Devices to measure Pressure
• In chemical and other industrial processing
plants it is often important to measure and
control the pressure in a vessel or process
and/or the liquid level in a vessel.
• Also, since many fluids are flowing in a pipe or
conduit, it is necessary to measure the rate at
which the fluid is flowing.
• Many of these flow meters depend upon devices
to measure a pressure or pressure difference.
• MANOMETERS are mainly used
MANOMETERS
• Simple U tube manometer
• Pressure pa is exerted on one arm
of the U tube and pb on the other
arm.
• Both pressures pa and pb are
pressure taps from a fluid meter
• The top of the manometer is filled
with liquid B, having a density of
rB, and the bottom with a more
dense fluid A, having a density of
rA
• Liquid A is immiscible with B.
• To derive the relationship between
pa and pb ………….
• We know, p2 = p3
pa  (Z  R) r B g  pb  Zr B g  Rr A g
pa  Rr B g  pb  Rr A g
 pa  pb  R( r A  r B ) g
• Write an assignment on Different types of
Manometers
Prob 1
• A manometer, as shown in Fig. is being
used to measure the pressure drop across
a flow meter. The heavier fluid is mercury,
with a density of 13.6 g/cm3, and the top
fluid is water, The reading on the
manometer is 32.7 cm. Calculate the
pressure difference in N/m2
• Ans: 4.04 x 104 N/m2
Prob 2
• A simple U tube manometer is installed
across an orifice meter. The manometer is
filled with Hg (specific gravity 13.6) & the
liquid above the Hg is CCl4 (s.g 1.6). The
manometer reads 200mm. What is the
pressure difference over the manometer.
• Ans: 23,544 N/m2
Prob 3
• A U-tube manometer filled with mercury is
connected between two points in a pipeline.
If the manometer reading is 26 mm of Hg,
calculate the pressure difference between
the points when (a) water is flowing through
the pipe (b) air at atmospheric pressure and
20ºC is flowing in the pipe.
Density of mercury = 13.6 gm/cc Density of
water = 1 gm/cc Molecular weight of air =
28.8
• (a) Water is flowing through the pipe:
Dp = (rm - r)gh = (13600 - 1000) x 9.812 x 0.026
= 3214.4 N/m2
• (b) Air at atmospheric pressure and 20ºC is
flowing in the pipe:
r = 28.8 x 101325/(8314 x 293) = 1.2 kg/m3
Dp = (rm - r)gh = (13600 - 1.2) x 9.812 x 0.026 =
3469.2 N/m2
From doran……..
Newtonian & Non-Newtonian fluids
• It has been found that the Shear stress for flow of fluid is directly
du
proportional to the velocity gradient (velocity/distance). 

• Introduce the proportionality constant “viscosity”…
“Newton’s law of viscosity”
dy
we get
du
 
dy
• A fluid obeys this law is Newtonian fluid....(i.e. constant viscosity)
• -----otherwise Non-Newtonian fluid
• All gases and most liquids which have simpler
molecular formula and low molecular weight
such as water, benzene, ethyl alcohol, CCl4,
hexane and most solutions of simple molecules
are Newtonian fluids.
• Generally non-Newtonian fluids are complex
mixtures: slurries, pastes, gels, polymer
solutions etc.,
Various non-Newtonian Behaviors
• Bingham-plastic: Resist a small shear stress but flow
easily under larger shear stresses. e.g. tooth-paste,
jellies, and some slurries.
• Pseudo-plastic: Most non-Newtonian fluids fall into this
group. Viscosity decreases with increasing velocity
gradient. e.g. polymer solutions, blood.
– Pseudo-plastic fluids are also called as Shear
thinning fluids. At low shear rates(du/dy) the shear
thinning fluid is more viscous than the Newtonian
fluid, and at high shear rates it is less viscous.
• Dilatant fluids: Viscosity increases with increasing
velocity gradient. They are uncommon, but
suspensions of starch and sand behave in this way.
– Dilatant fluids are also called as shear thickening
fluids.
Rheology of fermentation broth
• The fungus Aureobasidium pullulans is
used to produce an extra cellular
polysaccharide by fermentation of
sucrose. After 120 h fermentation, the
following measurements of shear stress
and shear rate were made with a rotating
cylinder viscometer. Plot the rheogram for
this fluid and name the fluid type.
Shear stress
(dyn/cm2)
Shear rate (s-1)
44.1
10.2
235.3
170
357.1
340
457.1
510
636.8
1020
Mechanism of Fluid Flow
• When a fluid flows through a pipe or channel, the character of the
flow can vary according to the conditions.
• The forms of flow can best be visualized by reference to a classical
experiment on the flow of water through a circular tube, first carried
out by Osborne Reynolds in 1883.
• Reynolds studied the effect of varying the conditions on the
character of flow and on the appearance of the thread of colored
liquid. This can be illustrated, for example, by varying the velocity of
the water through the tube.
• When the velocity is low, the thread of colored liquid remains
undisturbed in the centre of the water stream and moves steadily
along the tube, without mixing, this condition is known as s viscous,
or laminar flow.(Streamline flow)
Reynolds’ experiment
• At moderate velocities, a point is reached (the critical velocity) at
where the thread begins to waver, although no mixing occurs. This is
the phase of transitional flow.
• As the velocity is increased to high values eddies begin to occur in
the flow, so that the colored liquid mixes with the bulk of the water
immediately after leaving the jet. Since this is a state of complete
turbulence the condition is known as turbulent flow.
• As a result of his experiments Reynold found that flow conditions
were affected by four factors:
–
–
–
–
Diameter of pipe
Velocity of fluid
Density of fluid
Viscosity of fluid
These were connected together in a particular way and could be
grouped into a particular expression known now as Reynolds
Number: NRe or Re
NRe
• Reynolds number is given by….
N Re 
Dvr

• It can be seen that all the units cancel out; i.e.
Re is dimensionless.
N Re
(m)(m / s)(kg / m 3 )

kg / m s
Significance of Re
• For a straight circular pipe
Type of flow
Reynolds no.
Velocity
Laminar
<2100
Low
Turbulent
>4000
High
Transition
2100<Re<4000 Moderate
Prob 1
Reynolds Number in a Pipe
• Water at 303 K is flowing at the rate of 10 gal/min in a pipe
having an inside diameter (ID) of 2.067 in. Calculate the
Reynolds number. Given r=0.996 g/cc & =0.8007 Cp
• 1 gal = 3.785L
• Change all the units to SI
• D = (2.067)0.0254 m
• Since velocity = flowrate / CSA
v = {10(3.785x10-3) m3} / {(60 sec) (p/4)D2 m2}
• r=0.996 x103 kg/m3
• =0.8007 10-3 kg/m.s
• NRe = 19030 < 2100 (LAMINAR)
Prob 2
• NRe for milk flow:
• Whole milk @ 293K having a density of
1030 kg/m3 and viscosity 2.12 cP is
flowing at a rate of 0.605 kg/s in a glass
pipe having a dia of 63.5mm.
– Cal NRe
– Cal the flow rate needed in m3/s for Re=2100
and velocity in m/s
•
•
•
•
•
•
•
Vol.flow rate = Mass flow rate / density
Velocity = Vol.flow rate / CSA
v = (0.605 / 1030) / {(p/4) (63.5x10-3)2 }
NRe = 5722 Turbulent
If NRe =2100
v = 0.0608 m/s
Q = 2.155x10-4m3/s
Prob 3
• Pipe dia & Re:
An oil is being pumped inside a 10mm dia pipe @
Re of 2100. The oil density is 855 kg/m3 and
viscosity is 2.1x10-2 Pa-s.
a) What is the velocity in the pipe?
b) It is desired to maintain the same Re 2100 and
the same velocity as in part (a) using a second
fluid with a density of 925 kg/m3 and viscosity
1.5x10-2 Pa-s. What pipe dia should be used?
• a). Pa-s = kg/m-s
• v = 5.158 m/s
• b). D = 6.6 mm