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Economic diameter of tunnel

Short answer

The economic diameter minimizes total annual cost (construction plus value of energy lostdue to friction). It is found by expressing annual cost as a function of diameter, differentiating and solving, or by plotting total cost for trial diameters and selecting the minimum point.

What is meant by economic diameter of tunnel? How do you work out it?

The economic diameter of a tunnel (or penstock) is the diameter for which the total annual cost—the annual cost of construction (capital recovery + maintenance) plus the annual value of energy lost due to friction—is a minimum.

  • A larger diameter gives lower velocity and smaller head loss, so less energy is lost, but construction cost (excavation, lining) is higher.
  • A smaller diameter is cheaper to build but causes larger friction loss and loss of revenue every year.
  • The economic diameter is the best balance between these two opposing costs.

Working it out

  1. Annual construction cost. Cost of tunnel of length LL is roughly proportional to D2D^2 (excavation and lining). The annual cost is
C1=k1D2L×(capital recovery factor+O&M rate)\begin{aligned} C_1 = k_1 D^2 L \times (\text{capital recovery factor} + \text{O\&M rate}) \end{aligned}
  1. Annual cost of energy lost. Head loss by Manning (for a full circular section R=D/4R = D/4):
hf=n2V2LR4/3,V=4QπD2⇒hf=10.29 n2Q2LD16/3\begin{aligned} h_f &= \frac{n^2 V^2 L}{R^{4/3}}, \quad V = \frac{4Q}{\pi D^2} \\ \Rightarrow h_f &= 10.29\,\frac{n^2 Q^2 L}{D^{16/3}} \end{aligned}

Power lost PL=9.81 Q hf ηP_L = 9.81\,Q\,h_f\,\eta kW; annual energy lost =PL×8760×= P_L \times 8760 \times plant factor; its value at the energy rate rr gives

C2=k2D16/3C_2 = \frac{k_2}{D^{16/3}}
  1. Minimise total cost.
C=C1+C2=k1′D2+k2D16/3dCdD=2k1′D−163k2D19/3=0Deco=(8k23k1′)3/22\begin{aligned} C &= C_1 + C_2 = k_1' D^2 + \frac{k_2}{D^{16/3}} \\ \frac{dC}{dD} &= 2k_1' D - \frac{16}{3}\frac{k_2}{D^{19/3}} = 0 \\ D_{eco} &= \left(\frac{8 k_2}{3 k_1'}\right)^{3/22} \end{aligned}
  1. Graphical method (used in practice). Assume several trial diameters, compute for each the annual construction cost and annual loss cost, and plot them against DD. The diameter at the lowest point of the total cost curve is the economic diameter.
 Annual cost
   ^   \                       / construction
   |    \                    /    cost
   |     \     total cost  /
   |      \__       ___ /
   |        \ ''-.-'' /
   |         \  min /'.
   |          '-..-'   '-. energy loss cost
   |              |       '--.__
   +--------------+---------------> D
                D_eco
  1. Checks: velocity within the usual range (about 2–3 m/s in concrete-lined tunnels), minimum size for construction (about 2–2.5 m), and practical round size.

✓ Answer checked by Claude Opus 5.5 (high effort)

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