, Week 1 :S t a t i c fluids reading
I
phases states
-
of matter:
solias:definite shape, specific volume
liquids:definite volume, shape depends on container
gases:n e it h e r a definite shape for specific vo l u m e , molecules move to fill container
they are in
plasmas:neither definite shape nor wome
density:
per unit
v olume of substance
density object
is the mass a
or
defined as 1 *,
=
=density, m m a ss , V
=
volume,
=
units
vg/m3
=
Swater 19/cm3 =
100
=
kg/m3
of
density reveals phase st r u c t u re : density gases (liquids solids
+
+
pressure:
p E, F force applied to an A a re a thati s perpendicular to the force, units 1Pa =I N ma
=
pressure, area,
= -
P
=
=
WO mb (millibar) 1
=
x
10
4
Pa
fluid pressure has no direction:s c a l a r quantity
forces due to defined directions: perpendicular
pressure have well
always exerted to a surface
RO2
4 round
c1kg, legs dieg
=
m 4.8 c m
=
C) pressure each
leg exerts on carpet
41kg
=
4 =
1.2549/leg
F 10.25kg 9.8 MIS2 wo.4 5 N
=
mg
=
x
=
m2
M(2.4 100) 0.0018096
=
Acrae 12
=
=
=
p E
=
=
2
06 55510.8 pa
=
four
b) total pressure on the carpet
by all
legs
4 1 9.8 401.8N
=
F mg
=
C11kg
=
=
m =
A
=
Mr2 i(2.4 w0)2
=
=
0.0 0 1 8 0 9 6 m r
=
E 618096
=
p 5 5 5 0 9.5
=
=
pressure depth
+
p M
=
P pressure,
=
m mass,
=
g gravity,
=
A a re a
=
supporting
v vo l u m e
m eV, 2
density,
=
mass,
=
m
=
=
V A U,
=
v vo l u m e ,
=
c ross-sectional
A: area, U=depth
m e
=
An
p eA19
=
- > p =heg/pressure are to the weighto f any fluid)
atmospheric pressure:
1atmosphere (atm) Pa t m =
1.0 1
=
1 0 5 N/m2 l o1 k Pa
=
favg ig =
n g
=
pascal's principle:
force
pressure: per unita re a
Pascal's Principle:a change in pressure applied an
to enclosed said is transmitted undiminished to all portions of the fluid and to the walls of its container
hydraulic systems:p , P2, =
E =
work:f orce times distance moved
1100 49
⑤
R05 de 20d,
=
m 1160
=
bg d -zoc
i((2001) 2)2
9.8
Fz 1 1 6 0
=
=
Az
=
=
F211368 N
=
, *
z
-
168
F1 =
F1
=
2842
= =
100 28.4N
=
gauge pressure, absolute pressure, and pressure measurement
for it
garge pressure pressure
-
relative atmospheric
to
pressure positive for pressures above atmospheric pressure, and negative values below
absolute pressure:t h e of
atmospheric
sum
garge pressure and pressure
Pabs Pg =
+
Patm
in
* cases fluids
most absolute pressure of cannotb e negative
↳ smallesta bsolute pressure is 0, pulling --, pushing t
LD
Pg = -Patm (smahest)
manometer:U-shaped tube, both ends open to atmosphere (pressure pushes down on both sides
equally concerning out
it
Lo pressure
deeper on one side:
greater
↳is one side open
(Pg=egh)
to atm: ideal for measuring Page
1.0mmHg (millimetres mercury) 1 3 3 Pa
=
systolic pressure:maximum blood pressure
diastolic pressure:m i n i m u m blood pressure
example 11.7:
ff1 1.0g/mL
=
n?=
Paarge 18
mmHg 13342 239446
=
= =
Pg e 9 h
=
2394
100kg12 9.8 M/S2 h
=
=
=
h 0.244m
=
barometer:device thatm easures atmospheric pressure
Week 1 :Lec ture questions
L 1.1 :
Density
1) T
D
-
o
2) ice X
3)
*ice 11 mice m re a d =
> Me
a) rank
highestt o m a ss
lowest
M3
M1=
b) V 3 >Ve = V,
rank from highestt o
lowestvolume:
C) rank from highestt o
lowest
density:11782=13
11.2 water
density
One litre (= rocm = re m x wa m
100cm3)
=
of water has m=
lky
a) density of water in
ag,m3: 10o
bgim?
b) density of water in
glans:Iglan3
, 11.3 safe
Density a
of
Lorter = 1.2m
(thickness 4 a n
=
0.0 4
=
m
a) density of metal safe metal=7870 kg/m3
average
Linner -(0.04 x2) 1.1 2 m
=
1.2
=
Vinner (1.12)3
=
1.4 0 5 m3
=
Usage (1.2)
=
=
1.728m3
Parg mmair-Mrsafe-voir
=
Prietal (Usafe-Vair)
Parg
=
-
r safe
7870 yy1m3 =
(1728 1.405)
-
Parg
=
1.728
Parg 1471.1 Ug1M3
=
b) World the safe float?N o , itw i l l sinn because the
density is
greater than thato f water
11.4 Pressure
--
-
a)
I A
all
D
rank the pressures exerted
by the block:4a)Pb>4c
21.8 Uniform connected fluids
it---.* Im
P, Patm
=
Di P2
=
P, Patm =
+egh
P, W 1 0 0
=
1000
+
9.8
= 0.6 m
x
Pa 2 07.2
P, 207180 uPa
=
=
21.9 measuring gas pressure with a
mercury manometer
frg 13.6 g/m2
=
10
=
↳Iam
Pgas Patm
=
-fgU
Pgas 101300Pa-13.6 20" Rysm3 9.8m/S2 0.1 m
=
=
x =
Pgas=87972P0 8 8 uPa
=
the amounto f
we have to calculate pressure from the
the gas
acting
is on
mercury
Pgas Patm-Pe
=
11.10 non-uniform Gwids:
density of an unknown liquid in a manometer
eg egr 2V ?Vs
E->
->
= =
funknown=? mag
->
meg
:"Ira
=
=
ewater-woongim 1,(4) 1000
=
(5)
12 1250kg/m3
=
I
phases states
-
of matter:
solias:definite shape, specific volume
liquids:definite volume, shape depends on container
gases:n e it h e r a definite shape for specific vo l u m e , molecules move to fill container
they are in
plasmas:neither definite shape nor wome
density:
per unit
v olume of substance
density object
is the mass a
or
defined as 1 *,
=
=density, m m a ss , V
=
volume,
=
units
vg/m3
=
Swater 19/cm3 =
100
=
kg/m3
of
density reveals phase st r u c t u re : density gases (liquids solids
+
+
pressure:
p E, F force applied to an A a re a thati s perpendicular to the force, units 1Pa =I N ma
=
pressure, area,
= -
P
=
=
WO mb (millibar) 1
=
x
10
4
Pa
fluid pressure has no direction:s c a l a r quantity
forces due to defined directions: perpendicular
pressure have well
always exerted to a surface
RO2
4 round
c1kg, legs dieg
=
m 4.8 c m
=
C) pressure each
leg exerts on carpet
41kg
=
4 =
1.2549/leg
F 10.25kg 9.8 MIS2 wo.4 5 N
=
mg
=
x
=
m2
M(2.4 100) 0.0018096
=
Acrae 12
=
=
=
p E
=
=
2
06 55510.8 pa
=
four
b) total pressure on the carpet
by all
legs
4 1 9.8 401.8N
=
F mg
=
C11kg
=
=
m =
A
=
Mr2 i(2.4 w0)2
=
=
0.0 0 1 8 0 9 6 m r
=
E 618096
=
p 5 5 5 0 9.5
=
=
pressure depth
+
p M
=
P pressure,
=
m mass,
=
g gravity,
=
A a re a
=
supporting
v vo l u m e
m eV, 2
density,
=
mass,
=
m
=
=
V A U,
=
v vo l u m e ,
=
c ross-sectional
A: area, U=depth
m e
=
An
p eA19
=
- > p =heg/pressure are to the weighto f any fluid)
atmospheric pressure:
1atmosphere (atm) Pa t m =
1.0 1
=
1 0 5 N/m2 l o1 k Pa
=
favg ig =
n g
=
pascal's principle:
force
pressure: per unita re a
Pascal's Principle:a change in pressure applied an
to enclosed said is transmitted undiminished to all portions of the fluid and to the walls of its container
hydraulic systems:p , P2, =
E =
work:f orce times distance moved
1100 49
⑤
R05 de 20d,
=
m 1160
=
bg d -zoc
i((2001) 2)2
9.8
Fz 1 1 6 0
=
=
Az
=
=
F211368 N
=
, *
z
-
168
F1 =
F1
=
2842
= =
100 28.4N
=
gauge pressure, absolute pressure, and pressure measurement
for it
garge pressure pressure
-
relative atmospheric
to
pressure positive for pressures above atmospheric pressure, and negative values below
absolute pressure:t h e of
atmospheric
sum
garge pressure and pressure
Pabs Pg =
+
Patm
in
* cases fluids
most absolute pressure of cannotb e negative
↳ smallesta bsolute pressure is 0, pulling --, pushing t
LD
Pg = -Patm (smahest)
manometer:U-shaped tube, both ends open to atmosphere (pressure pushes down on both sides
equally concerning out
it
Lo pressure
deeper on one side:
greater
↳is one side open
(Pg=egh)
to atm: ideal for measuring Page
1.0mmHg (millimetres mercury) 1 3 3 Pa
=
systolic pressure:maximum blood pressure
diastolic pressure:m i n i m u m blood pressure
example 11.7:
ff1 1.0g/mL
=
n?=
Paarge 18
mmHg 13342 239446
=
= =
Pg e 9 h
=
2394
100kg12 9.8 M/S2 h
=
=
=
h 0.244m
=
barometer:device thatm easures atmospheric pressure
Week 1 :Lec ture questions
L 1.1 :
Density
1) T
D
-
o
2) ice X
3)
*ice 11 mice m re a d =
> Me
a) rank
highestt o m a ss
lowest
M3
M1=
b) V 3 >Ve = V,
rank from highestt o
lowestvolume:
C) rank from highestt o
lowest
density:11782=13
11.2 water
density
One litre (= rocm = re m x wa m
100cm3)
=
of water has m=
lky
a) density of water in
ag,m3: 10o
bgim?
b) density of water in
glans:Iglan3
, 11.3 safe
Density a
of
Lorter = 1.2m
(thickness 4 a n
=
0.0 4
=
m
a) density of metal safe metal=7870 kg/m3
average
Linner -(0.04 x2) 1.1 2 m
=
1.2
=
Vinner (1.12)3
=
1.4 0 5 m3
=
Usage (1.2)
=
=
1.728m3
Parg mmair-Mrsafe-voir
=
Prietal (Usafe-Vair)
Parg
=
-
r safe
7870 yy1m3 =
(1728 1.405)
-
Parg
=
1.728
Parg 1471.1 Ug1M3
=
b) World the safe float?N o , itw i l l sinn because the
density is
greater than thato f water
11.4 Pressure
--
-
a)
I A
all
D
rank the pressures exerted
by the block:4a)Pb>4c
21.8 Uniform connected fluids
it---.* Im
P, Patm
=
Di P2
=
P, Patm =
+egh
P, W 1 0 0
=
1000
+
9.8
= 0.6 m
x
Pa 2 07.2
P, 207180 uPa
=
=
21.9 measuring gas pressure with a
mercury manometer
frg 13.6 g/m2
=
10
=
↳Iam
Pgas Patm
=
-fgU
Pgas 101300Pa-13.6 20" Rysm3 9.8m/S2 0.1 m
=
=
x =
Pgas=87972P0 8 8 uPa
=
the amounto f
we have to calculate pressure from the
the gas
acting
is on
mercury
Pgas Patm-Pe
=
11.10 non-uniform Gwids:
density of an unknown liquid in a manometer
eg egr 2V ?Vs
E->
->
= =
funknown=? mag
->
meg
:"Ira
=
=
ewater-woongim 1,(4) 1000
=
(5)
12 1250kg/m3
=
