Selasa, 05 Mei 2015

TUGAS HIDROLIKA 1
Nama              : Jessica Ambar
NIM                : 110013062

1.      Dua buah kolam A dan B dihubungkan dengan pipa secara seri sebagai tergambar dibawah, L1=72 m, d1=1', f1=0,019, L2=52m, d2=8", f2=0,024, & L3=73m, d3=1', f3=0,020..Jika elevasi kolam A +100,00 dan debit pengaliran Q = 90 lt/dt berapa elevasi kolam B? Gambarkan garis energie dan tekanan (gravitsi bumi g = 9,81 m/dt2, minor loses diperhitungkan) !
2.      Kebutuhan air suatu daerah ditampung ke dalam bak Ditribusi dengan Debit 67,5 lt/dt dan mempunyai elevasi +125,50m diambilkan dari sebuah sungai didalam tanah dengan elevasi +25 (air tak terbatas). Pengambilan air dilakukan dengan pompa (elv + 50,00m) & sambungan pipa secara seri-pararel seperti gambar dibawah. Berapa tekanan manometer dititik sebelah kanan pompa P dan titik C ( g = 9,81 m/dt2 ) ?

1.       Penyelesaian :
diketahui :
L1
=
72
m
L2
=
52
m
L3
=
73
m
d1
=
1'
=
12
d2
=
8
d3
=
1'
=
12
f1
=
0.019
f2
=
0.024
f3
=
0.020
elev A
=
100
m
g
=
9.81
m/dt2
Q
=
90
lt/dt
ditanya :
gambar garis energie ?
elevasi B ?

penyelesaian :
D1
=
12
X
0.0254
=
0.3048
D2
=
8
X
0.0254
=
0.2032
D3
=
12
X
0.0254
=
0.3048

kehilangan energi antara kolam A dan B
hc1
=
0.5
V1^2
(dianggap kont.
=
0.5
V1^2
2g
sempurna)
2g
hf1
=
0.019
72
x
V1^2
=
4.48819
V1^2
0.3048
2g
2g
hf2
=
0.024
52
x
V1^2
=
6.14173
D1^4
x
V1^2
0.2032
2g
D2^4
2g
=
6.14173
0.008631
x
V1^2
0.0017049
2g
=
31.09252
V1^2
2g

hf3
=
0.020
73
x
V1^2
=
4.79003
V1^2
0.3048
2g
2g
hc2
=
0.50
V1^2
=
0.50
D1^4
x
V1^2
2g
D2^4
2g
=
0.50
0.008631
x
V1^2
0.0017049
2g
=
2.53125
V1^2
2g
he
=
(v2-v3)^2
=
A1
-
1
V1^2
2g
A2
2g
=
0.5
V1^2
2g

Ƹ h
=
Δh
=
hc1
+
hf1
+
hf2
+
hf3
+
hc2
+
he

=
43.90198
V1^2
2g

Q
=
90
lt/dt
=
0.90
m3/dt
Q
=
A1
X
V1
V1
=
Q
=
0.90
=
3.76147
m/dt
A1
0.23927
Δh
=
43.90198
V1^2
=
31.659273
m
2g
Jadi elevasi kolam B adalah
=
68.34072734
m




2.       Penyelesaian
Penyelesaian :
L1
=
200
m
L2
=
3000
m
L3
=
3000
m
L4
=
2750
m
L5
=
3000
m
D1
=
16
D2
=
13
D3
=
10
D4
=
8
D5
=
12
f1
=
0.015
f2
=
0.019
f3
=
0.021
f4
=
0.025
f5
=
0.020
Q
=
67.5
lt/dt
=
0.675
m3/dt
P
=
50
m
B
=
75
C
=
90
elev (A)
=
25
elev (D)
=
125.50
g
=
9.81
m/dt2
D1
=
16
X
0.0254
=
0.4064
D2
=
13
X
0.0254
=
0.3302
D3
=
10
X
0.0254
=
0.254
D4
=
8
X
0.0254
=
0.2032
D5
=
12
X
0.0254
=
0.3048

hf1
=
8 x 0.015 x 200
Q1^2
=
22.3828
Q1^2
3.14^2 x 9.81 x  0.4064^5
hf2
=
8 x 0.019 x 3000
Q2^2
=
1201.023
Q2^2
3.14^2 x 9.81 x  0.3302^5
hf5
=
8 x 0.020 x 3000
Q5^2
=
1886.419
Q5^2
3.14^2 x 9.81 x  0.3048^5
Q1
=
Q2
=
Q5
( hukum kontinuitas )
Q
=
Q3
+
Q4
A1
x
V1
=
A3 x V3
+
A4 x V4
………… 1
hf3
=
hf4
berawal dan berkumpul di satu batang
f3
L3
x
V3^2
=
f4
L4
x
V4^2
D3
2g
D4
2g
V3
=
( f4/f3 x L4/L3 x D3/D4 ) ^0.5 V4
…… 2

persamaan 2 masuk persamaan 1
A1
x
 V1
=
A3
( f4/ f3 x L4/L3 x D3/D4 ) ^ 0.5
V4 + A4 x V4
=
(A3( f4/ f3 x L4/L3 x d3/d4 ) ^0.5 + A4 )
V4
V4
=
A1
x
V1
A4 + ( f4/f3 x L4/L3 x D3/D4 ) ^ 0.5 A3
=
D1 ^ 2
x
V1
D4 ^ 2 +  ( f4/f3 x L4/L3 x D3/D4 ) ^ 0.5 D3 ^2
=
1.415975
V1
hf4
=
0.025
2750
x
V4^2
0.2032
2g
=
68.75
2.004985
V1^2
0.2032
=
678.359937
V1^2
2g
=
678.359937
A1^2
x
V1^2
A1^2
2g

hf4
=
678.359937
x
Q1^2
=
2056.871107
Q1^2
(1/4 x 3.14 (0.4064)^2)^2
2g
hf2
+
hf4
+
hf5
=
5144.312568
Q1^2
=
50
Q1^2
=
0.009719472
Q1
=
0.098587
m3/dt
=
Q2
=
Q5
V4
=
1.415975
V1
=
1.415975
Q1
A1
=
1.076708625
m/dt
Q4
=
A4
x
V4
=
0.034899183
m3/dt
Q3
=
Q1
-
Q4
=
0.063687817
m3/dt

hf3
=
8 x 0.021 x 3000
Q3^2
=
19.99155
m
3.14^2 x 9.81 x  0.2032^5
hf4
=
2056.871107
Q1^2
=
19.99155
m
hf2
=
1201.022641
Q2^2
=
11.67322
m
hf5
=
1886.41882
Q5^2
=
18.33485
m
hf2
+
hf4
+
hf5
=
49.99961
m
=
50 m
…….. OK !
hf1
=
22.382801
Q1^2
=
0.217547
m
Hp
=
hf1 + 25 + 125.50
=
150.72
m
DP
=
Q x h x Ɣw
=
283.024587
DK
ɲ x 75
PB
=
125.50 - 25 - hf2
=
88.8268
m
Ɣ
tekanan manometer PB     =
8.882678
kg/cm2
PC
=
125.50 - 40 - hf2 - hf4
=
53.8352
m
Ɣ
tekanan manometer PC    =
5.383524
kg/cm2
 tekanan manometer dititik sebelah kanan Pompa P
=
14.266202
kg/cm2