7 Dynamic Similitude
278
Inlet triangle
b
b
1
1
1
1
1
b
1
1 1 1 v
b
1
1
1 1
1
d Zt / cos
57 ,
0.860,
d
Q
v
20.40 m / s;
d b
u 30.37 m / s;v
v 17.68 m / s;w 35.14 m / s.
p
b
b
t
p
p
t h
t
°
−
= −
=
=
=
=
=
=
=
=
Outlet triangle:
b
b
2
2
2
2
2
b
b
2r
2r
2 2r
2 2 2 v
d Zt / cos
22 ,
0.963,
d
Q
v
16.32 m / s;v
v
15.72 m / s.
d b
p
b
b
t
p
t
p
t h
°
−
= −
=
=
=
=
=
=
Slip:
ψ λ
β
ψ
=
+
=
×
−
=
=
+
( .
) .
( .
.
) .
(
2 5 60
0 75 2 5 0 367
1 60
2
2
0
2 2
2
2 2
1 1
;
P f Z
b r
b r b r ) )(
)
.
(
)
.
.
r r
v
u v tg
m
u geo
r
b
b
2
1
2
2
2
2
0 191
61 74
1
1
0 840
−
=
= +
=
= +
=
,
,
ε
β
Pf
/ /
(
)
.
/ .
s v
v
m s
u
u geo
; 2
2
51 86
=
=
ε
Thus:
o
i
W 3544 J / kg; p
W 3614 Pa.
D
D
h rD
=
=
=
For the scroll, we first try a width leap of 2.5 (standard value): b 2 ' = 100 mm.
The
radial
velocity
is
then
(
)
'
'
2r
2 2
v
Q / d b
5.66 m / s
p
=
=
;
with
v
ms
u
2
2
51 86
83 77
=
=
°
.
:
. .
/ α
This angle is somewhat too small for a compact
scroll. The tangential angle is optimally around 5°, thus 2 85 .
a
°
≈
With α 2 = 85°:
The velocity at the outlet of the scroll is then on average
(
)
(
) : /
48.48 m / s.
'
2
3
3
3
b b Q h b
=
×
=
This velocity is too high. In order to reach the
same velocity level as at the entrance, the outlet section must be equivalent to a
circle with diameter 180 mm. This may be realised with b 3 = h 3 = 150 mm and then a
divergent part to h 4 = 170 mm, b 4 = 150 mm. We try b b 15 mm.
3
2
0
'
= =
Then: v = 3 77 m s,
85.84° :
'
3
2r
2
3
2
r
.
/
1.579 r 355 mm.
r
a =
=
→ =
With h 3 = 150 mm, we enlarge radius r 3 to 375 mm. The theoretical flow rate
at the scroll outlet, with
constant
u
v r=
, is
3
2
r
3
2u 2
3
r
v r b dr 0.894m / s
r
=
∫
. The actual
flow rate is about 10 % smaller due to boundary layer obstruction. So, the scroll is
correct.
3
3
3
3
3
2
2
2
2
r
r
2
ln
1.733 r 390 mm;h r r 165 mm.
r
tg
r
p
a
=
→ =
→ =
= − =
278
Inlet triangle
b
b
1
1
1
1
1
b
1
1 1 1 v
b
1
1
1 1
1
d Zt / cos
57 ,
0.860,
d
Q
v
20.40 m / s;
d b
u 30.37 m / s;v
v 17.68 m / s;w 35.14 m / s.
p
b
b
t
p
p
t h
t
°
−
= −
=
=
=
=
=
=
=
=
Outlet triangle:
b
b
2
2
2
2
2
b
b
2r
2r
2 2r
2 2 2 v
d Zt / cos
22 ,
0.963,
d
Q
v
16.32 m / s;v
v
15.72 m / s.
d b
p
b
b
t
p
t
p
t h
°
−
= −
=
=
=
=
=
=
Slip:
ψ λ
β
ψ
=
+
=
×
−
=
=
+
( .
) .
( .
.
) .
(
2 5 60
0 75 2 5 0 367
1 60
2
2
0
2 2
2
2 2
1 1
;
P f Z
b r
b r b r ) )(
)
.
(
)
.
.
r r
v
u v tg
m
u geo
r
b
b
2
1
2
2
2
2
0 191
61 74
1
1
0 840
−
=
= +
=
= +
=
,
,
ε
β
Pf
/ /
(
)
.
/ .
s v
v
m s
u
u geo
; 2
2
51 86
=
=
ε
Thus:
o
i
W 3544 J / kg; p
W 3614 Pa.
D
D
h rD
=
=
=
For the scroll, we first try a width leap of 2.5 (standard value): b 2 ' = 100 mm.
The
radial
velocity
is
then
(
)
'
'
2r
2 2
v
Q / d b
5.66 m / s
p
=
=
;
with
v
ms
u
2
2
51 86
83 77
=
=
°
.
:
. .
/ α
This angle is somewhat too small for a compact
scroll. The tangential angle is optimally around 5°, thus 2 85 .
a
°
≈
With α 2 = 85°:
The velocity at the outlet of the scroll is then on average
(
)
(
) : /
48.48 m / s.
'
2
3
3
3
b b Q h b
=
×
=
This velocity is too high. In order to reach the
same velocity level as at the entrance, the outlet section must be equivalent to a
circle with diameter 180 mm. This may be realised with b 3 = h 3 = 150 mm and then a
divergent part to h 4 = 170 mm, b 4 = 150 mm. We try b b 15 mm.
3
2
0
'
= =
Then: v = 3 77 m s,
85.84° :
'
3
2r
2
3
2
r
.
/
1.579 r 355 mm.
r
a =
=
→ =
With h 3 = 150 mm, we enlarge radius r 3 to 375 mm. The theoretical flow rate
at the scroll outlet, with
constant
u
v r=
, is
3
2
r
3
2u 2
3
r
v r b dr 0.894m / s
r
=
∫
. The actual
flow rate is about 10 % smaller due to boundary layer obstruction. So, the scroll is
correct.
3
3
3
3
3
2
2
2
2
r
r
2
ln
1.733 r 390 mm;h r r 165 mm.
r
tg
r
p
a
=
→ =
→ =
= − =