Dr. Todd Rayne of Hamilton College provided
these solutions.
1. A vertical water tank is 15 ft in diameter
and 60 ft. high. What is the volume of the tank in cubic feet?
V
= p r2
h
h
= 60 ft; d = 15 ft; r = d/2 = 7.5 ft
V
= p
x 7.5 ft2 x 60 ft
V
= 11,000 ft3 or 1.1 x 104
ft3
3. If the above tank were measured and found to
have an inside diameter of exactly 15.00 ft and a height of 60.00 ft, what
would be the volume in cubic feet?
V = p r2 h
h
= 60.00 ft; d = 15.00 ft; r = d/2 = 7.50 ft
V
= p
x 7.50 ft2 x 60.00 ft
V
= 10,600 ft3 or 1.06 x 104
ft3
5. If a well pumps at a rate of 8.4 gal per
minute, how long would it take to fill the tank described above?
V
= 300.3 m3 x 264.17 gal/m3 = 79,330.25 gal
t
= 79,330.25 gal/(8.4 gal/min) = 9400 min = 160 hr
7. If during the next week the pool still loses
2.35 in. of water to evaporation, even with 29 mm of rainfall, how many liters
of water must be added?
A
= 60.0 m2
Evaporation
= 2.35 in/week x 0.0254 in/m x 60.0 m2 = 3.58 m3/week
Precipitation
= 29 mm = 0.029 m x 60.0 m2 = 1.74 m3/week
Water
to be added is equal to evaporation less precipitation
Water
added = 3.58 m3/week – 1.74 m3/week
Water
added = 1.84 m3/week x 1000 L/m3 = 1800 L
9. The parking lot of the Spendmore Megamall has
an area of 128 ac. It is partially landscaped to provide some areas of grass.
Assume that an average of 63% of the water that falls on the parking lot will
flow into a nearby drainage ditch, and the rest either evaporates or soaks into
unpaved areas. If as summer thunderstorm drops 3.23 cm of ran, how many cubic
feet of water will flow into the drainage ditch?
A
= 128 acre x 43,560 ft2/acre
= 5.57568 x 106 ft2
Precipitation
= 3.23 cm x 0.394 in/cm x 0.833 ft/in = 0.106 ft
Volume
of rainfall = Precipitation x Area
Volume
of rainfall = 0.106 ft x 5.57567 x 106 ft2
= 5.91 x 105 ft3
Volume
of runoff = 63 % of volume of rainfall
Volume
of runoff = 5.91 x 105 ft3 x 0.63 = 3.72 x
105 ft 3
11. What mass of water at 15o C can
be cooled 1o C by the amount of heat needed to sublime (go from a
solid to a vapor state) 18 g of ice at 0o C?
18
g ice x 677 cal/g = 12,186 cal absorbed
12,000
cal x 1 g/cal = 12,000 g of water cooled
13. At a water elevation of 6391 ft, Mono Lake
has a volume of 2,939,000 ac-ft, and a surface area of 48,100 ac. Annual inputs
to the lake include 8.0 in. of direct precipitation, runoff from gauged streams
of 150,000 ac-ft per year, and ungauged runoff and groundwater inflow of 37,000
ac-ft per year. Evaporation is 45 in. per year.
First
convert all measurements to units of acre-feet per year
Precipitation
= 8 in/yr x 0.083 ft/in = 0.67
ft/yr
=
0.67 ft/yr x 48,100 ac = 32,067 ac-ft/yr
Evaporation
= 45 in/yr x 0.083 ft/in =
3.75 ft/yr
=
3.75 ft/yr x 48,100 ac = 180,375 ac-ft/yr
(A) Make a water budget showing inputs, in ac-ft
per year and outputs in ac-ft per year. Does the input balance the output?
INPUTS
(ac-ft/yr)
Precipitation 32,000
Streamflow 150,000
Ground
water inflow 37,000
Total 219,000
OUTFLOWS
(ac-ft/yr)
Evaporation 180,000
Total 180,000
Input
is greater than output
(B) Will the average lake level rise or fall
from the 6391-ft elevation over the long term?
The lake level will rise over the long term, as inputs are greater than
the only output, which is evaporation.
(C) What would be the lake surface area when the
inputs balance the outputs? (Assume
that the volume of gauged and ungauged runoff and ground-water inflows remain
constant with a change in lake surface area.)
In
order for the inputs and outputs to be equal, the lake will rise so that the
surface area is greater and more water evaporates as evaporation as well as the
precipitation input is a function of lake surface area.
The
excess input at an area of 48,100 acre surface area is 219,000 ac-ft/yr –
180,000 ac-ft/yr or 39,000 ac-ft/yr
The
net evaporation is total evaporation less precipitation
Net
evaporation = 45 in/yr – 8 in/yr = 37 in/yr = 3.08 in/yr
Increase
in area = (39,000 ac-ft/yr) / (3.08 ft/yr) = 12,649 ac
Total
area of lake = 12, 662 ac + 48,100 ac = 60,762 ac = 61,000 ac
(D) What
is the residence time for water in Mono Lake when the water surface is at 6391
ft?
Residence
time = (2,939,000 ac-ft) / (180,000 ac-ft/yr) = 16.3 yr