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Chapter 2 : Pressure Measurement 47

Intensity of pressure due to plunger,
F400
p = = 5.66 × 105 N/m2
7.068 u 10 – 4
a
Since the intensity of pressure will be equally transmitted (due to Pascal’s law), therefore the
intensity of pressure at the ram is also
= p = 5.66 × 105 N/m2
Weight W W
But intensity of pressure at the ram = = N/m 2
Area of ram A 0.0314
W
 ? = 5.66 × 105 or W = 0.0314 × 5.66 × 105 N = 17.77 × 103 N or 17.77 kN (Ans.)
0.0314
Example 2.4. For the hydraulic jack shown in Fig. 2.5 find the load lifted by the large piston
when a force of 400 N is applied on the small piston. Assume the specific weight of the liquid in the
ja is 9810 N/m3.
jack
Solution. Diam
So Diameter
met
e er
e ooff smal
small
alll pi
al ppiston,
ston
st on, d = 30 mm = 0.03 m
F = 400
400 N
40

Small
S
Smmall piston
10
100
1 00 mm
dia.
di
ia.
a

30 mm
mm
ddia.
dia
diia.
a. W Large piston
300 mm
m
L L

Liquid
L iqu
quid
id

Fig.. 2
2.5
.5
S 2 S
Area
Ar ea ooff sm
smal
small
all pi
al ppiston,
pist
ist
ston
ton
on,, a =
d .032 = 7.
× 0.0
00.03
0. 77.068
.06
.06 0–4
0688 × 10–4 2
m
4 
Diameter of the large piston, D = 100 mm = 0.1 m
S 2 S
Area of large piston, A = D = × 0.12 = 7.854 × 10–3 m2
 
Force on small piston, F = 400 N
Load lifted, W:
F 400
Pressure intensity on small piston, p = = 5.66 × 105 N/m2
a 7.068 u 10 – 4
Pressure intensity at section LL,
F
pLL = + Pressure intensity due to height of 300 mm of liquid
a
F 300
= + wh = 5.66 × 105 + 9810 ×
a 1000
= 5.66 × 105 + 2943 = 5.689 × 105 N/m2

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48 Fluid Mechanics
Pressure intensity transmitted to the large piston = 5.689 × 105 N/m2
Force on the large piston = Pressure intensity × area of large piston
= 5.689 × 105 × 7. 854 × 10–3 = 4468 N
Hence, load lifted by the large piston = 4468 N (Ans.)

2.4. ABSOLUTE AND GAUGE PRESSURES

Atmospheric pressure:
The atmospheric air exerts a normal pressure upon all surfaces with which it is in contact,
and it is known as atmospheric pressure. The atmospheric pressure is also known as ‘Barometric
pressure’.
The atmospheric pressure at sea level (above absolute zero) is called ‘Standard atmospheric
pressure’.
Note. The local atmospheric pressure may be a little lower than these values if the place under question is
higher than se
sea llevel,
vell, aand
eve nd higher values if the place is lower than sea level, due to the corresponding
decrease
de or increase
incr
crea column
e se of the colu mn ooff air standing, respectively.
lumn
Gauge
Gaug
Ga uge
ge ppres
pr
pressure:
ressu
surere::
Itt is the pressure,
pressu measured m with the help off pr ppressure
essure measuring
measuri ring instrument,
instrum
ument, in whichwhicichh the
thhe
atmospheric ic pressure
pres
pr e iss taken
t ken as datum. The atmospheric
ta atmosphher e ic pressure
preessure on tthe
h scale
he le is mark
markedked aass zero.
zerro.
ze
Gauges record ord
rd ppressure
ressure ab above or below the local al atm
atmospheric
mospheric pre pressure,
essure,e, since tthey
heyy me
he meas
measure
a ure
the difference in pressure ssure of of the
t e liquid
th liqu
qu to which they aare re cconnected
onnectcted
ed and tthat
h t off ssurrounding
ha urrounding air
ur r. If
air.
the pressure of the liquid is below bellow
b o thethe local
loc atmosph
atm heric press
atmospheric ssure, the
pressure, en thee ga
then gauge iiss ddesignated
esig
es i nated as
‘vacuum gauge’ and the recorded ed
d va
valueue ind
indicates
ica es tthe
he am mounnt by which tthe
amount he ppressure
ressure of the liqui
re uidd is
liquid
belo
be low
beloww local
loc
ocal
a atmosphericc ppressure,
ressure, i.e.
re i.e
. . negative
negati
ega
g ti
ttive
vee ppressure.
ressure.
ress
(Va
Vacucuum
(Vacuum um pressure is defined as the thhe pressure
essure below
b low the atmospheric
be os pressure).
Absolute
Abso
Ab olutee pressure:
pressssure:
Itt is nnecessary
eceess
ssarry to eestablish
stab
ablish
sh an ab
absolute
bsollute ppressure
ress
ssuree sca
sc
scale
aale
llee wh
whic
which
ich
ch iss indep
independent
ependent nt ooff th
the ch
cchanges
annge
g s in
a mo
atmosspheeric pr
atmospheric ressuure
pressure. re.. A pr
pressu
sure ooff ab
pressure bso
solutee zzero
absolute ero cann ex xist onl
exist nly iin
only n commppl
pleete vac
complete cuum
vacuum. m.
A
An
Anyy pressure
p esssu
pr surere measured
mea asure red
d above
abbovee the
th
he absolute
abbsolu ute
t zero
zeero off pressure
pre
ressur ure is
is termed
terme
meed as
as ann ‘‘absolute
ab
abs
bsolulu
ute pressure’.
preess
s urre’.
A schematic
schema
sc matic ddiagram
iag
agrram sh howwin
showingng th he gaug
the uge ppressure,
gauge ressuree, va acuum ppressure
vacuum ressuure and
annd the
the abbsso
solute
olute pre
absolute ess
s uree
pressure
is given in Fig. 22.6. .6.
6

Positive gauge Atmospheric

pressure pressure
Pressure

Negative gauge
pressure or vacuum

Absolute
pressure

Zero absolute pressure

Fig. 2.6. Relationship between pressures.

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Chapter 2 : Pressure Measurement 49

Mathematically:
1. Absolute pressure = Atmospheric pressure + gauge pressure
i.e., pabs = patm + pgauge
2. Vacuum pressure = Atmospheric pressure – absolute pressure
Units for pressure:
The fundamental S.I. unit of pressure is newton per square metre (N/m2). This is also known as
Pascal.
Low pressures are often expressed in terms of mm of water or mm of mercury. This is an
abbreviated way of saying that the pressure is such that will support a liquid column of stated height.
Note. When the local atmospheric pressure is not given in a problem, it is taken as 100 kN/m2 or 10 m of
water for simplicity of calculations.
Standard atmospheric pressure has the following equivalent values:
101.3 kN/m2 or 101.3 kPa; 10.3 m of water; 760 mm of mercury; 1013 mb (millibar)  1 bar
10 kPa = 105 N/m
100 N/m m2.
Example 2.5. Gi
Exam Given
G ive
ven ththat
that:
att:
BBarometer
arometer reading ding = 740
rreadi 7440 mm m of
of mercury;
m rc
me rcur
ury;
y;
SSpecific
Sppeccif i gravity ooff merc
ific
fic mercury
m erccur 13.6;
uryy = 13.6 6; Intensity of ppressure 40 kkPa.
ressure = 40 P .
Pa
Express
Expr
Ex prres
ess the intensity
th intens
int ty off pressure
sit
ity presssure in S.I.I units,
S.I. unitss, bothh gaugee and
and absolute.
abbssoolu
l te.
Solution. In IIntensity
nten
tens
tensit
nsit
ity
y of
of ppre
pressure,
rreessu
s ure, p = 40 kPa
ss kPa
Gauge pressure:
(i)
i p = 40 40 kPakPa
Pa = 4040 kN
kkN/m /m2 = 00.4
4 × 105 N/m
N/mm2 = 0.4 bar
bar ((Ans.)
ba Ans.)
bar = 11005 N
(1 bar N/m m2)
/m
p 0.4 u 10 5
(ii)
i h= = 4.077 m off water
wat
ater
ter
er (Ans.)
er (An
Ans.)
A s..)
w .81 u 100 3
99.81
p 0.4
0 .4 10 5
4 u 10
(iii)
(iii
i) h = 3
= 0.299
0.2299 m of
of mercury
merc
rcur
c ry (A
(Ans.)
Ans
ns.)
.))
w 8 u 100 u 13.6
9.81
9..81 13.6
6
ª Wher
Where,
re,, w spe specific
peci
pe ccifi
ifficc w
ifi weight;
eiigghhtt;; º
« 3 »
F water : w = 99.81
« For 811 kN
kkN/
N/m
kN/m / »
« 3»
¬ For
Fo r merc
me
mercury
rcur
rc ury
ur y : w 9.81
9.
9 . 811 u 13.6
13
3.6
3 .66 kN
kN/
kN/m
k N/
N /m
/ m ¼
Absolute
Abso
Ab te pressure:
sollute pre
resssurre:
Barometer reading (atmospheric pressure)
= 740 mm of mercury = 740 × 13.6 mm of water
740 u 13.6
= = 10.6 m of water
1000
Absolute pressure (pabs.) = Atmospheric pressure (patm.) + gauge pressure (pgauge).
 ? pabs = 10.06 + 4.077 = 14.137 m of water (Ans.)
= 14.137 × ( 9.81 × 103 ) = 1.38 × 105 N/m2 (Ans.) (p = wh)
= 1.38 bar (Ans.) (1 bar = 105 N/m2)
14.137
= = 1.039 m of mercury. (Ans.)
13.6
Example 2.6. Calculate the pressure at a point 5 m below the free water surface in a liquid that
has a variable density given by relation:

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50 Fluid Mechanics

U = (350 + Ay) kg/m3

where, A = 8 kg/m4 and y is the distance in metres measured from the free surface.
Solution. As per hydrostatic equation
dp = U.g.dy= g (350 + Ay)dy
Integrating both sides, we get:
5 5

³ dp = ³
0
g (350  Ay) dy ³
g (350  8 y) dy
0

2 5
y
p = g 350 y  8 u
2 0

52
= 9.81 (350 × 5 + 8 × ) = 18148 N/m2  18.15 kN/m2
2
(
(Ans.)
)
Example
Examp 2.7. Onn the the suction
suc
ucti
cti
tio
on side
on sid
i e of a pump
pum
u p a gauge shows a negative pressure of 0.35 bar.
E
Exp ess this pressure
pr
press
Express pres in terms
in ter
errms
ermms of:of:
((i)
i) InIntensity
Inte
nte
tensityy off ppressure,
reessur
ress
re ssur
ure,
ur e, kkPa,
Pa,
2
(ii)
i N/ N/m
N m absolute,
absolut
so tee,,
(iii)
i Metr
Metres
res ooff wa
w
wat
water
tteerr ga
gaug
gauge,
aug
ugge,
e
(iv) Met Metres
tres
es ooff oil (spe
(specific
ecific
cifi
ciific
fcg gr
gravity
ravi
avit
a ity
ty 00.82)) absolute,
a e and nd
d
((v)
v) Cent
Centimetres
C ent
ntiim
imet
etre
tre
ress off mmercury
ercu
er cury
uryy ggauge,
auge
a ge,
e,
Take
T kee atmospheric
ake atm
tmos
ossph
p er
pher ric pressure
eric pre
ress uree ass 76
ssur
ur 76 cm m ofof Hgg and
and relative
re densityy off mercuryy as 13.6.
Solution.
Solu
So luti
lu tio
tion. Given: Reading of the vacuum m gauge
ga = 0.35 barr
(i)
i Intensity
Intenssitityy of pressure,
presssur ure,, kkPa:
Paa:
Gauge
G
Gaug
Gaaugu e reading
r ad
re ad g = 0.35
ading 0 355 bar
0. barr = 00.35
.35
5 × 110 05 N/m
N/m2
= 0.
00.35 05 P
355 × 110 Paa = 35
3 kPa kPa
Pa (Ans
(Ans.)
n .)
2
(ii
ii)
i) N/m absolut
(ii) absolute:ute:
Atmospheric
Atmomoosp
sphe
heri
he ric pres
ri pressure,
ssu
suree, patm. = 76 ccm m of H Hg g
76
= (13.6 × 98 9810)
81010)) × 101396 N/m m2
100
Absolute
A
Ab bso
solu
solu
lutetee pressure
pre
ress
sssur
u e = At A
Atmospheric
tmosp
mospheerriic pressure
mo prreesssu
pres sure
re – Vacuum
re V cu
Vacuumumm pressure
pres
essu
esssuurree
pabs.
b = p atm
t – p vac.
= 101396 – 35000 = 66396 N/m2 absolute (Ans.)
(iii) Metres of water gauge:
p = Ugh = wh
p 0.35 u 10 5
  ? hwater (gauge) = 3.567m (gauge) (Ans.)
w 9810
(iv) Metres of oil (sp. gr. = 0.82) absolute:
66396
hoil (absolute) = = 8.254 m of water (absolute) (Ans.)
0.82 u 9810
(v) Centimetres of mercury gauge:
0.35 u 10 5
hmercury(gauge) = = 0.2623 m of mercury
13.6 u 9810
= 26.236 cm of mercury (Ans.)

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