conveyance structure

8/12/2019 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

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