Resolução do Capítulo 08 - Franco Brunetti

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Capítulo 8 
NOÇÕES DE INSTRUMENTAÇÃO PARA A MEDIDA DAS 
PROPRIEDADES DOS FLUIDOS 
 
 
Neste capítulo estabelecem-se apenas princípios para a medida de propriedades dos fluidos e dos 
escoamentos, baseados em seus conceitos, não sendo abordados instrumentos sofisticados de 
última geração. O objetivo é melhorar a compreensão das definições e das equações apresentadas 
nos capítulos anteriores. 
 
Exercício 8.1 
 
823,0
10
8235
m
kg840
8,9
8235
g
m
N8235
10170
4,1
V
EVE
N4,16,46EGGE
GEG
4OH
fl
fl
3
fl
fl
36flfl
ap
ap
2
r
==γ
γ=γ
==γ=ρ
=
×
==γ→γ=
=−=→−=
=−
− 
 
Exercício 8.2 
 
( )
( )
mm13347360hmm347m347,0h10181,8h1096,1
10
15,0
h1096,1G
m10181,8
6
025,0
6
D
V
h1096,1h
4
105,0h
4
D
V
EG
emersub
6
sub
5
4
sub
5
36
33
e
esf
sub
5
sub
22
sub
2
c
subcil
=−=⇒==⇒×+×=
+×γ=
×=×π=π=
×=×××π=π=
=
−−
−
−
−−
 
Exercício 8.3 
 
22
2
2
2
m
s.N1,13
101,025,00476,0
51005,02
s
m0476,0101,0
60
9Dnvcm05,0
2
101,10
vLD
M2
2
vLD
2
DDLv
2
DAM
=
×××π
×××=μ
=××π=π==−=ε
π
ε=μ
ε
πμ=πεμ=τ=
−
 
 
Exercício 8.4 
 
( ) ( )
( ) ( )
2
522
oe
2
oe
2
m
s.N93,2
05,018
600.81014,1105,0
v18
D
L18
tD
=×
−×××=μ
γ−γ=γ−γ=μ
− 
 
Exercício 8.5 
 
)2(9kk2025
45
k
45.k20,0
)1(7,43kk9025
95
k
95.k46,0
t
k
tk
21
2
1
21
2
1
2
1
=−→−=
=−→−=
−=ν
 
Fazendo-se (1) – (2) obtém-se: 00496,0k7,34k7000 11 =→= 
Da (1): 064,1k7,43k00496,09025 22 =→=−× 
 
t
04,1t00496,0 −=ν 
 
 
Exercício 8.6 
 
s
L2,23
s
m0232,0105061,86,09,0AvCCQ
kPa9,121029,1pm29,15
20
61,8p
s
m61,8
9,0
75,7
C
v
v
s
m75,7320vp
g2
v
g2
v
z
p
3
4
otcv
4
1
2
1
v
r
t
r
2
r
2
t
1
1
==××××==
−=×−=⇒−=−=γ
===
=×=⇒γ=
=+γ
−
 
Exercício 8.7 
 
g2
v
zp
HH
:eriorsupservatórioRe
2
1t
0
0
10
=+γ
= 
semt
St
s
cmem
2
=ν 
81,0
1,042,13
0853,04C
s
m42,13920gz2v
Dv
Q4
C
4
D
vCQ
Q
s
m0853,01422,06,0QCQ
s
m1422,0
4
09,036,22
4
D
vQ
s
m36,2215
10
101,020z
p
g2v
22D
22t
2
2o2t
2r
2D
2
2o
2t2D2r
2r
3
1t1D1r
322
1o
1t1t
4
6
0
0
1t
=×π×
×=
=×==
π=→
π=
==×==
=×π=π=
=⎟⎟⎠
⎞
⎜⎜⎝
⎛ +×=⎟⎟⎠
⎞⎜⎜⎝
⎛ +γ=
 
 
Exercício 8.8 
 
( )
( )
s
L40
s
m104AvQ
s
m438,320
pp
g2v
Pa000.30101062,0000.20p
p2,02,0p
8,3
p
g2
vp
3
2
1
10
1
44
1
2m1
0
2
11
=×==
=−×=⎟⎟⎠
⎞
⎜⎜⎝
⎛
γ−γ=
=−×+=
=×γ−×γ+
=γ=+γ
−
 
 
Exercício 8.9 
 
97,0
6,0
582,0
C
C
C;582,0
111,0
64,0
vA
QC
s
m1105,620gh2v
vACQ
s
m64,0
20
8,12
t
VQ
m8,128,044V
m8,0
000.10
000.81hhhAhA
VV
GE
c
D
v
to
D
t
toD
3
3
O2H
mad
subbasemadsubbaseO2H
cubomadsubO2H
====×==
=×==
=
===
=××=
=×=γ
γ=→γ=γ
γ=γ
=
 
Exercício 8.10 
 
( )
( ) Pa638201042,497,0
9375,010510p
9375,0
15
5,71
D
D
1
m1042,4
4
075,0
4
D
A
g2AC
D
D
1Q
p
pp
g2
D
D
1
AC
Q
232
234
44
1
2
23
22
2
2
2
2
2
D
4
1
22
21
4
1
2
2D
=
×××
×××=Δ
=⎟⎠
⎞⎜⎝
⎛−=⎟⎟⎠
⎞
⎜⎜⎝
⎛−
×=×π=π=
×
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛−γ
=Δ⇒⎟⎟⎠
⎞⎜⎜⎝
⎛
γ
−
⎟⎟⎠
⎞
⎜⎜⎝
⎛−
=
−
−
− 
 
 
Exercício 8.11 
 
66,0
96,0
634,0
C
C
C
634,0
0442,0
1028
Q
Q
C
s
m0442,0
4
075,010AvQ
96,0
10
6,9
v
v
C
s
m10520gh2v
s
m6,9
2,12
107,4
y2
gxv
v
D
c
3
t
r
D
32
ott
t
r
v
t
r
===
=×==
=×π×==
===
=×==
=××==
−
 
 
Exercício 8.12 
 
( )kconfirma106,5
10
15,074,3DvRe
s
m74,3
15,0
10664
D
Q4v
s
m106615,320
4
1,006,1Q
m15,31
10
106,1325,0
pp
p25,025,0p
06,1kseadota14.8figurada67,0
15
10
D
D
Com
pp
g2kAQ
5
6
11
1
2
3
2
1
1
3
3
2
4
4
21
2Hg1
1
2
21
2
×=×=ν=
=×π
××=π=
×=×××π×=
=⎟⎟⎠
⎞
⎜⎜⎝
⎛ −××=γ
−⇒=×γ−×γ+
=−==
⎟⎟⎠
⎞
⎜⎜⎝
⎛
γ
−=
−
−
−
 
 
 
Exercício 8.13 
 
s
m069,025,03084,1LH84,1Q
3
2
3
2
3 =××== 
 
Exercício 8.14 
 
467,0
06,0
028,0
Q
QC
s
m06,001,06AvQ
s
m68,120gh2v
s
m028,0
605
4,8
t
VQm4,81,222V
m1,2
210
1063
b
M3h
2
bhh
3
2hb
2
hh
3
2ApM
t
D
3
ottt
3
3
3
4
4
3
3
===
=×==⇒=×==
=×==⇒=××=
=×
××=γ=⇒
γ=×γ=×=
 
 
Exercício 8.15 
 
966,0
1210
10112
h
p
gh2
pg2
v
v
C
pg2v
g2
vp
gh2v
4
3
t
r
v
r
2
r
t
=
×
×=γ=
γ==
γ=→=γ
=
 
 
 
 
 
Exercício 8.16 
 
( )
2
2
8
v
v
d
d
4
d
v
4
d
v
s
m83202
gp2
vv
gp2
vv
g2
vp
g2
v
s
m232,320
pp
g2v
pp
g2
v
2
1
1
2
2
1
1
2
2
2
222
21
22
2
2
1
2
12
2
2
20
2
02
2
2
===⇒π=π
=×+=γ+=⇒γ+=⇒=γ+
=−×=⎟⎟⎠
⎞
⎜⎜⎝
⎛
γ
−=⇒γ=γ+
 
 
Exercício 8.17 
 
( )
s
m27,17,0
20
4
04,01
1058,120p
g2AC
Qg2v
g2
vp
g2ACQ
g2
vp
g2ACQ
2
42
2
23
2
o
2
D
2
1
2
112
o
2
D
2
2
1
oD
1
=
⎥⎥
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢⎢
⎢
⎣
⎡
−
××π×
π
×=⎟⎟⎠
⎞
⎜⎜⎝
⎛
γ−=
⎟⎟⎠
⎞
⎜⎜⎝
⎛
+γ=
⎟⎟
⎟
⎠
⎞
⎜⎜
⎜
⎝
⎛
+γ=
−
 
 
Exercício 8.18 
 
( ) ( )
4,0
8,075,7
2
D
D
75,7
2
4
D
75,7
4
D2
Q
QC
4
D
75,7
4
D
320
4
Dphg2Q
4
D2
4
DvQ
s
m22,020v2,0
g2
v)c
m5,32,15,02,05HHHH)b
kPa100Pa101051025p
kPa24Pa104,22,11022,1p)a
2
2
0
2
0
2
0t
D
2
0
2
0
2
0
0t
22
2
0,sp0s0,sp
44
s
44
0
=×=⎟⎟⎠
⎞
⎜⎜⎝
⎛=π×
π×
==
π×=π××=π×⎟⎟⎠
⎞
⎜⎜⎝
⎛
γ+=
π×=π=
=×=⇒=
=+−+=⇒−=
=×=××=×γ=
=×=××=×γ=