Hw2 Chapter 2 60f54

Nadia Karima Izzaty 1306369466 Dept. Teknik Sipil

2.2.1. A reservoir has the following inflows and outflows (in cubic meters) for the first three months of the year. If the storage at the beginning of January is 60 m 3, determine the storage at the end of March

n

S j =S 0 + ∑ I j−Q j i=1

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

S j =55 m

2.2.2. Compute the constant draft from a 500-hectare reservoir for a 30-day period during which the reservoir level dropped half a meter despite an average upstream inflow of 200,000 m3/day. During the period, the total seepage loss was 2 cm, the total precipitation was 10.5 cm, and the total evaporation was 8.5 cm. (1 hectare = 104 m2). 500 ha = 5.000.000 m2 Outflow Level drop = 0.5 * 5.000.000 = 2.5*106 m3

Inflow Precipitation = 0.105 * 5.000.000 = 525.000

Seepage = 0.02 * 5.000.000 = 100.000 m3

m3 Outflow = 200.000 m3/day = 6.000.000 m3/month

Evaporation = 0.085 *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 m3

2.2.4. From the hydrologic records of over 50 years on a drainage basin of area 500 km2, the average annual rainfall is estimated as 90 cm and the average annual runoff as 33 cm. A reservoir in the basin, having an average surface area of 1700 hectares, is planned at the basin outlet to collect available runoff for supplying water to a nearby community. The annual evaporation over the reservoir surface is estimated as 130 cm. There is no groundwater leakage or inflow to the basin. Determine the available average annual withdrawal from the reservoir for water supply. 1700 ha = 17.000.000 m2 Reservoir area = 500 km2 = 500.000.000 m2 Non-reservoir area = (500 – 17) = 483.000.000 m2 Outflow Evaporation = 0.13 * 17.000.000 =

Inflow Runoff = 0.33 * 483.000.000 = 159.390.000

22.100.000 m3

m3 Rainfall = 0.9 * 17.000.000 = 15.300.000 m3

S j =I −Q S j =( 159.390.000+15.300 .000 ) −(22.100 .000) S j =152.590.000 m3

2.5.2. Calculate the velocity and flow rate of a uniform flow 1 m deep in a 30-m wide stream with approximately rectangular cross section, bed slope 1 percent, and Manning's n of 0.035. Check that the criterion for fully turbulent flow is satisfied.

R=

R

v=

v=

(1∗30) A = =0.9375 P (1+30+1) 2 /3

. √S n

0.9375 2/3 . √ 0.01 0.035

v =2.737

m s 3

Q=v . A=82.11

m s

Fully turbulent flow criterion n6 . √ R . S ≥ 1.1∗10−13 0.0356 . √ 0.9375∗0.01 ≥1.1∗10−13 −10

1.78∗10

−13

≥1.1∗10

The flow is fully turbulent!

2.5.12. A rectangular open channel 12 m wide and 1 m deep has a slope of 0.001 and is lined with cemented rubble (n =0.025). Determine (a) its maximum discharge capacity, and (b) the maximum discharge obtainable by changing the crosssectional dimensions without changing the rectangular form of the section, the slope, and the volume of excavation. Hint: the best hydraulic rectangular section has a minimum wetted perimeter or a width-depth ratio of 2. a) Maximum discharge capacity R=

(1∗12) A = =0.857 P (1+12+1)

R2 /3 . √ S v= n 2/ 3

0.857 . √ 0.0 0 1 v= 0.02 5 v =1.141

m s 3

Q= A∗v=12∗1.141=13.692

m s

b) Maximum discharge capacity with best hydraulic section b=2 h 2

A=b∗h=2 h 2 h2=12

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

R=

v=

v=

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

2 /3

. √S n

1.2252 /3 . √ 0.001 0.025

v =1.45

m s

m3 Q= A∗v=12∗1.45=17.4 s

2.8.2 Solve Prob. 2.8.1 if the fluid is water. Assume Km =0.15 m 2/s and p =1000 kg/m 3. Calculate and compare the laminar and turbulent momentum fluxes at 20 cm elevation if v = 1.51 x 10 ~6 m2/s for water. For shear velocity (k = 0.4) u 1 Z = ln u ' k Z0 −6

1.51∗10 1 0.2 = ln u' 0.4 0.01 u' =2.016∗10−7 Velocity gradient du u' = dz k∗z −7

du 2.016∗10 = dz 0.4∗0.2

du =2.52∗10−6 dz Laminar

Turbulent

ρ∗v∗du τ= dz

ρ∗Km∗du τ= dz

τ =1000∗1.51∗10−6∗2.52∗10−6

τ =1000∗0.15∗2.52∗10−6

τ =3.81∗10−9

τ =3.78∗10−4

Ratio of laminar and turbulent 3.78∗10−4 R= =99212.6 3.81∗10−9 shows that the flow is rather turbulent.

2.8.5. The incoming radiation intensity on a lake is 200 W/m2. Calculate the net radiation into the lake if the albedo is a = 0.06, the surface temperature is 300C, and the emissivity is 0.97. ℜ=e∗σ∗T

4

−8

4

ℜ=0.97∗5.67∗10 ∗300

ℜ=445.492

W 2 m

Rn=Ri ( 1−α )−ℜ Rn=200∗( 1−0 0 .6 ) −445.492 Rn=−257.492

W m2