fluid_mechanics

222 Chapter 5 ■ Finite Control Volume Analysis
Section (1)
Fixed control volume
30°
W 2
T shaft
V 1
Section (2)
ω
D 2 = 2r 2 = 12 in.
D 1 = 2r 1 = 10 in.
h =
1 in.
ω
T shaft
(a)
F I G U R E E5.19
Fixed
control volume
W 2
W r2 V r2
U 2
V 2
30° V θ2
(b)
Then
m #
The rotor exit blade speed, U 2 , is
16 in.211725 rpm212 rad/rev2
U 2 r 2
112 in./ft2160 s/min2
90.3 ft/s
To determine the fluid tangential speed at the fan rotor exit, V 2 ,
we use Eq. 5.43 to get
The vector addition of Eq. 4 is shown in the form of a “velocity
triangle” in Fig. E5.19b. From Fig. E5.19b, we can see that
To solve Eq. 5 for V 2 we need a value of W 2 , in addition to the
value of U 2 already determined (Eq. 3). To get W 2 , we recognize
that
where V r2 is the radial component of either W 2 or V 2 . Also, using
Eq. 5.6, we obtain
or since
10.0766 lbm ft 3 21230 ft 3 min2
160 smin2
V 2 W 2 U 2
V 2 U 2 W 2 cos 30°
W 2 sin 30° V r 2
m # A 2 V r 2
A 2 2 r 2 h
0.294 lbms
(3)
(4)
(5)
(6)
(7)
(8)
where h is the blade height, Eqs. 7 and 8 combine to form
m # 2r 2 hV r 2
Taking Eqs. 6 and 9 together we get
m #
W 2
r2pr 2 h sin 30° rQ
r2pr 2 h sin 30°
Q
2pr 2 h sin 30°
Substituting known values into Eq. 10, we obtain
By using this value of W 2 in Eq. 5 we get
V 2 U 2 W 2 cos 30°
90.3 ft/s 129.3 ft/s210.8662 64.9 ft/s
Equation 1 can now be used to obtain
with BG units.
With EE units
W 2 1230 ft3 min2112 in.ft2112 in.ft2
160 smin22p16 in.211 in.2 sin 30°
29.3 fts
10.294 lbm/s2190.3 ft/s2164.9 ft/s2
W # shaft
332.174 1lbm ft21lbs 2 243550 1ft lb2/1hp s24
(9)
(10)
10.00912 slug/s2190.3 ft/s2164.9 ft/s2
W # m #
shaft U 2 V 2
31 1slug ft/s 2 2/lb43550 1ft lb2/1hp s24