Download - Conveyance Structure
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Chapter-4:
ConveyanceStructure
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Headrace/tailrace tunnel
Water should be lead from the intake to the
power station through the headrace tunnel,
and pressure shaft and away from the power
station through the tailrace tunnel
The water flow through the tunnel is
= 3/
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Headloss
Headloss during the transport is found from Manningformula:
=
2
2
Where,
Q = water flow (m3/s)
A = cross sectional area of tunnel (m2)
Q/A = average water velocity (m/s)
L = total tunnel length (m)
= , 30 < > 40 R = hydraulic radius (m)
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Head loss represents a loss in capacity (kW) which may be expressed as:
=9.81
2
2
= . x Q3= constant x Q3For Qmax,P =
Therefore,
=
Thus for one year load, the total annual energy loss as the integral of the capacity lossover one year:
= = / max
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The integral,
Equals the
area under the curve. The annual energy loss is
found by multipling this area with maximum
capacity loss i.e. P max
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When knowing the value of the energy i. e. V Rs. Per kWh.
Thus total economic loss =
[Rs. Per year]
In this equation, one parameter i.e.
Another parameter i.e. maximum capacity loss i.e. Qmax is
depends directly on the cross section.Thus, estimate Qmax for different cross section of the tunnel isone of the important part which is also directly affects onconstruction cost of tunnel.
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From experience we know that the maximum
water velocity through an unlined headrace
and tailrace tunnel should be somewhere
between 0.7 m per sec and 1.5 m per ssec.
Hence corresponding cross sections are
calculated of optimisation
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The tunnel cost
The cost caused by extending the tunnel crosssection with one square meter are
approximately 25 $ per stretch mete
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Example
In a proposed hydropower plant, it is decided the
turbine capacity to 66 MW and the net head to 415 m.then examine the optimize cross section of tunnel for
3000 m length. Take firm power value as 0.025 $ per
kWh
Here,
= 66..5 = 18
As, velocity should be inbetween 0.7 m/s to 1.5 m/s.
So, cross sectional area for Qmax=18 m3/s are
Velo. (m/s) 0.7 0.9 1.0 1.3 1.6
Area (m2) 25.7 20.0 18.0 13.8 11.2
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By graphical plot, the area is 2786 hours
With a depreciation time of 40 years, and a
discount rate of 10%, the annual costs will be
7650 $ per sq. m. per year.
Take M= 40 for properly blasted tunnel
Take
= 0 . 9for turbine/generator efficiency
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Tabular CalculationCross section (A-m2) 11.2 13.8 18.0 20.0 25.7
Pmax=.
$ -kW 899 521 256 193 99
70 70 70 70 70
Pmax x ($) 62900 36500 17900 13500 69007650 $ x (A-11.2) ($) 0 19900 52000 67300 110900
(4)+(5) -$ 62900 56400 69900 80800 117800
Note: - line 4 is economic loss due to head loss
- line 5 is the cost increase when extending the tunnel cross section above 11.2 m2
- Line is the sum of the economic head loss and the increased tunnel costs where we
have the cost minimum. Processing the fig, in line 6 in the graph as a function of the
tunnel cross section, we easily see how the costs are minimized for tunnel cross
section near 13-14 m2
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Thank you