SEMPRE LA STESSA EQUAZIONE…
scarica del condensatore
La scarica di un condensatore è governata dall’equazione
differenziale:
Q=−RC
dQ
dt
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyuaiabg2da9iabgkHiTiaadkfacaWGdbWaaSaaaeaacaqGKbGaamyuaaqaaiaabsgacaWG0baaaaaa@3E01@
ovvero, considerando che
Q=CV
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyuaiabg2da9iaadoeacaWGwbaaaa@396B@
:
V=−RC
dV
dt
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2da9iabgkHiTiaadkfacaWGdbWaaSaaaeaacaqGKbGaamOvaaqaaiaabsgacaWG0baaaaaa@3E0B@
.
(Alla prima equazione si arriva subito, partendo dalla legge
di Ohm
V=Ri
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2da9iaadkfacaWGPbaaaa@3992@
,
tenendo conto del fatto che
V=Q/C
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2da9maalyaabaGaamyuaaqaaiaadoeaaaaaaa@3981@
e ponendo
i=−
dQ
/
dt
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyAaiabg2da9iabgkHiTmaalyaabaGaaeizaiaadgfaaeaacaqGKbGaamiDaaaaaaa@3C80@
.
Il segno negativo dipende dal fatto che la carica sul condensatore diminuisce
in corrispondenza di una corrente che vogliamo assumere come positiva).
Le equazioni scritte si
integrano facilmente per separazione di variabili. Si ottiene:
−
dt
RC
=
dV
V
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0YaaSaaaeaacaqGKbGaamiDaaqaaiaadkfacaWGdbaaaiabg2da9maalaaabaGaaeizaiaadAfaaeaacaWGwbaaaaaa@3E1B@
−
t
RC
=lnV+lnk
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0YaaSaaaeaacaWG0baabaGaamOuaiaadoeaaaGaeyypa0JaciiBaiaac6gacaWGwbGaey4kaSIaciiBaiaac6gacaWGRbaaaa@40FC@
kV=
e
−t
RC
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4AaiaadAfacqGH9aqpcaWGLbWaaWbaaSqabeaadaWcaaqaaiabgkHiTiaadshaaeaacaWGsbGaam4qaaaaaaaaaa@3D69@
o, anche:
V=c
e
−t
RC
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2da9iaadogacaWGLbWaaWbaaSqabeaadaWcaaqaaiabgkHiTiaadshaaeaacaWGsbGaam4qaaaaaaaaaa@3D61@
.
La relazione ora trovata è
l’integrale generale dell’equazione (2).
Se si considera la condizione iniziale
V(0)=
V
0
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiaacIcacaaIWaGaaiykaiabg2da9iaadAfadaWgaaWcbaGaaGimaaqabaaaaa@3BA1@
,
si ottiene:
V=
V
0
e
−t
RC
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2da9iaadAfadaWgaaWcbaGaaGimaaqabaGccaWGLbWaaWbaaSqabeaadaWcaaqaaiabgkHiTiaadshaaeaacaWGsbGaam4qaaaaaaaaaa@3E44@
.
Analogamente si può scrivere:
Q=
Q
0
e
−t
RC
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyuaiabg2da9iaadgfadaWgaaWcbaGaaGimaaqabaGccaWGLbWaaWbaaSqabeaadaWcaaqaaiabgkHiTiaadshaaeaacaWGsbGaam4qaaaaaaaaaa@3E3A@
.
carica del condensatore
La carica è governata dall’equazione differenziale:
V=E−RC
dV
dt
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2da9iaadweacqGHsislcaWGsbGaam4qamaalaaabaGaaeizaiaadAfaaeaacaqGKbGaamiDaaaaaaa@3ED5@
,
dove E è la
tensione costante applicata al circuito.
Riconoscendo che l’equazione
incompleta corrispondente alla (7)
è la (2)
e che un integrale particolare della (7)
è dato da:
V=E
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2da9iaadweaaaa@3897@
è possibile scrivere l’integrale generale della (7)
come:
V=E+c
e
−t
RC
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2da9iaadweacqGHRaWkcaWGJbGaamyzamaaCaaaleqabaWaaSaaaeaacqGHsislcaWG0baabaGaamOuaiaadoeaaaaaaaaa@3F0D@
.
Tuttavia può essere utile
didatticamente effettuare una ricerca costruttiva della soluzione, integrando
l’equazione (7)
per derivazione. A tal fine poniamo:
dV
dt
=F
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaqGKbGaamOvaaqaaiaabsgacaWG0baaaiabg2da9iaadAeaaaa@3B6F@
.
Derivando la (7)
rispetto al tempo, si ha:
F=−RC
dF
dt
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOraiabg2da9iabgkHiTiaadkfacaWGdbWaaSaaaeaacaqGKbGaamOraaqaaiaabsgacaWG0baaaaaa@3DEB@
,
che è la stessa equazione della (2).
Pertanto possiamo scrivere:
dV
dt
=F=
c
1
e
−t
RC
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaqGKbGaamOvaaqaaiaabsgacaWG0baaaiabg2da9iaadAeacqGH9aqpcaWGJbWaaSbaaSqaaiaaigdaaeqaaOGaamyzamaaCaaaleqabaWaaSaaaeaacqGHsislcaWG0baabaGaamOuaiaadoeaaaaaaaaa@42FA@
e sostituire nella (7),
ottenendo:
V=E+c
e
−t
RC
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2da9iaadweacqGHRaWkcaWGJbGaamyzamaaCaaaleqabaWaaSaaaeaacqGHsislcaWG0baabaGaamOuaiaadoeaaaaaaaaa@3F0D@
(dove si è posto
c=−RC
c
1
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4yaiabg2da9iabgkHiTiaadkfacaWGdbGaam4yamaaBaaaleaacaaIXaaabeaaaaa@3C35@
).
Imponendo, ora, la condizione iniziale:
V(0)=0
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiaacIcacaaIWaGaaiykaiabg2da9iaaicdaaaa@3A9A@
,
si ha:
c=−E
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4yaiabg2da9iabgkHiTiaadweaaaa@3991@
e, in definitiva, la soluzione:
V=E(
1−
e
−t
RC
)
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2da9iaadweadaqadaqaaiaaigdacqGHsislcaWGLbWaaWbaaSqabeaadaWcaaqaaiabgkHiTiaadshaaeaacaWGsbGaam4qaaaaaaaakiaawIcacaGLPaaaaaa@407E@
.
caduta in un fluido
Un corpo pesante (omogeneo) di massa m cade in un fluido – per esempio nell’acqua: vogliamo studiare
l’andamento nel tempo della velocità v
del moto.
Impostiamo l’equazione differenziale di questo moto,
ammettendo che la forza di resistenza dell’acqua
F
R
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaaBaaaleaacaWGsbaabeaaaaa@37BA@
sia esprimibile con la formula:
F
R
=−Bv
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaaBaaaleaacaWGsbaabeaakiabg2da9iabgkHiTiaadkeacaWG2baaaa@3B79@
,
mentre la forza F
che accelera verso il basso il corpo è, ovviamente, data da:
F=V(μ−
μ
1
)g
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOraiabg2da9iaadAfacaGGOaGaeqiVd0MaeyOeI0IaeqiVd02aaSbaaSqaaiaaigdaaeqaaOGaaiykaiaadEgaaaa@4027@
,
(dove
μ
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiVd0gaaa@37A2@
e
μ
1
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiVd02aaSbaaSqaaiaaigdaaeqaaaaa@3889@
sono le densità del corpo e del liquido, V il volume del corpo e g l’accelerazione di gravità).
Per la seconda legge della
dinamica:
F+
F
R
=V(μ−
μ
1
)g−Bv=μV⋅a=μV
dv
dt
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOraiabgUcaRiaadAeadaWgaaWcbaGaamOuaaqabaGccqGH9aqpcaWGwbGaaiikaiabeY7aTjabgkHiTiabeY7aTnaaBaaaleaacaaIXaaabeaakiaacMcacaWGNbGaeyOeI0IaamOqaiaadAhacqGH9aqpcqaH8oqBcaWGwbGaeyyXICTaamyyaiabg2da9iabeY7aTjaadAfadaWcaaqaaiaabsgacaWG2baabaGaaeizaiaadshaaaaaaa@53C0@
,
ossia:
v=
V(μ−
μ
1
)g
B
−
μV
B
dv
dt
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamODaiabg2da9maalaaabaGaamOvaiaacIcacqaH8oqBcqGHsislcqaH8oqBdaWgaaWcbaGaaGymaaqabaGccaGGPaGaam4zaaqaaiaadkeaaaGaeyOeI0YaaSaaaeaacqaH8oqBcaWGwbaabaGaamOqaaaadaWcaaqaaiaabsgacaWG2baabaGaaeizaiaadshaaaaaaa@4955@
,
equazione che è formalmente identica alla (7).
Si può, quindi, risolvere l’equazione usando la relazione (11)
e operando le opportune sostituzioni. Se si ammette anche la condizione
iniziale:
v(0)=0
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamODaiaacIcacaaIWaGaaiykaiabg2da9iaaicdaaaa@3ABA@
,
si ottiene:
v=
V(μ−
μ
1
)g
B
(
1−
e
−Bt
μV
)
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamODaiabg2da9maalaaabaGaamOvaiaacIcacqaH8oqBcqGHsislcqaH8oqBdaWgaaWcbaGaaGymaaqabaGccaGGPaGaam4zaaqaaiaadkeaaaWaaeWaaeaacaaIXaGaeyOeI0IaamyzamaaCaaaleqabaWaaSaaaeaacqGHsislcaWGcbGaamiDaaqaaiabeY7aTjaadAfaaaaaaaGccaGLOaGaayzkaaaaaa@4ACE@
.
Si capisce facilmente che il la velocità v, al crescere di t, si approssima a un valore limite:
v
∗
=
V(μ−
μ
1
)g
B
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamODamaaCaaaleqabaGaey4fIOcaaOGaeyypa0ZaaSaaaeaacaWGwbGaaiikaiabeY7aTjabgkHiTiabeY7aTnaaBaaaleaacaaIXaaabeaakiaacMcacaWGNbaabaGaamOqaaaaaaa@4254@
.
Questo valore si trova anche ragionando sull’equazione (15)
scritta come:
a=
dv
dt
=−
Bv
μV
+(
1−
μ
1
μ
)g
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyyaiabg2da9maalaaabaGaaeizaiaadAhaaeaacaqGKbGaamiDaaaacqGH9aqpcqGHsisldaWcaaqaaiaadkeacaWG2baabaGaeqiVd0MaamOvaaaacqGHRaWkdaqadaqaaiaaigdacqGHsisldaWcaaqaaiabeY7aTnaaBaaaleaacaaIXaaabeaaaOqaaiabeY7aTbaaaiaawIcacaGLPaaacaWGNbaaaa@4B6C@
.
Infatti, la velocità limite si raggiunge quando la forza F è equilibrata dalla resistenza del
mezzo (che cresce con la velocità). In tale situazione, l’accelerazione a è nulla e da ciò si ricava subito:
v=
V(μ−
μ
1
)g
B
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamODaiabg2da9maalaaabaGaamOvaiaacIcacqaH8oqBcqGHsislcqaH8oqBdaWgaaWcbaGaaGymaaqabaGccaGGPaGaam4zaaqaaiaadkeaaaaaaa@412E@
.
conduzione termica (esterna)
Un corpo a temperatura T
è posto in un ambiente a temperatura fissa
T
a
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivamaaBaaaleaacaWGHbaabeaaaaa@37D7@
(per comodità, ammettiamo anche che sia
T>
T
a
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivaiabg6da+iaadsfadaWgaaWcbaGaamyyaaqabaaaaa@39B8@
). Sia C
la capacità termica del corpo considerato: in un intervallo di tempo
dt
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadshaaaa@37CC@
una quantità di calore
δQ=−CdT
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiTdqMaamyuaiabg2da9iabgkHiTiaadoeacaqGKbGaamivaaaa@3CE2@
passa dal corpo all’ambiente circostante e,
per essa, vale la relazione
−CdT=K(T−
T
a
)dt
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0Iaam4qaiaabsgacaWGubGaeyypa0Jaam4saiaacIcacaWGubGaeyOeI0IaamivamaaBaaaleaacaWGHbaabeaakiaacMcacaqGKbGaamiDaaaa@422B@
,
dove K dipende da
varie caratteristiche del corpo (superficie, spessore e conduttività termica
delle pareti). Se nella (21)
si sostituisce
T
∗
=T−
T
a
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivamaaCaaaleqabaGaey4fIOcaaOGaeyypa0JaamivaiabgkHiTiaadsfadaWgaaWcbaGaamyyaaqabaaaaa@3CA2@
,
si ottiene:
T
∗
=−
C
K
d
T
∗
dt
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivamaaCaaaleqabaGaey4fIOcaaOGaeyypa0JaeyOeI0YaaSaaaeaacaWGdbaabaGaam4saaaadaWcaaqaaiaabsgacaWGubWaaWbaaSqabeaacqGHxiIkaaaakeaacaqGKbGaamiDaaaaaaa@405C@
,
equazione che, ancora una volta, è formalmente identica alla
(2).
In analogia a quanto fatto con la (4),
scriveremo allora l’integrale della (22)
come:
T
∗
=
T
0
∗
e
−Kt
C
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivamaaCaaaleqabaGaey4fIOcaaOGaeyypa0JaamivamaaDaaaleaacaaIWaaabaGaey4fIOcaaOGaamyzamaaCaaaleqabaWaaSaaaeaacqGHsislcaWGlbGaamiDaaqaaiaadoeaaaaaaaaa@404F@
,
avendo posto
T
∗
(0)=
T
0
∗
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivamaaCaaaleqabaGaey4fIOcaaOGaaiikaiaaicdacaGGPaGaeyypa0JaamivamaaDaaaleaacaaIWaaabaGaey4fIOcaaaaa@3DB3@
.
Infine, tenendo conto della definizione della variabile
T
∗
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivamaaCaaaleqabaGaey4fIOcaaaaa@37E1@
,
possiamo sostituire:
T=(
T
0
−
T
a
)
e
−Kt
C
+
T
a
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivaiabg2da9iaacIcacaWGubWaaSbaaSqaaiaaicdaaeqaaOGaeyOeI0IaamivamaaBaaaleaacaWGHbaabeaakiaacMcacaWGLbWaaWbaaSqabeaadaWcaaqaaiabgkHiTiaadUeacaWG0baabaGaam4qaaaaaaGccqGHRaWkcaWGubWaaSbaaSqaaiaadggaaeqaaaaa@454B@
.
svuotamento di una buretta
Una buretta, riempita fino a un livello h, si svuota
lentamente (con gocciolamento continuo) attraverso il rubinetto d’efflusso. Lo
scorrimento del liquido può essere descritto attraverso la relazione di
Poiseuille:
dV
dt
=
π
r
4
8ηl
Δp
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaqGKbGaamOvaaqaaiaabsgacaWG0baaaiabg2da9maalaaabaGaeqiWdaNaamOCamaaCaaaleqabaGaaGinaaaaaOqaaiaaiIdacqaH3oaAcaWGSbaaaiabfs5aejaadchaaaa@4417@
dove V indica il
volume effluito (
dV
/
dt
MathType@MTEF@5@5@+=feaafeart1ev1aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSGbaeaacaqGKbGaamOvaaqaaiaabsgacaWG0baaaaaa@39A5@
è, quindi, la portata), r è il raggio
e l la lunghezza del capillare con il quale descriviamo il rubinetto, h
è la viscosità del liquido e Dp la differenza di pressione agli estremi del
capillare. Quest’ultimo termine può essere espresso in funzione dell’altezza
del liquido nella buretta:
Δp=dgh
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaamiCaiabg2da9iaadsgacaWGNbGaamiAaaaa@3C0F@
,
dove d rappresenta la densità del liquido e g
l’accelerazione di gravità.
Indichiamo, ora, il volume di liquido contenuto nella
buretta con Vh, mentre sia V0 il volume
iniziale. Chiaramente, vale la relazione:
V
h
=
V
0
−V
MathType@MTEF@5@5@+=feaafeart1ev1aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBaaaleaacaWGObaabeaakiabg2da9iaadAfadaWgaaWcbaGaaGimaaqabaGccqGHsislcaWGwbaaaa@3C84@
.
Inoltre, se
A=π
R
2
MathType@MTEF@5@5@+=feaafeart1ev1aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiabg2da9iabec8aWjaadkfadaahaaWcbeqaaiaaikdaaaaaaa@3B36@
è l’area della sezione della buretta, avremo:
V
h
=Ah
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBaaaleaacaWGObaabeaakiabg2da9iaadgeacaWGObaaaa@3AA3@
e sarà possibile, combinando le ultime tre relazioni,
esprimere Dp
come:
Δp=
dg
π
R
2
V
h
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaamiCaiabg2da9maalaaabaGaamizaiaadEgaaeaacqaHapaCcaWGsbWaaWbaaSqabeaacaaIYaaaaaaakiaadAfadaWgaaWcbaGaamiAaaqabaaaaa@40AD@
.
Sostituendo quest’ultima espressione nella (25),
otteniamo allora:
dV
dt
=
π
r
4
8ηl
⋅
dg
π
R
2
V
h
=
dg
r
4
8ηl
R
2
V
h
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaqGKbGaamOvaaqaaiaabsgacaWG0baaaiabg2da9maalaaabaGaeqiWdaNaamOCamaaCaaaleqabaGaaGinaaaaaOqaaiaaiIdacqaH3oaAcaWGSbaaaiabgwSixpaalaaabaGaamizaiaadEgaaeaacqaHapaCcaWGsbWaaWbaaSqabeaacaaIYaaaaaaakiaadAfadaWgaaWcbaGaamiAaaqabaGccqGH9aqpdaWcaaqaaiaadsgacaWGNbGaamOCamaaCaaaleqabaGaaGinaaaaaOqaaiaaiIdacqaH3oaAcaWGSbGaamOuamaaCaaaleqabaGaaGOmaaaaaaGccaWGwbWaaSbaaSqaaiaadIgaaeqaaaaa@5764@
e infine, poiché
dV=−d
V
h
MathType@MTEF@5@5@+=feaafeart1ev1aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeizaiaadAfacqGH9aqpcqGHsislcaqGKbGaamOvamaaBaaaleaacaWGObaabeaaaaa@3C7D@
(come si ottiene derivando la (27)),
possiamo scrivere:
V
h
=−
8ηl
R
2
dg
r
4
⋅
d
V
h
dt
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBaaaleaacaWGObaabeaakiabg2da9iabgkHiTmaalaaabaGaaGioaiabeE7aOjaadYgacaWGsbWaaWbaaSqabeaacaaIYaaaaaGcbaGaamizaiaadEgacaWGYbWaaWbaaSqabeaacaaI0aaaaaaakiabgwSixpaalaaabaGaamizaiaadAfadaWgaaWcbaGaamiAaaqabaaakeaacaWGKbGaamiDaaaaaaa@49FA@
,
equazione, anche questa volta, formalmente analoga alla (2).
Se procediamo all’integrazione nel modo già visto e,
sfruttando la (27),
ricaviamo poi V da Vh, otteniamo perciò, in
definitiva:
V=
V
0
(
1−
e
−
dg
r
4
8ηl
R
2
t
)
MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2da9iaadAfadaWgaaWcbaGaaGimaaqabaGcdaqadaqaaiaaigdacqGHsislcaWGLbWaaWbaaSqabeaacqGHsisldaWcaaqaaiaadsgacaWGNbGaamOCamaaCaaameqabaGaaGinaaaaaSqaaiaaiIdacqaH3oaAcaWGSbGaamOuamaaCaaameqabaGaaGOmaaaaaaWccaWG0baaaaGccaGLOaGaayzkaaaaaa@48CE@
.