Hydroelectric Power Plant Problems and Solutions

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Hydroelectric Power Plant Problems and Solutions Hydroelectric Problems 1.0 Find the torque and power transferred to the buckets of an impulse wheel with α= 00, β= 1600, κ= 0.44,

∅=0.46 , Cv = 0.98, a jet diameter of 10 inches and a pitch

diameter of 10ft. Find also, the hydraulic efficiency, and expressed as a percentage of the total head, find the head loss in bucket friction, the energy head loss at discharge and the head loss in the nozzle.

Solution: For operation under the purely artificial value of 1-ft net head, u=

∅ √ 2 g=8.02 ∅ = 3.68 fps,

V1 = 0.98

√ 2 g=¿ 7.86fps

yQ = yAV = (62.4)(0.545)(7.86) = 267lb/s Hence the expression for the torque exerted on the wheel by the water is: T = (r)(Fu) = 5(267/32.2)[ 1+ (0.940/1.2)](8.02)(Cv - ∅ ) where Cv = 0.98 and

∅=¿ 0.46,

T = 309 ft-lb The power transferred from the water to the bucket is FuU = Tω = T(u/r) = 309[(0.46)(8.02)/5] = 228 ft-lb/s The power input is: yQh = 267(1) = 228 ft-lb/s The hydraulic efficiency is nh =

228 267

= 0.85.

Note that the answer given here are for h=1ft. For the values of h, the flow rate, torque, and power can be found by adjusting the values for h = 1ft according to similarity laws.

Under 1-ft head υ1 = V1 – u = 7.86 – 3.68 = 4.18 fps and υ2 = υ1/

√ 1+k = 4.18/ √ 1.44=3.48 fps

Solving for the head loss in the bucket friction is 0.44(3.48)2/ 2g = 0.083 ft or 8.3% V2cosα2= u + V2cosβ2 = 3.68 + 3.48cos1600 = 3.68 – 3.27 = 0.41 fps V2sinα2 = V2sinβ2 = 3.48sin1600 = 1.19 fps Hence cotα2= 0.41/1.19 = 0.344 0r α2 = 710 V2 = 1.19/ 0.945 = 1.26 fps therefore the energy head loss at the discharge is: 1.262/2g = 0.025ft, 2.5 percent.

The head loss in the nozzle is approximately (1/C 2 – 1) 2g, about 4 percent; so the total hydraulic loss is 8.3 + 2.5 + 4.0 = 14.8%, which gives a close check on the computed hydraulic efficiency of 85%.

2.0 A tangential turbine develops 7200 hp at 200 rpm under a head of 790 ft at an efficiency of 82 percent. (a) If the speed factor is 0.46 compute the wheel diameter, the flow, the unit speed, unit power, unti flow, and specific speed. (b) For this turbine, what would be the speed, power and flow under a head of 529 ft? (c) For a turbine having the same design, what size of wheel should be used to develop 3800 hp under a 600 ft head and what would be its speed and rate of discharge? Assume no change in efficiency.

Solution: (a) Since Ø =

D

D1 = 1840

N / (1840

1

√H

√ 790 (0.46) / 200

D1 = 119 in From horsepower output = γQh /550 Q = (7200) (550) / [(62.4) (790) (0.82)] Q = 98 cfs Nu = ND1 /

√H

Nu= (200) (119) /

√ 790

Nu = 847 rpm Qu= Q / D12

√H

Qu = 98 / [(119)2

√ 790

Qu= 0.000246 cfs Pu = P / D12 H3/2 Pu = 7200 / [(119)2 (790)3/2] Pu = 0.0000229 hp

√ P / H5/4 Ns =200 √ 7200 / (790)5/4 Ns = N

Ns = 4.05 rpm (b)

Speed, N = Nu N = 847

/ D1

√H

√ 529 / 119

N = 164 rpm Power, P = PuD12H1/2 P = 0.0000229 (119)2 (529)3/2 P= 3946 hp Flow, Q = QuD12

√H

Q = 0.000246(119)2

√ 529

Q = 80.1 cfs The above three quantities might have been obtained by noting, for the same turbine (D1 unchanged), the speed varies as H 1/2, the power varies as H3/2, and Q varies as H1/2. Thus,

529 790

= 164 rpm

P = 7200(

529 ) 790

3/2

Q = 0.98

√

N = 200

√

529 790

= 3945 hp

= 80.2cfs.

(c) From P = PuD12H3/2 we obtain 3800 = 0.0000229(D 1)2(600)3/2 from which D12 = 11290 and D1 106 in.

N =NU

√ H / D1 N =847 √ 600 / 106 N =196 rpm

Q = QUD12

√H

Q = 0.000246(11290)

√ 600

Q = 68 cfs

3.0. A double-overhung impulse-turbine installation is to develop 20 000hp at 257rpm under a net head of 1120 ft. Determine N, wheel-pitch diameter and approximate jet diameter. And determine also the N and jet diameter of a single wheel with a single nozzle and in four nozzles.

For Double Over-hung: Solving for the Ns : Ns = NP0.5 / H5/4 = (257)(20 000/2)0.5/ 11205/4 = 3.97 Solving for the jet diameter (Dj): D =153.2 ∅

√ h N = (153.2)(0.45)( √ 1120 ¿/257 = 8.98 ft

Dj = D/12 = 8.98/12 = 0.7483 ft or 8.98 inches For Single wheel with a single nozzle: Solving for the Ns : Ns = NP0.5 / H5/4 = (257)(20 000/1)0.5/ 11205/4 = 5.61 Solving for the jet diameter (Dj): Note that the Dj is unchanged; hence Dj = 8.98 inches

For the single wheel with four nozzle:

Solving for the Ns : Ns = NP0.5 / H5/4 = (257)(20 000/4)0.5/ 11205/4 = 2.80 Solving for the jet diameter (Dj): Note that the Dj is unchanged; hence Dj = 8.98 inches 4.0 The Francis turbine which the test curves are shown in Fig 21-4a and b has a 27-indiameter runner and a maximum efficiency of 88% when discharging 38.8 cfs and developing 550 bhp at 600rpm under a net head of 141.8 ft. Compute Ns, Cr, assuming B/D =0.15.

Solution:

∅

, and

e

Solving for the value of Ns: Ns = NP0.5 / H5/4 = (600)(550)0.5 / 141.85/4 = 28.8

Solving for the ∅ =

∅ e:

πND √ 2 gh

using the derived formula ND = 153.2 ∅ √ H (600)( ∅

e

27 ) = (153.2)( ∅ 12

e

)

√ 141.8

= 0.740

Solving for the Cr: Cr = N2 / [ (63,800)(

∅

)2( B/D)(n)]

e

= (28.8)2 / [( 63 800)(0.740)2(0.15)(0.88)] Cr = 0.1799

5.0 A Kaplan turbine develops 10000 kW under a head of 12m when the following conditions prevail. Speed ratio=2, flow ratio=0.65, diameter of hub=0.3 times the external diameter of the vane and the overall efficiency=94 percent. Estimate (a) the speed (b)the diameter of the runner and (c) the specific speed.

Solution P= ρQgHŋ × 10−3 10000=Q(9.81)(12)(0.94) Q=90.37m3/s Flow ratio, ∅' =

Vn =0.65 √ 2 gH

Vf1=0.65

√ 2× 9.81× 12=9.97 m/ s

Area of flow, Ab=

90.37 =9.064 m2 9.97

Ab=

π 2 2 ( D −d h )=9.064 4

D2-(0.3D)2= D2=

9.064 × 4 =11.54 π

11.54 =12.682 0.91

Runner Diameter, D = 3.56 m.

Ans (b)

Speed ratio. ∅=¿

2=

Vb Vb = √2 gH √ 2 ×9.81 ×12

Vb=30.69 m.s= N=

πDN 60

30.69 × 60 =164.6∨165 rpm π ×3.56

Synchronous speed, N=

120 f p

If we take

120 f 120× 50 =165,then p= =36.36 p 165

Let 36 poles or 18 pairs of poles are taken.

Then, N= Ns=

N √P H

5 4

120 ×50 =166.7 rpm 36 =

166.7 √10000 5

25 4

Ans (a)

=746

Ans (c)

6.0 A turbine is to be installed at a point where the available head is 175 ft and the available flow will average 1000cfs. What type of turbine would you recommend?

Specify the operating speed and number of generator poles for 60-cycle electricity if a turbine with the highest tolerable specific speed to safeguard against cavitation is selected. Assume static draft head of 10ft and 90 percent turbine efficiency.

Approximately what size of turbine is required?

Solution Ns=NP0.5/H5/4 From Fig. A-42 for H=175ft and draft head = +10 ft, N s=≤48.

Bhp=ŋQ γH /550=¿ (0.90)(1000)(62.4)(175)/550=17 869 48=(N)(17869)0.5/1755/4 N=229rpm Npoles=7200/N=

7200 =31 ; use 32 poles 229

32=7200/N; N=225 rpm Ns=(225)(17 869)0.5/1755/4=47.3; Use a Francis Turbine.

From Fig. A-43 for Ns=47.3,

∅=0.77,

H ( 153.2 ) ( 0.77 ) ( √ 175 ) D=153.2 ∅ √ = =6.94 ft .

N 225

7.0 A Pelton wheel of 12-ft pitch diameter operates under a net head of 2000 ft.

Estimate the speed, power output and flow rate for best efficiency if the nozzle exit diameter is 4 in.

Solution: Vjet = Cu

√ 2gH

Take Cu = 0.94 Vjet = (0.94)

√ 2(32.2) (2000)

Vjet = 337 ft/s Ø = πDN/ (2gH)

1/2

Estimate Ø = 0.47 0.47 =

(π) (N/60) (12) ] / [(2) (32.2) (2000)] ¿

N = 268 rpm Q = AV Q = [ π) (4/12) ¿

2

/ 4] (337)

Q = 29.4 ft3/s Pw = ρQu(Vjet – u)(1-cosβ) Assume β = 165o u = Vjet/2 u=

337 ft/s 2

u = 168.5 fps

1/2

Pw = (1.94) (29.4) (168.5) (337-168.5) (1-cos165o) / 550 Pw = 5788 hp Ns = NP0.5/H5/4 = (268) (5788)0.5 / 20005/4 Ns= 1.52 Hence, η = 0.75 P= (0.75) (5788hp) P = 4341 hp 8.0 A Pelton wheel is to be selected to drive a generator at 600 rpm. The water jet is 75 mm in diameter and has a velocity of 100m/s. With the blade angle at 170°, the ratio of vane speed to initial jet speed at 0.47, and neglecting losses, determine (α) diameter of wheel to entreline of buckets (vanes), (b) power developed, and (c) kinetic energy per newton remaining in fluid.

Solution: (α) The peripheral speed of the wheel is u = 0.47(100) = 47m/s.

Then

600 [2π (D/2)] = 47m/s or D = 1.496m.

60

(b) The power, in kilowatts, is computed as P = Qu (V) (1-cosΘ) P= (1000) (π/4) (0.075)2(100) (47) (100 – 47) [1-(-0.9848) (

1 ) =2184kW. 1000

(c) From figure below, the absolute-velocity components leaving the vane are: Vx = (100-47) (-0.9848) +47= -5.2 m/s, Vy = (100-47) (0.1736) = 9.2m/s.

The kinetic energy remaining in the jet is (5.2 2+9.22) / [2(9.807)] = 5.69 m-N/N.

9.0 A turbine is to operate at 400 rpm under a net head of 1320 ft. If a single 6-inches diameter water jet is used, find the specific speed of this machine assuming C v = 0.98, ∅=0.45, and n = 0.85. Find the required pitch diameter of the wheel.

Solution: Solving for velocity: V=

Cv √ 2 gh

= 0.98

√ ( 64.4 ) ( 1320 )=286 fps

Solving for the value of Q : Q = AV = (0.196)(286) = 56.0 cfs Solving for Bhp: Bhp = n (yQh/550) = 0.85[( 62.4)(56)(1320)/550] = 7130 Solving for Ns: Ns = N

√ Bhp / h5/4 = 400 √ 7130 /(1320)5/4 = 4.25

Solving for u:

u=

∅ √ 2 gh

= 0.45

√ ( 64.4 ) ( 1320 )

= 131.2fps

Solving for the pitch diameter N=

u (60) πD

;D=

u (60) πN

=

131.2(60) = 6.26ft or 75.2 inches π ( 400)

10.0 A Pelton type turbine was installed 30m below the head gate of the penstock. The head loss due to friction is 15 percent of the given elevation. The length of the penstock is 80m and the coefficient of friction is 0.00093. Determine the power output in KW. Solution: Head Loss, hL hL = 0.15( 30m ) hL = 4.5m Net Head, h h = 30m - 4.5m h = 25.5m v=

√ 2 gh

v=

9.81m ( √ sec )( 25.5m) 2

2

v = 22.368 m/ sec 2 hL =

2 fL V 2 gD 2

4.5m =

2 ( 0.00093 ) (80 m)(22.368) (9.81 m/sec 2) D

D = 1.686 m

Q=AxV Q = ( π/4 )( 1.686 m )( 22.368 m2 ) Q = 49.94 m3 /sec Power = w Q h Power = (9.81m/s2)(49.94 m3 /sec)(25.5m) Power = 12,492.74 KW