Lecture Packet 1. Course Introduction Water Balance Equation

  1.72, Groundwat er Hydrology Prof. Charles Harvey

  

Le ct u r e Pa ck e t # 1 : Cou r se I n t r odu ct ion , W a t e r Ba la n ce Equ a t ion

W h e r e doe s w a t e r com e fr om ?

  I t cycles. The t ot al supply doesn’t change m uch.

  39 Moist ur e over 100 land Pr ecipit at ion on land

  385 Pr ecipit at ion on ocean 424

  61 Evaporat ion from Evaporat ion t he ocean fr om t he land

  Surface runoff I nfilt rat ion Evapot ranspirat ion and evaporat ion Groundw at er

  38 sur face recharge discharge Groundw at er flow

  1 Gr oundw at er discharge

Low per m eabilit y

st rat a

  H y dr ologic cy cle w it h y e a r ly flow volu m e s ba se d on a n n u a l su r fa ce ₃ pr e cip it a t ion on e a r t h , ~ 1 1 9 ,0 0 0 k m / y e a r .

  3000 BC – Ecclesiast es 1: 7 ( Solom on) “ All t he rivers run int o t he sea; yet t he sea is not full; unt o t he place from whence t he rivers com e, t hit her t hey ret urn again.” Greek Philosophers ( Plat o, Arist ot le) em braced t he concept , but m echanism s w ere not underst ood. _{t h}

  17 Cent ury – Pierre Perrault show ed t hat rainfall w as sufficient t o explain flow of t he Seine.

  Th e e a r t h ’s e n e r gy ( r a dia t ion ) cy cle

  Solar ( short w ave) radiat ion T errest rial ( long- w ave) radiat ion

  Reflect ed Out going I ncom ing

  Space

  99.998

  6

  18

  6

  4

  39

  27 Backscat t er ing Net radiant by air em ission by greenhouse Reflect ion gases by clouds

  Net radiant em ission by

  At m osphere

  11 Net clouds 4 absor pt ion by

  Absor pt ion greenhouse by clouds gasses & clouds

  20 Absor pt ion by Reflect ion at m osphere Net

  Net radiant Net lat ent by surface sensible em ission by heat flux heat flux surface

  15

  7

  24

  46 Ocean and Absor pt ion by Heat ing of surface

  Land surface

  46 0.002 Cir cu la t ion r e dist r ibu t e s e n e r gy r) ₂ /y m c ₂₀ l a ₁ c _{- 20}

  1 ( x lu

• _{- 60} f n o ti ia
^4.0 d ra t e _{3.0 Tot al Flux} N _2.0 r) Lat ent y _Heat l/ ^1.0 a _{2 2} c (1 r _Ocean fe s - 1.0 Cur r ent s _Sensible n _Heat ra

  T y

• _{- 2.0} rg e n E _{- 4.0 - 3.0}

_{90 N}

_{60 30 30 60 90 S} Lat it ude

  Th e Ea r t h ’s En e r gy ( Ra dia t ion) Cycle : Spa t ia l dist r ibu t ion of e ne r gy a n d t e m pe r a t u r e dr ive s cir cu la t ion – bot h globa l a n d loca l:

  Polar High Polar east erlies

  Polar front / subarct ic low West erlies

  W e st e r lie s

  Subt ropical high Trade w inds

  Tr a de W in ds

  Equit orial low Trade w inds

  Tr a de W in ds

  Subt ropical high West erlies

  Polar front / subarct ic low Polar east erlies

  Polar High

  Cou ple d Ea r t h Cy cle s H y dr ologic Cy cle M a t e r ia l Cycle En e r gy Cycle

  Solar Radiat ion

  Eart h Erosion

  Radiat ion Dissolut ion

  Evapo- t ranspirat ion Precipit at ion

  Tect onics

  9 Est im at es from river out flow s indicat e 17 x 10 Tons/ Year of m at erial is t ransport ed ₉ int o t he Ocean. Anot her 2 or 3 x 10 Tons/ Year is t rapped in reservoirs.

  80% of m at erial t ransport ed is part iculat e 20% is dissolved 0.05 m m / yr ( m ay have been accelerat ed by m an, and act ual rat es m ay be m uch larger) Local rat es depend on relief, precipit at ion, and rock t ype.

  At m ospher e ( 12,900) 71,000 Neg. 2,700 Neg. 1,000

  116,000 Biom ass ( 1, 120 ) 505,000 71,000

  Glaciers Riv ers ( 24,000,000) ( 2,120)

  Soil Moist ur e Lakes & m ar shes ( 16,500) ( 102,000)

  458,000 2,700 46,000 44,700

  

43,800 Ground Wat er Oceans 2,200 ( 10,800,000) ( 1,338,000,000) Glaciers Fresh Wat er At m osphere Oceans Re la t iv e v olu m e s of w a t e r in gla cie r s, fr e sh w a t e r , a t m osph e r e a n d oce a n s. Est im a t e of t h e W or ld W a t e r Ba la n ce Pa r a m e t e r Su r fa ce V olu m e V olu m e Equ iva le n t Re side n ce a r e a ( k m 3 ) X % de pt h ( m ) Tim e _{2 6} ( k m ) X 1 0 ₆ 1 0

  Oceans and seas 361 1370 94 2500 ~ 4000 years Lakes and 1.55 0.13 < 0.01 0.25 ~ 10 years reservoirs Sw am ps < 0.1 < 0.01 < 0.01 0.007 1- 10 years River channels < 0.1 < 0.01 < 0.01 0.003 ~ 2 w eeks Soil m oist ure 130 0.07 < 0.01 0.13 2 w eeks – 1 year Groundw at er 130 60 4 120 2 – w eeks

  10,000 years I cecaps and glaciers

  17.8

  30

  2 60 10- 1000 years At m ospheric w at er 504 0.01 < 0.01 0.025 ~ 10 days Biospheric w at er < 0.1 < 0.01 < 0.01 0.001 ~ 1 w eek ₃ Am azon is 6,000 km / yr ( ~ 5x m ore t han Zaire- Congo) 0.0003% is pot able and available.

  What are w at er needs for hum ans? Prim it ive condit ions – 3 t o 5 gallons/ day Urban use – 150 gallons/ day US Fresh Wat er Use – 1,340 gallons/ day

  Where does t he w at er go? To m ake t hings…t o clean t hings….

  I t e m Ga llon s

  1 pound of cot t on 2,000 1 pound of grain- fed beef 800 1 loaf of bread 150 1 car 100,000 1 kilow at t of elect ricit y

  80 1 pound of rubber 100 1 pound of st eel

  25 1 gallon of gasoline 10 1 load of laundry 60 1 t en- m inut e show er 25- 50

  Som e point s:

• A lot of water is used for agriculture _o

  56% ( 76 BGD) is consum pt ive use _o 20% ( 28 BGD) is lost in conveyance _o 24% ( 33 BGD) is ret urn flow

• A lot of water is used for therm oelectric power generation _o

  87% of all indust rial wat er use

• A sm all savings in either of these categories would free up significant quant it ies for public supplies

  W a t e r Use in t h e US in 1 9 9 0 Fr e sh W a t e r Use 1 9 9 0 Tot a l 3 3 9 BGD

  California 35.1 Texas 20.1 I daho 19.7 I llinois 18.0 Colorado 12.7 Louisiana 11.7 Michigan 11.6 New York

  10.5 Pennsylvania 9.8 I ndiana 9.4 Mont ana 9.3 Four st at es account for 27% 11 st at es account for 50% RFWS land ₃ ( 110,300 k m / year ) Tot al

  Tot al r unoff ₃ evapot ranspirat ion ( 40,700 k m / y ear) on land ₃ ( 69,600 k m / y ear)

  Rem ot e flow ₃ ( 7,774 k m / year) Uncapt ur ed floodw at er ₃ ( 20,426 k m / y ear)

  Geogr aphically and t em por ally accessible runoff ( AR) ₃ ( 12,500 k m / y ear)

  Wit hdr aw s I nst r eam uses _{3 3} [ 4,430 k m / year ( 35% ) ] [ 2,350 k m / year ( 19% ) ] Hum an appropriat ion of ET Hum an appropriat ion of AR _{3 3} [ 18,200 k m / y ear ( 26% ) ] [ 6,780 k m / year ( 54% ) ]

  Hum an appropriat ion of accessible RFWS land ₃ [ 24,980 k m / year ( 30% ) ] Hum an appropriat ion of t ot al RFWS land ₃ [ 24,980 k m / y ear ( 23% ) ]

  Analy sis of hum an appropr iat ion of renew able freshw at er supply ( RFSW) on land. Figur e adapt ed fr om Post el et al., Science, ( 271) p. 758, Feb. 9, 1996

  Post e l e t a l.’s Ca lcu la t ion s

1) Calculat ion of appropriat ed ET – indirect • 132 billion tons of biom ass produced a year (Vitousek, 1992).

• 30% used by people.
• Approxim ate people’s use of ET from the proportion of biom ass – 30% .
3<
• Subtract agricultural irrigation (2,000 km
₃ / year) and assum e half t he w at er. from parks and law ns is irrigat ion ( 80 km / year) .

  2) I naccessible runoff

• 95 of the Am azon, Half of the Zaire-Congo and all of the Polar rivers 7774
₃ .km / year ( did not consider ot her nort hern rivers – conservat ive?) . 3<
• Flood water. 11,100 km / year of runoff is base flow ( 6) .
3<
• Capacity of m an-m ade reservoirs is 5,500 km / year.
• Accessible runoff = Baseflow + reservoir capacity = 12,500.

  adj ust ed for accessible

  3) Wit hdraw als ( consum ed) ₃

• 12,000 m / ha average irr igat ion wat er applicat ion.
• 240,000,000 ha of world irrigated area.
3<
• 2,880 km / year.
₃ • 65% consum ed = 1,870 km / year. _{3 3} • I ndustrial use: 975 km / year, 9% consum ed, 90 km / year. ₃ 3<
• Municipal use: 300 km / year, 17% consum ed, 50 km / year.
₃ • Reservoir losses – 5% loss, 275 km / year. ₃ • Total consum ed = 2,285 km / year.

  4) I nst ream use ₃

• 28.3 L/ s per 1,000 population, applied to 1990 population, 4,700 km
₃ /
• Assum ing 50% of waste gets treatm ent – 2,350 km / year.
• Neglects dispersed pollution (agricultural) and flood waters.

  Est im at ed w at er use in t he Unit ed St at es, billion gallons per day ( bgd) , for dom est ic and com m ercial purposes. Adapt ed from Solley, Pierce, and Perlm an, 1993.

  

Re la t ive M e r it s of Su r fa ce a n d Su bsu r fa ce Re se r v oir s

  Surface Reservoirs Subsurface Reservoirs Disadvant ages Advant ages

  Few new sit es free ( in USA) Many large- capacit y sit es available High evaporat ive loss, even w here hum id Pract ically no evaporat ive loss clim at e prevails Need large areas of land Need very sm all areas of land May fail cat ast rophically Pract ically no danger of failure Varying wat er t em perat ure Wat er t em perat ure uniform Easily pollut ed Usually high biological purit y, alt hough pollut ion can occur Easily cont am inat ed by radioact ive fallout Not rapidly cont am inat ed by radioact ive fallout Wat er m ust be conveyed Act as conveyance syst em s, t hus obviat ing t he need for pipes or canals

  W a t e r sh e d H y dr ologic Bu dge t s

  Delineat ion of a w at ershed ( drainage basin, river basin, cat chm ent )

• Area that topographically appears to contribute all the water that flows t hrough a given cross sect ion of a st ream . I n ot her w ords, t he area over w hich w at er flow ing along t he surface w ill event ually reach t he st ream , upst ream of t he cross- sect ion.
• Horizontal projection of this area is the drainage area.
• The boundaries of a watershed are called a divide, and can be traced on a t opographic m ap by st art ing at t he locat ion of t he st ream cross- sect ion t hen draw ing a line aw ay from t he st ream t hat int ersect s all cont our lines at right angles. I f you do t his right , t he lines draw n from bot h sides of t he st ream should int ersect . Moving t o eit her side

  P ET G in S Q G ou t

  W a t e r Ba la n ce Equa t ion ∂S = P − ( ) + + G Q ET + G _{in out} ∂t

  ∂S

  At st eady- st at e: = 0

  ∂t P G Q ET G _{in out} = ( ) + + +

  Hydrologic Product ion Runoff, or Hydrologic Cir culat ion

  Q and P are t he only quant it ies t hat w e can t ry t o m easure direct ly. I f st eady st at e is assum ed, t hese m easurem ent s can be used t o calibrat e m odels of evapot ranspirat ion and groundw at er flow . From and engineering point of view w e are int erest ed in underst anding w hat cont rols Q.

  ) Q = P + (G − G − ET

_{in out}

  How m uch of Q is available t o hum an use? How can w e increase Q?

  La k e Pa r a bola

  ( A sim ple quadrat ic m odel of t he cross sect ion of a circular lake)

  4

  3

  2 H = cr

  2 π

  2 = π = H

  

H A r

c

1 H H

  π π

  2 ( ) = ( ) = =

  V H A h dh Hdh H ∫ ∫ c 2c

• - 2 - 1

  1

  

2

  D isch a r ge in t o La k e Pa r a bola ₃ I f t here is const ant volum e flux Q m / day int o t he lake, how does t he dept h depend

  on t im e? ( At t im e 0, t he dept h is H o m ) Since t here is no out flow , t he charge in st orage m ust equal t he inflow

  Mass- balance ( w at er balance) equat ion ₂

  dV ( H ) d ( H ) dH π π

  Q = = = H dt 2c dt c dt

  This different ial equat ion can be solved by separat ion of variables

  T H cQ dt = hdh

  ∫ ∫ π

  H _o

  2

  2 cQ H H ²

  T = − H = T H _o

• o 2cQ

  2 2 π π

  3

  2 H [ m ] ₃ Q= 10 [ m / day] c= 0.1 [ 1/ m ] 1 2 4 6

  8 10 14

  12 T [ days]

  D isch a r ge in t o a n d Ev a por a t ion ou t of La k e Pa r a bola

  e = rat e of evaporat ion [ L/ T]

  ( π

  H dV ) =

  

Ae Q = Q − He −

dt c

dH

  π dH

  π H = Q − He = Qc − e c dt c dt

  πH

  10

  8 H [ m ] 6 Q= 10 [ m ³ / day]

  4 c= 0.1 [ 1/ m ] e= 0.03 [ m / day]

  2 200 400 600 800 1000 T [ days]

  dt dH Qc St e a dy St a t e Solu t ion = − e

  πH Qc = (10)( 1.

  0 ) H = = 10 6 . m

  π e

  π ( . 0 03)

  D ischa r ge , e va por a t ion a n d se e pa ge

  4

  3

2 H

  1 - 2 - 1

  1

  2 dV (H e π k π ² Q − eA − kHA = Q − H − H

  ) =

dt c c

  π ² H dh = Q − eπ H − kπ H dh = Qc − e − kH c dt c c dt πH

  2.75

  2.5

  2.25 H [ m ]

  2 ₃ Q= 10 [ m / day] c= 0.1 [ 1/ m ] 1.75 e= 0.03 [ m / day] 1.50 k= 0.03 [ 1/ day]

  1.25 200 400 600 800 1000 T [ days]

  dh Qc St e a dy St a t e Solu t ion = − e − kH dt ₂ πH

  Qc − eH − kH = 0 π

  Quadrat ic form ula: ₂

  ) − e ± (e − 4 ck Q π π π

• - 3.8 m , 2.8 m

  H = = 2k π

  Re side n ce Tim e ( st e a dy st a t e , com ple t e m ix in g)

  Very useful concept for degradat ion and chem ical react ions No m ixing

  Q Here at T Here at T R

  V = QT T = V/ Q Exact ly _{R R} Com plet e m ixing

  Q Q

  T R = V/ Q On Average

  For t h is pr oble m : Consider discharge and evaporat ion out of Lake Parabola. Assum e no seepage. ₃ Q = input = out put = 10 [ m / day] π ^{2 π 2 3} ) 10 6 .

  V (H = H = = , 1 765m 2( 1 0 ) .

  2 c ₃ , 1 765m /

  T = _R

  V Q = = 176 5 . days ₃ 10 m / day

  Consider yourself: Quant it y Men Wom en Wat er w eight [ Kg]

  60

  50 Average I nt ake [ Kg/ day]

  3

  2.1 Residence Tim e [ day]

  14

  14