EAS 345 Lab #4 Solutions
HEAT AND HYDROLOGY: EVAPORATION AND SNOWMELT
1. Calculate the depth of water, Dz that will evaporate in 8 hours if the average sunlight striking level ground is
1000 W m-2 and ⅓ of the sunlight is used to evaporate water
I
Dz 
1 dQ
DT
Dz
 cz
 L
A dt
Dt
Dt
1 IDt 1 1000 8  3600

 0.00384m
3 L 3 2.5 106 1000
2. Find the total sunlight striking level ground on 23 November at latitude 41N and at 41S.
dSE  149.5 2.5 cos(#days fromJune 21)  149.5 2.5 cos(152) 147.3
  23.5  cos(# days fromJune 21)  23.5  cos(152)  20.75
H  cos1[ tan( ) tan( )]  cos1[ tan( ) tan(20.75 )] 
 cos1[ tan(41 ) tan(20.75 )]  71  1.24
  41
 cos1[ tan(41 ) tan(20.75 )]  109  1.91   41
 Q  S  149.5 
A
0
 d 
 SE 
2
H sin( ) sin( )  cos( ) cos( ) sin(H )
86400

Qtot = ____1.47(10)7 J m-2
Qtot = 4.16(10)7 J m-2
b: Assuming 25% of this heat is used to evaporate water, calculate the depth of water evaporated per day
and per year at this rate.
7
Dz  0.25
Q
1.4710
 0.25
 0.00147m
L
2.5 106 1000
Daily evaporation = 0.00147 m
Annual evaporation = 0.537 m
0.00416 m
1.52 m
3. Calculate the depth of snow melted if 1 cm of rain with T = 5C falls on the snow. Assume that the snow is initially
at 0C and that snow = 100 kg m-3.
Dz snow
1
dQ  0  czDT  LDz
A
c liq zDT
4186 1000 0.01 (5)


 0.00625m
L f  snow
3.35 105  100
4. Calculate the rates of heat transport and evaporation from a lake with C = 1.9(10)-3 if v = 5 m s-1,
A: T0 = 280 K, RH0 = 100%,
B: T2 = 270 K, RH2 = 30%.
Assume air = 1.2 kg m-3
1 dQ
= Cc( T 0 - T 2 )v  1.9 10 3 1.2 1004  (280  270 )  5  114
A dt
e  RH  610 2
7
10
 1 610 2  991
T C
10
 0.3  610 2
3
10
 149
1 dQ CL( e0 - e2 )v 1.9 103  2.5 106  (991 149)  5
=

 158
A dt
461.7  275
Rv T
Dz 
Q
158

 6.24108 m  s -1  1.96 m  yr-1
6
L 2.5 10 1000
5. Recalculate the rates of heat transport and evaporation in Problem 4 if both temperatures are raised
10C and all other conditions remain the same
Sensible heat fluxes remain the same but evaporation rates double because saturated vapor
pressure doubles with every increase of 10C.
6. Calculate the rates of sensible heat transfer and condensation onto a snow surface if T2 = 5C, RH2 = 100% and v =
5 m s-1. Then compute the depth of melted snow as a result of each of these processes if the snow has a density, snow =
100 kg m-3 and if the process continues for 5 hours. Compare this to your answer in problem 3.
Assume that the snow surface is at T0 = 0C and since
the atmosphere is stable (cold air below), Cturb = 10-3
1 dQ
= Cc( T 0 - T 2 )v  10 3 1.2 1004  (273  278 )  5  30
A dt
1 dQ CL( e0 - e2 )v 103  2.5 106  (610 863)  5
=

 25
A dt
461.7  275.5
Rv T
All downward heat fluxes contribute to melting the snow.
1 dQ
30  25
Dz  A dt Dt 
 5  3600  .030 m
L
3.35 10 5 100
9. Use the 4-panel graph to calculate daily potential evaporation, given that T = 70ºF, Td = 60ºF, Iavg = 300 W m-2 
600 cal cm-2 day-1, v = 5 m s-1  10 mph
PE = 0.23 in day-1
A:Calculate all required quantities in the sheet for the Ring Infiltrometer Data (see red below).
B: Plot the data using Excel with f on the y axis (max value = 8 on graph) and time in min on the x axis
C: From the graph, estimate f0 = _______________ cm hr-1 and fc = ____________cm hr-1.
D: Use Horton's equation together with the values in the table and fo and fc to calculate k.
3
time min
Dt hr
Vol H2O (cm )
F (cm)
0
0.0000
0
0.000
1
0.0167
63
0.110
0.11
0.5
6.59
2
0.0167
120
0.210
0.100
1.5
5.99
5
0.0500
269
0.470
0.260
3.5
5.21
10
0.0833
436
0.762
0.292
7.5
3.50
1
20
0.1667
681
1.191
0.428
15
2.57
0
30
0.1667
862
1.507
0.316
25
1.90
60
0.5000
1153
2.016
0.509
45
1.02
90
0.5000
1298
2.269
0.253
75
0.51
120
0.5000
1440
2.517
0.248
105
0.50
DF (cm)
Tavg
f (cm/hr)
8
7
6
f
5
4
3
2
0
20
40
60
80
100
120
time (min)
f0  7.0, fc  0.4
8
7
f  fc  ( fo  fc)e
 kt
1  fo  fc  60  (7  0.4) 
k  ln 
 ln
 2.33
t  f  fc  25  (1.9  0.4) 
6
5
f
Once f0 and fc have been determined from the graphs to the
right, rearrange Horton’s solution for infiltration, f, to solve for k
and substitute one set of values from the table above.
4
3
2
1
0
0
2
4
time (min)
6
8
2. Given initial and final infiltration rates, fo = 1.25 cm hr-1, fc = 0.05 cm hr-1, and the maximum storage, Smax = 5 cm
t
S(t)
R
fc + (f0 - fc)(Smax-S)/Smax
fin
Rxs
S(t+Dt)
0
0
0.5
1.25
0.50
0.45
0.00
1
0.45
0.3
1.14
0.30
0.25
0.00
2
0.70
0.2
1.08
0.20
0.15
0.00
3
0.85
0.7
1.05
0.70
0.65
0.00
4
1.50
0.9
0.89
0.89
0.84
0.01
5
2.34
1.1
0.69
0.69
0.64
0.41
6
2.98
0.6
0.54
0.54
0.49
0.06
7
3.46
0.3
0.42
0.30
0.25
0.00
8
3.71
0.5
0.36
0.36
0.31
0.14
9
4.02
0.4
0.28
0.28
0.23
0.12
10
4.26
0.4
0.23
0.23
0.18
0.17
11
4.44
0.6
0.19
0.19
0.14
0.41
12
4.57
0.5
0.15
0.15
0.10
0.35
13
4.67
0.4
0.13
0.13
0.08
0.27
14
4.75
0.4
0.11
0.11
0.06
0.29
15
4.81
0.2
0.10
0.10
0.05
0.10
4.86
5.66
2.34
EAS 345 Lab #6
GROUNDWATER
1. Calculate the hydraulic conductivity, K for an unconfined aquifer with bottom at h = 0 where two piezometers
40 meters apart record heads (or water table heights), h1=50 m and h2 = 35 m. The discharge, q/L = 0.01 m3s-1/m is
uniform and flows from h1 to h2.
q
h
= - KH
L
x
K
q Dx
1 40
 0.01
 6.27 10  4
LH Dh
42.5 15
2. A confined aquifer 50 m thick with hydraulic conductivity, K = 10-3 and head with slope h/x = 0.1 empties
into a river 1 km long. Find the contribution of ground water flow into the river.
q = - KHL
h
 10 3  50 1000  2  0.1  10
x
The factor, 2 occurs because
every river has two banks!
3. Find the total and available volume of water (to a well) in an aquifer 100 m thick with an area of 1 km 2 if porosity,
P = 0.4 and specific yield, Sy = 0.15.
P
S
VH 2O
V
VA( H 2O )
V
 VH 2O  P  V  0.4 108  4 107
 VA( H 2O )  S  V  0.15108  1.5 107
4. Calculate how long water in the aquifer of problem 3 would last if it is not replenished and if it is pumped at the
rate, q = 0.01 m3 s-1
q
DVAH2O
Dt
 Dt 
q
0.1

 6.67105 s
7
DVAH2O 1.5 10
5. If 2 cm of rain falls but 0.8 cm runs off on the surface, calculate the rise of the water table if the specific yield, Sy =
0.2 and the soil was at field capacity
Sy 
V A( H 2O )
V

DH A( H 2O )
DH
 DH 
DH A( H 2O )
Sy

(2  0.8)
 6 cm
0.2
6. Find the rate of change of height of the water table, h/t, in a rectangular area x = 200 m long and y = 100 m
wide of an unconfined aquifer with bottom at h = 0 if the water table has a height H = 35 m on the left side and H =
20 m on the right and the slope of the water table h/x = -.25 on the left and h/x = -.10 on the right. Assume the
porosity is P = 0.25, specific yield, Sy = 0.15 and hydraulic conductivity, K = 10-4. To do this you must calculate the
discharge at the left (inflow) and right (outflow) sides of the aquifer, take the difference to find the volume
accumulation rate and divide by the surface area of the region.
35 m
100 m
35 m
h
q
 10 4  35 (0.25)  8.7510 4
  = - KH
x
 L  left
h
q
 10 4 15 (0.10)  1.5 10 4
  = - KH
x
 L  right
DVH 2O LDxDH H 2O DxS y DH 200 0.15 DH
q
q



    
LDt
LDt
Dt
Dt
 L  left  L  right
 q 
DH
1
q 
5
1

       2.0810 m  s
Dt 200 0.15  L  left  L  right 
In 24 hours the water table will rise 1.8 m
7. Find the speed of water moving through an aquifer with K = 10-5 and porosity, P = 0.4 if the slope of the water
table, h/x = 1.
q
K h
105
v=


0.1  2.5 106 m  s 1
PA
P x
0.4
Porosity reduces the effective cross section area available to the flowing water.
Therefore the velocity through and aquifer is increased by a factor P-1.
8. Observations show that the total length of all tributaries is related to the watershed's area by, L = 1.72A0.94,
where the length is in km and the area in km2. If the same unconfined aquifer as in Problem 7 empties into all the
tributaries in an area, A = 100 km2, find the total length of the tributaries, and the total base flow that will result.
Assume the aquifer is H = 35 m thick.
L = 1.72A0.94  1.721000.94  130.5 km
q = - KHL
h
 10 5  35 1.305 10 5  2  0.1  4.57 m3  s 1
x
EAS 345 HYDROLOGYLab 7
WELL FLOW AND GROUNDWATER
1. Water is pumped from a well in an unconfined aquifer at a rate q = 0.01 m3 s-1. A piezometer (test well) 100 m
away (r = 100) records a steady height, H = 60 m. Find H at the edge of the well (r = 0.5 m) if K = 10-5.
 r2 
ln
q
 
u
K ( H 22 - H 12 )
 r 1   602  0.01 ln(200)
 H 12  H 22 
qu =
K
105 
 r2 
ln  
 r1 
H1  43.7
2. Calculate K for a steady well with q = 0.02 m3 s-1 in an unconfined aquifer, given that, r1=50 m, H1=50.0 m:
r2=20 m, H2=45.5 m
r 
qu ln 2 
K
 r 1   0.02 ln(2.5)  1.36105 m  s 1
K 
qu =
 ( H 22 - H 12 )  (2500 2070)
 r2 
ln  
 r1 
3. Find the maximum sustainable pumping rate for a well of r = 0.5 m in an unconfined aquifer with h = 60 m
at a distance, r = 100 m if K = 10-4.



2
2 

K
(
)  104  ( 6022 - 0 )
H
H
2
1


 0.213 m3  s 1
max(qu ) = max

ln200
 r2  
ln




 r1  

To maximize q, subtract
as little as possible, thus
H12 = 0. This is curious
because then there is no
water in the well.
4. For a confined aquifer, find K for a well pumping water at a rate q=0.01 m3s-1 if, H = 35 and r1=40 m, h1=40 m:
r2=20 m, h2=15 m
r 
qc ln 2 
2KH( h2 - h1 )
 r 1   0.01 ln(0.5)  1.26106 m  s 1
K 
qc =
2H( h2 - h1 ) 2  35 (15  40)
 
ln r 2 
 r1 
5. Calculate the drawdown, Zr for an unsteady well of radius r = 0.5 m in a confined aquifer with q = 0.001, S =
2(10)-4, H = 40, and K = 9(10)-5 that has been pumping for 10 hours.
q  e-u
q
du 
W (u )
Zr=
 r2 S
4T 4Tt u
4T
r 2S
0.25 2  10 4
u

 3.85 106
5
4Tt 4  9  10  36000
W (u )  .5772 ln(u )  u  2.72
W (u )  .5772 ln(u )  11.9
Zr= 
Zr= 
r2 S
2500 2 104

 3.8510 2
5
4Tt 4  9 10  36000
2
u2 
.001
(2.7)  2.38 m
4T
.001
(11.9)  10.5 m
4T
6, 7. Use the method of images to recalculate the drawdown for the well in Problem 5 if a: a river is located 50 m
away from the well, b: the aquifer ends 50 m away.
z well river  z1  z 2  10.5  2.4  8.1
z wellend  z1  z 2  10.5  2.4  12.9
7. Given a well with T = 10-3, q = 0.002, S = 2(10)-4, calculate the drawdown at a piezometer 25 m from the well at the
times listed below.
8. Water is pumped from a well in a confined aquifer at a rate, q = 0.003 m3 s-1. The drawdown at a piezometer a
distance r = 20 m away as a function of time is given in the table below. Complete the table and use the graphs of
u vs W(u) to find the transmissibility, T, and the specific yield, Sc.
t
1 min=
5 min=
30 min=
1 h=
2 h=
5 h=
Zr
t (s)
0.737
2.16
4.04
4.78
5.53
6.52
60
300
1800
3600
7200
18000
Points are plotted on the Zr-T curve then copied to the W(u)-u curve and
overlain until they fit. The pair chosen at the red cross mark is u=0.1, W(u)
= 1.84. The points together with the cross mark are then returned to the ZrT curve and overlain with the original points. The cross mark then yields t =
210, Zr = 1.9. With these values, we solve for T and S below.
T=
S
q
4 Z r
4Ttu
r
2
W(u) 
0.003
1.84  2.3 104
4 (1.9)
4  2.3 104  210 0.1

 4.83105
2
20
Matching pair Zr = 1.9, t = 210
101
102
103
104
t (sec)
105
106
Matching pair u=0.1, W(u) = 1.84
Lab 8
STREAM DISCHARGE AND RATING CURVES
1. Find the rating curve for a stream with the following values of q and z,
q(m3 s-1)
7.9
36.2
71.6
111.7
z(m)
0.5
1
1.5
2
Determine z0 by graphing points
and extrapolating curve to q = 0.
This yields z0  0.3
Then use 2 (z,q) pairs in rating curve
equation to solve for K and b as below.
ln(q)= ln(K) + b ln(z - z 0 )
ln(36.2)  ln(K )  b ln(1.0  0.3)
ln(111.7)  ln(K )  b ln(2.0  0.3)
4.72  3.59  b(0.531 0.357)
b  1.27 : K  57
2.5
2
1.5
Excel’s Solver
gives slightly
different results
1
0.5
0
0
Var
z0
K
b
20
40
60
80
100
120
Value
0.275781
55.0582
1.298567
2. Find q when z = 3.5
q = K(z - z0 )b  60(3.2)1.3  272 m3  s 1
3. On a rainy day (24 hours) 10 cm of precipitation falls and all runs down the stream. If the basin area is, A = 109 m2
a. Calculate the mean runoff or discharge. b. Will the stream flood? Flood stage is z = 3.5 m. Explain.
q = ERA = 0.10/86400109 = 1157 m3 s-1
Since this is much larger than q at flood stage (assuming values from the last problem) stream will definitely flood.
4. Use the Manning Formula to find the average current speed and discharge in a river with a rectangular bed that is
40 m wide, 4 m deep, has a slope, s = .002, and a very rough stream bed with n = .15.
A
40  4

 3.33
Pwet (40  2  4)
1
2
1
1
v = A R3 S 2 
 3.33 0.002  0.99 m  s 1
n
0.15
q  Av  158.9 m 3  s 1
R
5. Once every several years a natural stream or river
overflows its banks. The water spreads out onto the
floodplain. For the accompanying cross section of the
Chemung River (a tributary of the Susquehanna River) at
Elmira, NY, indicate the bounds of the floodplain on the
contour map and on the cross section assuming that water
level can rise 10 m above flood stage.
Click to see water rise on this and on the next slide.