ππ£
ππ₯+ π£
ππ£
ππ¦+ π€
ππ£
ππ§
= πππ€
ππ‘+ π’
ππ€
ππ₯+ π£
ππ€
ππ¦+ π€
ππ€
ππ§
Equação de Euler
Equação de Euler:
Escoamento invΓscido (sem βatritoβ)
πππ₯ βππ
ππ₯+
πππ₯π₯
ππ₯+
πππ¦π₯
ππ¦+
πππ§π₯
ππ§= π
ππ’
ππ‘+ π’
ππ’
ππ₯+ π£
ππ’
ππ¦+ π€
ππ’
ππ§
πππ¦ βππ
ππ¦+
πππ₯π¦
ππ₯+
πππ¦π¦
ππ¦+
πππ§π¦
ππ§= π
ππ£
ππ‘+ π’
ππ£
ππ₯+ π£
ππ£
ππ¦+ π€
ππ£
ππ§
πππ§ βππ
ππ§+
πππ₯π§
ππ₯+
πππ¦π§
ππ¦+
πππ§π§
ππ§= π
ππ€
ππ‘+ π’
ππ€
ππ₯+ π£
ππ€
ππ¦+ π€
ππ€
ππ§
β πππ = 0
Equação de Euler:
Escoamento invΓscido (sem βatritoβ)
πππ₯ βππ
ππ₯= π
ππ’
ππ‘+ π’
ππ’
ππ₯+ π£
ππ’
ππ¦+ π€
ππ’
ππ§
πππ¦ βππ
ππ¦= π
ππ£
ππ‘+ π’
ππ£
ππ₯+ π£
ππ£
ππ¦+ π€
ππ£
ππ§
πππ§ βππ
ππ§= π
ππ€
ππ‘+ π’
ππ€
ππ₯+ π£
ππ€
ππ¦+ π€
ππ€
ππ§
π π β π»π = πππ
ππ‘
β πππ = 0
Equação de Euler:
Exemplo: Um campo de escoamento permanente,
incompressΓvel e sem atrito Γ© dado por π = 2π₯π¦ π + π¦2 π em
unidades arbitrΓ‘rias. Seja a massa especΓfica 0 = constante e
despreze a gravidade. Encontre uma expressΓ£o para o
gradiente de pressão na direção x.
Οgxββpβx
=Οβuβt+u
βuβx+v
βuβy
+wβuβz
Οgyββpβy
=Οβvβt+u
βvβx+v
βvβy
+wβvβz
Οgzββpβz=Ο
βwβt
+uβwβx
+vβwβy
+wβwβz
Οgββp = ΟdV
dt
Equação de Euler:
Exemplo: Um campo de escoamento permanente,
incompressΓvel e sem atrito Γ© dado por π = 2π₯π¦ π β π¦2 π em
unidades arbitrΓ‘rias. Seja a massa especΓfica 0 = constante e
despreze a gravidade. Encontre uma expressΓ£o para o
gradiente de pressão na direção x.
Οgxββpβx=Ο
βuβt+u
βuβx+v
βuβy
+wβuβzΟgββp = Ο
dV
dt
π = u π + π£ π + π€ π0
0 0 0
β ββpβx
= Ο0 2xyβ 2xyβx
βy2β 2xyβy
2y 2x
= Ο04xy2β2xy2 = 2Ο
0xy2
ββpβx
= β 2Ο0xy2
Eq. de Navier-Stokes
Equação da quantidade de movimento linear
πππ₯ βππ
ππ₯+
πππ₯π₯
ππ₯+
πππ¦π₯
ππ¦+
πππ§π₯
ππ§= π
ππ’
ππ‘+ π’
ππ’
ππ₯+ π£
ππ’
ππ¦+ π€
ππ’
ππ§
πππ¦ βππ
ππ¦+
πππ₯π¦
ππ₯+
πππ¦π¦
ππ¦+
πππ§π¦
ππ§= π
ππ£
ππ‘+ π’
ππ£
ππ₯+ π£
ππ£
ππ¦+ π€
ππ£
ππ§
πππ§ βππ
ππ§+
πππ₯π§
ππ₯+
πππ¦π§
ππ¦+
πππ§π§
ππ§= π
ππ€
ππ‘+ π’
ππ€
ππ₯+ π£
ππ€
ππ¦+ π€
ππ€
ππ§
Equação da quantidade de movimento linear
ππ₯π¦ = πππ£
ππ₯+
ππ’
ππ¦
ππ₯π§ = πππ€
ππ₯+
ππ’
ππ§
ππ¦π§ = πππ€
ππ¦+
ππ£
ππ§
ππ₯π₯ = 2πππ’
ππ₯
ππ¦π¦ = 2πππ£
ππ¦
ππ§π§ = 2πππ€
ππ§
πππ₯ βππ
ππ₯+
πππ₯π₯
ππ₯+
πππ¦π₯
ππ¦+
πππ§π₯
ππ§= π
ππ’
ππ‘+ π’
ππ’
ππ₯+ π£
ππ’
ππ¦+ π€
ππ’
ππ§
πππ¦ βππ
ππ¦+
πππ₯π¦
ππ₯+
πππ¦π¦
ππ¦+
πππ§π¦
ππ§= π
ππ£
ππ‘+ π’
ππ£
ππ₯+ π£
ππ£
ππ¦+ π€
ππ£
ππ§
πππ§ βππ
ππ§+
πππ₯π§
ππ₯+
πππ¦π§
ππ¦+
πππ§π§
ππ§= π
ππ€
ππ‘+ π’
ππ€
ππ₯+ π£
ππ€
ππ¦+ π€
ππ€
ππ§
π = πππ
ππ‘
ππ
ππ‘ Newton:
ππ
ππ‘
πππ = πππππ
ππ‘
ππππ
ππ‘=
ππ’π
ππ₯π+
ππ’π
ππ₯π
πππ = πππ’π
ππ₯π+
ππ’π
ππ₯π
Equação da quantidade de movimento linear
ππ₯π¦ = πππ£
ππ₯+
ππ’
ππ¦
ππ₯π§ = πππ€
ππ₯+
ππ’
ππ§
ππ¦π§ = πππ€
ππ¦+
ππ£
ππ§
ππ₯π₯ = 2πππ’
ππ₯
ππ¦π¦ = 2πππ£
ππ¦
ππ§π§ = 2πππ€
ππ§
πππ₯ βππ
ππ₯+
πππ₯π₯
ππ₯+
πππ¦π₯
ππ¦+
πππ§π₯
ππ§= π
ππ’
ππ‘+ π’
ππ’
ππ₯+ π£
ππ’
ππ¦+ π€
ππ’
ππ§
πππ¦ βππ
ππ¦+
πππ₯π¦
ππ₯+
πππ¦π¦
ππ¦+
πππ§π¦
ππ§= π
ππ£
ππ‘+ π’
ππ£
ππ₯+ π£
ππ£
ππ¦+ π€
ππ£
ππ§
πππ§ βππ
ππ§+
πππ₯π§
ππ₯+
πππ¦π§
ππ¦+
πππ§π§
ππ§= π
ππ€
ππ‘+ π’
ππ€
ππ₯+ π£
ππ€
ππ¦+ π€
ππ€
ππ§
πππ₯π₯
ππ₯+
πππ¦π₯
ππ¦+
πππ§π₯
ππ§=
π
ππ₯2π
ππ’
ππ₯+
β’ Considerando fluido newtoniano: constante
= π 2π2π’
ππ₯2+
π2π£
ππ¦ππ₯+
π2π’
ππ¦2+
π2π€
ππ§ππ₯+
π2π’
ππ§2
π2π’
ππ₯2
π2π’
ππ₯2
π2π£
ππ¦ππ₯
π2π’
ππ¦2
π2π€
ππ§ππ₯
π2π’
ππ§2
= ππ2π’
ππ₯2+
π2π’
ππ¦2+
π2π’
ππ§2+
π2π’
ππ₯2+
π2π£
ππ¦ππ₯+
π2π€
ππ§ππ₯
π
ππ¦π
ππ£
ππ₯+
ππ’
ππ¦+
π
ππ§π
ππ€
ππ₯+
ππ’
ππ§
2πππ’
ππ₯
πππ£
ππ₯+
ππ’
ππ¦
πππ€
ππ₯+
ππ’
ππ§
Equação da quantidade de movimento linear
ππ₯π¦ = πππ£
ππ₯+
ππ’
ππ¦
ππ₯π§ = πππ€
ππ₯+
ππ’
ππ§
ππ¦π§ = πππ€
ππ¦+
ππ£
ππ§
ππ₯π₯ = 2πππ’
ππ₯
ππ¦π¦ = 2πππ£
ππ¦
ππ§π§ = 2πππ€
ππ§
πππ₯ βππ
ππ₯+
πππ₯π₯
ππ₯+
πππ¦π₯
ππ¦+
πππ§π₯
ππ§= π
ππ’
ππ‘+ π’
ππ’
ππ₯+ π£
ππ’
ππ¦+ π€
ππ’
ππ§
πππ¦ βππ
ππ¦+
πππ₯π¦
ππ₯+
πππ¦π¦
ππ¦+
πππ§π¦
ππ§= π
ππ£
ππ‘+ π’
ππ£
ππ₯+ π£
ππ£
ππ¦+ π€
ππ£
ππ§
πππ§ βππ
ππ§+
πππ₯π§
ππ₯+
πππ¦π§
ππ¦+
πππ§π§
ππ§= π
ππ€
ππ‘+ π’
ππ€
ππ₯+ π£
ππ€
ππ¦+ π€
ππ€
ππ§
πππ₯π₯
ππ₯+
πππ¦π₯
ππ¦+
πππ§π₯
ππ§=
π
ππ₯2π
ππ’
ππ₯+
π
ππ¦π
ππ£
ππ₯+
ππ’
ππ¦+
π
ππ§π
ππ€
ππ₯+
ππ’
ππ§
= ππ2π’
ππ₯2+
π2π’
ππ¦2+
π2π’
ππ§2+
π2π’
ππ₯2+
π2π£
ππ¦ππ₯+
π2π€
ππ§ππ₯
π
ππ₯
ππ’
ππ₯+
ππ£
ππ¦+
ππ€
ππ§
β’ Considerando fluido incompressΓvel: constanteπ» β π
β π» β π = 0
β’ Considerando fluido newtoniano: constante
= ππ2π’
ππ₯2+
π2π’
ππ¦2+
π2π’
ππ§2
Equação da quantidade de movimento linear
ππ₯π¦ = πππ£
ππ₯+
ππ’
ππ¦
ππ₯π§ = πππ€
ππ₯+
ππ’
ππ§
ππ¦π§ = πππ€
ππ¦+
ππ£
ππ§
ππ₯π₯ = 2πππ’
ππ₯
ππ¦π¦ = 2πππ£
ππ¦
ππ§π§ = 2πππ€
ππ§
πππ₯ βππ
ππ₯+
πππ₯π₯
ππ₯+
πππ¦π₯
ππ¦+
πππ§π₯
ππ§= π
ππ’
ππ‘+ π’
ππ’
ππ₯+ π£
ππ’
ππ¦+ π€
ππ’
ππ§
πππ¦ βππ
ππ¦+
πππ₯π¦
ππ₯+
πππ¦π¦
ππ¦+
πππ§π¦
ππ§= π
ππ£
ππ‘+ π’
ππ£
ππ₯+ π£
ππ£
ππ¦+ π€
ππ£
ππ§
πππ§ βππ
ππ§+
πππ₯π§
ππ₯+
πππ¦π§
ππ¦+
πππ§π§
ππ§= π
ππ€
ππ‘+ π’
ππ€
ππ₯+ π£
ππ€
ππ¦+ π€
ππ€
ππ§
πππ₯π₯
ππ₯+
πππ¦π₯
ππ¦+
πππ§π₯
ππ§
β’ Considerando fluido incompressΓvel: constante β π» β π = 0
= ππ2π’
ππ₯2+
π2π’
ππ¦2+
π2π’
ππ§2
β’ Considerando fluido newtoniano: constante
πππ₯π¦
ππ₯+
πππ¦π¦
ππ¦+
πππ§π¦
ππ§
πππ₯π§
ππ₯+
πππ¦π§
ππ¦+
πππ§π§
ππ§
= ππ2π£
ππ₯2+
π2π£
ππ¦2+
π2π£
ππ§2
= ππ2π€
ππ₯2+
π2π€
ππ¦2+
π2π€
ππ§2
= ππ2π’
ππ₯2+
π2π’
ππ¦2+
π2π’
ππ§2
= ππ2π£
ππ₯2+
π2π£
ππ¦2+
π2π£
ππ§2
= ππ2π€
ππ₯2+
π2π€
ππ¦2+
π2π€
ππ§2
Equação da quantidade de movimento linear
ππ₯π¦ = πππ£
ππ₯+
ππ’
ππ¦
ππ₯π§ = πππ€
ππ₯+
ππ’
ππ§
ππ¦π§ = πππ€
ππ¦+
ππ£
ππ§
ππ₯π₯ = 2πππ’
ππ₯
ππ¦π¦ = 2πππ£
ππ¦
ππ§π§ = 2πππ€
ππ§
πππ₯ βππ
ππ₯+
πππ₯π₯
ππ₯+
πππ¦π₯
ππ¦+
πππ§π₯
ππ§= π
ππ’
ππ‘+ π’
ππ’
ππ₯+ π£
ππ’
ππ¦+ π€
ππ’
ππ§
πππ¦ βππ
ππ¦+
πππ₯π¦
ππ₯+
πππ¦π¦
ππ¦+
πππ§π¦
ππ§= π
ππ£
ππ‘+ π’
ππ£
ππ₯+ π£
ππ£
ππ¦+ π€
ππ£
ππ§
πππ§ βππ
ππ§+
πππ₯π§
ππ₯+
πππ¦π§
ππ¦+
πππ§π§
ππ§= π
ππ€
ππ‘+ π’
ππ€
ππ₯+ π£
ππ€
ππ¦+ π€
ππ€
ππ§
πππ₯π₯
ππ₯+
πππ¦π₯
ππ¦+
πππ§π₯
ππ§
β’ Considerando fluido incompressΓvel: constante β π» β π = 0
β’ Considerando fluido newtoniano: constante
πππ₯π¦
ππ₯+
πππ¦π¦
ππ¦+
πππ§π¦
ππ§
πππ₯π§
ππ₯+
πππ¦π§
ππ¦+
πππ§π§
ππ§
ππ2π’
ππ₯2+
π2π’
ππ¦2+
π2π’
ππ§2
ππ2π£
ππ₯2+
π2π£
ππ¦2+
π2π£
ππ§2
ππ2π€
ππ₯2+
π2π€
ππ¦2+
π2π€
ππ§2
Equação de Navier-Stokes
πππ₯ βππ
ππ₯+ π
π2π’
ππ₯2 +π2π’
ππ¦2 +π2π’
ππ§2 = πππ’
ππ‘+ π’
ππ’
ππ₯+ π£
ππ’
ππ¦+ π€
ππ’
ππ§
πππ¦ βππ
ππ¦+ π
π2π£
ππ₯2 +π2π£
ππ¦2 +π2π£
ππ§2 = πππ£
ππ‘+ π’
ππ£
ππ₯+ π£
ππ£
ππ¦+ π€
ππ£
ππ§
πππ§ βππ
ππ§+ π
π2π€
ππ₯2 +π2π€
ππ¦2 +π2π€
ππ§2 = πππ€
ππ‘+ π’
ππ€
ππ₯+ π£
ππ€
ππ¦+ π€
ππ€
ππ§
β’ e incompressΓvel: constante
β’ para fluido newtoniano: constante
π π β π»π + ππ»2π = πππ
ππ‘
IncΓ³gnitas:
π, π’, π£ π π€
π»2 =
π2
ππ₯2 +π2
ππ¦2 +π2
ππ§2
1
π
π
πππ
π
ππ+
1
π2
π2
ππ2 +π2
ππ₯2
com a eq. da continuidade: sistema de 4 incógnitas e 4 equaçáes
Equação de Navier-Stokes
Exemplo 1:
Um fluido viscoso de massa especΓfica e
viscosidade dinΓ’mica constantes escorre devido a
gravidade entre duas placas distantes 2h uma da
outra, conforme figura abaixo. O fluxo estΓ‘ totalmente
desenvolvido, com uma ΓΊnica componente de
velocidade w = w(x). NΓ£o hΓ‘ gradientes de pressΓ£o
aplicados, somente a gravidade. Resolva a equação
de Navier-Stokes para o perfil de velocidade entre as
placas.
z
x
h h
Οgxββpβx
+ΞΌβ2u
βx2+β2u
βy2+β2u
βz2=Ο
βuβt+u
βuβx+v
βuβy
+wβuβz
Οgyββpβy
+ΞΌβ2v
βx2+β2v
βy2+β2v
βz2=Ο
βvβt+u
βvβx+v
βvβy
+wβvβz
Οgzββpβz+ΞΌ
β2w
βx2+β2w
βy2+β2w
βz2=Ο
βwβt
+uβwβx
+vβwβy
+wβwβz
Equação de Navier-Stokes
Exemplo 1:
Um fluido viscoso de massa especΓfica e
viscosidade dinΓ’mica constantes escorre devido a
gravidade entre duas placas distantes 2h uma da
outra, conforme figura abaixo. O fluxo estΓ‘ totalmente
desenvolvido, com uma ΓΊnica componente de
velocidade w = w(x). NΓ£o hΓ‘ gradientes de pressΓ£o
aplicados, somente a gravidade. Resolva a equação
de Navier-Stokes para o perfil de velocidade entre as
placas.
Οgzββpβz+ΞΌ
β2w
βx2+β2w
βy2+β2w
βz2=Ο
βwβt
+uβwβx
+vβwβy
+wβwβz
0 0 0 0 0 0 0
Οg + ΞΌβ2w
βx2= 0 β
β2w
βx2= β
ΟgΞΌ
= βk ββwβx
= βkx + C1 β w x = βk2x2 + C1x + C2
w βh = 0w +h = 0
z
x
h h
β w x =k2
h2 β x2 β w x =Οg2ΞΌ
h2 β x2β C1= 0
C2= βkh2 2
Equação de Navier-Stokes Exemplo 2: Para um escoamento laminar e permanente de um fluido
incompressΓvel e newtoniano de massa especΓfica e viscosidade , no interior de uma tubulação horizontal de seção circular, com diΓ’metro D e comprimento L:
a) considerando um gradiente de pressΓ£o constante e ππ
ππ₯=
βπ
πΏ= πΎ
βπ
πΏ, calcule o perfil
de distribuição de velocidades;
b) calcule a vazΓ£o volumΓ©trica;
c) calcule a velocidade mΓ©dia; e
d) expresse a perda de carga unitΓ‘ria ( βπ πΏ) em função dos demais parΓ’metros
xr
u(r)
Equação de Navier-Stokes Exemplo 2:
a) considerando um gradiente de pressΓ£o constante e ππ
ππ₯=
βπ
πΏ= πΎ
βπ
πΏ, calcule o perfil
de distribuição de velocidades;
π π β π»π + ππ»2π = ππ·π
π·π‘
β πππ₯ βππ
ππ₯+ π
1
π
π
πππ
ππ’
ππ+
1
π2
π2π’
ππ2 +π2π’
ππ₯2 = π π’ππ’
ππ₯+ π£
ππ£
ππ+ π€
ππ€
ππ
βπ
π
π
πππ
ππ’
ππ=
ππ
ππ₯=
βπ
πΏ= πΎ
βπ
πΏ
β π’ π =βπΟπ
4πΏππ 2 β π2
βπ
πππ
ππ’
ππ= π
πΎβπ
ππΏ
β 0
π π
πππ
ππ’
ππππ =
0
π
ππΎβπ
ππΏππ
β πππ’
ππ=
π2
2
πΎβπ
ππΏβ
π
π ππ’
ππππ =
π
π π
2
πΎβπ
ππΏππ
xr
u(r)
Equação de Navier-Stokes Exemplo 2:
a) considerando um gradiente de pressΓ£o constante e ππ
ππ₯=
βπ
πΏ= πΎ
βπ
πΏ, calcule o perfil
de distribuição de velocidades;
b) calcule a vazΓ£o volumΓ©trica;
c) calcule a velocidade mΓ©dia; e
π’ π =βπΟπ
4πΏππ 2 β π2
xr
u(r)
π = π΄
πππ ππ΄ = π΄
π’ ππ΄ = 0
π βπΟπ
4πΏππ 2 β π2 2πππ π =
πβπΟπ
2πΏπ 0
π
π 2 β π2 π ππ
=πβπΟπ
2πΏπ π 2
π2
2β
π4
40
π
=πβπΟπ
2πΏππ 2
π 2
2β
π 4
4=
πβπΟπ
2πΏπ
π 4
4β π =
πβπΟππ·4
128 πΏπ
π = ππ π΄ β ππ =π
π΄=
4π
ππ·2
4
=
πβππππ·4
128 πΏπ
ππ·2
4
β ππ =βπΟππ·2
32 πΏπ
Equação de Navier-Stokes Exemplo 2:
a) considerando um gradiente de pressΓ£o constante e ππ
ππ₯=
βπ
πΏ= πΎ
βπ
πΏ, calcule o perfil
de distribuição de velocidades;
b) calcule a vazΓ£o volumΓ©trica;
c) calcule a velocidade mΓ©dia; e
d) expresse a perda de carga unitΓ‘ria ( βπ πΏ) em função dos demais parΓ’metros
π’ π =βπΟπ
4πΏππ 2 β π2
xr
u(r)
π =πβπΟππ·4
128 πΏπ
ππ =βπΟππ·2
32 πΏπβ
βπ
πΏ=
32 πΏπππ
Οππ·2
Equação da continuidade:
Equação de Euler: (escoamento invΓscido)
Equação de Navier-Stokes: (fluido newtoniano e
incompressΓvel)
ππ
ππ‘+ π» β ππ = 0
π π β π»π = πππ
ππ‘
π π β π»π + ππ»2π = πππ
ππ‘
Aula 4 β Equaçáes Diferenciais
Equação da continuidade
CinemΓ‘tica
Equação da quantidade de movimento linear
Equação de Euler
Equação de Navier-Stokes
BIBLIOGRAFIA:
WHITE, Frank. M. MecΓ’nica dos Fluidos. 6Βͺ ed. McGraw-
Hill, 2010.
WHITE, Frank. M. Viscous Fluid Flow. 3Βͺ ed. MacGraw-
Hill, 2006.
FOX Robert W.; MCDONALD Alan T. Introdução Γ
MecΓ’nica dos FluΓdos. 8Βͺ ed. John Wiley and Sons, N.Y.,
Tradução: LTC, 2014.
www.HidroUff.uff.br
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