PELTON WHEEL (LOSSES AND EFFICIENCIES)

Head losses occur in the pipelines conveying the water to the nozzle due to friction and bend. Losses also occur in the nozzle and are expressed by the velocity coefficient,C_v.

The jet efficiency (h_j) takes care of losses in the nozzle and the mechanical efficiency (hm) is meant for the bearing friction and windage losses. The overall efficiency (ho) for large Pelton turbine is about 85 – 90%. Following efficiency is usually used for Pelton wheel.

Pipeline transmission efficiency¼ Energy at end of the pipe Energy available at reservoir Figure 3.3 shows the total headline, where the water supply is from a reservoir at a headH₁above the nozzle. The frictional head loss,h_f, is the loss as the water flows through the pressure tunnel and penstock up to entry to the nozzle.

Then the transmission efficiency is

h_trans^{¼ ðH}12hfÞ/H₁ ¼H/H₁ ð3:5Þ

The nozzle efficiency or jet efficiency is h_j¼Energy at nozzle outlet

Energy at nozzle inlet ¼C²₁/2gH ð3:6Þ

Nozzle velocity coefficient

Cv¼ Actual jet velocity

Theoretical jet velocity¼C1

=

^{pffiffiffiffiffiffiffiffiffi2gH}

Therefore the nozzle efficiency becomes

h_j^¼C²₁

=

^2gH¼C2v ð3:7Þ The characteristics of an impulse turbine are shown in Fig. 3.4.

Figure 3.4 shows the curves for constant head and indicates that the peak efficiency occurs at about the same speed ratio for any gate opening and that

Figure 3.3 Schematic layout of hydro plant.

Figure 3.4 Efficiency vs. speed at various nozzle settings.

the peak values of efficiency do not vary much. This happens as the nozzle velocity remaining constant in magnitude and direction as the flow rate changes, gives an optimum value ofU/C₁at a fixed speed. Due to losses, such as windage, mechanical, and friction cause the small variation. Fig. 3.5 shows the curves for power vs. speed. Fixed speed condition is important because generators are usually run at constant speed.

Illustrative Example 3.1:A generator is to be driven by a Pelton wheel with a head of 220 m and discharge rate of 145 L/s. The mean peripheral velocity of wheel is 14 m/s. If the outlet tip angle of the bucket is 1608, find out the power developed.

Solution:

Dischargerate;Q¼145 L/s Head;H¼220 m

U1¼U2¼14 m/s b₂ ^¼18021608¼208 Refer toFig. 3.6

Using Euler’s equation, work done per weight mass of water per sec.

¼ ðC_w1U₁2C_w2U₂Þ

But for Pelton wheelC_w2is negative

Figure 3.5 Power vs. speed of various nozzle setting.

Therefore

Work done / s¼ ðC_w1U₁þC_w2U₂Þ Nm / s From inlet velocity triangle

Cw1¼C1 and C²₁ 2g¼H Hence,C1¼ ffiffiffiffiffiffiffiffiffi

p2gH

¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2£9:81£220

p ¼65:7 m/s

Relative velocity at inlet is

V₁¼C₁2U₁¼65:7214¼51:7 m/s From outlet velocity triangle

V₁¼V₂¼51:7 m/s(neglecting friction) and cos b2¼^U2þC_V ^w2

2 or

cosð20Þ ¼14þCw2

51:7 Therefore

Cw2¼34:58 m/s

Hence, work done per unit mass of water per sec.

¼ ð65:7Þð14Þ þ ð34:58Þð14Þ ¼1403:92 Nm Power developed¼ð1403:92Þð145Þ

1000 ¼203:57 kW Figure 3.6 Inlet and outlet velocity triangles.

Design Example 3.2:A Pelton wheel is supplied with 0.035 m³/s of water under a head of 92 m. The wheel rotates at 725 rpm and the velocity coefficient of the nozzle is 0.95. The efficiency of the wheel is 82% and the ratio of bucket speed to jet speed is 0.45. Determine the following:

1. Speed of the wheel 2. Wheel to jet diameter ratio

3. Dimensionless power specific speed of the wheel Solution:

Overall efficiencyho¼Power developed Power available [ P¼rgQHhoJ/s¼^rgQHh₁₀₀₀^o kW

¼9:81ð0:035Þð92Þð0:82Þ ¼25:9kW

Velocity coefficient C_v¼ C1

ffiffiffiffiffiffiffiffiffi p2gH orC1¼Cv

ffiffiffiffiffiffiffiffiffi p2gH

¼0:95½ð2Þð9:81Þð92Þ^1/2 ¼40:36 m/s 1. Speed of the wheel is given by

U¼0:45ð40:36Þ ¼18:16 m/s 2. IfDis the wheel diameter, then

U¼vD

2 or D¼2U

v ^¼

ð2Þð18:16Þð60Þ

725ð2p^{Þ ¼0:478m} Jet area A¼ Q

C₁¼0:035

40:36¼0:867£10²³m² and Jet diameter,d, is given by

d¼ 4A p

1/2

¼ ð4Þð0:867£10²³Þ p

1/2

¼0:033m

Diameter ratio D

d ¼0:478

0:033¼14:48

3. Dimensionless specific speed is given by Eq. (1.10) Nsp¼ NP^1/2

r^1/2ðgHÞ^5/4

¼ 725 60

£ ð25:9Þð1000Þ 10³

1/2

£ 1

ð9:81Þ£ð92Þ

5/4

¼ ð12:08Þð5:09Þð0:0002Þ

¼0:0123rev

¼ ð0:0123Þð2p^Þrad

¼0:077rad

Illustrative Example 3.3:The speed of Pelton turbine is 14 m/s. The water is supplied at the rate of 820 L/s against a head of 45 m. If the jet is deflected by the buckets at an angle of 1608, find the hP and the efficiency of the turbine.

Solution:

Refer to Fig. 3.7 U₁ ¼ U₂ ¼ 14 m/s Q ¼ 820 L/s ¼ 0.82 m³/s H ¼ 45 m

b₂ ¼ 18021608 ¼ 208 Velocity of jet

C₁¼Cv ffiffiffiffiffiffiffiffiffi p2gH

, assuming Cv¼0:98

¼0:98 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð2Þð9:81Þð45Þ

p ¼29:12 m/s

Figure 3.7 Velocity triangle for Example 3.3.

Assuming b1¼1808

b2¼18021608¼208 Cw1¼C1¼29:12 m/s

V1¼C12U1¼29:12214¼15:12 m/s From outlet velocity triangle,

U1¼U2(neglecting losses on buckets) V2¼15:12m / s and U2 ¼14 m/s

Cw2¼V2cosa22U2¼15:12 cos 208 214

¼0:208 m/s

Work done per weight mass of water per sec

¼ ðC_w1þC_w2ÞU

¼ ð29:12þ0:208Þ£ð14Þ ¼410:6Nm / s [ Power developed¼ ð410:6Þð0:82£10³Þ

1000 ¼336:7kW

¼451hP Efficiencyh1¼Power developed

Available Power

¼ ð1000Þð336:7Þ

ð1000Þð9:81Þð0:82Þð45Þ ¼0:930 or 93:0%

Illustrative Example 3.4:A Pelton wheel develops 12,900 kW at 425 rpm under a head of 505 m. The efficiency of the machine is 84%. Find (1) discharge of the turbine, (2) diameter of the wheel, and (3) diameter of the nozzle. Assume C_v ¼ 0.98, and ratio of bucket speed to jet speed ¼ 0.46.

Solution:

Head,H ¼ 505 m.

Power,P ¼ 12,900 kW Speed,N ¼ 425 rpm Efficiency,ho ¼ 84%

1. LetQbe the discharge of the turbine Using the relationho ¼ P

9:81QH

or

0:84¼ 12;900

ð9:81Þð505ÞQ¼2:60 Q or

Q¼3:1m³/s 2. Velocity of jet

C¼Cv ffiffiffiffiffiffiffiffiffi p2gH

ðassume Cv¼0:98Þ or

C¼0:98 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð2Þð9:81Þð505Þ

p ¼97:55 m/s

Tangential velocity of the wheel is given by U¼0:46C¼ ð0:46Þð97:55Þ ¼44:87 m/s and

U¼pDN

60 ; hence wheel diameter is D¼60U

pN ¼ð60Þð44:87Þ

ðpÞð425Þ ¼2:016m 3. Letdbe the diameter of the nozzle

The discharge through the nozzle must be equal to the discharge of the turbine. Therefore

Q¼^p₄£d²£C

3:1¼ ð^p₄Þðd²Þð97:55Þ ¼76:65d² [ d¼ ffiffiffiffiffiffiffiffiffiffiffi

3:1 76:65

q ¼0:20m

Illustrative Example 3.5:A double Overhung Pelton wheel unit is to operate at 12,000 kW generator. Find the power developed by each runner if the generator is 95%.

Solution:

Output power ¼ 12,000 kW Efficiency,h¼95%

Therefore, power generated by the runner

¼12;000

0:95 ¼12;632kW

Since there are two runners, power developed by each runner

¼12;632

2 ¼6316kW

Design Example 3.6: At the power station, a Pelton wheel produces 1260 kW under a head of 610 m. The loss of head due to pipe friction between the reservoir and nozzle is 46 m. The buckets of the Pelton wheel deflect the jet through an angle of 1658, while relative velocity of the water is reduced by 10%

due to bucket friction. The bucket/jet speed ratio is 0.46. The bucket circle diameter of the wheel is 890 mm and there are two jets. Find the theoretical hydraulic efficiency, speed of rotation of the wheel, and diameter of the nozzle if the actual hydraulic efficiency is 0.9 times that calculated above. Assume nozzle velocity coefficient,C_v ¼ 0.98.

Solution:

Refer toFig. 3.8.

Hydraulic efficiency h_h¼ Power output

Energy available in the jet¼ P 0:5mC²₁

At entry to nozzle

H¼610246¼564m Using nozzle velocity coefficient

C1¼Cv ffiffiffiffiffiffiffiffiffi p2gH

¼0:98 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð2Þð9:81Þð564Þ

p ¼103:1 m/s

Now

Wm¼U1Cw12U2Cw2

¼U Ufð þV₁Þ2^½U2V₂cos 180^ð 8 2a^Þg

¼U C½ð 12UÞð12kcosaÞwhereV2¼kV1

Therefore,W/m¼0.46C₁(C₁20.46C₁)(120.9 cos 1658) Substitute the value ofC₁

W/m¼5180:95

Theoretical hydraulic efficiency¼ Power output Energy available in the jet

¼ 5180:95

0:5£103²¼98%

Actual hydraulic efficiency¼ ð0:9Þð0:98Þ ¼0:882 Wheel bucket speed¼ ð0:46Þð103Þ ¼47:38 m/s Wheel rotational speed¼N¼ð47:38Þð60Þ

ð0:445Þð2p^Þ¼1016rpm Actual hydraulic efficiency¼ Actual power

energy in the jet¼ð1260£10³Þ 0:5mC²₁ Therefore,m¼ ð1260£10³Þ

ð0:882Þð0:5Þð103²Þ¼269 kg/s For one nozzle,m¼134.5 kg/s

For nozzle diameter, using continuity equation,m¼rC1A¼rC₁pd² 4 Hence,d¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð134:5Þð4Þ ðp^Þð103£10³Þ r

¼0:041m¼41mm

Illustrative Example 3.7:A Pelton wheel has a head of 90 m and head lost due to friction in the penstock is 30 m. The main bucket speed is 12 m/s and the nozzle discharge is 1.0 m³/s. If the bucket has an angle of 158at the outlet andC_v ¼ 0.98, find the power of Pelton wheel and hydraulic efficiency.

Figure 3.8 Velocity triangle for Example 3.6.

Solution:(Fig. 3.9) Head ¼ 90 m

Head lost due to friction ¼ 30 m

Head available at the nozzle ¼ 90230 ¼ 60 m Q¼ 1 m³/s

From inlet diagram C₁¼Cv

ffiffiffiffiffiffiffiffiffi p2gH

¼0:98£ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð2Þð9:81Þð60

p Þ ¼33:62 m/s

Therefore,V₁¼C₁2U₁¼33.62212¼21.62 m/s From outlet velocity triangle

V₂¼V₁¼21:16 m/s(neglecting losses) U₂¼U₁¼12 m/s

Cw2¼V2cosa₂U2¼21:62 cos 158 212¼8:88 m/s Figure 3.9 Velocity triangle for Example 3.7.

and

Cr2¼V2sina¼21:62 sin 158¼5:6 m/s Therefore,

C2¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C²_w2þCr²₂

q ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð8:88Þ²þ ð5:6Þ² q

¼10:5 m/s [ Work done¼C²₁2C²₂

2 ¼ð33:62Þ²2ð10:5Þ²

2 ¼510 kJ/kg

Note Work done can also be found by using Euler’s equation (C_w1U₁þ C_w2U₂)

Power¼510kW Hydraulic efficiency

h_h^{¼ work done}

kinetic energy¼ð510Þð2Þ

ð33:62Þ²¼90:24%

Design Example 3.8: A single jet Pelton wheel turbine runs at 305 rpm against a head of 515 m. The jet diameter is 200 mm, its deflection inside the bucket is 1658and its relative velocity is reduced by 12% due to friction. Find (1) the waterpower, (2) resultant force on the bucket, (3) shaft power if the mechanical losses are 4% of power supplied, and (4) overall efficiency. Assume necessary data.

Solution:(Fig. 3.10)

Velocity of jet,C1¼Cv ffiffiffiffiffiffiffiffiffi p2gH

¼0:98 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð2Þð9:81Þð515

p Þ ¼98:5 m/s

Discharge,Qis given by

Q¼Area of jet£Velocity¼p

4£ð0:2Þ²ð98:5Þ ¼3:096 m³/s 1. Water power is given by

P¼rgQH¼ ð9:81Þð3:096Þð515Þ ¼15641:5 kW 2. Bucket velocity,U₁, is given by

U₁¼Cv ffiffiffiffiffiffiffiffiffi p2gH

¼0:46 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð2Þð9:81Þð515Þ

p ¼46 m/sðassumingCv ¼0:46Þ Relative velocity,V₁, at inlet is given by

V₁¼C₁2U₁ ¼98:5246¼52:5 m/s

and

V₂¼0:88£52:5¼46:2 m/s From the velocity diagram

Cw2¼U22V2cos 15¼46246:2£0:966¼1:37 m/s Therefore force on the bucket

¼r^QðCw12Cw2Þ ¼1000£3:096ð98:521:37Þ

¼300714N

3. Power produced by the Pelton wheel

¼ð300714Þð46Þ

1000 ¼13832:8kW Taking mechanical loss¼4%

Therefore, shaft power produced¼0.96£13832.8¼13279.5 kW 4. Overall efficiency

h_o ¼13279:5

15641:5¼0:849 or 84:9%

Figure 3.10 Velocity triangles for Example 3.8.