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Direct runoff hydrograph analysis, 4250-acre basin

Short answer

Excess release time is 3 h, the time of concentration. At t=4 h, 100% of the basin contributes. Peak discharge Qp = (0.20/12) ft/hr × 4250 acre × 43560 ft²/acre ÷ 3600 ≈ 857 cfs. The plateau occurs because after t_c = 3 h the entire basin contributes at equilibrium, keeping flow constant until rain stops.

A hydrograph for a 4,250-acre basin is shown in the accompanying sketch. The given hydrograph actually appeared as a direct runoff hydrograph from the basin, caused by net rain falling at an intensity of 0.20 in./hr for a duration of 5 hr, beginning at t = 0.

[Figure: direct surface runoff rate Q (cfs) versus time t (hr), t = 0 to 8. The rising limb passes about 200 cfs near t = 1 and about 700 cfs near t = 3, then the hydrograph flattens to a plateau at the peak QpQ_p between t = 3 and t = 5. The falling limb passes about 400 cfs near t = 6 and about 200 cfs near t = 7 and reaches zero at t = 8.]

Note: in = inches; cfs = cubic feet/sec; 12 inches = 1 foot; 1 acre = 43560 sq.ft.

(a) Determine the excess release time of the basin. (b) What percentage of the drainage basin was contributing to direct runoff 4 hr after rain began (t = 4)? (c) Use your response to part (b) to determine QpQ_p as shown in the sketch. Do not scale QpQ_p from the drawing. (d) Note that rain continued to fall between t = 3 and t = 5. Why did the hydrograph form a plateau between t = 3 and t = 5, rather than continue to rise during these 2 hours?

The hydrograph is a trapezoid produced by constant net rain of 0.20 in/hr for 5 hr. The rise stops when the whole basin contributes (equilibrium), and the recession starts when rain stops.

 Q
 Qp |        ________
    |      /|       |\
    |    /  |       |  \
    |  /    |       |    \
    |/______|_______|______\___ t (hr)
    0       3       5       8
         rain 0 -> 5 hr

(a) Excess release time

Rain stops at t = 5 hr and the direct runoff ends at t = 8 hr. The time taken for the last excess water to leave the basin is

trelease=8−5=3 hrt_{release} = 8 - 5 = 3 \text{ hr}

This equals the time of concentration, tc=3t_c = 3 hr (also the duration of the rising limb, 0 to 3 hr).

(b) Contributing area at t = 4 hr

Since t=4 hr>tc=3 hrt = 4\text{ hr} > t_c = 3\text{ hr}, water from every part of the basin, including the farthest point, has reached the outlet. So

AcontributingA=100%\frac{A_{contributing}}{A} = 100\%

(At t < 3 hr the contributing fraction is t/3, e.g. 33% at t = 1 hr.)

(c) Peak discharge QpQ_p

With the whole basin contributing, QpQ_p = net rain intensity × area (equilibrium discharge).

Qp=i A=0.2012 ft/hr×4250×43560 ft2=3,085,500 ft3/hr=3,085,5003600=857 cfs\begin{aligned} Q_p &= i\,A = \frac{0.20}{12}\ \text{ft/hr} \times 4250 \times 43560\ \text{ft}^2 \\ &= 3,085,500 \text{ ft}^3/\text{hr} \\ &= \frac{3,085,500}{3600} = 857 \text{ cfs} \end{aligned}

Answer: QpQ_p ≈ 857 cfs (the sketch values are only approximate readings).

(d) Why the plateau between t = 3 and t = 5 hr

After t = 3 hr (= tct_c), the entire basin already contributes. Rain continues at the same constant intensity, so the inflow to the outlet from the basin is constant: the rate of runoff leaving equals the rate of net rain falling (equilibrium). No new area can be added to increase the flow, so the hydrograph stays flat at QpQ_p until the rain stops at t = 5 hr. It then recedes to zero in another tct_c = 3 hr.

✓ Answer checked by Claude Sonnet 5.5

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