Post on 06-Apr-2018
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Counts Analysis of the Uniform Steady Flow Experiments
The first experiment1. Calculate the flow discharge (Qmeasured)
Given :
hVnotch = 9,15 cm
p = 27 cm
g = 981 cm/s2
h/p = 0,339
= 35
Kv = 0,183 cm (obtained from graph)
Ce = 0,5821 (obtained from graph)
Solution:
Qmeasured = Ce15
82g tan
2
( hVnotch + Kv)
2,5
= 0,5821
15
89812 tan
2
35
0,183+9,15 2,5
= 1153,766 cm3/s
2. Calculate the sectional area of flow (A)Given:
h ( h average) = 0,983 cm
b = 30 cm
Solution:
A = b x h= 30 x 0,983
= 29,5 cm2
3. Calculate the flow velocity (v)Given:
Q = 1153,766 cm3/s
A = 29,5 cm2
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Solution:
v = A
Q
=29,5
1153,766
= 39,111 cm/s
4. Calculate Viscocity at T = Given:
cm
2/s
cm2/s
Solution:
By using Linier Interpolation Method
()
( )
x 310 cm2/s
5. Calculate the number Renold (Re)Given:
v = 39,111 cm/s
h average = 0,983 cm
=3
10446,8 cm2/s
Solution:
Re =
averagehv
=3
10446,8
983,0111,39
= 4553,499
The flow included the type of turbulent flow for Re> 1000
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6. Calculating the circumference of wet (P)Given:
b = 30 cm
h ( h average) = 0,983 cm
Solution:
P = b + 2h
= 30 + (2 x 0,983)
= 31,967 cm
7. Calculate the hydraulic radius (R)Given:
A = 29,5 cm2
P = 31,967 cm
Solution:
R =P
A
=
31,967
29,5
= 0,923 cm
8. Calculating empirical formula based on coefficientGiven:
R = 0,923 cm
Sf = So = 0,0047 cm
v = 39,111 cm/s
Solution:
a. Manning Coefficient
nv
SfR 21
32
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=
39,111
0,0047923,0 21
32
= 0,00166
b. Strickler Coefficient
ks =2
13
2
SfR
v
=2
13
2
0,0047983,0
39,111
= 601,862
c. Chezy Coefficient
C = SfR
v
=0047,0983,0
39,111
= 593,860
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The second experiment1. Calculate the flow rate (Qmeasured)
Given :
hVnotch = 12 cm
p = 27 cm
g = 981 cm/s2
h/p = 0,444
= 35
Kv = 0,183 cm (obtained from graph)
Ce = 0,5919 (obtained from graph)
Solution:
Qmeasured = Ce15
82g tan
2
( hVnotch + Kv)
2,5
= 0,591915
89812 tan
2
35
0,183+12 2,5
= 2283,954 cm3/s
2. Calculate the sectional area of flow (A)Given:
h ( h average) = 1,333 cm
b = 30 cm
Solution:
A = b x h
=30 x 1,333
= 40 cm2
3. Calculate the flow velocity (v)Given:
Q = 2283,954 cm3/s
A = 40 cm2
Solution:
v =
A
Q
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=40,0
2283,954
= 57,099 cm/det
4. Calculate Viscocity at T = Given:
cm2/s
cm2/s
Solution:
By using Linier Interpolation Method
()
( )
x 310 cm2/s
5. Calculate the number Renold (Re)Given:
v = 57,099 cm/s
h aerage = 1,333 cm
=3
10446,8 cm2/s
Solution:
Re =
averagehv
=3
10446,8
333,1099,57
= 9013,946
The flow included the type of turbulent flow for Re> 10006. Calculating the circumference of wet (P)
Given:
b = 30 cm
h ( h ataverage) = 1,333 cm
Solution:
P = b + 2h
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= 30 + (2 x 1,333 )
= 32,667 cm
7. Calculate the hydraulic radius (R)Given:
A = 40 cm2
P = 32,667 cm
Solution:
R =P
A
=32,667
40,0
= 1,223 cm
8. Calculating empirical formula based on coefficientGiven:
R = 1,224 cm
Sf = So = 0,0047 cm
v = 57,099 cm/s
Solution:
a. Manning Coefficientn
v
SfR 21
32
=57,099
0,0047224,1 21
32
= 0,00137
b. Strickler Coefficientks
21
32
SfR
v
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=2
1
3
2
0047,0224,1
099,57
= 727,682
c. Chezy Coefficient
C = SfR
v
=0047.0224,1
099,57
= 59,860
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The third experiment1. Calculate the flow rate (Qmeasured)
Given :
hVnotch = 12,30 cm
p = 27 cm
g = 981 cm/s2
h/p = 0,456
= 35
Kv = 0,183 cm (obtained from graph)
Ce = 0,5936 (obtained from graph)
Solution:
Qmeasured = Ce15
82g tan
2
( hVnotch + Kv)
2,5
= 0,593615
89812 tan
2
35
0,183+12,30 2,5
= 2434,044 cm3/s
2. Calculate the sectional area of flow (A)Given:
h ( h average) = 1,683 cm
b = 30 cm
Solution:
A = b x h
= 30 x 1,683
= 50,5 cm2
3. Calculate the flow velocity (v)Given:
Q = 2434,044 cm3/s
A = 50,5 cm2
Solution:
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v =A
Q
=50,5
2434,044
= 48,199 cm/s
4. Calculate Viscocity at T = Given:
cm2/s
cm2/s
Solution:
By using Linier Interpolation Method
()
( )
x 310 cm2/s
5. Calculate the number Renold (Re)Given:
v = 48,199 cm/s
R = 1,513 cm
=3
10446,8 cm2/s
Solution:
Re =
Rv
=3
10446,8
513,148,199
= 9606,299
The flow included the type of turbulent flow for Re> 10006. Calculating the circumference of wet (P)
Given:
b = 30 cm
h ( h ataverage) = 1,683 cm
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Solution:
P = b + 2h
= 30 + (2 x 1,683)
= 33,367 cm
7. Calculate the hydraulic radius (R)Given:
A = 50,5 cm2
P = 33,367 cm
Solution:
R =P
A
=33,367
50,5
= 1,513 cm
8. Calculating empirical formula based on coefficientGiven:
R = 1,513 cm
Sf = So = 0,0047 cm
v = 48,199 cm/s
Solution:
a. Manning Coefficientn
v
SfR 21
32
=48,199
0,00471,513 21
32
= 0,00187
b. Strickler Coefficient
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ks =
21
32
SfR
v
=2
1
32
0047,0513,1
48,199
= 533,338
c. Chezy Coefficient
C = SfR
v
=0047.0513,1
48,199
= 571,477
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The fourth experiment1. Calculate the flow rate (Qmeasured)
Given :
hVnotch =13,50 cm
p = 27 cm
g = 981 cm/s2
h/p = 0,500
= 35
Kv = 0,183 cm (obtained from graph)
Ce = 0,6015 (obtained from graph)
Solution:
Qmeasured = Ce15
82g tan
2
( hVnotch + Kv)
2,5
= 0,601515
89812 tan
2
35
0,18313,50 2,5
= 3102,841 cm3/s
2. Calculate the sectional area of flow (A)Given:
h ( h average) = 1,700 cm
b = 30 cm
Solution:
A = b x h
= 30 x 1,700
= 51,0 cm2
3. Calculate the flow velocity (v)Given:
Q = 3102,841 cm3/s
A = 51,0 cm2
Solution:
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v =A
Q
=51,0
3102,841
= 60,840 cm/s
4. Calculate Viscocity at T = Given:
cm2/s
cm2/s
Solution:
By using Linier Interpolation Method
()
( )
x 310 cm2/s
5. Calculate the number Renold (Re)Given:
v = 60,840 cm/s
R = 1,527 cm
=3
10446,8 cm2/s
Solution:
Re =
Rv
=3
10446,8
527,160,840
= 12245,800
The flow included the type of turbulent flow for Re> 10006. Calculating the circumference of wet (P)
Given:
b = 30 cm
h ( h ataverage) = 1,700 cm
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Solution:
P = b + 2h
= 30 + (2 x 1,700)
= 33,400 cm
7. Calculate the hydraulic radius (R)Given:
A = 51,00 cm2
P = 33,400 cm
Solution:
R =P
A
=33,400
51,0
= 1,527 cm
8. Calculating empirical formula based on coefficient
Given:
R = 1,527 cm
Sf = So = 0,0047 cm
v = 60,840 cm/s
Solution:
a. Manning Coefficient
nv
SfR 21
32
=60,840
0,00471,527 21
32
= 0,00149
b. Strickler Coefficient
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ks =
21
32
SfR
v
=2
13
2
0047,01,527
60,840
= 669,254
c. Chezy CoefficientC
SfR
v
=0047.0527,1
60,840
= 718,172
9. Calculate the number Renold (Re)Given:
v = 60,840 cm/s
R = 527,1 cm
=3
10446,8 cm2/s
Solution:
Re =
Rv
=3
10446,8
527,160,840
x
= 12245
The flow included the type of turbulent flow for Re> 1000
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The fifth Experiment1. Calculate the flow rate (Qmeasured)
Given :
hVnotch =14,80 cm
p = 27 cm
g = 981 cm/s2
h/p = 0,548
= 35
Kv = 0,183 cm (obtained from graph)
Ce = 0,6123 (obtained from graph)
Solution:
Qmeasured = Ce15
82g tan
2
( hVnotch + Kv)
2,5
= 0,612315
89812 tan
2
35
0,18314,80 2,5
= 3963,309 cm3/s
2. Calculate the sectional area of flow (A)Given:
h ( h average) = 2,000 cm
b = 30 cm
Solution:
A = b x h
= 30 x 2,000
= 60,0 cm2
3. Calculate the flow velocity (v)Given:
Q = 3963,309 cm3/s
A = 60,0 cm2
Solution:
v =
A
Q
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=60,0
3963,309
= 66,055 cm/s
4. Calculate Viscocity at T = Given:
cm2/s
cm2/s
Solution:
By using Linier Interpolation Method
()
( )
x 310 cm2/s
5. Calculate the number Renold (Re)Given:
v = 66,055 cm/s
R = 1,765 cm
=3
10446,8 cm2/s
Solution:
Re =
Rv
=3
10446,8
765,1055,66
x
= 15641,759
The flow included the type of turbulent flow for Re> 10006. Calculating the circumference of wet (P)
Given:
b = 30 cm
h ( h ataverage) = 2,000 cm
Solution:
P = b + 2h
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= 30 + (2 x 2,000)
= 34,000 cm
7. Calculate the hydraulic radius (R)Given:
A = 60,0 cm2
P = 34,000 cm
Solution:
R =P
A
=34,000
60,0
= 1,765cm
8. Calculating empirical formula based on coefficientGiven:
R = 1,765 cm
Sf = So = 0,0047 cm
v = 66,055 cm/s
Solution:
a. Manning Coefficient
n
v
SfR 21
32
=66,055
0,00471,765 21
32
= 0,00152
b. Strickler Coefficient
ks =2
13
2
SfR
v
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=2
13
2
0047,0765,1
055,66
= 659,254
c. Chezy CoefficientC
SfR
v
=0047.0765,1
055,66
= 725,306
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DISCUSSION
1. Tail gate function is to equalize or control the water level is raised orelevated water level at every turn of the discharge. But from an experiment
conducted each height h is not the same. This is caused because each turn
of the discharge will affect the water level.
2. From the experiments have been conducted Manning coefficient valuesobtained do not approach the price of 0.01 m -1 / 3 sec or 0.00215 cm-1
/ 3 dt. This is because the slope of the basic channels that are too big
which causes a higher flow rate so that the value of Manning coefficient
become large.
3. Renold Numbers 1000, included turbulent flow.
4. Heights h1, h2, h3 on one trial is not the same as the channel bottom slopeis too large and lack of precision when measuring height.
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CONCLUSION
1. From the steady uniform flow experiments can be known that in the uniformflow discharge, muka air,kecepatan dan berat jenis yang tetap sepanjang
waktu dan seragam diberbagai tempat yang di tinjau.
2. Setelah melakukan lima kali percobaan didapatkan nilai rata-rataManning,Strickler, dan Chezy sebesar:
Manning : 0,0030 Strickler : 339,0172 Chezy : 362,254
3. Nilai koefisienManning mempunyai nilai yang berkebalikan dengan nilaikoefisien Strickler ( n =
Ks
1)