HW2 Chapter 2

 Nadia Karima Izzaty 1306369466 Dept. Teknik Sipil

2.2.1. 2.2.1. A reserv reservoir oir has the the following following inflows inflows and outflows outflows (in cubic cubic meters) meters) for the first first three months of the year. If the storage at the beginning of January is 60 m ! determine the storage at the end of "arch

n

S j =S 0 +

 I  −Q ∑ =  j

i

 j

1

S j =60 +[ ( 4− 8 ) + ( 6 −11) +( 9−5 )] 3

S j =55 m

2.2.2. 2.2.2. #om$u #om$ute te the consta constant nt draft from from a %00&he %00&hect ctar aree reser reservoi voirr for for a 0&da 0&day y $er $eriod iod during which the reservoir level dro$$ed half a meter des$ite an average u$stream inflow of 200!000 m 'day. uring the $eriod! the total see$age loss was 2 cm! the total $reci$itation was 10.% cm! and the total eva$oration was .% cm. (1 hectare * 10 + m2).

500 ha = 5.000.000 m 2 ,utflow e!el dr"p = 0.5 # 5.000.000 = 2.5#10 6 m3

Inflow $re%ipitati"n = 0.105 # 5.000.000 = 525.000

Seepa&e = 0.02 # 5.000.000 = 100.000 m 3

m3 '(t)l"* = 200.000 m 3+day = 6.000.000 m3+m"nth

,!ap"rati"n = 0.0-5 #5.000.000 = 425.000 m3

S j = I −Q S j =( 525.000 + 6.000 .000)−( 2.500 .000+ 100.000 + 425.000 ) S j =3.500.000 m

3

2.2.+. -rom the hydrologic records of over %0 years on a drainage basin of area %00 m2! the average annual rainfall is estimated as /0 cm and the average annual runoff as  cm. A reservoir in the basin! having an average surface area of 100 hectares! is $lanned at the basin outlet to collect available runoff for su$$lying water to a nearby community. he annual eva$oration over the reservoir surface is estimated as 10 cm. here is no groundwater leaage or inflow to the basin. etermine the available average annual withdrawal from the reservoir for water su$$ly.

100 ha = 1.000.000 m 2 /eer!"ir area = 500 km2 = 500.000.000 m 2  N"nreer!"ir area = 500  1 = 4-3.000.000 m 2 ,utflow ,!ap"rati"n = 0.13 # 1.000.000 =

Inflow /(n")) = 0.33 # 4-3.000.000 = 159.390.000

22.100.000 m 3

m3 /ain)all = 0.9 # 1.000.000 = 15.300.000 m 3

S j = I −Q S j =( 159.390.000 + 15.300 .000 ) −(22.100 .000 ) S j =152.590.000 m

3

2.%.2. #alculate the velocity and flow rate of a uniform flow 1 m dee$ in a 0&m wide stream with a$$roimately rectangular cross section! bed slo$e 1 $ercent! and "anning3s n of 0.0%. #hec that the criterion for fully turbulent flow is satisfied.

 R =

( 1∗30 )  A  = =0.9375  P ( 1 + 30 + 1 )

 R

2 /3

. √ S n

v=

v=

0.9375

2/ 3

. √ 0.01

0.035

v =2.737

m s 3

 m Q= v . A =82.11 s

(lly t(r(lent )l"* %riteri"n − 13

n . √  R . S ≥ 1.1∗10 6

6

0.035

− 13

. √ 0.9375 ∗0.01 ≥ 1.1∗10 − 10

1.78∗10

−13

≥ 1.1∗10

The )l"* i )(lly t(r(lent7

2.%.12. A rectangular o$en channel 12 m wide and 1 m dee$ has a slo$e of 0.001 and is lined with cemented rubble (n *0.02%). etermine (a) its maimum discharge ca$acity! and (b) the maimum discharge obtainable by changing the crosssectional dimensions without changing the rectangular form of the section! the slo$e! and the volume of  ecavation. 4int5 the best hydraulic rectangular section has a minimum wetted $erimeter or a width&de$th ratio of 2.

a 8aim(m di%har&e %apa%ity  R =

( 1∗12 )  A  = =0.857  P ( 1 + 12 + 1 )

 R

2 /3

. √ S n

v=

v=

0.857

2/ 3

. √ 0.001

0.02 5

v =1.141

m s 3

m Q= A ∗v =12∗1.141 = 13.692 s

  8aim(m di%har&e %apa%ity *ith et hydra(li% e%ti"n b =2 h  A = b∗ h=2 h 2h

2

2

=12

h =2.45 m; b =2 h= 4.9 m

 R =

( 2.45∗4.9 )  A  = =1.225  P ( 2.45 + 4.9 + 2.45 )

 R

v=

v=

2 /3

. √ S n

1.225

v =1.45

2 /3

. √ 0.001

0.025

m s

 m Q= A ∗v =12∗1.45 = 17.4 s

3

2..2 olve 7rob. 2..1 if the fluid is water. Assume 8m *0.1% m 2's and $ *1000 g'm . #alculate and com$are the laminar and turbulent momentum flues at 20 cm elevation if v * 1.%1  10 96 m2's for water.

 "r hear !el"%ity k = 0.4 u 1 Z   =  ln u '  k  Z 0 −6

1.51 ∗10

u' 

=

1 0.4

 ln

0.2 0.01

−7

' 

u =2.016∗10

:el"%ity &radient ' 

du u  = dz k ∗ z −7

du 2.016∗10 = 0.4∗ 0.2 dz

du =2.52∗10−6 dz :aminar

urbulent

 ρ∗v ∗du τ = dz

 ρ ∗ Km∗du τ = dz −6

τ =1000∗1.51∗10

∗2.52∗10−

6

−9

−4

τ =3.81∗10

τ =3.78∗10

/ati" ") laminar and t(r(lent −4

 R =

3.78∗10

−9

3.81∗10

−6

τ =1000∗0.15∗2.52∗10

=99212.6

h"* that the )l"* i rather t(r(lent.

2..%. he incoming radiation intensity on a lae is 200 ;'m2. #alculate the net radiation into the lae if the albedo is a * 0.06! the surface tem$erature is 00#! and the emissivity is 0./. 4

ℜ= e∗σ ∗T 

ℜ= 0.97∗5.67∗10− ∗300 8

ℜ= 445.492

4

 W  m

2

 Rn = Ri ( 1− α )−ℜ  Rn =200∗( 1−0 0 . 6 ) −445.492

 Rn =−257.492

 W  m

2