Capitolo 3
MODELLAZIONE ACUSTICA
Scopo principale di questo capitolo è quello di dedurre un’equazione d’onda, simile alle equazioni delle onde acustiche, che incorpori tutte le caratteristiche acustiche del fenomeno della combustione.
3.1 – Equazioni di base per l’instabilità acustica di
combustione dei gas
Stabilito come la fisica del flusso governi il processo, in questo capitolo se ne traccerà il modello deducendo un’equazione d’onda simile all’equazione delle onde acustiche che incorpori tutte le caratteristiche del fenomeno in questione.
Si partirà dalle consuete equazioni di bilancio della quantità di moto, della massa e dell’energia riferite ad un volume infinitesimo di fluido presente in camera.
(equazione di continuità della massa)
( )
w t +∇⋅ = ∂ ∂ u ρ ρ r (3.1) 94dove:
ρ= densità media della miscela aria-metano
u= vettore velocità
w=sorgente di massa distribuita nel fluido (equazione della quantità di moto)
ρ u +
(
ρu⋅∇)
u=−∇ +∇⋅τ +m +ρg ∂ ∂ e v p t r r r (3.2) dove: p = pressione em = quantità di moto immessa nel fluido da una distribuzione di sorgenti
v
τ =tensore delle tensioni g = azioni di massa (equazione dell’energia)
( )
(
e)
(
)
Q t e v⋅ −∇⋅ + ⋅ ∇ = ⋅ ∇ + ∂ ∂ q u τ u r r r 0 0 ρ ρ (3.3) dove:Q = calore prodotto dalla combustione
q = calore trasmesso
0
e = energia interna totale definita come:
= + u⋅u 2 1
0 c T
e v
Dove c è il calore specifico a volume v costante medio della miscela:
1( ) 4 4 vCH CH aria c aria v c c c ρ ρ ρ + =
In primo luogo si ricava una forma dell’equazione dell’energia (3.3) che riporti a primo membro la derivata totale della pressione. A tal fine si inserisce l’espressione di nella (3.3), ottenendo così:
0 e
( )
( )
( )
( )
( )
(
)
(3.6) 0 0 0 0 0 e e e Q t e t e v ⋅ −∇⋅ + ⋅ ∇ = ⋅ ∇ ⋅ + ∇ ⋅ + ∇ ⋅ + ∂ ∂ + ∂ ∂ q u τ u u u r ρ ρ r ρ r r r ρ ρ Raccogliendo e : 0( )
( )
( )
( )
( )
(
)
(3.7) 0 0 0 e Q t e t e v ⋅ −∇⋅ + ⋅ ∇ = ∇ ⋅ + + ⋅∇ + ∇⋅ ∂ ∂ + ∂ ∂ q u τ u u u r ρ ρr ρ r r r ρ ρRicordando che: ∇r ⋅
( )
ρu =u⋅∇r( )
ρ +ρ∇r ⋅( )
u , si ha:( )
( )
( )
( )
(
)
(1.7) 0 0 0 e Q t e t e v⋅ −∇⋅ + ⋅ ∇ = ∇ ⋅ + +∇⋅ ∂ ∂ + ∂ ∂ q u τ u u r r r r ρ ρ ρ ρRicordando l'equazione di continuità (1.1), si nota subito come il termine tra parentesi non sia altro che l’espressione a destra dell’equazione di continuità stessa; ne segue quindi che:
( )
( )
(
)
(3.8) 0 0 0 e w e Q t e v ⋅ −∇⋅ + ⋅ ∇ = ∇ ⋅ + + ∂ ∂ q u τ u r r r ρ ρSostituendo ora la l’espressione dell’energia interna, si ottiene:
(
)
(3.9) 2 1 2 1 2 1 Q T c w T c t T c v v v v + ⋅ ∇ − ⋅ ⋅ ∇ = + ⋅ ∇ ⋅ + + ⋅ + ∂ + ⋅ ∂ q u τ u u u u u u u r r r ρ ρCon l'ipotesi che non dipenda dal tempo: cv
( )
(
)
(3.10) 2 1 2 2 1 Q T c w T wc t t T c v v v v + ⋅ ∇ − ⋅ ⋅ ∇ = ⋅ ∇ ⋅ + ∇ ⋅ + ⋅ + + ∂ ⋅ ∂ + ∂ ∂ q u τ u u u u u u u u r r r r ρ ρ ρ ρRaccogliendo il primo e il quinto termine della (1.10) si ottiene:
( )
(
)
(3.11) 2 1 2 2 1 Q w T wc t T t T c v v v + ⋅ ∇ − ⋅ ⋅ ∇ = ⋅ ∇ ⋅ + ⋅ + + ∂ ⋅ ∂ + + ⋅∇ ∂ ∂ q u τ u u u u u u u u r r r r ρ ρ ρChe è pari, in forma compatta a:
( )
(
)
(3.12) 2 1 2 2 1 Q w T wc t Dt T D c v v v + ⋅ ∇ − ⋅ ⋅ ∇ = ⋅ ∇ ⋅ + ⋅ + + ∂ ⋅ ∂ + q u τ u u u u u u u r r r ρ ρ ρIl sistema di equazioni di partenza è quindi ora pari a:
( )
(
)
( )
(
)
+ ⋅ ∇ − ⋅ ⋅ ∇ = ⋅ ∇ ⋅ + ⋅ + + ∂ ⋅ ∂ + + + ⋅ ∇ + ∇ − = ∇ ⋅ + ∂ ∂ = ⋅ ∇ + ∂ ∂ (3.12) 2 1 2 2 1 2 3 1 3 Q w T wc t Dt T D c ) . ( p t ) . ( w t v v v e v q u τ u u u u u u u g m τ u u u u r r r r r r r ρ ρ ρ ρ ρ ρ ρ ρSi può ora ipotizzare che la miscela si comporti come un gas perfetto, segua cioè la legge:
p=ρRT (3.13)
dove:
p = pressione
R = costante della miscela = pm Ru
Ru=costante universale del gas 8.314 [j/kmol K°] pm = peso molare della miscela
Differenziando l’espressione (3.13) si deriva la relazione tra la derivata totale della pressione e quella della temperatura e della densità ( sempre nell’ipotesi che R sia costante), a partire dalla vecchia formulazione dell’equazione di conservazione dell’energia (3.3):
( )
(
)
(
)
Q e t e v⋅ −∇⋅ + ⋅ ∇ = ⋅ ∇ + ∂ ∂ q u τ u r r r 0 0 ρ ρ (3.3)( )
( )
e( )
( )
e e( )
(
)
Q t e t e v⋅ −∇⋅ + ⋅ ∇ = ⋅ ∇ + ∇ ⋅ + ∇ ⋅ + ∂ ∂ + ∂ ∂ q u τ u u u r r 0 0r r r 0 0 0 ρ ρ ρ ρ ρ( )
( )
( )
e( )
e( )
(
)
Q t e e t e v ⋅ −∇⋅ + ⋅ ∇ = ⋅ ∇ + ∇ ⋅ + ∂ ∂ + + ⋅∇ ∂ ∂ q u τ u u u r 0 0 0 r 0r r r 0 ρ ρ ρ ρ( )
( )
( )
( )
( )
(
)
Q t e e t e v⋅ −∇⋅ + ⋅ ∇ = + ⋅∇ + ∇⋅ ∂ ∂ + + ⋅∇ ∂ ∂ q u τ u u u r ρ r ρ ρr r r ρ 0 0 0( )
( )
( )
( )
(
)
Q t e e t e v ⋅ −∇⋅ + ⋅ ∇ = +∇⋅ ∂ ∂ + + ⋅∇ ∂ ∂ q u τ u u r ρ r ρ r r ρ 0 0 0( )
(
)
Q w e Dt e D v⋅ −∇⋅ + ⋅ ∇ = + 0 r τ u r q 0 ρ( )
(
)
w e Q Dt e D v 0 0 =∇r ⋅ τ ⋅u −∇r ⋅q+ − ρSostituendo ora l’espressione di nella derivata totale, si ottiene: e0
(
)
Q e w Dt T c D v v 0 2 1 − + ⋅ ∇ − ⋅ ⋅ ∇ = + ⋅ q u τ u u r r ρ(
)
Q e w Dt D Dt DT cv 2 v 0 1 − + ⋅ ∇ − ⋅ ⋅ ∇ = ⋅ + τ u q u u r r ρ ρ(
)
Q e w Dt D Dt DT cv +ρu⋅ u =∇r ⋅ τv⋅u −∇r ⋅q+ − 0 ρ 98Considerando ora l’equazione della quantità di moto (3.2), ripetuta di seguito:
(
)
p ( . ) t u u τv me g 32 u ρ ρ ρ + ⋅∇ =−∇ +∇⋅ + + ∂ ∂ r r r si ottiene che:(
p)
(
)
Q e w Dt DT cv +u⋅ −∇r +∇r ⋅τv +me +ρg =∇r ⋅ τv⋅u −∇r ⋅q+ − 0 ρIsolando quindi la derivata totale della temperatura a primo membro:
(
p)
(
)
Q e w Dt DT cv =−u⋅ −∇r +∇r ⋅τv +me +ρg +∇r ⋅ τv ⋅u −∇r ⋅q+ − 0 ρ(
)
(
)
Q e w p Dt DT cv =u⋅∇r −u⋅ ∇r ⋅τv +me +ρg +∇r ⋅ τv⋅u −∇r ⋅q+ − 0 ρApplicando quindi l’operatore gradiente all’equazione dei gas perfetti, si ottiene:
(
ρ)
= ρ ∇( )
+ ∇( )
ρ∇ =
∇rp r RT Rr T RTr
Sostituendo quindi l’espressione ottenuta nell’equazione precedente, si ha che:
( )
( )
(
R T RT) (
)
(
)
Q e w Dt DT cv =u⋅ ρ ∇r + ∇r ρ −u⋅ ∇r ⋅τv +me +ρg +∇r ⋅ τv ⋅u −∇r ⋅q+ − 0 ρ( )
T RT( )
(
)
(
)
Q e w R Dt DT cv = u⋅∇ + u⋅∇ −u⋅ ∇⋅τv +me + g +∇⋅ τv ⋅u −∇⋅q+ − 0 r r r r r ρ ρ ρ ρDall’equazione di conservazione di massa (3.1), si può ricavare l’espressione di u⋅∇r
( )
ρ :( )
w t +∇⋅ = ∂ ∂ u ρ ρ r (3.1)( )
( )
w t + ⋅∇ + ∇⋅ = ∂ ∂ u u r ρ ρr ρ( )
( )
t w ∂ ∂ − ⋅ ∇ − = ∇ ⋅ ρ ρ u ρ u r rSostituendolo nell’equazione precedente, si ottiene:
( )
( )
(
)
(
)
Q e w t w RT T R Dt DT c v e v v 0 − + ⋅ ∇ − ⋅ ⋅ ∇ + + + ⋅ ∇ ⋅ − ∂ ∂ − ⋅ ∇ − + ∇ ⋅ = q u τ g m τ u u u r r r r r ρ ρ ρ ρ ρ Quindi:( )
( )
(
)
(
)
Q e w t RT RTw T R RT Dt DT c v e v v 0 − + ⋅ ∇ − ⋅ ⋅ ∇ + + + ⋅ ∇ ⋅ − ∂ ∂ − + ∇ ⋅ + ⋅ ∇ − = q u τ g m τ u u u r r r r r ρ ρ ρ ρ ρRicordando che p=ρRT, si ha che:
( )
( )
(
)
(
)
Q e w t RT RTw T R p Dt DT c v e v v 0 − + ⋅ ∇ − ⋅ ⋅ ∇ + + + ⋅ ∇ ⋅ − ∂ ∂ − + ∇ ⋅ + ⋅ ∇ − = q u τ g m τ u u u r r r r r ρ ρ ρ ρDato che, la derivata totale della pressione è per definizione pari a:
p t p Dt Dp + ⋅∇ ∂ ∂ = u r
sostituendo a tale definizione l’espressione della pressione, si ottiene:
(
)
(
)
RT t RT Dt Dp ρ ρ ∇ ⋅ + ∂ ∂ = u r( )
( )
(
RT)
t RT t T R Dt Dp ρ ρ + ⋅∇ ρ ∂ ∂ + ∂ ∂ = u r( )
( )
ρ ρ( )
( )
ρ ρ + ⋅∇ + ⋅∇ ∂ ∂ + ∂ ∂ = Ru r T RTu r t RT t T R Dt Dp( )
( )
( )
R( )
T t RT t T R Dt Dp + ⋅∇ + ⋅∇ ∂ ∂ + ∂ ∂ =ρ ρ u r ρ ρ u rricordando l’espressione del bilancio della massa si ha:
( )
w t +∇⋅ = ∂ ∂ u ρ ρ r( )
( )
w t + ⋅∇ + ∇⋅ = ∂ ∂ u u r ρ ρr ρ( )
( )
u u⋅∇ = − ∇⋅ + ∂ ∂ρ r ρ ρr w tsi può sostituire il termine tra parentesi ottenendo:
( )
RT(
w( )
)
R( )
T t T R Dt Dp + − ∇⋅ + ⋅∇ ∂ ∂ =ρ ρr u ρ u r( )
( )
(
( )
)
u u + − ∇⋅ + ⋅∇ ∂ ∂ =ρ r T RT w ρr t T R Dt Dp( )
+(
− ∇⋅( )
u)
=ρ RT w ρr Dt T D R Dt Dp( )
+(
− ∇⋅( )
u)
= ρ RT w ρr Dt T D c c R Dt Dp v v( )
+(
− ∇⋅( )
u)
= ρ RT w ρr Dt T D c c R Dt Dp v vSi può ora sostituire l’espressione del bilancio energetico precedentemente trovata nell’equazione ottenendo:
{
( )
( )
(
)
(
τ u)
q}
(
( )
u)
g m τ u u u ⋅ ∇ − + − + ⋅ ∇ − ⋅ ⋅ ∇ + + + ⋅ ∇ ⋅ − ∂ ∂ − + ∇ ⋅ + ⋅ ∇ − = r r r r r r ρ ρ ρ ρ w RT w e Q t RT RTw T R p c R Dt Dp v e v v 0( )
{
( )
(
)
(
)
Q e w}
RTw RT( )
( . ) t RT RTw T R c R c R p Dt Dp v e v v v 19 3 0 u q u τ g m τ u u u ⋅ ∇ − + − + ⋅ ∇ − ⋅ ⋅ ∇ + + + ⋅ ∇ ⋅ − ∂ ∂ − + ∇ ⋅ + ⋅ ∇ − = r r r r r r ρ ρ ρ ρ( )
( )
{
( )
(
)
(
)
Q e w}
( . ) t RT RTw T R c R RT RTw c R p Dt Dp v e v v v 20 3 0 − + ⋅ ∇ − ⋅ ⋅ ∇ + + + ⋅ ∇ ⋅ − ∂ ∂ − + ∇ ⋅ + ⋅ ∇ − + ⋅ ∇ − = q u τ g m τ u u u u r r r r r r ρ ρ ρ ρ Dato che R = c −c =γ −1 v v p v c c( )(
)
( ) (
){
( )
(
)
(
)
Q e w}
( . ) t RT RTw T R RT RTw p Dt Dp v e v 321 1 1 0 − + ⋅ ∇ − ⋅ ⋅ ∇ + + + ⋅ ∇ ⋅ − ∂ ∂ − + ∇ ⋅ − + ⋅ ∇ − + − ⋅ ∇ − = q u τ g m τ u u u u r r r r r r ρ ρ ρ γ ρ γ( )(
)
( ) (
){
( )
(
)
(
)
Q e w}
( . ) t RT RTw T R p RTw p Dt Dp v e v 322 1 1 0 − + ⋅ ∇ − ⋅ ⋅ ∇ + + + ⋅ ∇ ⋅ − ∂ ∂ − + ∇ ⋅ − + ⋅ ∇ − + − ⋅ ∇ − = q u τ g m τ u u u u r r r r r r ρ ρ ρ γ γ( )
( )
( ) (
){
( )
(
)
(
)
Q e w}
( . ) t RT RTw T R p RTw p p Dt Dp v e v 323 1 0 − + ⋅ ∇ − ⋅ ⋅ ∇ + + + ⋅ ∇ ⋅ − ∂ ∂ − + ∇ ⋅ − + ⋅ ∇ − + ⋅ ∇ + ⋅ ∇ − = q u τ g m τ u u u u u r r r r r r r ρ ρ ρ γ γ In definitiva si ha:( )
(
)(
) (
){
( )
(
)
(
)
Q e w}
( . ) t RT T R RTw RTw p Dt Dp v e v 324 1 1 0 − + ⋅ ∇ − ⋅ ⋅ ∇ + + + ⋅ ∇ ⋅ − ∂ ∂ − ∇ ⋅ − + − + + ⋅ ∇ − = q u τ g m τ u u u r r r r r ρ ρ ρ γ γ γ( )
(
){
( )
(
)
(
)
Q e w}
( . ) t RT T R RTw RTw RTw p Dt Dp v e v 325 1 0 − + ⋅ ∇ − ⋅ ⋅ ∇ + + + ⋅ ∇ ⋅ − ∂ ∂ − ∇ ⋅ − + − + + ⋅ ∇ − = q u τ g m τ u u u r r r r r ρ ρ ρ γ γ γ( )
(
){
( )
(
)
(
)
Q e w}
( . ) t RT T R RTw p Dt Dp v e v 326 1 0 − + ⋅ ∇ − ⋅ ⋅ ∇ + + + ⋅ ∇ ⋅ − ∂ ∂ − ∇ ⋅ − + + ⋅ ∇ − = q u τ g m τ u u u r r r r r ρ ρ ρ γ γ γ 102Il sistema di equazioni che verrà utilizzato in seguito assume quindi la seguente forma:
( )
(
)
( )
(
){
( )
(
)
(
)
}
= − + ⋅ ∇ − ⋅ ⋅ ∇ + + + ⋅ ∇ ⋅ − ∂ ∂ − ∇ ⋅ − + + ⋅ ∇ − = + + ⋅ ∇ + ∇ − = ∇ ⋅ + ∂ ∂ = ⋅ ∇ + ∂ ∂ ) . ( RT p ) . ( w e Q t RT T R RTw p Dt Dp ) . ( p t ) . ( w t v e v e v 13 3 26 3 1 2 3 1 3 0 ρ ρ ρ ρ γ γ γ ρ ρ ρ ρ ρ q u τ g m τ u u u g m τ u u u u r r r r r r r r r3.2 – Sviluppo dell’equazione d’onda acustica
A partire dal sistema di equazioni precedentemente ricavato, ed in particolare dall’equazione dell’energia, è possibile ottenere un’equazione d’onda simile a quella delle onde acustiche che in più ha un termine forzante a membro destro.
Al fine di derivarne l’espressione è opportuno considerare tutte le grandezze e le variabili che compaiono nelle equazioni di conservazione e svilupparle in due termini:
- componente media della grandezza (funzione solo dello spazio)
- componente fluttuante della grandezza (funzione dello spazio e del tempo) Si ottiene così: ) , ( ) (r u r t u u = + ′ T =T(r)+T′(r,t) me =me(r)+me′(r,t) ) , ( ) (r p r t p p = + ′ w=w(r)+w′(r,t) q=q(r)+q′(r,t) ) , ( ) (r ρ r t ρ ρ = + ′ Q=Q(r)+Q′(r,t) τv =τv(r)+τ′v(r,t)
Sviluppando ora le grandezze presenti nell’equazione di conservazione dell’energia nella forma (3.23), si ottiene:
(
)
(
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)
(
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(
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(
){ (
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(
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(
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(
Q(r) Q(r,t))
e(
w(r) w(r,t)) }
( . ) ) t , r ( ) r ( ) t , r ( ) r ( ) t , r ( ) r ( ) t , r ( ) r ( ) t , r ( ) r ( t ) t , r ( ) r ( ) t , r ( T ) r ( T R ) t , r ( T ) r ( T ) t , r ( ) r ( R ) t , r ( ) r ( ) t , r ( w ) r ( w ) t , r ( T ) r ( T R ) t , r ( ) r ( ) t , r ( p ) r ( p Dt ) t , r ( p ) r ( p D e e v v v 24 3 1 0 + ′ − ′ + + ′ + + ′ + + ′ + ⋅ ∇ ⋅ ′ + − ′ + ⋅ ∇ − ⋅ ⋅ ∇ + ∂ ′ + ∂ ′ + − ′ + ∇ ⋅ ′ + ′ + − + ′ + ′ + + ′ + ⋅ ∇ ′ + − = ′ + g m m τ τ u u q q u τ u u u u ρ ρ ρ ρ ρ ρ γ γ γ r r r r rSostituendo l’espressione dell’energia = + u⋅u
2 1
0 c T
e v , e sviluppandone le grandezze presenti nelle due componenti, si ottiene:
(
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2 1 ) , ( ) ( ) , ( ) ( ) , ( ) ( ) , ( ) ( ) , ( ) ( ) , ( ) ( ) , ( ) ( ) , ( ) ( ) , ( ) ( ) , ( ) ( ) , ( ) ( ) , ( ) ( 1 ) , ( ) ( ) , ( ) ( ) , ( ) ( ) , ( ) ( ) , ( ) ( t r w r w t r r t r r t r T r T c t r Q r Q t r r t r r t r r t r r t r r t t r r t r T r T R t r T r T t r r R t r r t r w r w t r T r T R t r r t r p r p Dt t r p r p D v e e v v v ′ + + ′ + + ′ ⋅ + ′ − ′ + + ′ + + ′ + + ′ + ⋅ ∇ ⋅ ′ + − ′ + ⋅ ∇ − ⋅ ⋅ ∇ + ∂ ′ + ∂ ′ + − ′ + ∇ ⋅ ′ + ′ + − + ′ + ′ + + ′ + ⋅ ∇ ′ + − = ′ + u u u u g m m τ τ u u q q u τ u u u u ρ ρ ρ ρ ρ ρ γ γ γ r r r r rRiordinando i termini, in modo da evidenziare i vari gruppi di componenti, si ha che:
104
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2 1 ) , ( ) ( ) , ( ) ( 1 ) , ( ) ( ) , ( ) ( ) , ( ) ( ) , ( ) ( 1 ) , ( ) ( ) , ( ) ( ) , ( ) ( ) , ( ) ( ) , ( ) ( ) , ( ) ( ) , ( ) ( ) , ( ) ( 1 ) , ( ) ( ) , ( ) ( ) , ( ) ( ) , ( ) ( ) , ( ) ( t r w r w t r r t r r t r T r T c t r Q r Q t r r t r r t r r t r r t r r t r r t r r t t r r t r T r T R t r T r T t r r R t r r t r w r w t r T r T R t r r t r p r p Dt t r p r p D v e e v v v v ′ + + ′ + + ′ ⋅ + ′ − ′ + − + ′ + + ′ + + ′ + ⋅ ∇ ⋅ ′ + − − ′ + ⋅ ∇ − ′ + ⋅ ′ + ⋅ ∇ + ∂ ′ + ∂ ′ + − ′ + ∇ ⋅ ′ + ′ + − + ′ + ′ + + ′ + ⋅ ∇ ′ + − = ′ + u u u u g m m τ τ u u q q u u τ τ u u u u γ ρ ρ γ ρ ρ ρ ρ γ γ γ r r r r r(
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2 1 ) , ( ) ( ) , ( ) ( ) ( ) , ( 2 1 ) , ( ) , ( ) ( 2 1 ) , ( ) ( ) ( ) ( ) ( ) ( 2 1 1 ) , ( ) , ( ) ( ) , ( ) , ( ) ( ) ( ) ( ) , ( ) ( 1 ) , ( ) , ( ) ( ) , ( ) , ( ) ( ) ( ) ( 1 ) , ( ) , ( ) ( ) , ( ) , ( ) ( ) ( ) ( 1 ) , ( ) , ( ) ( ) , ( ) , ( ) ( ) ( ) ( 1 ) , ( ) ( 1 ) , ( ) , ( ) ( ) , ( ) , ( ) ( ) ( ) ( 1 ) , ( ) , ( ) , ( ) ( 1 ) , ( ) , ( ) , ( ) ( ) , ( ) , ( ) , ( ) ( ) , ( ) ( ) ( ) , ( ) , ( ) , ( ) ( ) ( ) , ( ) ( ) , ( ) ( ) ( ) ( ) ( ) ( 1 ) , ( ) , ( ) ( ) , ( ) , ( ) ( ) ( ) ( 1 ) , ( ) , ( ) ( ) , ( ) , ( ) ( ) ( ) ( ) , ( ) , ( ) ( ) , ( ) , ( ) ( ) ( ) ( ) , ( t r t r t r w t r r t r w r r t r w t r t r r w t r r r w r r r w t r w t r T r w t r T t r w r T r w r T c t r Q r Q t r t r r t r t r r r r t r t r r t r t r r r r t r t r r t r t r r r r t r r t r t r r t r t r r r r t t r t r T t t r r T R t r T t r t r r T t r t r t r T r t r r T r t r t r T t r r r T t r r t r T r r r T r r R t r w t r T r w t r T t r w r T r w r T c t r t r p r t r p t r r p r r p t r p t r r p t r t r p r r p r t t r p v e e e e v v v v v v v v V u u u u u u u u u u u u g u g u g u g u m u m u m u m u τ u τ u τ u τ u q q u τ u τ u τ u τ u u u u u u u u u u u u u u u u ′ ⋅ ′ ′ + ′ ⋅ ′ + ⋅ ′ + ′ ⋅ ′ + ′ ⋅ + ⋅ − − ′ ′ + ′ + ′ + − ′ + − + ⋅ ′ ′ + ⋅ ′ + ⋅ ′ + ⋅ − + ′ ′ + ′ + ′ + − + ′ ⋅ ∇ ⋅ ′ + ⋅ ∇ ⋅ ′ + ′ ⋅ ∇ ⋅ + ⋅ ∇ ⋅ − − ′ ⋅ ∇ + ⋅ ∇ − − ′ ⋅ ′ ⋅ ∇ + ⋅ ′ ⋅ ∇ + ′ ⋅ ⋅ ∇ + ⋅ ⋅ ∇ − + ∂ ′ ∂ ′ + ∂ ′ ∂ − − ′ ∇ ⋅ ′ ′ + ∇ ⋅ ′ ′ + + ′ ∇ ⋅ ′ + ∇ ⋅ ′ + ′ ∇ ⋅ ′ + + ∇ ⋅ ′ + ′ ∇ ⋅ + ∇ ⋅ − + ′ ′ + ′ + ′ + − + ′ ⋅ ∇ ′ + ⋅ ∇ ′ + ′ ⋅ ∇ + ⋅ ∇ − = ′ ∇ ⋅ ′ + ∇ ⋅ ′ + ′ ∇ ⋅ + ∇ ⋅ + ∂ ′ ∂ γ γ ρ ρ ρ ρ γ γ γ γ γ ρ ρ γ ρ ρ ρ ρ ρ ρ ρ ρ γ γ γ r r r r r r r r r r r r r r r r r r r r r r}
r r r rConsiderando l’equazione per le condizioni medie, che si ottiene dalla precedente eliminando le fluttuazioni, si ha:
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τ u u u q u τ u u u ) ( ) ( ) ( ) ( 1 ) ( ) ( 1 ) ( ) ( ) ( 2 1 1 ) ( ) ( ) ( ) ( ) ( ) ( 1 ) ( ) ( ) ( 1 ) ( ) ( ) ( ) ( ) ( ) ( r r r r r r r r r w r w r T c r Q r r r r T r r R r w r T R r r p r p r e v v v ρ γ γ γ γ ρ γ γ γ + ⋅ − + ⋅ ∇ ⋅ − − ⋅ − − − + ⋅ ∇ − ⋅ ⋅ ∇ − + ∇ ⋅ − + + ⋅ ∇ − = ∇ ⋅ r r r r r rSottraendo quest’ultima alla precedente, si ottiene l’equazione dell’energia per le fluttuazioni di pressione:
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( . ) ) t , r ( ) t , r ( ) r ( w ) t , r ( ) r ( ) r ( w ) t , r ( w ) t , r ( T ) r ( w ) t , r ( T ) t , r ( w ) r ( T c ) t , r ( Q ) t , r ( ) t , r ( ) r ( ) t , r ( ) t , r ( ) r ( ) t , r ( ) t , r ( ) r ( ) t , r ( ) t , r ( ) r ( ) t , r ( ) t , r ( ) r ( ) t , r ( ) t , r ( ) r ( ) t , r ( ) t , r ( ) t , r ( ) r ( ) t , r ( ) t , r ( ) r ( t ) t , r ( ) t , r ( T t ) t , r ( ) r ( T R ) t , r ( T ) t , r ( ) t , r ( ) r ( T ) t , r ( ) t , r ( ) t , r ( T ) r ( ) t , r ( ) r ( T ) r ( ) t , r ( ) t , r ( T ) t , r ( ) r ( ) r ( T ) t , r ( ) r ( ) t , r ( T ) r ( ) r ( R ) t , r ( w ) t , r ( T ) r ( w ) t , r ( T ) t , r ( w ) r ( T c ) t , r ( ) t , r ( p ) r ( ) t , r ( p ) t , r ( ) r ( p ) t , r ( p ) t , r ( ) r ( p ) t , r ( ) t , r ( p ) r ( t ) t , r ( p v e e e v v v v v v V 26 3 2 1 2 1 2 1 1 1 1 1 1 1 1 1 1 1 u u u u u u u u u u g u g u g u m u m u m u τ u τ u τ u q u τ u τ u τ u u u u u u u u u u u u u ′ ⋅ ′ ′ + ′ ⋅ ′ + ⋅ ′ + ′ ⋅ ′ + ′ ⋅ − − ′ ′ + ′ + ′ − ′ − + ⋅ ′ ′ + ⋅ ′ + ⋅ ′ − + ′ ′ + ′ + ′ − + ′ ⋅ ∇ ⋅ ′ + ⋅ ∇ ⋅ ′ + ′ ⋅ ∇ ⋅ − − ′ ⋅ ∇ − − ′ ⋅ ′ ⋅ ∇ + ⋅ ′ ⋅ ∇ + ′ ⋅ ⋅ ∇ − + ∂ ′ ∂ ′ + ∂ ′ ∂ − − ′ ∇ ⋅ ′ ′ + ∇ ⋅ ′ ′ + + ′ ∇ ⋅ ′ + ∇ ⋅ ′ + ′ ∇ ⋅ ′ + + ∇ ⋅ ′ + ′ ∇ ⋅ − + ′ ′ + ′ + ′ − + ′ ⋅ ∇ ′ + ⋅ ∇ ′ + ′ ⋅ ∇ − = ′ ∇ ⋅ ′ + ∇ ⋅ ′ + ′ ∇ ⋅ + ∂ ′ ∂ γ γ ρ ρ ρ γ γ γ γ γ ρ ρ γ ρ ρ ρ ρ ρ ρ ρ γ γ γ r r r r r r r r r r r r r r r r r r r rConsiderando ora l’equazione di conservazione della quantità di moto (3.2) , e sviluppando le sue variabili nei termini di fluttuazione e media si ottiene:
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g m m τ τ u u u u u u ) , ( ) ( ) , ( ) ( ) , ( ) ( ) , ( ) ( ) , ( ) ( ) , ( ) ( ) , ( ) ( ) , ( ) ( ) , ( ) ( t r r t r r t r r t r p r p t r r t r r t r r t t r r t r r e e v v ρ ρ ρ ρ ρ ρ ′ + + ′ + + ′ + ⋅ ∇ + ′ + ∇ − = = ′ + ∇ ⋅ ′ + ′ + + ∂ ′ + ∂ ′ + r r r Effettuando i calcoli: g g m m τ τ u u u u u u u u u u u u u u u u u u ) , ( ) ( ) , ( ) ( ) , ( ) ( ) , ( ) ( ) , ( ) , ( ) , ( ) , ( ) ( ) , ( ) , ( ) , ( ) ( ) , ( ) ( ) ( ) ( ) , ( ) , ( ) ( ) ( ) , ( ) ( ) , ( ) ( ) ( ) ( ) ( ) , ( ) , ( ) , ( ) ( t r r t r r t r r t r p r p t r t r t r t r r t r t r t r r t r r r r t r t r r r t r r t r r r r r t t r t r t t r r e e v v ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ′ + + + ′ + + ′ ⋅ ∇ + ⋅ ∇ + ′ ∇ − ∇ − = ′ ∇ ⋅ ′ ′ + ′ ∇ ⋅ ′ + ′ ∇ ⋅ ′ + + ′ ∇ ⋅ + ∇ ⋅ ′ ′ + ∇ ⋅ ′ + + ∇ ⋅ ′ + ∇ ⋅ + ∂ ′ ∂ ′ + ∂ ′ ∂ r r r r r r r r r r r rConsiderando l’equazione per le condizioni medie della quantità di moto, che si ottiene dalla precedente eliminando le fluttuazioni, si ha:
g m τ u u( ) ( ) ( ) ( ) ( ) ( ) ) (r r r p r v r e r ρ r ρ ⋅∇r =−∇r +∇r ⋅ + +
Sottraendo alla penultima equazione scritta, l’ultima, si ottiene
) . ( ) t , r ( ) t , r ( ) t , r ( ) t , r ( p ) t , r ( ) t , r ( ) t , r ( ) t , r ( ) r ( ) t , r ( ) t , r ( ) t , r ( ) r ( ) t , r ( ) r ( ) r ( ) r ( ) t , r ( ) t , r ( ) r ( ) r ( ) t , r ( ) r ( ) t , r ( ) r ( t ) t , r ( ) t , r ( t ) t , r ( ) r ( e v m g 330 τ u u u u u u u u u u u u u u u u ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ′ + ′ + ′ ⋅ ∇ + ′ ∇ − = ′ ∇ ⋅ ′ ′ + ′ ∇ ⋅ ′ + ′ ∇ ⋅ ′ + ′ ∇ ⋅ + ∇ ⋅ ′ ′ + ∇ ⋅ ′ + ∇ ⋅ ′ + ∂ ′ ∂ ′ + ∂ ′ ∂ r r r r r r r r r
Complessivamente, il sistema di equazioni a cui ci si è riportati e da cui si ricaverà l’equazione d’onda è quello formato dall’equazione (3.26) ed (3.30)
3.2.1 – Equazione d’onda
Per derivare esplicitamente l’equazione d’onda si derivi rispetto al tempo l’equazione dell’energia (3.26):
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( . ) ) t , r ( ) t , r ( ) r ( w ) t , r ( ) r ( ) r ( w t ) t , r ( w ) t , r ( T ) r ( w ) t , r ( T ) t , r ( w ) r ( T c ) t , r ( Q t ) t , r ( ) t , r ( ) r ( ) t , r ( ) t , r ( ) r ( t ) t , r ( ) t , r ( ) r ( ) t , r ( ) t , r ( ) r ( t ) t , r ( ) t , r ( ) r ( ) t , r ( ) t , r ( ) r ( t ) t , r ( t ) t , r ( ) t , r ( ) r ( ) t , r ( ) t , r ( ) r ( t t ) t , r ( ) t , r ( T t ) t , r ( ) r ( T t R ) t , r ( T ) t , r ( ) t , r ( ) r ( T ) t , r ( ) t , r ( ) t , r ( T ) r ( ) t , r ( ) r ( T ) r ( ) t , r ( ) t , r ( T ) t , r ( ) r ( ) r ( T ) t , r ( ) r ( ) t , r ( T ) r ( ) r ( t R ) t , r ( w ) t , r ( T ) r ( w ) t , r ( T ) t , r ( w ) r ( T t c ) t , r ( ) t , r ( p ) r ( ) t , r ( p ) t , r ( ) r ( p t t ) t , r ( p ) t , r ( ) t , r ( p t ) t , r ( ) r ( p t ) t , r ( t ) t , r ( p ) r ( t ) t , r ( p v e e e v v v v v v V 31 3 2 1 2 1 2 1 1 1 1 1 1 1 1 1 1 1 2 2 u u u u u u u u u u g u g u g u m u m u m u τ u τ u τ u q u τ u τ u τ u u u u u u u u u u u u u u ′ ⋅ ′ ′ + ′ ⋅ ′ + ⋅ ′ + ′ ⋅ ′ + ′ ⋅ ∂ ∂ − − ′ ′ + ′ + ′ − ′ ∂ ∂ − + ⋅ ′ ′ + ⋅ ′ + ⋅ ′ ∂ ∂ − + ′ ′ + ′ + ′ ∂ ∂ − + ′ ⋅ ∇ ⋅ ′ + ⋅ ∇ ⋅ ′ + ′ ⋅ ∇ ⋅ ∂ ∂ − − ′ ⋅ ∇ ∂ ∂ − − ′ ⋅ ′ ⋅ ∇ + ⋅ ′ ⋅ ∇ + ′ ⋅ ⋅ ∇ ∂ ∂ − + ∂ ′ ∂ ′ + ∂ ′ ∂ ∂ ∂ − − ′ ∇ ⋅ ′ ′ + ∇ ⋅ ′ ′ + + ′ ∇ ⋅ ′ + ∇ ⋅ ′ + ′ ∇ ⋅ ′ + + ∇ ⋅ ′ + ′ ∇ ⋅ ∂ ∂ − + ′ ′ + ′ + ′ ∂ ∂ − + ′ ⋅ ∇ ′ + ⋅ ∇ ′ + ′ ⋅ ∇ ∂ ∂ − = ∂ ′ ∂ ∇ ⋅ ′ + ′ ∇ ⋅ ∂ ′ ∂ + + ∇ ⋅ ∂ ′ ∂ + ∂ ′ ∂ ∇ ⋅ + ∂ ′ ∂ γ γ ρ ρ ρ γ γ γ γ γ ρ ρ γ ρ ρ ρ ρ ρ ρ ρ γ γ γ r r r r r r r r r r r r r r r r r r r r rSi applichi la divergenza all’equazione (3.30), ottenendo così:
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( . ) ) t , r ( p ) t , r ( ) t , r ( ) t , r ( ) t , r ( ) r ( ) t , r ( ) t , r ( ) t , r ( ) r ( ) t , r ( ) r ( ) r ( ) r ( ) t , r ( ) t , r ( ) r ( ) r ( ) t , r ( ) r ( ) t , r ( ) r ( t ) t , r ( ) t , r ( ) r ( t ) t , r ( t ) t , r ( ) r ( e v 332 2 τ m g u u u u u u u u u u u u u u u u u ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ′ ⋅ ∇ + ′ ⋅ ∇ + ′ ⋅ ∇ ⋅ ∇ + ′ −∇ = ′ ∇ ⋅ ′ ′ ⋅ ∇ + ′ ∇ ⋅ ′ ⋅ ∇ + ′ ∇ ⋅ ′ ⋅ ∇ + ′ ∇ ⋅ ⋅ ∇ + ∇ ⋅ ′ ′ ⋅ ∇ + ∇ ⋅ ′ ⋅ ∇ + ∇ ⋅ ′ ⋅ ∇ + ∂ ′ ∂ ′ ⋅ ∇ + ∇ ⋅ ∂ ′ ∂ + ∂ ′ ∂ ⋅ ∇ r r r r r r r r r r r r r r r r r r r r r Riorganizzando i termini:(
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( . ) ) t , r ( ) t , r ( ) t , r ( ) t , r ( ) r ( ) t , r ( ) t , r ( ) t , r ( ) r ( ) t , r ( ) r ( ) r ( ) r ( ) t , r ( ) t , r ( ) r ( ) r ( ) t , r ( ) r ( ) t , r ( ) r ( t ) t , r ( ) t , r ( ) r ( t ) t , r ( t ) t , r ( ) r ( ) t , r ( p e v 333 2 g m τ u u u u u u u u u u u u u u u u u ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ′ ⋅ ∇ + ′ ⋅ ∇ + ′ ⋅ ∇ ⋅ ∇ + ′ ∇ ⋅ ′ ′ ⋅ ∇ − ′ ∇ ⋅ ′ ⋅ ∇ − ′ ∇ ⋅ ′ ⋅ ∇ − ′ ∇ ⋅ ⋅ ∇ − ∇ ⋅ ′ ′ ⋅ ∇ − ∇ ⋅ ′ ⋅ ∇ − ∇ ⋅ ′ ⋅ ∇ − ∂ ′ ∂ ′ ⋅ ∇ − ∇ ⋅ ∂ ′ ∂ − = ∂ ′ ∂ ⋅ ∇ + ′ ∇ r r r r r r r r r r r r r r r r r r r r rDividendo l’equazione (3.33) per ρ:
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( . ) ) t , r ( ) t , r ( ) t , r ( ) t , r ( ) r ( ) t , r ( ) t , r ( ) t , r ( ) r ( ) t , r ( ) r ( ) r ( ) r ( ) t , r ( ) t , r ( ) r ( ) r ( ) t , r ( ) r ( ) t , r ( ) r ( t ) t , r ( ) t , r ( ) r ( t ) t , r ( ) t , r ( p t ) t , r ( e v 335 1 1 1 1 1 1 1 1 1 1 1 1 1 2 g m τ u u u u u u u u u u u u u u u u u ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ′ ⋅ ∇ + ′ ⋅ ∇ + ′ ⋅ ∇ ⋅ ∇ + ′ ∇ ⋅ ′ ′ ⋅ ∇ − ′ ∇ ⋅ ′ ⋅ ∇ − ′ ∇ ⋅ ′ ⋅ ∇ − ′ ∇ ⋅ ⋅ ∇ − ∇ ⋅ ′ ′ ⋅ ∇ − ∇ ⋅ ′ ⋅ ∇ − ∇ ⋅ ′ ⋅ ∇ − ∂ ′ ∂ ′ ⋅ ∇ − ∇ ⋅ ∂ ′ ∂ − + ′ ∇ − = ∂ ′ ∂ ⋅ ∇ r r r r r r r r r r r r r r r r r r r r rSi riorganizzi la (3.31) in questo modo:
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w(r,t) (r) (r,t)(
w(r,t) (r,t) (r,t)) }
( . ) ) t , r ( ) t , r ( ) r ( w ) t , r ( ) r ( ) r ( w t p ) t , r ( w ) t , r ( T ) r ( w ) t , r ( T ) t , r ( w ) r ( T c ) t , r ( Q t p ) t , r ( ) t , r ( ) r ( ) t , r ( ) t , r ( ) r ( t p ) t , r ( ) t , r ( ) r ( ) t , r ( ) t , r ( ) r ( t p ) t , r ( ) t , r ( ) r ( ) t , r ( ) t , r ( ) r ( t p ) t , r ( t p ) t , r ( ) t , r ( ) r ( ) t , r ( ) t , r ( ) r ( t p t ) t , r ( ) t , r ( T t ) t , r ( ) r ( T t R p ) t , r ( T ) t , r ( ) t , r ( ) r ( T ) t , r ( ) t , r ( ) t , r ( T ) r ( ) t , r ( ) r ( T ) r ( ) t , r ( ) t , r ( T ) t , r ( ) r ( ) r ( T ) t , r ( ) r ( ) t , r ( T ) r ( ) r ( t R p ) t , r ( w ) t , r ( T ) r ( w ) t , r ( T ) t , r ( w ) r ( T t R p t ) t , r ( ) t , r ( p ) t , r ( t ) t , r ( p ) r ( t ) t , r ( p p t ) t , r ( p ) t , r ( ) t , r ( p t ) t , r ( ) r ( p t ) t , r ( t ) t , r ( p ) r ( p t ) t , r ( t ) t , r ( p p v e e e v v v v v v 37 3 2 1 2 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 u u u u u u u u u u g u g u g u m u m u m u τ u τ u τ u q u τ u τ u τ u u u u u u u u u u u u u u u ′ ⋅ ′ ′ + ′ ⋅ ′ + ⋅ ′ + ′ ⋅ ′ + ′ ⋅ ∂ ∂ − − ′ ′ + ′ + ′ − ′ ∂ ∂ − + ⋅ ′ ′ + ⋅ ′ + ⋅ ′ ∂ ∂ − + ′ ′ + ′ + ′ ∂ ∂ − + ′ ⋅ ∇ ⋅ ′ + ⋅ ∇ ⋅ ′ + ′ ⋅ ∇ ⋅ ∂ ∂ − − ′ ⋅ ∇ ∂ ∂ − − ′ ⋅ ′ ⋅ ∇ + ⋅ ′ ⋅ ∇ + ′ ⋅ ⋅ ∇ ∂ ∂ − + ∂ ′ ∂ ′ + ∂ ′ ∂ ∂ ∂ − − ′ ∇ ⋅ ′ ′ + ∇ ⋅ ′ ′ + + ′ ∇ ⋅ ′ + ∇ ⋅ ′ + ′ ∇ ⋅ ′ + ∇ ⋅ ′ + ′ ∇ ⋅ ∂ ∂ − + ′ + ′ + ′ ′ ∂ ∂ + ∂ ′ ∂ ⋅ ∇ ′ − ′ ⋅ ∇ ∂ ′ ∂ − ⋅ ∇ ∂ ′ ∂ − + + ∂ ′ ∂ ∇ ⋅ ′ + ′ ∇ ⋅ ∂ ′ ∂ + ∇ ⋅ ∂ ′ ∂ + ∂ ′ ∂ ∇ ⋅ − = ∂ ′ ∂ ⋅ ∇ + ∂ ′ ∂ γ γ γ γ ρ ρ ρ γ γ γ γ γ γ γ γ γ γ ρ ρ γ γ ρ ρ ρ ρ ρ ρ ρ γ γ γ γ γ γ γ γ γ γ r r r r r r r r r r r r r r r r r r r r r rRiorganizzando ancora l’equazione (3.37) si ottiene:
