Tải bản đầy đủ

Bài tập Turbine chap 12

Chapter 12 / Turbomachinery

CHAPTER 12
Turbomachinery
Elementary Theory

ω = 800 × π / 30 = 83.8 rad/s
u1 = ωr1 = 83.8 × 0.04 = 3.35 m/s
u 2 = ω r2 = 83.8 × 0.125 = 10.48 m/s

Q = 2π r1b1Vn1 , but Vn1 = u1 since β1 = 45D

12.2

∴Q = 2π × 0.04 × 0.05 × 3.35 = 0.0421 m3 / s
Vn2 =

Q
0.0421
=
= 2.14 m/s

2π r2b2 2π × 0.125 × 0.025

Vt2 = u2 −

Vt1 = 0

Vn2

tan β 2

= 10.48 −

2.14
= 6.77 m/s
tan 30D

(α1 = 90D under ideal conditions)

(

)

∴T = ρQ r2Vt2 − rV
1 t1 = 1000 × 0.0421(0.125 × 6.77 − 0) = 35.6 N ⋅ m
W P = ωT = 83.8 × 35.6 = 2980 W

H t = ωT / γ Q = 2980 /(9810 × 0.0421) = 7.22 m
Q = 7.6 ×10−3 m3/s

12.4

ω = 2000 ×

π
30

= 209 rad/s

175
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.



Chapter 12 / Turbomachinery
Q
7.6 ×10−3
=
= 1.94 m/s
2π r2b2 2π × 0.0625 × 0.01

Vn2 =

u2 = ωr2 = 209 × 0.0625 = 13.06 m/s
∴ Ht =

13.06
u2
(13.06 − 1.94cot 60°) = 15.9 m
(u2 − Vn2 cot β 2 ) =
9.81
g

 = γ QH = 9810 × 0.8 × 7.6 ×10−3 ×15.9 = 948 W
W
t

 = 948/ 746 = 1.21 hp
W
Compute loss in suction pipe:
2
⎛ L
⎞ Q
hL = ⎜ f + Σ K ⎟
⎝ D
⎠ 2 gA 2

11


= ⎜ 0.015 ×
+ 2 × 0.19 + 0.8⎟


0.1

12.6

0.05 2
2

⎛ π⎞
2 × 9.81 × ⎜ ⎟ × 0.1 4
⎝ 4⎠

= 5.85 m

Water at 20°C: γ = 9792 N/m3 , pv = 2340 Pa. Substitute known data into NPSH
relation, solving for Δz:
∴Δ z =

patm − pv

γ

101×103 − 2340
− 5.85 − 3 = 1.23 m
− hL − NPSH =
9792

Dimensional Analysis and Similitude
Q1 =

N
ω1
970
Q = 1Q =
× 0.8 = 0.65 m 3 / s
ω 2 2 N 2 2 1200

∴from Fig. 12.9, H1 ≅ 11.5 m and W 1 ≅ 91 kW.
2

⎛ω ⎞
⎛ 1200 ⎞
∴ H 2 = ⎜ 2 ⎟ H1 = ⎜
⎟ ×11.5 = 17.6 m
⎝ 970 ⎠
⎝ ω1 ⎠

12.8

2

3

⎛ 1200 ⎞
 = ⎛ ω2 ⎞ W

∴W

⎟ 1=⎜
2
⎟ × 91 = 172 kW
⎝ 970 ⎠
⎝ ω1 ⎠

12.10

CQ =

3

Q
Q / 3600
=
= 1.061× 10 −4 Q
3
ω D 304 × 0.2053

(Q in m 3 /h)

176
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 12 / Turbomachinery

CH =

gH
9.81H
=
= 2.526 × 10−3 H (H in m)
2
2
2
2
ω D 304 × 0.205

Tabulate CQ and CH using selected values of Q and H from Fig. 12.6:
CQ×10 −3
0
0.53
1.06
1.59
2.12
2.65
3.18

Q (m3/h)
0
50
100
150
200
250
300

H (m)
54
53
52
50
47
41
33

CH ×10−1
1.36
1.34
1.31
1.26
1.19
1.04
0.83

The dimensionless curve shown in Fig. 12.12 is for the 240-mm impeller. Since
the impellers are not the same (240 mm versus 205 mm), dynamic similitude
does not exist, and thus the curves are not the same.

Compute the specific speed: Ω P =
12.12

ω Q
( gH P )3 / 4

=

1800 ×

π

× 0.15
30
= 1.30 ,
(9.81 × 22)3 / 4

hence use a mixed flow pump. As an alternate, since Ω P is close to unity, a
radial flow pump could be employed.
Fig. 12.13: At η = 0.75 (best eff. ), CQ ≅ 0.048 ,
C H ≅ 0.018, C W ≅ 0.0011, C NPSH ≅ 0.023.

ω = 750 ×

π
30

= 78.5 rad/s

1/3

(a)
12.14

⎛ 1240 ×10−3 ⎞
D = ⎜⎜
⎟⎟
⎝ 0.049 × 78.5 ⎠
H=

= 0.685 m

0.018 × 78.52 × 0.6852
= 5.3 m
32.2

H NPSH =

0.023 × 78.52 × 0.6852
= 6.78 m
9.81

 = 0.0011×1000 × 78.53 × 0.6855 = 80,250 W = 80.25 kW
W

or
12.16

 = 82.25/0.746 = 107.6 hp
W

ω = 600 ×

π
30

= 62.8 rad /s,

Q = 22.7 / 60 = 0.378 m 3 / s,

177
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 12 / Turbomachinery

ΩP =

ω Q
( gH P )

=

3/ 4

2

62.8 0.378
= 0.751
(9.81×19.5)3/ 4
2

H 2 ⎛ ω 2 ⎞ ⎛ D2 ⎞
=⎜
⎟ ⎜
⎟ =2
H1 ⎝ ω1 ⎠ ⎝ D1 ⎠
2

3

Q2 ⎛ ω 2 ⎞⎛ D2 ⎞
=⎜
⎟⎜
⎟ =2
Q1 ⎝ ω1 ⎠⎝ D1 ⎠

2

⎛ ω ⎞ ⎛ D ⎞ ⎛ ω ⎞⎛ D ⎞
∴ ⎜ 2 ⎟ ⎜ 2 ⎟ = ⎜ 2 ⎟⎜ 2 ⎟
⎝ ω1 ⎠ ⎝ D1 ⎠ ⎝ ω1 ⎠⎝ D1 ⎠

12.18

∴Use a radial flow pump.

3

or

ω 2 D2
=
ω1 D1

4

⎛ω ⎞
∴ ⎜ 2 ⎟ = 2, ω2 = 4 2 ω1 = 1.19 ω1 and D2 = 4 2 D1 = 1.19 D1
⎝ ω1 ⎠

Assume a pump speed N = 2000 rpm or ω = 2000 ×

π
30

= 209 rad/s

 /(γ Q) = 200 ×103 /(8830 × 0.66) = 34.3 m
HP = W
f

∴Ω P =

ω Q
( gH P )

3/ 4

=

209 0.66
= 2.16
(9.81× 34.3)3/ 4

The specific speed suggests a mixed-flow pump. However, if N = 1000 rpm, a
radial-flow pump may be appropriate. Consider both possibilities.
Mixed flow: from Fig. 12.14, at best η :
CW ≅ 0.0117, CQ ≅ 0.148, CH ≅ 0.067

Use
12.20

CQ =

gH
Q
and CH = 2 P2
3
ωD
ωD

Combining and solving for D and ω
Q / CQ

D=

gHP / CH

ρ=

Q
CQD3

0.66 / 0.148
= 0.251 m
9.81× 34.3/ 0.067

∴D =

ω=

and ω =

0.66
= 282 rad/s or N = 282 × 30 / π = 2674 rpm
0.148 × 0.2513

γ
g

=

8830
= 900 kg/m3
9.81

178
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 12 / Turbomachinery
 = C  ρω 3D5 = 0.0117 × 900 × 2823 × 0.2515
∴W
P
W

= 2.35 × 105 W or 235 kW
Radial flow: from Fig. 12.12, at best η :
CW ≅ 0.027, CQ ≅ 0.0165, C H ≅ 0.125
∴D =

ω=

0.66 / 0.0165
= 0.878 m
9.81× 34.3/ 0.125

0.66
59.1× 30
= 59.1 rad/s or N =
= 564 rpm
3
π
0.0165 × 0.878

 = 0.0027 × 900 × 59.13 × 0.8785 = 2.62 × 105 W
W
P

or 262 kW

Hence, a mixed-flow pump is preferred.
Use of Turbopumps

12.22

The intersection of the system demand curve with the head-discharge curve
yields Q = 2.75 m3 / min, H P = 12.6 m, W P = 7.2 kW.
N 1 = 1350 rpm, Q 1 = 2.75 m 3 / min, H 1 = 12.6 m,

12.24

W 1 = 7.2 kW, N 2 = 1200 rpm, D 2 = D1

179
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 12 / Turbomachinery
3

⎛ N ⎞⎛ D ⎞
⎛ 1200 ⎞
3
∴Q2 = Q1 ⎜ 2 ⎟⎜ 2 ⎟ = 2.75 ⎜
⎟ = 2.44 m /min
1350
N
D


⎝ 1 ⎠⎝ 1 ⎠
2

2

3

5

⎛N ⎞ ⎛D ⎞
⎛ 1200 ⎞
H 2 = H1 ⎜ 2 ⎟ ⎜ 2 ⎟ = 12.6 ⎜
⎟ = 9.96 m
⎝ 1350 ⎠
⎝ N1 ⎠ ⎝ D1 ⎠
2

⎛N ⎞ ⎛D ⎞
⎛ 1200 ⎞
W 2 = W 1 ⎜ 2 ⎟ ⎜ 2 ⎟ = 7.2 ⎜
⎟ = 5.06 kW
⎝ 1350 ⎠
⎝ N1 ⎠ ⎝ D1 ⎠
3

Efficiency will remain approximately the same.
Compute system demand:
2
2
14
⎛ L
⎞V

⎞ 3
= ⎜ 0.01 ×
+ 4 × 0.1⎟
= 0.40 m
H P = ⎜ f + ΣK ⎟
⎝ D
⎠ 2g ⎝
0.3
⎠ 2 × 9.81

12.26

Q = VA = 3 ×

∴Ω P =

π
4

× 0.32 = 0.212 m2

ω Q
( gH P )

3/ 4

=

31.4 0.212
= 5.19
(9.81 × 0.4)3/ 4

ω = 300 × π / 30 = 31.4 rad/s
∴Axial pump is appropriate.

(a)

12.28

The intersection of the pump curve with the demand curve yields H P ≅ 64 m
and Q ≅ 280 m3 / h. ∴ From Fig. 12.6, W P ≅ 64 kW and NPSH ≅ 8.3 m

180
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 12 / Turbomachinery

Use energy eqn. to establish system demand:
e⎤

f ≅ 1.325 ⎢ln 0.27 ⎥
D⎦

H P = Δz + f

−2

0.000255 ⎤

= 1.325 ⎢ln 0.27 ×
0.45 ⎥⎦


−2

= 0.017

0.017 × 5000
L Q2
= 192 +
Q2 = 192 + 381Q2
2
5
2
D 2 gA
2 × 9.81× (π / 4) × (0.45)

From Fig. 12.6, at best ηP , Q ≅
12.30

240 m3
= 0.067 m3/s, and HP ≅ 65 m
3600 s

Assume three pumps in series, so that HP = 3× 65 = 195 m. Then the demand
head is
H P = 192 + 381× (0.067) 2 = 193.7 m

Hence three pumps in series are appropriate. The required power is
WP = γ QHP/ηP = 9810× 0.86× 0.067 ×195/0.75 = 146,965 W
W P = 146,965/746 = 197 hp

or

(a) For water at 80°C, pv = 46.4 × 103 Pa, and ρ = 9553 kg/m3. Write the energy
equation from the inlet (section i) to the location of cavitation in the pump:
pi

γ

12.32

+

Vi2 pv
=
+ NPSH
2g γ

∴ NPSH =

pi − pv

γ

Vi2 (83 − 46.4) ×103
62
+
=
+
= 5.67 m
2g
9533
2 × 9.81

(b) NPSH1 = 5.67 m, N1 = 2400 rpm, N2 = 1000 rpm, D2/D1 = 4
2

2

⎛N ⎞ ⎛D ⎞
⎛ 1000 ⎞
∴ NPSH 2 = NPSH1 ⎜ 2 ⎟ ⎜ 2 ⎟ = 5.67 ⎜

⎝ 2400 ⎠
⎝ N1 ⎠ ⎝ D1 ⎠

2

( 4 )2 = 15.8 m

Given:
12.34

L = 200 m , D = 0.05 m , Δz = z 2 − z1 = −3 m , V = 3 m/s , ν = 6 × 10 −7 m 2 /s ,

ρ = 1593 kg/m 3 , pv = 86.2 × 103 Pa , pa = 101 × 103 Pa

Compute the pump head:

181
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 12 / Turbomachinery
Re =

3 × 0.05
6 × 10

−7

= 2.5 × 105

(

)

f = 1.325 ⎡ ln 5.74Re-0.9 ⎤



−2

= 1.325 ⎡ ln 5.74 × (2.5 × 105 ) −0.9 ⎤



−2

= 0.015

2
fL ⎞ V 2

⎛ 0.015 × 200 ⎞ 3
= −3 + ⎜ 1 +
= 25.0 m
H P = Δz + ⎜ 1 + ⎟

0.05
D ⎠ 2g

⎠ 2 × 9.81


(a) Choose a radial-flow pump. Use Fig. P12.35 to select the size and speed:
C H = 0.124, CQ = 0.0165, η = 0.75
Q = 3×
∴D =
∴ω =

π

4

4

× 0.052 = 0.00589 m 2

CHQ 2
CQ2 gH P
Q

CQ D

3

=

=

4

0.124 × 0.00589 2
0.01652 × 9.81 × 25.0

0.00589
0.0165 × 0.090

3

= 0.090 m

= 490 rad/s, or N = 490 ×

30

π

= 4680 rpm

(b) Available net positive suction head:
NPSH =

pa − pv
101×103 − 86.2 ×103
− Δz =
+ 3 = 3.95 m
1593 × 9.81
ρg

(c)

182
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 12 / Turbomachinery
Turbines
ω = 120 × π / 30 = 12.6 rad/s

α1 = cot −1 (2π r12b1ω / Q + cot β1 )
= cot −1 (2π × 4.52 × 0.85 × 12.6 /150 + cot 75D ) = 6.1D

Vt1 = u1 + Vn1 cot β1 = ω r1 +
= 12.6 × 4.5 +

12.36

Q
cot β1
2π r1b1

150
cot 75D = 58.37 m/s
2π × 4.5 × 0.85

Vt2 = u2 + Vn2 cot β 2 = ω r2 +
= 12.6 × 2.5 +

Q
cot β 2
2π r2b2

150
cot100D = 29.52 m/s
2π × 2.5 × 0.85

∴ T = ρQ(rV
1 t1 − r2Vt2 )
= 1000 ×150(4.5 × 58.37 − 2.5 × 29.52) = 2.83 × 107 N ⋅ m

 = ωT = 12.6 × 2.83 × 107 = 3.57 × 108 W or 357 MW
W
T

 =W
 , hence
Under ideal conditions ηT = 1, and W
T
f
HT = W T / γ Q = 3.57 × 108 /(9810 × 150) = 243 m
N 2 = 240 rpm , W 2 = 2200 kW , D 2 = 0.9 m, W 1 = 9kW , H1 = 7.5 m
3
5
2
2
W 2 ⎛ N 2 ⎞ ⎛ D2 ⎞
H 2 ⎛ N 2 ⎞ ⎛ D2 ⎞
From the similarity rules  = ⎜
=⎜
⎟ ⎜
⎟ and
⎟ ⎜

H1 ⎝ N1 ⎠ ⎝ D1 ⎠
W1 ⎝ N1 ⎠ ⎝ D1 ⎠

12.38

Substitute second eqn. into the first to eliminate ( N 2 / N 1 ), and solve for D 1 :
1/ 2
3/ 4
 ⎞1/ 2 ⎛ H ⎞3/ 4
⎛W
⎛ 9 ⎞ ⎛ 45 ⎞
1
2
∴ D1 = D2 ⎜  ⎟ ⎜
⎟ = 0.9 ⎜
⎟ ⎜
⎟ = 0.221 m
⎝ 2200 ⎠ ⎝ 7.5 ⎠
⎝ W2 ⎠ ⎝ H1 ⎠
1/ 2

⎛H ⎞
N 2 = N1 ⎜ 1 ⎟
⎝ H2 ⎠

⎛ D2 ⎞
⎛ 7.5 ⎞

⎟ = 240 ⎜

D
⎝ 45 ⎠
⎝ 1⎠

1/ 2

⎛ 0.9 ⎞

⎟ = 399 rpm
⎝ 0.221 ⎠

183
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 12 / Turbomachinery

Write energy eqn. from upper reservoir (loc. 1) to surge tank (loc. 2) and solve
for Q:
1/ 2



D
Q = ⎢ 2 gA2
( z1 − z2 ) ⎥
fL



1/ 2

⎡ 2 × 9.81 × (π / 4) 2 × 0.855

=⎢
(650 − 648.5) ⎥
0.025 × 2000



= 0.401 m 3 / s

Apply energy eqn. from loc. 2 to lower reservoir (loc. 3) and determine H T :
2
⎛ L
⎞ Q
HT = z2 − z3 − ⎜ f + K v ⎟
⎝ D
⎠ 2 gA2

12.40

⎛ 0.02 × 100 ⎞
= 648.5 − 595 − ⎜
+ 1⎟
⎝ 0.85


0.4012
⎛π ⎞
2 × 9.81× ⎜ ⎟ × 0.855
⎝4⎠
2

= 53.4 m

∴ W T = γ QHTηT = 9810 × 0.401 × 53.4 × 0.9 = 1.89 × 105 W or 189 kW

∴ From Fig. 12.32, use a Francis turbine.
A representative value of the specific speed is 2 (Fig. 12.20):
ΩT ( gHT )5 / 4
2(9.81× 53.3)5 / 4
∴ω =
=
= 306 rad/s
 / ρ )1/ 2
(W
(2.67 × 105 /1000)1/ 2
T

or

N = 306 × 30 / π = 2920 rpm

ω = 200 × π / 30 = 20.94 rad/s and from Fig. P12.42 at best efficiency

(ηT = 0.8), φ ≈ 0.42, cv = 0.94
V1 = cv 2 gHT = 0.94 2 × 9.81× 120 = 45.6 m/s

∴Q =

12.42

WT
4.5 × 10 6
=
= 4.78 m 3 / s
γ H T η T 9810 × 120 × 0.8

This is the discharge from all of the jets.
Determine the wheel radius r : r =

φ 2 gHT 0.42 2 × 9.81× 120
=
= 0.973 m
20.94
ω

Hence, the diameter of the wheel is 2r = 2 × 0.973 = 1.95 m
Compute diameter of one jet: D j = 2r / 8 = 1.95 / 8 = 0.244 m, or 244 mm
Let N j = no. of jets. Then each jet has a discharge of Q / N j and an area

184
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 12 / Turbomachinery
Q / Nj
V1
Nj =

=

π
4

D2j ∴Solving for N j :

Q
1
4.78
=
= 2.24
V1 ⎛ π 2 ⎞
⎛π ⎞
2
D
45.6
0.244
×
×

⎜ ⎟
j ⎟
⎝4

⎝4⎠

∴Use three jets

ΩT =

ω (W T / ρ )1/ 2
( gHT )5 / 4

=

20.94(4.5 × 106 /1000)1/ 2
= 0.204
(9.81 × 120)5 / 4

Q = 2082 / 6 = 347 m3 / sec (one unit)

HT =

WT
427,300 × 746
=
= 110.1 m
γ QηT 9810 × 347 × 0.85

Write energy eqn. from upper reservoir (loc. 1) to lake (loc. 2):
2
⎛ L
⎞ Q
+ HT + z2
z1 = ⎜ f + ΣK ⎟
2
⎝ D
⎠ 2gA

12.44

⎛ 0.01 × 390

∴ 309 = ⎜
+ 0.5 ⎟
D



which reduces to 2.4 −

347 2
⎛π ⎞
2 × 9.81 × ⎜ ⎟ D 4
⎝4⎠
2

+ 110.1 + 196.5

38,840 4980
− 4 =0
D5
D

∴ Solving, D ≅ 8 m

∴From Fig. 12.32, a Francis or pump/turbine unit is indicated.
(a) Let H be the total head and Q the discharge delivered to the turbine; then
HT = 0.95H = 0.95 × 305 = 289.8 m
and
12.46

Q=


W
10.4 × 10 6
T
=
= 4.30 m 3 / s.
γHT ηT 9810 × 289.8 × 0.85

Write energy eqn. from reservoir to turbine outlet:
2
⎛ L
⎞ Q
H = HT +⎜ f +∑K⎟
2
⎝ D
⎠ 2gA

4.32
⎛ 0.02 × 3000

∴ 305 = 289.8 + ⎜
+ 2⎟
D

⎠ 2 × 9.81× (π / 4)2 D 4

185
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 12 / Turbomachinery
which reduces to 15.2 −

91.67 3.06
− 4 =0
D5
D

∴Solving D = 1.45 m

(b) Compute jet velocity: V1 = cv 2 gHT = 0.98 2 × 9.81× 289.8 = 73.9 m/s
The flow through one nozzle is Q/4 and the jet area is πD 2j / 4. Hence

π D 2j
4

=

Q / 4 4.3/ 4
=
= 0.0146 m 2
V1
73.9
4 × 0.0146

∴ Dj =

π

= 0.136 m
2

12.48

⎡ 4.15 × (9.81 × 3.7)5 / 4 ⎤
6
WT = 1000 ⎢
⎥ = 4.99 × 10 W (one unit).
×
50
/
30
π



Total power developed is 9.21 × 10 6 W. Hence, required number of units is

9.21/ 4.99 = 1.8 ∴ Use two turbines.
(a)

H T = z1 − z 2 −

fL Q 2
D 2 gA 2

0.015 × 350 × 0.252
= 915 − 892 −
= 11.8 m
0.3 × 2 × 9.81× (0.7854 × 0.32 ) 2

 = γ QH η = 9810 × 0.25 ×11.8 × 0.85 = 2.46 ×104 W or 24.6 kW
W
T
T T
(b) Compute the specific speed of the turbine:
12.50

ω=N

π
30

∴ΩT =

= 1200 ×

 /ρ
ω W
T

( gHT )

5/4

π
30
=

= 126 rad/s

126 2.46 × 104 /1000

( 9.81×11.8)

5/4

= 1.65

Hence, from Fig. 12.20, a Francis turbine is appropriate.
(c) From Fig. 12.24, the turbine with ΩT = 1.063 is chosen:
CH = 0.23, CQ = 0.13, and ηT = 0.91.

186
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 12 / Turbomachinery

D=

ω=

4

CHQ2
=
CQ2 gHT

4

0.23 × 0.252
= 0.293 m or approximately 0.30 m
0.132 × 9.81 × 11.8

Q
0.25
30
=
= 71.2 rad/s or N = 71.2 ×
= 680 rpm
3
3
π
CQ D
0.13 × 0.30

W T = γ QHTηT = 9810 × 0.25 × 11.8 × 0.91 = 2.63 × 10 4 W, or 26.3 kW

Calculate a new specific speed based on the final design data:
ΩT =

71.2 × 2.63 ×104 /1000
(9.81×11.8)5/4

= 0.96

An acceptable value according to Fig. 12.20.

187
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 12 / Turbomachinery

188
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.



x

Tài liệu bạn tìm kiếm đã sẵn sàng tải về

Tải bản đầy đủ ngay

×