Economic Diameter of Penstock
The weight and first cost of the penstock increases with increasing diameter, and the output in electrical energy also increases owing to the reduction in frictional headloss. The economic diameter for a penstock required to carry a discharge \(Q\) is the one at which the annual cost due to higher investment does not exceed the annual value of the increment in energy output. Mathematically:
\[
\frac{dC_1}{dD} \le \frac{dC_2}{dD}
\]
where \(C_1\) = annual cost due to investment for pipe diameter \(D\), and \(C_2\) = value of energy produced.
—
Let \(H\) = design head, and \(\sigma\) = permissible stress in steel. Then shell thickness:
\[
b = \frac{0.1\,H\,D}{2\,\sigma}
\]
Weight of the penstock per unit length (with 20% allowance) is:
\[
\text{weight per meter} = 1.2 \times 7850 \pi D b
= \frac{1.2 \times 7850 \pi \times 0.1 H D^2}{2 \sigma}
= \frac{1480 H D^2}{\sigma}
\]
For pipe length \(L\):
\[
\text{total weight} = \frac{1480 H D^2 L}{\sigma}
\]
Assuming the annual cost is proportional to first cost:
\[
C_1 = k_1 \frac{1480 H D^2 L}{\sigma}
\]
So:
\[
\frac{dC_1}{dD} = 2 \cdot 1480\,k_1 \frac{H L}{\sigma}\,D
\]
—
Now consider the frictional head loss. Using Darcy‐Weisbach kind of relation:
\[
\text{head loss} = \frac{\lambda L v^2}{2gD}
\]
Expressing \(v\) in terms of discharge \(Q\) and diameter \(D\), one gets:
\[
\text{head loss} = \frac{\lambda L Q^2}{12.1\,D^5}
\]
Total potential power (neglecting losses) is \(9.81\,Q\,H\) [kW]. With an overall efficiency of 80%, the power lost due to friction becomes:
\[
9.81 \times 0.80 \times \frac{\lambda L Q^2}{12.1\,D^5} \times Q
\]
If the plant runs \(t\) hours per year, energy lost is:
\[
E = 0.65 \frac{\lambda L Q^3 t}{D^5} \quad [\text{kWh}]
\]
If \(k_2\) = value of energy per kWh at generator terminals, then:
\[
C_2 = 0.65 \frac{\lambda L Q^3 t}{D^5} \cdot k_2
\]
Differentiating \(C_2\) with respect to \(D\):
\[
\frac{dC_2}{dD} = -5 \cdot 0.65 \, \frac{\lambda L Q^3 t}{D^6} \cdot k_2
\]
—
Setting the economic criterion \(\frac{dC_1}{dD} = \frac{dC_2}{dD}\), we get:
\[
2 \cdot 1480 \frac{k_1 H L}{\sigma} D = 5 \cdot 0.65 \frac{\lambda L Q^3 t}{D^6} \, k_2
\]
Rearrange:
\[
D^7 \le \frac{5 \cdot 0.65}{2 \cdot 1480} \frac{\lambda \sigma k_2}{k_1} \frac{Q^3}{H} t
\]
Simplified form often used:
\[
D \le \left( \frac{1}{1100} \frac{\lambda \sigma k_2}{k_1} \frac{Q^3}{H} t \right)^{1.7}
\]
—
**Where:**
– \(Q\) = discharge in m³/s (cumecs)
– \(H\) = design head in m
– \(\sigma\) = allowable stress in steel in kg/cm²
– \(k_1\) = annual cost of penstock per kg
– \(k_2\) = value of 1 kWh at generator terminals
– \(t\) = annual operating hours
– \(\lambda\) = friction coefficient (take ≈ 0.02 for preliminary estimates)
