1.72, Groundwat er Hydrology
Th e e a r t h ’s e n e r gy ( r a dia t ion ) cy cle
surface Heat ing of surface
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:
W e st e r lie s
Polar east erlies
West erlies
West erlies
Polar east erlies Tr a de W in ds
Tr a de W in ds Polar High
Polar front / subarct ic low
Subt ropical high
Trade w inds Equit orial low Trade w inds Subt ropical high
Polar front / subarct ic low
Cou ple d Ea r t h Cy cle s
Local rat es depend on relief, precipit at ion, and rock t ype.
Glaciers Fresh Wat er At m osphere Oceans 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
Where does t he w at er go? To m ake t hings…t o clean t hings….
• A sm all savings in eit her of t hese cat egories w ould free up significant quant it ies for public supplies
3 t em por ally accessible
runoff ( AR) ( 12,500 k m / y ear)
Hum an appropriat ion of accessible RFWSland
[ 24,980 k m / year ( 30% ) ] Hum an appropriat ion of t ot al
RFWSland
[ 24,980 k m / y ear ( 23% ) ] Hum an appropriat ion of ET
[ 18,200 k m / y ear ( 26% ) ]
Wit hdr aw s [ 4,430 k m / year ( 35% ) ]
I nst r eam uses [ 2,350 k m / year ( 19% ) ]
Hum an appropriat ion of AR [ 6,780 k m / year ( 54% ) ]
3) Wit hdraw als ( consum ed)
• 12,000 m3/ ha average irr igat ion wat er applicat ion. • 240,000,000 ha of world irrigat ed area.
• 2,880 km3/ year.
• 65% consum ed = 1,870 km3/ year.
• I ndust rial use: 975 km3/ year, 9% consum ed, 90 km3/ year. • Municipal use: 300 km3/ year, 17% consum ed, 50 km3/ year. • Reservoir losses – 5% loss, 275 km3/ year.
• Tot al consum ed = 2,285 km3/ year. 4) I nst ream use
• 28.3 L/ s per 1,000 populat ion, applied t o 1990 populat ion, 4,700 km3/ year. • Assum ing 50% of w ast e get s t reat m ent – 2,350 km3/ year.
• Neglect s dispersed pollut ion ( agricult ural) and flood w at ers.
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
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 t hat t opographically appears t o cont ribut e all t he w at er t hat flow s 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.
• Horizont al proj ect ion of t his area is t he drainage area.
• The boundaries of a w at ershed are called a divide, and can be t raced 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
S Gin
Gou t
W a t e r Ba la n ce Equa t ion
evapot ranspirat ion and groundw at er flow .
D isch a r ge in t o La k e Pa r a bola
This different ial equat ion can be solved by separat ion of variables
(
H
π
dV
)
=
Q
−
Ae
=
Q
−
He
dt
c
dH
π
dH
π
H
=
Q
−
He
=
Qc
−
e
c
dt
c
dt
π
H
10 8
H [ m ] 6
Q= 10 [ m3/ day]
4 c= 0.1 [ 1/ m ]
e= 0.03 [ m / day] 2
0
200 400 600 800 1000 T [ days]
dt
dH
Qc
St e a dy St a t e Solu t ion
=
−
e
π
H
Qc
=
(10)( 1
0 )
.
H
=
π
=
10 6
.
mRe 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 T0 Here at TR
V = QTR TR = V/ Q Exact ly
Com plet e m ixing
Q Q
TR = V/ Q On Average
For t h is pr oble m :
Consider discharge and evaporat ion out of Lake Parabola. Assum e no seepage. Q = input = out put = 10 [ m3/ day]
(
H
V
)
=
π
H
2=
π
10 6
.
2=
,
1 765
m
3T
2
c
2( 1
0 )
.
R
=
V
Q
=
,
1 765
m
3=
176 5
.
days
/
10
m
3/
day
Consider yourself:
Quant it y
Wat er w eight [ Kg] Average I nt ake [ Kg/ day] Residence Tim e [ day]
Men 60
3 14
Wom en 50 2.1
