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contoh soal mekanika fluida
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Nama : Hendri Andrian
NIM : 12010033
Kelas : Teknik perminyakan A
Ujian : Mekanika Fluida
1. Diketahui:
D = 200 mm = 200.10-3m = 2.10-1m
p = 800 mm = 800.10-3m = 8.10-1m
A = 20 mm = 20.10-3m = 2.10-2m
Ρ = 6
Ρ = 0.6
Ditanya:
Dapatkah silinder mengapung secara vertikal?
Jawab:
a. Rprl silinder = ��
��
0,6 =��
����
=600 kg/m3
b. Rprl base = ���
��
6 =��
����
=6000 kg/m3
c. V silinder = π.r2.t silinder
= 3,14 (0,1)2.0,8
=0,025 m3
d. V base = π.r2.t base
= 3,14 (0,1)2.0,02
=0,00628
e. W base = Wf yang dipindahkan
Ws + W base = ρ fluida.J (V silinder tercelup + V benda)
(ρs.g.Vs)+(ρb.g.Vb) = 9810 (A silinder tercelup.htercelup + V benda)
(600.9,81.0,025)+(6000.9,81.0,00628)= 9810 (0,314.h tercelup + 0, 00628)
147,15 + 36,96 =308,34h tercelup + 6,161
184,11 – 6,161 = 308,34 h tercelup
h tercelup = �� ,��–�,���
���,�
=0,578 m
f. OB = ½ h tercelup
= ½ 0,578
= 0,289 m
g. OG = ½ h silinder
= ½ 0,8
= 0,4
h. BG = OG – OB
= 0,4 – 0,928
= 0,111 m
i. Bm =���
�
=
�.�.�.�.�
��
.������ !"
=
#,$�.%,$.%,$.%,$.%,$
��
&�,� .�,�.�,�'.�,()�
= 0,0437 m
j. GM = BM – BG
= 0,00437 – 0,111
= - 0,107 m
Jadi, silinder tersebut tidak stabil (terendam) didalam air.
2. Diketahui:
Pm = 2 bar = 2.105 Pa
µm = 50 kali udara
ρm = 80 kali udara
ρp = udara
µp = udara
Ditanya:
Pp =...?
Jawab:
Rn = �*.�*.+*
,* =
�".�".+"
,"
�*
�" =
�"
�* x
+"
+* x
,*
,"
= -
���- x
.(
� x
(�-
-
= .(
��
Eu = �*.�*.
/* =
�".�"
/"
/"
/* =
�"
�*x(
�*
�")2
/"
.�����/0 =
-
���-x(
��
.()2
/"
.�����/0 = 5,12.10-4
Pp = 5,12.10-4. 200000
= 102,4 Pa
Jadi Pp nya adalah 102,4 Pa.