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2-hr UH from sketch, net rain 5 in/hr

Short answer

For 5 in/hr for2hr, net rain is10 in, so peak discharge =10×600 =6000 cfs. Total direct runoff equals rainfall depth,10 in. For 0.5 in/hr for4 hr, net rain 2 in gives two 1-in blocks; at 8 p.m. (t=4 hr) Q =400+400=800 cfs.

A 2-hr unit hydrograph for a basin is shown in the sketch.

[Figure: 2-hr unit hydrograph, discharge (cfs) versus time (hr), t = 0 to 9. The curve rises from 0 at t = 0, passes about 200 cfs near t = 1 and 400 cfs near t = 2, peaks at 600 cfs at t = 3, then falls through about 400 cfs near t = 4 and 200 cfs near t = 5, and tails off to near zero by about t = 9. Note: in = inches; 12 inches = 1 feet; cfs = cubic feet/sec; a unit hydrograph is the response of catchment due to 1 inch effective rainfall over it.]

(a) Determine the peak discharge (in cfs) for a net rain of 5.00 in./hr and a duration of 2 hr. (b) What is the total direct surface runoff (in inches) for the storm described in part (a)? (c) A different storm with a net rain of 0.50 in./hr lasts for 4 hr. What is the discharge at 8 p.m. if the rainfall started at 4 p.m.?

A 2-hr UH gives the response to 1 inch of net rain in 2 hr (intensity 0.5 in/hr). Ordinates are directly proportional to the net rain depth, and responses of successive blocks are added with a time lag (principle of proportionality and superposition).

Read from the sketch (t, hr: Q, cfs): 0: 0; 1: 200; 2: 400; 3: 600 (peak); 4: 400; 5: 200; tail to about 0 at 9.

(a) Peak discharge for 5.00 in/hr for 2 hr

Duration is the same as the UH duration (2 hr), so just scale.

net rain=5.00×2=10 inQpeak=10×600=6000 cfs\begin{aligned} \text{net rain} &= 5.00 \times 2 = 10 \text{ in} \\ Q_{peak} &= 10 \times 600 = 6000 \text{ cfs} \end{aligned}

Answer: 6000 cfs (at t = 3 hr).

(b) Total direct surface runoff

Direct runoff depth equals net (effective) rain depth.

Runoff=5.00 in/hr×2 hr=10 in\text{Runoff} = 5.00 \text{ in/hr} \times 2 \text{ hr} = 10 \text{ in}

Answer: 10 inches over the basin.

(c) Storm of 0.50 in/hr for 4 hr, discharge at 8 p.m.

Net rain is 0.5 × 4 = 2 in, which is two successive 2-hr blocks of 0.5 in/hr × 2 hr = 1 in each. Each block gives one UH (multiplier 1); the second is lagged by 2 hr.

Rain starts at 4 p.m., so 8 p.m. is t = 4 hr:

Q(4)=1×U(4)+1×U(4−2)=400+400=800 cfsQ(4) = 1\times U(4) + 1\times U(4-2) = 400 + 400 = 800 \text{ cfs}

Answer: 800 cfs at 8 p.m.

✓ Answer checked by Claude Sonnet 5.5

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