7.6 Turbomachine Design Example: Centrifugal Fan
277
Then:
b
b
b
1
1
1
1
tg
u / v :
57 .
b
b
°
= −
≈ −
This value then differs somewhat from the
theoretically optimum value (− 55°). With i 0.85 : W 3432 J / kg;
h
D
≈
=
then
2
2
W 0.735
u
D
y =
=
and
2 2u
W u v ,
D =
thus: 2u
v
50.22 m / s.
=
With an estimate of the
work reduction factor of 0.80, we obtain (
)
.
/ .
v
m s
u geo
2
62 78
=
Suppose that we can realise the deceleration ratio w 2 /w 1 = 0.70. Then:
2
2u
2u
2r
w
24.60 m / s, v
50.22 m / s, w
18.11 m / s, w
16.65 m / s.
=
=
= −
=
These
results seem acceptable, since with
/
1
2r
1
v
17.68 m / s : v v = 0.94
=
.
Geometrically:
b
2u geo
2r
rot
2 2 2
( w )
5.55 m / s,v
Q / ( d b ) 17.47m / s,
p
t
= −
=
=
thus
b
2
17.6 .
b
°
≈ −
The rotor blades can only be swept moderately backward. We knew this
already, as
0.735
y =
is quite high. With straight blades, to
b
1
57
b
°
= −
corresponds
b
2
22
b
°
= −
(see Fig. 7.21). So, we can opt for straight backward inclined blades.
The fan might then realise somewhat less total pressure rise than the target value,
because the blade outlet angle is somewhat too big in magnitude. If it later turns out
to be necessary, we may enlarge the outlet diameter somewhat.
Width at outlet: rotor
2 2 2r
2
Q
d b v : b 38 mm.
p
=
=
We take b
mm
2
40
=
.
2r 2u
M
2
M 2
w v
C
1.0
w
s
=
, which gives M 1.38
s
, with
st
M
2 2 2
Z M .
2 r b r
s
p
=
M
brdr
b r b r r r Z
st =
=
+
−
=
∫
1 1
2 2
2
1
2
16 5
(
):
. . We take Z = 16. Choice of plate
thickness: t = 3 mm.
We can now correct the velocity triangles, using improved values of the obstruction factors. At this stage, we still take
0.90
v
h =
and
0.85.
i
h =
2
2
0
p 2917 m / s .
D
r
=
Fig. 7.21 Meridional and orthogonal sections of the rotor
277
Then:
b
b
b
1
1
1
1
tg
u / v :
57 .
b
b
°
= −
≈ −
This value then differs somewhat from the
theoretically optimum value (− 55°). With i 0.85 : W 3432 J / kg;
h
D
≈
=
then
2
2
W 0.735
u
D
y =
=
and
2 2u
W u v ,
D =
thus: 2u
v
50.22 m / s.
=
With an estimate of the
work reduction factor of 0.80, we obtain (
)
.
/ .
v
m s
u geo
2
62 78
=
Suppose that we can realise the deceleration ratio w 2 /w 1 = 0.70. Then:
2
2u
2u
2r
w
24.60 m / s, v
50.22 m / s, w
18.11 m / s, w
16.65 m / s.
=
=
= −
=
These
results seem acceptable, since with
/
1
2r
1
v
17.68 m / s : v v = 0.94
=
.
Geometrically:
b
2u geo
2r
rot
2 2 2
( w )
5.55 m / s,v
Q / ( d b ) 17.47m / s,
p
t
= −
=
=
thus
b
2
17.6 .
b
°
≈ −
The rotor blades can only be swept moderately backward. We knew this
already, as
0.735
y =
is quite high. With straight blades, to
b
1
57
b
°
= −
corresponds
b
2
22
b
°
= −
(see Fig. 7.21). So, we can opt for straight backward inclined blades.
The fan might then realise somewhat less total pressure rise than the target value,
because the blade outlet angle is somewhat too big in magnitude. If it later turns out
to be necessary, we may enlarge the outlet diameter somewhat.
Width at outlet: rotor
2 2 2r
2
Q
d b v : b 38 mm.
p
=
=
We take b
mm
2
40
=
.
2r 2u
M
2
M 2
w v
C
1.0
w
s
=
, which gives M 1.38
s
, with
st
M
2 2 2
Z M .
2 r b r
s
p
=
M
brdr
b r b r r r Z
st =
=
+
−
=
∫
1 1
2 2
2
1
2
16 5
(
):
. . We take Z = 16. Choice of plate
thickness: t = 3 mm.
We can now correct the velocity triangles, using improved values of the obstruction factors. At this stage, we still take
0.90
v
h =
and
0.85.
i
h =
2
2
0
p 2917 m / s .
D
r
=
Fig. 7.21 Meridional and orthogonal sections of the rotor