Solve Problem of Hydroelectric Power Plant

HYDROELECTRIC POWER PLANT PROBLEMS AND SOLUTION

1. A Pelton wheel is required to develop 4500 kW at 400 rpm operating under an available head of360 m. There are two equal jets and the bucket angle is 170 0. The bucket pitch circle diameter is 1.82 m. Taking k for the buckets as 0.85, determine (a) the efficiency of the runner and (b) the diameter of each jet. Given:

Required: P = 4500 kW N = 400 rpm H = 300 m N=2 ϴ = 1700 D = 1.82 m k 0.8

(a) ηD = ? (b) d = ?

Solution: v 1= √ 2 gH

√(

¿ 2 9.81

¿ √ 2 gH k v m ( 360 m )( 0.85 ) s2

)

¿ 77.4836757

m s

vb =

¿ 38.11799086

(a)

η D=

πDN = 60

m s

1−kcosθ 2 2g v 1 v b −v b ) 2 ( g v1

(

π ( 1.82 m ) 40 60

m min

)

0

¿

[

η D=0 . 9183

(b) Power develop, P=

4500 kW =4900.36 kW 0.9183

Power develop per jet, P=

4900.36 kW =2450.18 kW 2

P=γQH =γA V 1 H =γ

π d2 V1H 4

( )

2

4 P=γπ d V 1 H 2

d=

d=

4P γπ V 1 H

√

( )]

1−0.85 cos 170 m (77.48 )( 78.12 )−( 38.12 )2 m s 9.81 2 s

4 ( 2950.18 kw ) m ( 9.81 )( π ) 77.4836757 (360 m ) s

(

d=0.106773422 m

)

2

[ ] m 2 s 2 m 77.48 s

2 9.81

(

)

(

)

2. A Pelton wheel develops 8 MW under a net head of 130 m at a speed of 200 rpm. Assuming cv = 0.98, hydraulic efficiency = 87 percent, speed ratio = 0.46 and the ratio of jet-to-wheel diameter = 1/9, determine (a) the flow required (b) the diameter of the wheel (c) the diameter and number of jets of jets needed. Given: P = 8 MW H = 130 m N = 200 rpm Cv = 0.98 η D=87 ϕ=0.46 m=

1 4

a.) Q = ? b.) D = ? c.) d = ? ; n = ? Solution: a.) Q = ? P=γQH η D Q=

P = γH η D

(

8000 kw =7.21 m3 /s kN 9.81 3 ( 130 m ) (0.87) m

)

b.) D = ? V b=

πDN 60

√(

V b=ϕ √ 2 gH =( 0.46 ) 2 9.81

m m ( 130 m ) =23.231 s s

)

( ms ) =2.22m rev π (200 min )

( 60 ) 23.231 D=

c.) d=? m=

D d

D 1 = d 9 d=

2.22 m =0.2467 m 9 n=?

n=

7.21 π ( 0.2467 )2 0.98 √ 2 ( 9.81 )( 130 ) ( ) 4

=3

3. A Pelton wheel has to develop 12 MW under a head of 300 m at a speed of 500 rpm. If the diameter of the jets is not to exceed 1/9 of the wheel diameter, estimate the number and diameter of the bucket wheel and the quantity of flow. Assume overall efficiency = 88 percent, cv = 0.97 and Ф = 0.45.

Given: P = 12 MW H = 300 rpm N = 500 rpm d 1 = D 9 ηo =88

C v =0.97 ϕ=0.45

Solution: Q=? V 1=C v √ 2 gH =0.97 √ 2 ( 9.81 ) (300 m)=74.42 m/s Q= A V 1 P=QγH η0

√(

V b=ϕ √ 2 gH =0.45 2 9.81

Q=

P = γH η o

(

D=

)

πDN 60

60(34.52 mls) =1.32 m π ( 500 rpm )

d=? 1 d = 9 D d=

)

12000 kw =4.63 m3 / s kN 9.81 3 ( 300 m) (0.88) m

D=? V b=

m m ( 300 m )=34.52 2 s s

1.32 =0.147 m 9

n=? Q= A V 1 n 3

n=

[

4.63 m /s =3.67∨4 2 π (0.147) ( 74.42 m/s ) 4

]

4. A Pelton wheel to be designed is to run at 300 m under an effective head of 150 m. The ratio of the nozzle diameter to the pitch circle diameter is 1/12. Assuming efficiency = 84%, cv = 0.98 and speed ratio = 0.45, determine (a) diameter of the wheel, (b) diameter of the jet (c) the quantity of water flow (d) the minimum number of buckets required; and (e) the power developed. Given: N = 300 rpm H = 150 m d 1 = D 12 η1=84 C v =0.98 ϕ=0.45 Solution:

√(

V 1=C v √2 gH =0.98 2 9.81

√(

V b=0.45 2 9.81

a.) D = ?

m m ( 150 m )=53.16 2 s s

)

m (150 m)=24.41mls s2

)

V b=

D=

πDN 60

60(24.41 mls) =1.55 m π (300 rpm)

b.) d =? d 1 = D 12 d=

1.55 =0.129 m 12

c.) Q = ?

[

2

]

π (0.129) Q= ( 53.16 mls )=0.695 m3 /s 4 d.) z = ? z=

m 12 +15= + 15=21 2 2

e.) P = ?

(

P= 9.81

kN m3 0.695 ( 150 m ) ( 0.84 )=859.0617 k 3 s m

)(

)

5. A francis turbine with an overall efficiency of 76 percent is required to produce 180kW. It is working under a head of 8 m. The peripheral velocity is 0.25 (2gH) 1/2 and the radial velocity of flow is 0.95 (2gH)1/2. The wheel runs at 150 rpm and the hydraulic losses in the turbine are 20

percent of the available energy. Assuming radial discharge, determine (a) the guide vane angle (b) the wheel vane angle at inlet (c) the diameter of the wheel at inlet. Given: ηo = 76%

Peripheral velocity = 0.25 (2gH)1/2 = Vf = 11.902 m/s

P = 180 kW

Radial velocity = 0.95 (2gH)1/2 = Vb = 3.1321m/s

H=8m

N = 150 rpm

Solution:

o 

shaft power 180 kW   0.76 water power QgH

o 

180 kW  0.76 kg m 1 kN Q (1000 3 ) (9.81 2 ) ( ) (8 m) m s 1000 N

180 kW kg m 1 kN  o (1000 3 ) (9.81 2 ) ( ) (8 m) m s 1000 N 180 kW  kg m 1 kN (0.76) (1000 3 ) (9.81 2 ) ( ) (8 m) m s 1000 N

Q

Q  3.018

m3 s

P  QgH h

h 

P  QgH

 h  0.76

180 kW m kg m 1 kN (3.018 ) (1000 3 ) (9.81 2 ) ( ) (8 m) s m s 1000 N 3

Vw Vb gH gH h (9.81)(8)(. 76) Vw   Vb 3.1321

h 

Vw  19.043

tan  

m s

Vf Vw  Vf  11.902    tan     19.043   Vw 

  tan  

  32.00565   32 0' 20.35"

ans (a)

Vb  DN D

Vb 3.1321  N  150     60 

D  0.399 m

ans (c)

Vw = Vb + X X = Vw + Vb = 19.043 m/s – 3.1321 m/s X = 15.911 m/s

tan  

Vf

X  Vf   tan    X

 

  11.902   tan     15.911    36.7978

  36 47 ' 52.08 "

ans (b)

6. An inward flow reaction turbine having an overall efficiency of 75 percent delivers 132 kW. The head H is 9 m, velocity of the periphery of the wheel is 54 m/s and the radial velocity is 18 m/s. The wheel makes 120 rpm. The hydraulic losses in the turbine are 20 percent of the available energy. Determine the discharge at the inlet, the guide blade angle, the wheel blade angle and the diameter and width of the wheel. Assume radial discharge.

Given: Vf = 18 m/s ηo = 25 %

N = 120 rpm

P = 132 kW

Vb = 54 m/s

H=9m Solution: a)

Q=?

ηl = 20 %

P  QH o Q

P H o

Q

132 kW 9.81 m s 2  9 m  0.75

Q  1.99 m

s

∝1=?

b)

∝1=tan−1

ηH=

3

vf1 vw 1

vw 1 vb1 gH o

0.8=

( v w1 ) ( 54 m/s ) ( 9.81m/s 2 ) ( 9 m )

v w 1=1.308 m/ s α 1=tan −1

vf 1 vw 1

α 1=tan −1

18 m/s 1.31 m/s

α 1=850 50'

c) β1 = ?

η H =1−η L =1−0.2=0.8

β 1=tan −1

β 1=tan

−1

0

vf 1 v b 1−v w 1 18 m/ s (54−18 ) m/ s

β 1=26 33

'

d) D = ? v b 1=πDN D=

54 m/ s π ( 120 rev /min ) ( 1min / sec )

D=8.60 m e) B1 = ? Q= AV Q=πD B1 v f 1

B 1=

1.99 m3 /s π ( 8.6 m/s ) (18 m/ s )

B 1=0.0041 m

7. From the performance curves of turbines it is seen that a turbine of one meter diameter under a head of one meter develops a speed of 25 rpm. What diameter should the prototype have it is to develop 1000 kW, working under a head of 200 m with a specific speed of 150? Given: Nm = 400 m

Q = 4.5 m3/s ηt = 82% Ns = 60 rpm

Solution: Dp = diameter for prototype = ?

(√ )

Dm = DP

D p=1 m

H p Nm Hm Np

25 rpm (√ 2001 mm ) 150 rpm

D p=2.36 m

8) The following data relates to a hydroelectric plant, Head = 400 m, Discharge = 4.5 m3/s. Turbine efficiency = 82%, Specific speed = 60, Determine (a) the power develop (b) the type of turbine (c) the speed of the turbine. Given: H = 400 m Q = 4.5 m3/s ηt = 82% Ns = 60 rpm Solution:

a) P = ? P=QγH ηt P=( 4.5 m3 /s ) ( 9.81m/ s2 ) ( 400 m )( 0.82 ) P=14479.56 kW

b) The Pelton turbine which has a capacity from 300 m up to 2000 m of heads.

c) N = ? N√P

NS=

H

60=

5 4

N √ 14479.56 kW 5

400 4

N=891.97 rpm

9) A runner blade is to be designed for a Francis turbine, 1.5 m outer diameter and 0.75 m inner diameter, to operate under a head of 120 m with a specific speed of 150 and generate 14000 kW. Assume hydraulic efficiency is 92 percent. What should be the inlet and outlet blade angles if the water has to enter a wheel an angle of 11020’ and leave the blade with no velocity of whirl? Given: Outer diameter = 1.5 m Inner diameter = 0.75 m H = 120 m Ns = 150 rpm P = 14000 kW ηH = 92% α1 = 11020’ Solution:

NS=

N √P H

N=

5 4

150 ( 120 m ) √ 14000 kW

N=503.505 rpm v w =π Do N v w =π ( 1.5 m )( 503.505 rev /min ) v w 1=39.545

tan α 1=

( 601 minsec )

m s

vf1 vw 1

v f 1=v w1 tanα 1 v f 1=( 39.545 m/s )( tan 11.333 ) v f 1=7.926 m/s

η D=

v w1 v b 1 gH

v b 1=

( 9.81 m/s 2 ) ( 120m ) ( 0.92 ) 39.545 m/s

v b 1=27.387 m/s

x=v w1 −v b x=12.16 m/s

β 1=tan

−1

vf 9.23 m/ s = x ❑ 12.16 m/ s

0

0

β 1=33.10 =33 6

β 2=tan −1

[

'

vf 1 v 7.23 m/s = f1 = v b 1 πDN π ( 0.75 m )( 503.51 rev /min )( 1 min/60 sec ) β 2 = 21.840 =210 50'

]