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Instantaneous unit hydrograph
Department of Civil Engineering...New Mexico State University
Contents
- 1 Instantaneous unit hydrograph (IUH)
- 2 S-hydrograph
- 3 Lagged S-hydrographs
- 4 Lagged S-hydrographs: Rainfall depth
- 5 Finite-duration unit hydrograph
- 6 Finite-duration unit hydrograph: Reference depth
- 7 From unit hydrograph to IUH
- 8 From unit hydrograph to IUH: Derivation
- 9 Dimensional analysis of the IUH
- 10 IUH for a reference rainfall depth
- 11 IUH for a reference rainfall depth: Dimensions
- 12 Runoff-volume-normalized IUH
- 13 Runoff-volume-normalized IUH: Derivation
- 14 Relationship between the two IUHs
- 15 Relationship between the two IUHs: Discharge
- 16 IUH normalization and conservation of mass
- 17 IUH normalization: Rainfall depth
- 18 Deriving the S-hydrograph from the IUH
- 19 Deriving the S-hydrograph from the IUH: Verification
- 20 IUH and the convolution integral
- 21 IUH and the convolution integral: Dimensions
- 22 Instantaneous rainfall and the Dirac delta function
- 23 Instantaneous rainfall: Convolution
- 24 Numerical example: Given data
- 25 Numerical example: IUH normalization
- 26 Numerical example: Discharge
- 27 Numerical example: Discharge hydrograph
- 28 Numerical example: Runoff volume
- 29 Numerical example: Conservation of mass
1 Instantaneous unit hydrograph (IUH)
- The IUH describes the direct runoff response to an instantaneous pulse of effective rainfall.
- It is the limiting form of a finite-duration unit hydrograph as the rainfall duration approaches zero.
- The IUH can be derived from an S-hydrograph.
- We will distinguish two definitions:
- Rainfall-depth-normalized IUH: \(U(t)\), with dimensions \(L^2/T\).
- Runoff-volume-normalized IUH: \(u(t)\), with dimensions \(T^{-1}\).
2 S-hydrograph
- An S-hydrograph is produced by continuous effective rainfall at a constant intensity \(I\).
- It can be constructed by summing successive unit hydrographs at equal time intervals.
- Let
- \(S(t)\): S-hydrograph discharge (\(L^3/T\)).
- \(I\): effective rainfall intensity (\(L/T\)).
- \(A\): watershed area (\(L^2\)).
- The S-hydrograph approaches a constant discharge when the watershed reaches equilibrium. \[ S(\infty)=AI \]
3 Lagged S-hydrographs
- Consider two identical S-hydrographs:
- \(S(t)\): original S-hydrograph, beginning at \(t=0\).
- \(S(t-\Delta t)\): S-hydrograph lagged by \(\Delta t\).
- Their difference represents runoff caused by uniform effective rainfall lasting only \(\Delta t\). \[ \Delta Q(t)=S(t)-S(t-\Delta t) \]
4 Lagged S-hydrographs: Rainfall depth
- The effective rainfall intensity is \(I\).
- The rainfall pulse lasts for \(\Delta t\).
- The corresponding effective rainfall depth is \[ D=I\Delta t \]
- Therefore, \(\Delta Q(t)\) represents the direct runoff hydrograph from a rainfall pulse of depth \(I\Delta t\).
5 Finite-duration unit hydrograph
- The rainfall pulse has duration \(\Delta t\) and depth \(I\Delta t\).
- Dividing the resulting discharge by the rainfall depth gives a rainfall-depth-normalized unit hydrograph. \[ U_{\Delta t}(t)= \frac{S(t)-S(t-\Delta t)}{I\Delta t} \]
- Dimensional analysis: \[ [U_{\Delta t}] =\frac{L^3/T}{(L/T)T} =\frac{L^2}{T} \]
- The resulting ordinates are discharge per unit rainfall depth.
6 Finite-duration unit hydrograph: Reference depth
- Let \(D_0\) be a reference effective rainfall depth.
- Multiplying the rainfall-depth-normalized unit hydrograph by \(D_0\) gives the corresponding discharge hydrograph. \[ Q_{D_0,\Delta t}(t)=D_0U_{\Delta t}(t) \]
- Substituting the S-hydrograph difference: \[ \boxed{ Q_{D_0,\Delta t}(t)= \frac{D_0}{I\Delta t} \left[S(t)-S(t-\Delta t)\right] } \]
- The dimensions of \(Q_{D_0,\Delta t}(t)\) are \(L^3/T\).
7 From unit hydrograph to IUH
- As \(\Delta t\) approaches zero, the rainfall pulse becomes instantaneous.
- Start with the rainfall-depth-normalized unit hydrograph: \[ U_{\Delta t}(t)= \frac{1}{I} \frac{S(t)-S(t-\Delta t)}{\Delta t} \]
- Taking the limit: \[ U(t)=\lim_{\Delta t\to0}U_{\Delta t}(t) \]
8 From unit hydrograph to IUH: Derivation
- Substituting the S-hydrograph difference into the limit: \[ U(t)=\frac{1}{I} \lim_{\Delta t\to0} \frac{S(t)-S(t-\Delta t)}{\Delta t} \]
- The difference quotient becomes the derivative of \(S(t)\). \[ \boxed{ U(t)=\frac{1}{I}\frac{dS(t)}{dt} } \]
- This is the rainfall-depth-normalized IUH.
9 Dimensional analysis of the IUH
- The S-hydrograph has dimensions of discharge: \[ [S]=\frac{L^3}{T} \]
- Its derivative has dimensions \[ \left[\frac{dS}{dt}\right]=\frac{L^3}{T^2} \]
- Since rainfall intensity has dimensions \(L/T\), \[ [U]=\frac{L^3/T^2}{L/T} =\boxed{\frac{L^2}{T}} \]
- Thus, \(U(t)\) represents discharge per unit rainfall depth, not discharge itself.
10 IUH for a reference rainfall depth
- Let \(D_0\) be the reference effective rainfall depth, commonly 1 in. or 1 cm.
- Multiplying \(U(t)\) by \(D_0\) gives the discharge hydrograph for that rainfall depth. \[ Q_{D_0}(t)=D_0U(t) \]
- Substituting the S-hydrograph derivative: \[ \boxed{ Q_{D_0}(t)= \frac{D_0}{I}\frac{dS(t)}{dt} } \]
11 IUH for a reference rainfall depth: Dimensions
- Dimensional verification: \[ [Q_{D_0}] =\frac{L}{L/T}\frac{L^3}{T^2} =\boxed{\frac{L^3}{T}} \]
- Alternatively, define a normalized rainfall intensity: \[ I^*=\frac{I}{D_0}, \qquad [I^*]=T^{-1} \]
- Then, \[ Q_{D_0}(t)=\frac{1}{I^*}\frac{dS(t)}{dt} \]
12 Runoff-volume-normalized IUH
- Another common IUH definition normalizes runoff by the total runoff volume rather than rainfall depth.
- Define \(u(t)\) as the runoff-volume-normalized IUH.
- Since the total effective rainfall volume is \(AD_0\), \[ u(t)=\frac{Q_{D_0}(t)}{AD_0} \]
- Substituting the previous expression: \[ u(t)= \frac{1}{AD_0} \frac{D_0}{I}\frac{dS(t)}{dt} \]
13 Runoff-volume-normalized IUH: Derivation
- Canceling the reference rainfall depth \(D_0\): \[ \boxed{ u(t)=\frac{1}{AI}\frac{dS(t)}{dt} } \]
- Its dimensions are \[ [u]=\frac{L^3/T^2}{L^2(L/T)} =\boxed{T^{-1}} \]
- Unlike \(U(t)\), the volume-normalized IUH \(u(t)\) has dimensions of inverse time.
14 Relationship between the two IUHs
- The rainfall-depth-normalized IUH is \[ U(t)=\frac{1}{I}\frac{dS(t)}{dt} \]
- The runoff-volume-normalized IUH is \[ u(t)=\frac{1}{AI}\frac{dS(t)}{dt} \]
- Therefore, \[ \boxed{U(t)=Au(t)} \]
15 Relationship between the two IUHs: Discharge
- The discharge hydrograph for an instantaneous rainfall pulse of depth \(D_0\) is \[ Q_{D_0}(t)=D_0U(t) \]
- Substituting \(U(t)=Au(t)\): \[ \boxed{Q_{D_0}(t)=AD_0u(t)} \]
- These are equivalent representations of the same runoff response.
- Both produce discharge with dimensions \(L^3/T\).
16 IUH normalization and conservation of mass
- The total direct runoff volume equals the effective rainfall volume. \[ \int_0^\infty Q_{D_0}(t)\,dt=AD_0 \]
- Substituting \(Q_{D_0}(t)=AD_0u(t)\): \[ AD_0\int_0^\infty u(t)\,dt=AD_0 \]
- Therefore, \[ \boxed{\int_0^\infty u(t)\,dt=1} \]
- The volume-normalized IUH integrates to unity.
17 IUH normalization: Rainfall depth
- Since the two IUHs are related by \[ U(t)=Au(t) \]
- Integrating both sides: \[ \int_0^\infty U(t)\,dt =A\int_0^\infty u(t)\,dt \]
- Therefore, \[ \boxed{\int_0^\infty U(t)\,dt=A} \]
- Both normalization conditions follow from conservation of mass, assuming all effective rainfall becomes direct runoff.
18 Deriving the S-hydrograph from the IUH
- Consider continuous effective rainfall of constant intensity \(I\), beginning at \(t=0\).
- The runoff response is the accumulation of infinitesimal rainfall pulses.
- Using the volume-normalized IUH: \[ S(t)=A\int_0^t Iu(t-\tau)\,d\tau \]
- Since \(I\) is constant, change the integration variable: \[ \boxed{ S(t)=AI\int_0^t u(\tau)\,d\tau } \]
19 Deriving the S-hydrograph from the IUH: Verification
- Differentiate the S-hydrograph expression: \[ \frac{dS(t)}{dt}=AIu(t) \]
- Rearranging: \[ \boxed{ u(t)=\frac{1}{AI}\frac{dS(t)}{dt} } \]
- This independently confirms the S-hydrograph derivation.
- As \(t\to\infty\), the S-hydrograph approaches \[ S(\infty)=AI \] because \(u(t)\) integrates to unity.
20 IUH and the convolution integral
- For an arbitrary effective rainfall intensity \(i(t)\), the direct runoff hydrograph is obtained by convolution.
- Using the rainfall-depth-normalized IUH: \[ \boxed{ Q(t)=\int_0^t i(\tau)U(t-\tau)\,d\tau } \]
- Substituting \(U(t)=Au(t)\): \[ \boxed{ Q(t)=A\int_0^t i(\tau)u(t-\tau)\,d\tau } \]
21 IUH and the convolution integral: Dimensions
- The volume-normalized convolution equation is \[ Q(t)=A\int_0^t i(\tau)u(t-\tau)\,d\tau \]
- Dimensional verification: \[ [Q]=L^2\frac{L}{T}\frac{1}{T}T =\boxed{\frac{L^3}{T}} \]
- The IUH serves as the response kernel for a linear, time-invariant watershed model.
22 Instantaneous rainfall and the Dirac delta function
- An instantaneous effective rainfall pulse of depth \(P_e\) can be represented using the Dirac delta function. \[ i(t)=P_e\delta(t) \]
- The delta function satisfies \[ \int_{-\infty}^{\infty}\delta(t)\,dt=1, \qquad [\delta]=T^{-1} \]
- The rainfall depth is recovered by integrating the rainfall intensity: \[ \int_{-\infty}^{\infty}i(t)\,dt=P_e \]
23 Instantaneous rainfall: Convolution
- Substituting the instantaneous rainfall pulse into the convolution integral: \[ Q(t)=A\int_{-\infty}^{\infty} P_e\delta(\tau)u(t-\tau)\,d\tau \]
- Using the sifting property: \[ \boxed{Q(t)=AP_eu(t)} \]
- Equivalently, from the S-hydrograph: \[ \boxed{ Q(t)=\frac{P_e}{I}\frac{dS(t)}{dt} } \]
- These two expressions give the same direct runoff hydrograph.
24 Numerical example: Given data
- A watershed has the following properties:
- Watershed area: \(A=7.2\ \mathrm{km^2}\).
- Instantaneous effective rainfall depth: \(P_e=15\ \mathrm{mm}\).
- Time interval: \(\Delta t=1\ \mathrm{hr}\).
- The volume-normalized IUH ordinates are:
Time (hr) \(u(t)\) (hr\(^{-1}\)) 0 0 1 0.25 2 0.50 3 0.25 4 0
25 Numerical example: IUH normalization
- Verify the IUH normalization by trapezoidal integration: \[ \int_0^4 u(t)\,dt \approx \frac{\Delta t}{2} \left[u_0+2u_1+2u_2+2u_3+u_4\right] \]
- Substituting \(\Delta t=1\ \mathrm{hr}\) and the IUH ordinates: \[ =\frac{1}{2} [0+2(0.25)+2(0.50)+2(0.25)+0] =\boxed{1} \]
- Therefore, the IUH satisfies the unit-integral condition.
26 Numerical example: Discharge
- For instantaneous effective rainfall, \[ Q(t)=AP_eu(t) \]
- The total effective rainfall volume is \[ AP_e=(7.2\times10^6)(0.015) =108{,}000\ \mathrm{m^3} \]
- Since \(u(t)\) is given in hr\(^{-1}\), discharge in m\(^{3}\)/s is \[ Q(t)= 108{,}000\,u(t)\frac{1\ \mathrm{hr}}{3600\ \mathrm{s}} \]
- Numerically, with \(u(t)\) expressed in hr\(^{-1}\), \[ \boxed{Q(t)=30u(t)} \]
- The factor 3600 converts hours to seconds.
27 Numerical example: Discharge hydrograph
- Using \(Q(t)=30u(t)\), with the specified units:
Time (hr) IUH (hr\(^{-1}\)) Discharge (m\(^{3}\)/s) 0 0 0 1 0.25 7.5 2 0.50 15 3 0.25 7.5 4 0 0 - The peak discharge is \[ \boxed{Q_p=15\ \mathrm{m^3/s}} \]
- The peak occurs at \(t=2\ \mathrm{hr}\).
28 Numerical example: Runoff volume
- Verify runoff volume by trapezoidal integration: \[ V=\int_0^4 Q(t)\,dt \]
- Using \(\Delta t=1\ \mathrm{hr}=3600\ \mathrm{s}\): \[ V\approx 3600\left[ \frac{0+7.5}{2} +\frac{7.5+15}{2} +\frac{15+7.5}{2} +\frac{7.5+0}{2} \right] \] \[ \boxed{V=108{,}000\ \mathrm{m^3}} \]
29 Numerical example: Conservation of mass
- The effective rainfall volume is \[ AP_e=(7.2\times10^6)(0.015) =108{,}000\ \mathrm{m^3} \]
- The integrated direct runoff volume is \[ V=108{,}000\ \mathrm{m^3} \]
- Therefore, \[ \boxed{V=AP_e} \]
- The computed hydrograph conserves the effective rainfall volume.