--- title: "UH derivation from complex storms" subject: "Engineering Hydrology" chapter: 5 key: "semester_5/engineering_hydrology:uh-complex-storms-procedure" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/uh-complex-storms-procedure site: BCT Engineering --- # UH derivation from complex storms Engineering Hydrology · Chapter 5: Hydrograph Analysis - Asked in: 2072 Chaitra, 8 marks - Video: [4-Hour Unit Hydrograph Derivation | Complex Storm Hydrograph Numerical Explained #civilengineering](https://www.youtube.com/watch?v=mMmzhQtRPmo) — Class 1 to B.Tech(Civil) - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/uh-complex-storms-procedure - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-5 ## Question Describe the procedure of derivation of unit hydrograph from complex storms using appropriate expressions. ## Short answer Separate baseflow to obtain DRH; compute net rainfall for each period. Solve the convolution equations forward: U₁ = Q₁/P₁, Uₙ = (Qₙ − Σ Pₘ Uₙ₋ₘ₊₁ for m>1) / P₁. Check that the volume equals 1 cm runoff; smooth or use least-squares if oscillations occur. ## Answer A unit hydrograph (UH) derived from a complex storm uses a direct runoff hydrograph (DRH) that results from several successive periods of rainfall excess of different intensity. Because the UH ordinates cannot be read off directly as in a single burst, the DRH is **deconvoluted** using the convolution (discrete) equation. #### Basic equation If the net rain in successive periods of duration $D$ is $P_1, P_2, \dots, P_M$ (cm), the UH of duration $D$ has ordinates $U_1, U_2, \dots, U_N$, and the DRH has ordinates $Q_1, Q_2, \dots, Q_n$, then $$ Q_n = \sum_{m=1}^{n\le M} P_m\, U_{n-m+1}, \qquad n = M+N-1 $$ Writing it out for each ordinate: $$ \begin{aligned} Q_1 &= P_1U_1 \\ Q_2 &= P_1U_2 + P_2U_1 \\ Q_3 &= P_1U_3 + P_2U_2 + P_3U_1 \\ &\ \ \vdots \end{aligned} $$ #### Procedure 1. Select a storm with several periods of rain of uniform duration $D$ over the whole basin, and its stream-flow record. 2. Separate base flow from the storm hydrograph to get the DRH ordinates $Q_n$ at intervals $D$. 3. Compute the rainfall excess $P_m$ of each period by deducting losses (φ-index or a loss model). A period with no excess stays in the sequence as $P_m=0$, because it only shifts the later periods in time. 4. Number of UH ordinates $N = n - M + 1$ (n = number of DRH ordinates, M = number of excess-rain periods). 5. Solve the equations successively (forward substitution): $$ U_1=\frac{Q_1}{P_1},\quad U_2=\frac{Q_2-P_2U_1}{P_1},\quad U_3=\frac{Q_3-P_2U_2-P_3U_1}{P_1},\ \dots $$ 6. Check: the area of the derived UH must equal 1 cm of runoff over the basin: $\sum U_i \cdot D\cdot 3600 = A\times 0.01$ m. Adjust (smooth) the ordinates if needed. 7. Plot the UH and, if necessary, re-compute the DRH from it to confirm that it reproduces the observed DRH. #### Remarks - Forward substitution accumulates errors, so ordinates may oscillate or become negative. In that case the **least-squares** (matrix) form $\mathbf{Q}=[\mathbf{P}]\mathbf{U}$, with $\mathbf{U}=(\mathbf{P}^T\mathbf{P})^{-1}\mathbf{P}^T\mathbf{Q}$, or a trial-and-error / optimisation approach with a non-negative constraint is used (Collins' method: assume a UH, compute the DRH, correct the UH, and repeat). - The UH obtained from the average of several storms is more reliable. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With M excess-rain periods and DRH ordinates n, the number of UH ordinates N is: A. n - M + 1 B. n + M - 1 C. n x M D. n - M Answer: A. n = M + N - 1, so N = n - M + 1. (Key point: Equation) 2. U3 is computed as: A. (Q3 - P2 U2 - P3 U1)/P1 B. Q3/P3 C. (Q3 - P1 U2)/P2 D. Q3/P1 Answer: A. Q3 = P1U3 + P2U2 + P3U1. (Key point: Forward substitution) 3. If forward substitution gives oscillating or negative ordinates, one should: A. Use least-squares or Collins' trial-and-error method B. Ignore C. Increase P D. Remove baseflow again Answer: A. Least-squares with U = (P^T P)^-1 P^T Q reduces error accumulation. (Key point: Error accumulation) 4. The derived UH is verified by checking that: A. Its area equals 1 cm over the basin B. Peak equals Q1 C. Sum equals P D. Base equals 0 Answer: A. Sum U x D x 3600 = A x 0.01 m. 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