92
2 Basic Components
A: The momentum balance (no pressure force in axial direction) into the axial
direction on the streamtube results in
(2.32)
Momentum into the axial direction locally results in
Combination gives
(2.33)
The work equations within the streamtube upstream and downstream of the rotor
are
From that:
(2.34)
Combination of (Eq. 2.33) and (Eq. 2.34) results in
2
2
0
3a
2u
3u
1a 0
3a
0
3a
v v
v
v
v ( v v )
( v v )
.
2
2
+
−
−
=
−
+
Due to the radius increase within the streamtube downstream the rotor, v u decreases.
Thus, v 3u is smaller than v 2u . We ignore the difference between v 3u and v 2u here. We
accept that in (Eq. 2.34) p 1  − p 2 becomes somewhat smaller. This is equivalent to the
assumption that some losses occur within the flow downstream of the rotor. With
this simplification follows
1
1a
2a
0
3a
2
v
v
( v v ).
=
=
+
The flow deceleration is then, as with the one-dimensional analysis, the same upstream and downstream of the rotor.
We set 1 a
2a
0
3a
0
2u
v
v
v ( 1 a ), v
v ( 1 2a ), v
u( 2b ).
=
=
−
=
−
=−
The energy extraction in the streamtube according to (Eq. 2.34), ignoring the
difference between v u
2 and v u
3 , is
(2.35)
a
1a 3a
0
L
sv ( v
v ).
r
− =
−
a
1
2
L ( p p )s.
=
−
1
2
1a 0
3a
p p
v ( v v ).
r
−
=
−
2
2
2
2
a
0
a
3
1
1
2
2
p
v
p
v
p v
p
v
,
.
2
2
2
2
r
r
r
r
+
=
+
+
=
+
2
2
2
2
2
2
2
2
0
3
2u
3u
0
3a
1
2
2
1
v v
v
v
v v
p p
v v
.
2
2
2
2
r
−
−
−
−
−
=
+
=
+
2
2
2
2
2
2
2
2
2
0
3a
2u
0
1
2
1
2
v v
v
v
p p
v v
u
( 4a 4a )
( 4b ).
2
2
2
2
2
r
−
−
−
+
=
−
=
−
−