Homework 1 answers
Note: These answers ignore significant figures.
1.1 Four inches of runoff result from a storm on a drainage area of 50 mi2. Convert this
amount to acre-feet and cubic meters. (Note: 1 mi2 = 640 acres.)
Volume = Depth times Area:
4in * 50mi 2 *
640ac 1 ft
*
 10,667 ac  ft
1mi 2 12in
2
0.0254m  1609.344m 
3
4in * 50mi *
*
  13,157,139m
in
mi


2
1.2 Assume you are dealing with a vertical walled reservoir having a surface area of
400,000 m2 and that an inflow of 0.5 m3/sec occurs. How many hours will it take to
raise the reservoir level by 30 cm?
Volume = Rate times Time so: Time = Volume divided by Rate (with Volume =
Area times Depth):
400,000m 2 * 30cm *
0.5
m3
s
1m
100cm * 1h  66.7h
3600s
1.4 During a 24-hr time period, the inflow to a 500-acre vertical walled reservoir was
100 cfs. During the same interval evaporation was 1 in. Was there a rise or fall in
surface water elevations? How much was it? Give the answer in inches.
To get the volume of water added to the reservoir, multiply the time (24 hours) by
the flow rate (100 ft3/s). Divide this volume by the area (500 acres) to get what the
increase in depth would be if there had been no evaporation. Subtract the
evaporation (1 inch) from this to get net change. A positive result indicates a rise:
ft 3 3600s
24h *100
*
s
h *12 in  4.76in potential rise
ft
43,560 ft 2
500ac *
ac
4.76in  1in  3.76in actual rise
1.10 The evaporation rate from the surface of a 3650-acre lake is 100 acre-ft/day.
Determine the depth change (feet) in the lake during a 365-day year if the inflow to
the lake is 25.2 cfs. Is the change in lake depth an increase or decrease?
This one is very much like 1.4:
ft 3 86400s
*
s
d
 5 ft potential rise
43,560 ft 2
3650ac *
ac
ac  ft
365d *100
d  10 ft potential fall
3650ac
Change : 5 ft  10 ft  5 ft Since this is negative, it' s a decrease.
365d * 25.2
1.11 One and one-half inches of runoff are equivalent to how many acre-feet if the
drainage area is 25 mi2? (Note: 1 acre = 43,560 ft2.)
2
1.5in * 25mi 2 *
1 ft  5280 ft 
1ac
*
 2000ac  ft
 *
12in  mi  43,560 ft 2
1.12 One-half inch of rain per day is equivalent to an average rate of how many cubic
feet per second if the area is 500 acres?
Rate = Volume divided by Time:
0.5in
1 ft 43,560 ft 2
d
ft 3
* 500ac *
*
 10.5
d
12in
ac
86400s
s