Trichterauslegung - Lehrstuhl Mechanische Verfahrenstechnik

2a ⋅
=
2a ⋅
2
b + b − 4ac
2
( b − b − 4ac )
Das Integral ergibt:
v
∫
0
av
2
dv
=
+ bv + c
2 2
⋅ v + b − b + 4ac 2a ⋅
=
2 2
⋅ v + b − b + 4ac 2a ⋅
b
2
1
⎢ ⎢ ⎡
⋅ ln
− 4ac ⎣
2
b + b −
2
( b − b − 4ac)
⋅ v + 4ac
=
⋅ v + 4ac
2
b + b − 4ac
⋅ v + 2c
( ) ⎥ ⎥ ⎤
2
b − b − 4ac ⋅ v + 2c⎦
Beide Seiten der Integralgleichung 4.290) ergeben zusammen:
1 ⎡
2
b + b −
⋅ v + 2c⎤
⋅ ln⎢
⎥ = t
2
2
b − 4ac ⎢⎣
( b − b − 4ac)
⋅ v + 2c⎥⎦
Auflösen nach v mit dem Parameter f
⎡( b + f ) ⋅ v + 2c⎤
ln⎢
= t ⋅ f
( b f ) v 2c
⎥
⎣ − ⋅ + ⎦
( b + f ) ⋅ v + 2c
= exp( t ⋅ f )
b − f ⋅ v + 2c
160
2
b + b −
⋅ v + 2c
2
( b − b − 4ac) ⋅ v + 2c
= b
2 − 4ac
nach Gl.( 4.296):
( )
( b + f ) ⋅ v + 2c = [( b − f ) ⋅ v + 2c] ⋅ exp( t ⋅ f )
( b + f ) ⋅ v − ( b − f ) ⋅ v ⋅ exp( t ⋅ f ) = 2c ⋅ exp( t ⋅ f ) − 2c
v ⋅ [( b + f ) − ( b − f ) ⋅ exp( t ⋅ f )] = 2c ⋅ [ exp( t ⋅ f ) −1]
2c ⋅ [ exp( t ⋅ f ) −1]
v =
( b + f ) − ( b − f ) ⋅ exp( t ⋅ f )
2c ⋅[ exp( t ⋅ f ) −1]
2c ⋅[ exp( t ⋅ f ) −1]
v =
=
f + f ⋅ exp( t ⋅ f ) + b − b ⋅ exp( t ⋅ f ) f ⋅[ exp( t ⋅ f ) + 1] + b ⋅[ 1−
exp( t ⋅ f )]
−1
[ exp( t ⋅ f ) + 1]
Zweckmäßiges Erweitern des Bruches mit
liefert:
−
[ exp( t ⋅ f ) + 1] 1
exp( t ⋅ f ) −1
exp( t ⋅ f ) −1
2c ⋅
2c ⋅
exp( t ⋅ f ) + 1 exp( t ⋅ f ) + 1
v =
1 exp ( t f )
=
− ⋅ exp( t ⋅ f ) −1
f + b ⋅
− b ⋅
+ f
exp( t ⋅ f ) + 1 exp( t ⋅ f ) + 1
Mit der tanh-Funktion Gl.(4.232) und ihrem Argument t ⋅ f / 2
( 2x)
( 2x)
exp −1
= tanh( x)
exp + 1
exp( t ⋅ f ) −1
2c ⋅
exp( t ⋅ f ) + 1
v =
exp( t ⋅ f ) −1
− b ⋅
+ f
exp( t ⋅ f ) + 1
c ⋅ tanh( t ⋅ f / 2)
v =
b
f
− ⋅ tanh( t ⋅ f / 2)
+
2
2
( t ⋅ f / 2)
( t ⋅ f / 2)
folgt (4.232)
2c ⋅ tanh
=
− b ⋅ tanh
( t ⋅ f / 2)
( t ⋅ f / 2)
+
c ⋅ tanh
=
f b
− ⋅ tanh
2
( t ⋅ f / 2)
( t ⋅ f / 2)
c ⋅ tanh
v = ( 4.307)
b
f
− ⋅ tanh +
2
2
Jetzt müssen wiederum die Koeffizienten f, b und c
+
f
2
Schüttec_4 VO Partikelmechanik und Schüttguttechnik, Trichterauslegung Prof. Dr. Jürgen Tomas, 04.06.2013