EAS 345 Lab #4 Solutions HEAT AND HYDROLOGY: EVAPORATION AND SNOWMELT

1. Calculate the depth of water, z 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

t

zL

t

Tzc

dt

dQ

AI

1

m00384.01000105.2

360081000

3

1

3

16

L

tIz

2. Find the total sunlight striking level ground on 23 November at latitude 41 N and at 41S.

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.

)sin()cos()cos()sin()sin(864005.149

2

0 HHd

SA

Q

SE

4191.1109)]75.20tan()41tan([cos4124.171)]75.20tan()41tan([cos)]75.20tan()tan([cos)]tan()tan([cos

1

1

11

H

3.147cos(152)2.5149.521) June from days cos(#2.5149.5dSE 75.20)152cos(5.23)21Junefromdays#cos(5.23

m00147.01000105.2

1047.125.025.0

6

7

L

Qz

Daily evaporation = 0.00147 m 0.00416 mAnnual evaporation = 0.537 m 1.52 m

m00625.01001035.3

)5(01.010004186

01

5

snowf

liqsnow L

Tzcz

zLTczdQA

3. Calculate the depth of snow melted if 1 cm of rain with T = 5C falls on the snow. Assume that the snow is initially at 0C and that snow = 100 kg m-3.

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

1145)270280(10042.1109.1 3 )vT-Tc(C = dt

dQ

A

120

1582757.461

5)149991(105.2109.1 63

TR

)ve-eCL( =

dt

dQ

A

1

v

20

14926103.099126101

2610

10

3

10

7

10

CT

RHe

1186

yrm96.1sm1024.61000105.2

158 --

L

Qz

5. Recalculate the rates of heat transport and evaporation in Problem 4 if both temperatures are raised 10C 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 10C.

6. Calculate the rates of sensible heat transfer and condensation onto a snow surface if T 2 = 5C, 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 = 0C and since the atmosphere is stable (cold air below), Cturb = 10-3

305)278273(10042.110 3 )vT-Tc(C = dt

dQ

A

120

All downward heat fluxes contribute to melting the snow.

255.2757.461

5)863610(105.210 63

TR

)ve-eCL( =

dt

dQ

A

1

v

20

m030.360051001035.3

25305

t

LdtdQ

A1

z

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

time min t hr Vol H2O (cm3) F (cm) DF (cm) Tavg f (cm/hr)

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

20 0.1667 681 1.191 0.428 15 2.57

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

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 axisC: 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.

f0 7.0, fc 0.4

0

1

2

3

4

5

6

7

8

0 20 40 60 80 100 120

time (min)

f

0

1

2

3

4

5

6

7

8

0 2 4 6 8

time (min)

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.

33.2)4.09.1(

)4.07(ln

25

60ln

1

)(

fcf

fcfo

tk

efcfofcf kt

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 S(t+Dt)

Rxs

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.

41027.615

40

5.42

101.0

h

x

LH

qK

x

hKH- =

L

q

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.

101.0210005010 3

x

hKHL- = q

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.

78)(

)(

78

105.11015.0

104104.0

2

2

2

2

VSVV

VS

VPVV

VP

OHAOHA

OHOH

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

sV

qt

t

Vq

OAH

OAH 57

1067.6105.1

1.0

2

2

The factor, 2 occurs because every river has two banks!

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

cm62.0

)8.02()()()( 222

y

OHAOHAOHAy S

HH

H

H

V

VS

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.

44

right

44

left

105.1)10.0(1510

1075.8)25.0(3510

x

hKH- =

L

qx

hKH- =

L

q

35 m

35 m100 m

15

rightleft

rightleft

sm1008.215.0200

1

15.020022

L

q

L

q

t

H

t

H

t

HxS

tL

HxL

tL

V

L

q

L

q yOHOH

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.

165

105.21.04.0

10

smx

h

P

K

PA

q = v

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.

kmA1.72 = L 0.940.94 5.13010072.1

1355 57.41.0210305.13510

smx

hKHL- = q

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.

7.43

10

)200ln(01.060

ln

ln

1

52

22

H

Krrq

HH

rr

)H-H( K = q 1

2u

221

1

2

21

2

u

15

2

1036.1)20702500(

)5.2ln(02.0ln

ln

sm

)H-H(rrq

K

rr

K = q

21

2

1

2u

1

2u

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

132

42 213.0

200ln

06010

ln

max)max(

sm

)-(

r

r

)H-H( K = q

2

1

2

21

2

u

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.

To maximize q, subtract as little as possible, thus H1

2 = 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

161026.1)4015(35

)5.0ln(01.0ln

ln

sm

2)h-hH(2rrq

K

rr

)h-hKH(2 = q

12

1

2c

1

2

12c

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.

mT4

= Z

uuW

4Tt

Sru

uWT4

qdu

ue

T4

q = Z

r

-u

4Tt

Srr 2

5.10)9.11(001.

9.11)ln(5772.)(

1085.3360001094

10225.0

)(

65

42

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.

mT4

= Z

uuuW

4Tt

Sru

r 38.2)7.2(001.

72.2)ln(5772.)(

1085.3360001094

1022500 25

422

2

9.124.25.10211.84.25.1021

zzzzzz

endwell

riverwell

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 Z r t (s)

1 min= 0.737 60

5 min= 2.16 30030 min= 4.04 1800

1 h= 4.78 3600

2 h= 5.53 7200

5 h= 6.52 18000 4103.284.1)9.1(

003.0 4

W(u) Z4

q =T

r

52

4

1083.420

1.0210103.24

r

4TtuS

2

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 Zr-T 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.

101 102 103 104 105 106

t (sec)

Matching pair Zr = 1.9, t = 210

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) z(m)7.9 0.536.2 171.6 1.5111.7 2

0

0.5

1

1.5

2

2.5

0 20 40 60 80 100 120

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.

Kbb

bKbK

)z -(z b + (K) = (q) 0

57:27.1)357.0531.0(59.372.4

)3.00.2ln()ln()7.111ln()3.00.1ln()ln()2.36ln(

lnlnln

Var Value

z0 0.275781

K 55.0582

b 1.298567

Excel’s Solver gives slightly different results

2. Find q when z = 3.5133.1 272)2.3(60 sm)z -K(z = q b

0

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 = 0.10/86400109 = 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.

13

1

wet

9.158

99.0002.033.315.0

1

33.3)4240(

440

smAvq

smSRAn

1 = v

P

AR

2

1

3

2

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.