Development of probabilistic inundation mapping for dam failure induced floods

Abstract

A primary function of flood inundation mapping is to forecast flood hazards and assess potential losses. However, uncertainties limit the reliability of inundation hazard assessments. Major sources of uncertainty should be taken into consideration by an optimal flood management strategy. This study focuses on the 20 km reach downstream of the Shihmen Reservoir in Taiwan. In the modified dam failure model proposed in this study, the surface area of the water in the reservoir and water elevation are allowed to vary with the volume of the reservoir. A dam failure induced flood herein provides the upstream boundary conditions of flood routing. The two major sources of uncertainty that are considered in the hydraulic model and the flood inundation mapping herein are uncertainties in the dam break model and uncertainty of the roughness coefficient. The perturbance moments method is applied to a dam break model and the hydrosystem model to develop probabilistic flood inundation mapping. Uncertain variables such as roughness coefficient, breach width and reservoir surface area can be considered in these models and the variability of outputs can be quantified. The probabilistic flood inundation mapping for dam break induced floods can be developed with consideration of the variability of output using a hydraulic model. Two different probabilistic flood inundation mappings are discussed and compared. Probabilistic flood inundation mappings are hoped to provide improved insights in support of the evaluation of concerning reservoir flooded areas.

Appendix

We derive the formula of dam failure.

• Governing equations

  $$\frac{{d\forall_{L} }}{dt} = A_{L} (Z_{L} )\frac{{dZ_{L} }}{dt} = Q_{u} - Q_{B}$$

  (27)
  $$\frac{{d\forall_{B} }}{dt} = KQ_{B} S_{D} ,\quad \, \forall_{B} = \lambda_{B} \frac{{b_{B} \delta^{2} }}{{S_{D} }}$$

  (28)
  $$Q_{B} (t) = \sqrt {\frac{8}{27}} g^{{\frac{1}{2}}} b_{B} \eta^{{\frac{3}{2}}}$$

  (29)

• Derivation

  • Step (1)

    $A_{L}$ is constant.

    Initial condition: $Z_{C} (0) = Z_{D} ,\;\delta (0) = 0,\;\eta (0) = 0$

    From Eq. (27)

    $$\begin{aligned} & \left\{ \begin{array}{l} Z_{L} (t) = Z_{D} + \eta (t) - \delta (t) \hfill \\ \frac{{dZ_{L} }}{dt} = \frac{{Q_{u} - Q_{B} }}{{A_{L} }} \hfill \\ \end{array} \right. \, \Rightarrow \, \frac{{d(Z_{D} + \eta (t))}}{dt} = \frac{d\delta (t)}{dt} + \frac{{Q_{u} - Q_{B} }}{{A_{L} }} \\ & \because \frac{{dZ_{D} }}{dt} = 0 \, and \, Q_{u} = 0 \, \therefore \, \frac{d\eta (t)}{dt} = \frac{d\delta (t)}{dt} - \frac{{Q_{B} }}{{A_{L} }} \\ \end{aligned}$$

    (30)

  • Step (2)

    From Eq. (28)

    $$\begin{aligned} & \left\{ \begin{aligned} \frac{{d\forall_{B} }}{dt} = KQ_{B} S_{D} \hfill \\ \forall_{B} = \lambda_{B} \frac{{b_{B} \delta^{2} }}{{S_{D} }} \hfill \\ \end{aligned} \right. \, \Rightarrow \, \frac{{d\left( {\lambda_{B} \frac{{b_{B} \delta^{2} }}{{S_{D} }}} \right)}}{dt} = KQ_{B} S_{D} \\ & \Rightarrow \, \frac{{d\delta^{2} }}{dt} = \frac{{KQ_{B} S_{D} }}{{\lambda_{B} \frac{{b_{B} }}{{S_{D} }}}} \, \Rightarrow { 2}\delta \frac{d\delta }{dt} = \frac{{KQ_{B} S_{D}^{2} }}{{\lambda_{B} b_{B} }} \, \Rightarrow \, \frac{d\delta }{dt} = \frac{{KS_{D}^{2} }}{{2\lambda_{B} b_{B} }}\frac{{Q_{B} }}{\delta } \\ \end{aligned}$$

    (31)

    From Eq. (29) and Eq. (31)

    $$\frac{d\delta }{dt} = \frac{{KS_{D}^{2} }}{{2\lambda_{B} b_{B} }}\frac{{\sqrt {\frac{8}{27}} g^{{\frac{1}{2}}} b_{B} \eta^{{\frac{3}{2}}} }}{\delta } = \alpha \frac{{\eta^{{\frac{3}{2}}} }}{\delta }$$

    (32)

  • Step (3)

    With Eqs. (29), (30) and (32)

    $$\frac{d\eta (t)}{dt} = \frac{d\delta (t)}{dt} - \frac{{Q_{B} }}{{A_{L} }} = \alpha \frac{{\eta^{{\frac{3}{2}}} }}{\delta } - \frac{{\sqrt {\frac{8}{27}} g^{{\frac{1}{2}}} b_{B} }}{{A_{L} }}\eta^{{\frac{3}{2}}} = \alpha \frac{{\eta^{{\frac{3}{2}}} }}{\delta } - \beta \eta^{{\frac{3}{2}}}$$

    (33)

  • Step (4)

    Take Eq. (33) over (6)

    $$\begin{aligned} & \frac{d\eta (t)}{d\delta (t)} = \frac{{\alpha \frac{{\eta^{{\frac{3}{2}}} }}{\delta } - \beta \eta^{{\frac{3}{2}}} }}{{\alpha \frac{{\eta^{{\frac{3}{2}}} }}{\delta }}} = 1 - \frac{\beta }{\alpha }\delta \Rightarrow \eta (t) = \delta - \frac{\beta }{2\alpha }\delta^{2} + C \\ & \because \delta (0) = 0,\;\eta (0) = 0\,\therefore\, C = 0 \\ & \eta (t) = \delta - \frac{\beta }{2\alpha }\delta^{2} \\ \end{aligned}$$

    (34)

  • Step (5)

    With Eqs. (32) and (34)

    $$\begin{aligned} \frac{d\delta }{dt} & = \alpha \frac{{\eta^{{\frac{3}{2}}} }}{\delta } = \alpha \frac{{\left( {\delta - \frac{\beta }{2\alpha }\delta^{2} } \right)^{{\frac{3}{2}}} }}{\delta } \\ & \Rightarrow \int {\delta^{{\frac{1}{2}}} \left( {1 - \frac{\beta }{2\alpha }\delta } \right)^{{\frac{3}{2}}} } d\delta = \alpha t + C \\ & \Rightarrow \frac{{2\delta^{{\frac{1}{2}}} }}{{\left( {1 - \frac{\beta }{2\alpha }\delta } \right)^{{\frac{1}{2}}} }} = \alpha t + C \, \because t = 0,\;\delta (0) = 0\therefore C = 0 \\ & \Rightarrow \frac{4\delta }{{\left( {1 - \frac{\beta }{2\alpha }\delta } \right)}} = (\alpha t)^{2} \\ & \Rightarrow \delta (t) = \frac{{\alpha^{2} }}{4}\frac{{t^{2} }}{{1 + \frac{1}{8}\alpha \beta t^{2} }} \\ \end{aligned}$$

    (35)

  • Step (6)

    From Eq. (32)

    $$\eta (t) = \frac{{\alpha^{2} }}{4}\frac{{t^{2} }}{{\left( {1 + \frac{1}{8}\alpha \beta t^{2} } \right)^{2} }}$$

    (36)

  • Step (7)

    With Eqs. (29) and (36)

    $$Q_{B} (t) = \sqrt {\frac{8}{27}} g^{{\frac{1}{2}}} b_{B} \eta^{{\frac{3}{2}}} = \sqrt {\frac{8}{27}} g^{{\frac{1}{2}}} b_{B} \left( {\frac{{\alpha^{2} }}{4}\frac{{t^{2} }}{{\left( {1 + \frac{1}{8}\alpha \beta t^{2} } \right)^{2} }}} \right)^{{\frac{3}{2}}} = \sqrt {\frac{8}{27}} g^{{\frac{1}{2}}} b_{B} \left( {\frac{{\alpha^{3} }}{8}\frac{{t^{3} }}{{\left( {1 + \frac{1}{8}\alpha \beta t^{2} } \right)^{3} }}} \right)$$

    (37)

    By application of parallel axis theorem, we derive the first, second, third and fourth moments of the function $Y$, respectively.

• Variance deviation

  $$\begin{aligned} & \int {(x + d)^{2} dm = } \int {x^{2} dm + 2d\int {xdm + d^{2} } } \int {dm} \\ & \because \int {xdm} = 0 \\ & \int {(x + d)^{2} dm = } \int {x^{2} dm + d^{2} } \int {dm} \\ & \Rightarrow E\left[ {d^{2} } \right] = E\left[ {d\mu_{i}^{2} } \right] + P_{i} e_{i}^{2} \\ & \Rightarrow E\left[ {d\mu_{i}^{2} } \right] = E\left[ {d^{2} } \right] - P_{i} e_{i}^{2} \\ \end{aligned}$$

• Skewness coefficient

  $$\begin{aligned} & \int {(x + d)^{3} dm = } \int {x^{3} dm + 3d\int {x^{2} dm + 3d^{2} } } \int {xdm} + d^{3} \int {dm} \\ & \because \int {xdm} = 0 \\ & \int {(x + d)^{3} dm = } \int {x^{3} dm + 3d\int {x^{2} dm} } + d^{3} \int {dm} \\ & \Rightarrow E\left[ {d^{3} } \right] = E\left[ {d\mu_{i}^{3} } \right] + 3e_{i} E\left[ {d\mu_{i}^{2} } \right] + P_{i} e_{i}^{3} \\ & \Rightarrow E\left[ {d\mu_{i}^{3} } \right] = E\left[ {d^{3} } \right] - 3e_{i} E\left[ {d\mu_{i}^{2} } \right] - P_{i} e_{i}^{3} \\ & \gamma = \frac{{E\left[ {(Y - \mu )^{3} } \right]}}{{\left( {E\left[ {(Y - \mu )^{2} } \right]} \right)^{1.5} }} = \frac{{\sum\limits_{1}^{N} {E\left[ {d\mu_{i}^{3} } \right]} }}{{\left( {\sum\limits_{1}^{N} {E\left[ {d\mu_{i}^{2} } \right]} } \right)^{1.5} }} \\ \end{aligned}$$

• Kurtosis coefficient

  $$\begin{aligned} & \int {(x + d)^{4} dm = } \int {x^{4} dm + 4d\int {x^{3} dm + 6d^{2} } } \int {x^{2} dm} + 4d^{3} \int {xdm} + d^{4} \int {dm} \\ & \because \int {xdm} = 0 \\ & \int {(x + d)^{4} dm = } \int {x^{4} dm + 4d\int {x^{3} dm + 6d^{2} } } \int {x^{2} dm} + d^{4} \int {dm} \\ & \Rightarrow E\left[ {d^{4} } \right] = E\left[ {d\mu_{i}^{4} } \right] + 4e_{i} E\left[ {d\mu_{i}^{3} } \right] + 6e_{i}^{2} E\left[ {d\mu_{i}^{2} } \right] + P_{i} e_{i}^{4} \\ & \Rightarrow E\left[ {d\mu_{i}^{4} } \right] = E\left[ {d^{4} } \right] - 4e_{i} E\left[ {d\mu_{i}^{3} } \right] - 6e_{i}^{2} E\left[ {d\mu_{i}^{2} } \right] - P_{i} e_{i}^{4} \\ & \kappa = \frac{{E\left[ {(Y - \mu )^{4} } \right]}}{{\left( {E\left[ {(Y - \mu )^{2} } \right]} \right)^{2} }} = \frac{{\sum\limits_{1}^{N} {E\left[ {d\mu_{i}^{4} } \right]} }}{{\left( {\sum\limits_{1}^{N} {E\left[ {d\mu_{i}^{2} } \right]} } \right)^{2} }} \\ \end{aligned}$$