Economic diameter of penstock

 Economic diameter of penstock

The weight and first cost of the penstock increases with increasing diameter, the output in electrical energy is also increased owing to the reduction in frictional headloss. The economic diameter for a penstock required to carry a discharge Q is the one at which annual costs due to greater investment do not exceed the annual value of resulting increment energy output. Mathematically, this criterion may be expressed by the relation`\frac{dC_1}{dD}≤\frac{dC_2}{dD}`    where C₁ is the annual cost due to the investment for a pipe of a diameter D₁ and C₂ is the value of energy that can be produced at the same diameter.

Let H be the design head at any section. Shell thickness b =`\frac{0.1HD}{2σ}`     , where σ = permissible stress in steel

Weight of penstock section of unit length say 1 meter, adding 20% additional allowance neglecting joint efficiency = 1.2 ×7850πDb

                =`\frac{1.2×7850π×0.1H×D^2}{2σ}`    =`\frac{1480HD^2}{σ}` 

Weight of pipe of length L meter will be `\frac{1480HD^2L}{σ}`       kg

If annual operating charges including depreciation and maintenance is taken proportional to first cost

C₁=`k_1\frac{1480HD^2L}\sigma`   

Therefore `\frac{dC_1}{dD}=2\times1480k_1\frac{HL}{σ}D`

Let us now consider the influence of the change in diameter on the annual energy output and its value.

Friction head loss =`\frac{λLv^2}{2gD}`  =`\frac{λLv^2}{2g(\frac{πD^2}{4})^2D}` =`\frac{λLQ^2}{12.1×D^5}` 

With a head H and discharge Q, power potential is 9.81QH [kW]

Taking overall efficiency as 80%, the potential power corresponding to frictional loss will be

9.81×80% `(\frac{λLv^2}{12.1×D^5})Q`     [kW]

If t is annual duration of operation in hours the energy generated will be

`\frac{9.81×0.80}{12.1}`×`\frac{λLQ^3 t}{D^5}`    [kWh]

=0.65×`\frac{λLQ^3 t}{D^5}`        [kWh]

If k₂ is the value of energy at generator terminals

C₂= 0.65×`\frac{λLQ^3 t}{D^5}`    ×k₂

Therefore `\frac{dC_2}{dD} = -5×0.65×\frac{λLQ^3 t}{D^6}×k_2`     

So, 2×1480`\frac{k_1 HDL}{σ}`≤ 3.25𝝀L`\frac{Q^3t}{D^6} `×`k_2t`

D⁷≤`\frac{3.25}{2\times1480}\times\frac{λσk_2}{k_1}×\frac{Q^3}{H}t`

Now, D ≤ `({\frac1{1100}\times\frac{\lambda\sigma k_2}{k_1}\times\frac{Q^3}Ht})^{1.7}` [m]

Where,

Q = discharge in cumecs

H= design head in meters

σ= allowable stress in steel in kg/cm²

K₁= Annual cost of penstock per kg

K₂= value of one kWh at generator terminals in the same units

t = annual duration of operation in hours

𝝀= friction coefficient, value of which may be taken as 0.02 for preliminary estimates