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Transcript of PmF4
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8/19/2019 PmF4
1/11
UNIVERSIDAD NACIONAL
AUTÓNOMA DE MÉXICO
FACULTAD DE INGENIERÍA
Departamento de Termo f!"do#
La$orator"o de Me%&n"%a de F!"do# '
In(en"er)a Me%&n"%a
*ra%t"%a + ,Cuerpos Sumergidos”
*rofe#or-
ALVARE. SANC/E. MILTON CARLOS M0I0
Gr!po1 '2
A!mno-
Mart)ne3 V"ramonte# F""$erto
45'674
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Objetivo:
Determinar la posición del Centro de Presiones sobre unasuperficie plana parcialmente y totalmente sumergida en un
líquido en reposo
!ntroducción:
Lo# %!erpo# #8"do# #!mer("do# en !n )9!"do 9!e e:per"mentan !nemp!;e a #! nom$re= todo %!erpo #!mer("dotota o par%"amente en !n )9!"do e:per"menta !n emp!;e >ert"%a ?ae de
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En e e9!""$r"o am$a# f!er3a# ap"%ada# #o$re p!nto# d"ferente#e#tar&n a"neada#
E;empo na>"o-
S" por efe%to de !na f!er3a atera= %omo a prod!%"da por !n (ope demar= e e;e >ert"%a de na>)o #e "n%"nara )o=e# de%"r= a %apa%"dad para re%!perar a >ert"%a"dad0 Eo #e %on#"(!ed"#eando %on>en"entemente e %a#%o ? repart"endo a %ar(a de modo9!e re$a;e a po#"%"8n de %entro de (ra>edad= %on a 9!e #e %on#"(!ea!mentar e $ra3o de par0
Sa$emo# 9!e a pre#"8n
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amp"o0 Con#"dere !na #!perf"%"e pana >ert"%a= %!?o e:tremo #!per"or%o"n%"de %on a #!perf"%"e "$re de )9!"do0 La pre#"8n >ar"ar& de#de%ero en M=
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#$%&' (cm)
*$+, (cm)
-$+,. (cm)
D$/+ (cm)
0esultados
1vento "asa2gr3 42mm3 455e6p (m) 4557eo (m) 18
+ %' .+ ,%,&& ,+9% &;
% ', ; ,+&9/ ,+&/ +++
. /' '/ ,+&'. ,+&; ,/
; +,, / ,+/&& ,+&,; ,&&
' +%' /. ,+&&. ,+/& ';.+
+', &+ ,+&.' ,+/ '%&
/ +/' &9 ,+//. ,+/
& %,, 9 ,+/;% ,+/+ ,'+9
9 %%' +,% ,+/. ,+9 %/%
+, %', ++, ,+' ,+ ,.
++ %/' ++ ,++ ,+'% ,';+% .,, +%% ,+'& ,+; +,&'
+. .%' +%9 ,+.; ,+%9 ,,,.
+; .', +. ,++; ,++9 ,,,.
+' ./' +;% ,++; ,++ ,,,%
+ ;,, +;& ,++' ,+'&& +/
+/ ;%' +'; ,++' ,+ ,9%
+& ;', +% ,+'&/ ,+';' %;
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Parcialmente Sumergido
h´ ´ exp= 2mL
ρbh
2 h´´teo=a+d−h
3
1vento +
h´ ´ exp= 2(0.25)(0.285)
(1000)(0.071)(0.031)2 =0.2088[m ]
h´´teo=0.1+0.103−0.031
3=0.1926[m]
1vento %
h´ ´ exp= 2(0.05)(0.285)
(1000)(0.071)(0.046)2 =0.1897 [m ]
h´´teo=0.1+0.103−0.046
3=0.1876[m ]
1vento .
h´ ´ exp= 2(0.075)(0.285)
(1000)(0.071)(0.057)2=0.1853[m ]
h´´teo=0.1+0.103−0.075
3=0.184[m ]
1vento ;
h´ ´ exp= 2(0.1)(0.285)
(1000)(0.071)(0.067)2 =0.1788[m ]
h´´teo=0.1+0.103−0.067
3=0.1806[m ]
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1vento '
h´ ´ exp= 2(0.125)(0.285)
(1000)(0.071)(0.073)2 =0.1883[m ]
h´´teo=0.1+0.103−0.073
3=0.1786[m ]
1vento
h´ ´ exp= 2(0.150)(0.285)
(1000)(0.071)(0.081)2 =0.1835[m]
h´´teo=0.1+0.103−0.081
3 =0.176[m]
7otalmente sumergido
h´ ´ exp= mL
ρb d (h−d
2)❑
h´´teo=a+d−h+h´ h´ =(h−
d
2)2
+ d
2
12
h−d
2
1vento +,
h´ ´ exp= (0.250)(0.285)
(1000)(0.071)(0.103)(0.11−(0.103)
2)❑=0.1665[m ]
h´ =(0.11−
0.103
2)2
+0.103
2
12
0.11−0.103
2
=0.0736[m]
h´´teo=0.1+0.103−0.11+0.0736=0.1666 [m]
1vento ++
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h´ ´ exp= (0.275)(0.285)
(1000)(0.071)(0.103)(0.116−(0.103)
2)❑=0.1661[m ]
h´ =(0.116−0.1032 )
2
+ 0.1032
12
0.116−0.103
2
=0.0782[m]
h´´teo=0.1+0.103−0.116+0.0782=0.1652[m ]
1vento +%
h´ ´ exp= (0.300)(0.285)
(1000)(0.071)(0.103)(0.122− (0.103)2
)❑=0.1658[m]
h´ =(0.122−
0.103
2)2
+0.103
2
12
0.122−0.103
2
=0.083[m]
h´´teo=0.1+0.103−0.122+0.083=0.164[m ]
1vento +.
h´ ´ exp= (0.325)(0.285)
(1000)(0.071)(0.103)(0.129−(0.103)
2)❑ =0.1634[m ]
h´ =(0.129−
0.103
2)2
+0.103
2
12
0.129−0.103
2
=0.0889[m]
h´´teo=0.1+0.103−0.129+0.0889=0.1629 [m]
1vento +;
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h´ ´ exp= (0.350)(0.285)
(1000)(0.071)(0.103)(0.136−(0.103)
2)❑ =0.1614 [m ]
h´ =(0.136−0.1032 )
2
+ 0.1032
12
0.136−0.103
2
=0.0949[m]
h´´teo=0.1+0.103−0.136+0.0949=0.1619 [m]
1vento +'
h´ ´ exp= (0.375)(0.285)
(1000)(0.071)(0.103)(0.142− (0.103)2
)❑=0.1614[m]
45$,+,(m)
h´´teo=0 .1+0 .103−0 .142+0.10=0 .161[m ]
1vento +
h´ ´ exp= (0.400)(0.285)
(1000)(0.071)(0.103)(0.148−(0.103)2
)❑ =0.1615[m]
4
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1vento +&
h´ ´ exp= (0.450)(0.285)
(1000)(0.071)(0.103)(0.162− (0.103)
2)❑=0.1587[m ]
4
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C!ando a #!perf"%"e e#t& totamente #!mer("da e %entro de pre#"one# %am$"apa!at"namente0
La# %on%!#"one# anter"ore# me pare%en totamente demo#trat">a# a a rea%"8n9!e a pre#"8n a!menta #e(n a prof!nd"dad en !n f!"do= en e#te %a#o a(!a0 A#)
%omo a#e(!ro
