---
title: "Linear reservoir and Clark UH from time-area"
subject: "Engineering Hydrology"
chapter: 7
key: "semester_5/engineering_hydrology:linear-reservoir-clark-uh"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/linear-reservoir-clark-uh
site: BCT Engineering
---

# Linear reservoir and Clark UH from time-area

Engineering Hydrology · Chapter 7: Flow Routing

- Asked in: 2069 Chaitra, 2+6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/linear-reservoir-clark-uh
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-7

## Question

What is linear reservoir? Explain the procedure to obtain Clark UH from time area method.

## Short answer

A linear reservoir has S=KO. Clark UH from time‑area: (1) compute translation inflow Iᵢ=2.78Aᵢ/Δt for 1 cm rain; (2) obtain K from recession; (3) route through linear reservoir via O₂=C₀(I₁+I₂)+C₂O₁ with C₀=Δt/(2K+Δt), C₂=(2K−Δt)/(2K+Δt). IUH ordinates are O; average successive ordinates gives Δt‑h UH.

## Answer

#### Linear reservoir

A **linear reservoir** is a reservoir (or storage) in which the storage is directly proportional to the outflow:

$$
S = K\,O
$$

where $K$ is the storage constant (units of time). With continuity $I - O = dS/dt$, the response is $K\,dO/dt + O = I$, and its IUH is $u(t) = \dfrac{1}{K}e^{-t/K}$.

#### Procedure for Clark UH from the time-area method

1. **Isochrones and areas.** Divide the catchment by isochrones at interval $\Delta t$ up to $t_c$ and find the strip areas $A_1, A_2, \dots, A_N$.
2. **Time-area (translation) inflow.** For 1 cm excess rain:
$$
I_i = \frac{2.78\,A_i}{\Delta t}\ \text{m}^3/\text{s}
$$
This is the inflow hydrograph to an imaginary linear reservoir at the outlet.
3. **Storage constant.** Get $K$ from the recession of an observed hydrograph, or by a regional formula.
4. **Route through the linear reservoir:**
$$
O_2 = C_0 (I_1 + I_2) + C_2 O_1,\quad C_0 = \frac{\Delta t}{2K + \Delta t},\quad C_2 = \frac{2K - \Delta t}{2K + \Delta t}
$$
starting with $O_1 = 0$. The outflow ordinates are the IUH.
5. **$\Delta t$-hour unit hydrograph.** The ordinates are the average of successive IUH ordinates, $U_n = (O_n + O_{n-1})/2$.
6. **Longer-duration UH.** For a unit hydrograph of duration $D$ (a multiple of $\Delta t$), use the S-curve or average successive ordinates.

```text
 Time-area        Linear reservoir      IUH / UH
 inflow I(t) ---> S = K O  ----------> O(t), U(t)
 (translation)    (storage effect)
```

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. A linear reservoir is one in which:
   A. Outflow is constant
   B. Storage is directly proportional to outflow (S = K O)
   C. Storage is proportional to the square of outflow
   D. Inflow equals outflow at all times
   Answer: B. A linear reservoir has S = K O, where K is the storage constant with units of time. (Key point: Definition of linear reservoir)

2. In Clark's procedure, what is the inflow to the imaginary linear reservoir?
   A. The observed streamflow
   B. The time-area (translation) hydrograph I = 2.78 A/dt for 1 cm excess rain
   C. The rainfall hyetograph directly
   D. The baseflow
   Answer: B. Strip areas converted to discharge by 2.78 A/dt form the translation hydrograph that is routed through the reservoir. (Key point: Role of translation inflow)

3. Which is the correct C2 in O2 = C0(I1 + I2) + C2 O1?
   A. (2K - dt)/(2K + dt)
   B. dt/(2K + dt)
   C. (K - dt)/(K + dt)
   D. 2K/(2K + dt)
   Answer: A. C0 = dt/(2K+dt) and C2 = (2K-dt)/(2K+dt) from the linear reservoir continuity equation. (Key point: Routing coefficients)

4. How is the dt-hour unit hydrograph obtained from the routed outflow ordinates?
   A. Multiply ordinates by dt
   B. Take the average of successive IUH ordinates, (On + On-1)/2
   C. Take the maximum of successive ordinates
   D. Subtract successive ordinates
   Answer: B. Averaging two consecutive IUH ordinates gives the dt-hour UH ordinate. (Key point: Unit hydrograph from IUH)

## Related in Engineering Hydrology

- [Procedure of deriving Clark UH](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aclark-uh-procedure) — Engineering Hydrology, chapter 7, asked 2×
- [2-hr UH from figure: area, flood, Tc, two storms](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Auh-2hr-figure-four-parts) — Engineering Hydrology, chapter 5
- [DAD curve procedure and isohyetal area data](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Adad-curve-procedure-isohyetal) — Engineering Hydrology, chapter 2
- [Principle and procedure of slope-area method](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aslope-area-principle-procedure) — Engineering Hydrology, chapter 4
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