unit v turbines 2 [bhaskar]
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9/11/2012 Dr. L. Bhaskar 1
• The draft tube is a pipe of gradually increasing area which connects the outlet of the runner with the tailrace. One end of the draft tube is connected to the outlet of the runner while the other end is submerged below the level of water in the tail race.
• It creates a negative head at the outlet of the runner thereby increasing the net head on the turbine.
• It converts a large proportion of rejected kinetic energy into useful pressure energy
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Turbine Specific Speed (rpm)
Pelton 10 – 30Turgo 20 – 70Cross flow 20 – 200Francis 30 – 400Propeller and Kaplan 200 - 1000
Generally, the turbine manufacturer specifies thespecific speed of its turbine.
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Specific Speed (rpm) Turbine
MKS SI
10 to 35 8.5 to 30 Pelton wheel with single jet
35 to 60 30 to 51 Pelton wheel with 2 or more jets
60 to 300 51 to 225 Francis turbine
300 to 1000 255 to 860 Kaplan or Propeller
Note: Generally, the turbine manufacturer specifies thespecific speed of its turbine.
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MORE ADAPTED TYPE OF TURBINE AS FUNCTION OF THE SPECIFIC SPEED.
Specific Speed in r.p.m. Turbine type Jump height in m
Until 18 Pelton of an injector 800
From 18 to 25 Pelton of an injector 800 to 400
From 26 to 35 Pelton of an injector 400 to 100
From 26 to 35 Pelton of two injectors 800 to 400
From 36 to 50 Pelton of two injectors 400 to 100
From 51 to 72 Pelton of four injectors 400 to 100
From 55 to 70 Very slow Francis 400 to 200
From 70 to 120 Slow Francis 200 to 100
From 120 to 200 Normal Francis 100 to 50
From 200 to 300 Quick Francis 50 to 25
From 300 to 450 Extra-quick Francis 25 to 15
From 400 to 500 Extra-quick helix 15
From 270 to 500 Slow Kaplan 50 to 15
From 500 to 800 Quick Kaplan 15 to 5
From 800 to 1100 Extra-quick Kaplan Less than 5
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1) Gross Head
Difference Between the Head race level and Tail race level
Static (No water flow) / Total Head – H1
2) Net or Effective Head
Head available at the entrance of the turbine: H = H1 - hf
a) Net Head for a Reaction TurbineH = {(P1/w) + (V1
2/2g) + Z1} – {Z2 + V22/2g)}
b) Net Head for Impulse TurbineH = {(P1/w) + (V1
2/2g) + Z1} – Z2
Head of Hydraulic TurbinesHead of Hydraulic Turbines
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Efficiencies of hydraulic Turbines
• Hydraulic efficiency
• Mechanical efficiency
• Volumetric efficiency
• Overall efficiency
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Efficiencies of Hydraulic TurbinesEfficiencies of Hydraulic Turbines1) Hydraulic Efficiency – due to hydraulic losses
Power developed by the runnerNet power supplied at the turbine entrance
SI Unit: kWMetric Unit : Horse Power/Water Horse Power
(W.H.P)2) Mechanical Efficiency – Due to mechanical losses (
bearing friction)
Power available at the turbine shaft (P)Power developed by the runner
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3) Volumetric Efficiency – due to amt of water slips directly to the tail race
Amount of water striking the runnerAmount of water supplied to the turbine
4) Overall Efficiency
Power available at the turbine shaft (P)Net power supplied at the turbine entrance
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Hydraulic efficiency
• Hydraulic efficiency =• Power delivered to runner/power
supplied at inlet• = R.P/W.P
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Efficiencies of a Turbine• R.P = For Pelton Turbine
= For a radial flow Turbine
• W.P = Power supplied at inlet of turbine and also called water power
• W. P =• Where W is weight of water striking the vanes
of the turbine per second=ρ g Q • Vw1 = Velocity of whirl at inlet• Vw1 = Velocity of whirl at outlet
[ ] kwuVVg
W ww
100021 ×±
[ ]kwuVuVg
W ww
10002211 ±
kwHQwkwHW10001000
××=
×
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Efficiencies of a Turbine• Mechanical efficiency• Power at the shaft of the turbine/power
delivered by water to the runner• = S.P/R.P• Volumetric efficiency• Volume of water actually striking the
runner/Volume of water supplied to the turbine
• Overall efficiency• Power available at the shaft of the
turbine/Power supplied at the inlet of the turbine=Shaft power/water power
• Overall Eff. = Mechanical eff. x Hydraulic eff.
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• R.P → W. P → S.P• Hydraulic efficiency= ηh = W.P/R.P• Mechanical efficiency= ηm = S.P/R.P• Overall efficiency= ηo = S.P/W.P• R.P = Power delivered by water to
runner• W.P = Power supplied at inlet of
turbine also called as water power• S.P = Power available at the shaft of
the turbine
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• It is the operation by which the speed of the turbine is kept constant under all conditions of working load. This is done automatically by a governor which regulates the rate flow through the turbines according to the changing load conditions on the turbine.
• Governing of a turbine is absolutely necessary if the turbine is coupled to an electric generator which is required to run at constant speed under all fluctuating load conditions.
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Problems based on Specific speed• Q1. An hydraulic turbine develops 8.5 MW
power under a head of 30 meters at 142 rpm. Determine the specific speed of the turbine and state the type of the turbine for this application.
• Given: P = 8.5 MW = 8.5 x 10 6 W.h = 30 m.N = 142 rpm.
• To Find: Specific speed of turbine, NS and type of turbine
• Solution:
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Problems based on Specific speed• Q2. A turbine is to operate under a head of
30 m at 250 rpm. The discharge is 10.5 cumec. The efficiency of the turbine is 89%. Determine the following. (i) Specific speed of the turbine. (ii) Power generated by the turbine and (iii) Type of turbine.
• Given: h = 30 m.N = 250 rpm.Q = 10.5 cumec = 10.5 m3/secηo = 89% = 0.89
• To Find: (i). NS (ii). P and (iii). type of turbine
• Solution:
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Unit quantities• The unit operating conditions for a turbine
are those under which that particular turbine would run when working under a head of one ft. ( or unit head in any other system)assuming there to be no change in efficiency.
• This allows the performance of a given turbine to be compared when working under different heads and enables the characteristic curves to be drawn which show the efficiency at all running conditions.
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Unit Speed
• It is defined as the speed of a turbine working under a head of 1 m. It is denoted by Nu .
• Unit speed =HNNu =
2
2
1
1
HN
HNNu ==
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Unit Discharge
HQQu =
1
2
1
1
HQ
HQQu ==
It is defined as the discharge passing through a turbine, which is working under a head of one metre. It is denoted by Qu . The expression for unit discharge is given by
Unit discharge =
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Unit Power
• It is defined as the power developed by a turbine working under a head of 1 m. It is denoted by Pu .
• Unit power = 2/3HPPu =
2/32
22/3
1
1
HP
HPPu ==
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Problems based on unit quantities• Q3. An hydraulic turbine develops 10,000 kW
power under a head of 30 meters at 110 rpm. Determine the speed and power developed by the turbine when the head on the turbine is reduced to 20 m.
• Given: P1 = 10,000 kWh1 = 30 mN1 = 110 rpmh2 = 20 m
• To Find: Speed, N2 and power, P2 of the turbine
• Solution:
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Characteristic curves of a TurbineThese are curves which are characteristic of a particular turbine which helps in studying the performance of the turbine under various conditions. These curves pertaining to any turbine are supplied by its manufacturers based on actual tests.The data that must be obtained in testing a turbine are the following:
1. The speed of the turbine N2. The discharge Q3. The net head H4. The power developed P5. The overall efficiency ηo6. Gate opening (this refers to the percentage of the
inlet passages provided for water to enter the turbine)
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Characteristic curves of a TurbineThe characteristic curves obtained are the following:
a) Constant head curves or main characteristic curvesb) Constant speed curves or operating characteristic
curvesc) Constant efficiency curves or Muschel curves
Constant head curves: Maintaining a constant head, the speed of the turbine is varied by admitting different rates of flow by adjusting the percentage of gate opening. The power P developed is measured mechanically. From each test the unit power Pu, the unit speed Nu, the unit discharge Qu and the overall efficiency ηo are determined. The characteristic curves drawn are
a) Unit discharge vs unit speedb) Unit power vs unit speedc) Overall efficiency vs unit speed
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Characteristic curves of a TurbineConstant speed curves: In this case tests are conducted at a constant speed varying the head H and suitably adjusting the discharge Q. The power developed P is measured mechanically. The overall efficiency is aimed at its maximum value.The curves drawn are
P vs Qηo vs Qηo vs Puηo max vs % Full load
Constant efficiency curves: These curves are plotted from data which can be obtained from the constant head and constant speed curves. The object of obtaining this curve is to determine the zone of constant efficiency so that we can always run the turbine with maximum efficiency.This curve also gives a good idea about the performance of the turbine at various efficiencies.
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Main Characteristic curves of a Pelton turbine
• N is constant100% GO
75% GO
50% GO
25% GO
Nu
Qu
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Main Characteristic curves of a Pelton turbine
• N is constant
100% GO
75% GO50% GO
25% GO
Nu
ηo
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Main Characteristic curves of a Pelton turbine
• N is constant
100% GO
75% GO
50% GO
25% GO
Nu
Pu
Main Characteristic curves of a Pelton turbine
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Main Characteristic curves of a Kaplan turbine
• N is constant
100% GO75% GO
50% GO
25% GO
Nu
Qu
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Main Characteristic curves of a Kaplan turbine
• N is constant
100% GO
75% GO
50% GO25% GO
Nu
ηo
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Main Characteristic curves of a Kaplan turbine
• N is constant
100% GO
75% GO50% GO
25% GO
Nu
Pu
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Main Characteristic curves of a Francis turbine
• N is constant
100% GO75% GO
50% GO
25% GO
Nu
Qu
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Main Characteristic curves of a Francis turbine
• N is constant
100% GO
75% GO
50% GO25% GO
Nu
Pu
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Main Characteristic curves of a Francis turbine
• N is constant
100% GO
75% GO50% GO
25% GO
Nu
ηo
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Operating Characteristic curves of a turbine
Q
P, ηo
ηo
P
N is constant
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Operating Characteristic curves of a turbine
P
ηo
N is constant
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Operating Characteristic curves of a turbine
% Full load
ηo
N is constant
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Constant Efficiency curves for Reaction turbine
N
P
Constant Efficiency curves for Reaction turbine
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12,600 MW Year 1983
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HYDRO-POWER POTENTIAL IN INDIA
BASINS/RIVERS POWER POTENIAL IN MW AT
60% L.F.
Indus 19988
Ganges 10715
Central India Rivers 2740
West flowing Rivers 6149
East flowing Rivers 9532
Brahamaputra 34920
TOTAL 84044
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Anatomy of A Hydroelectric Power Plant
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