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 B, and the bottom with a more dense fluid A, having a density of

• Liquid A is immiscible with B. • To derive the relationship between

pa and pb ………….

gRpgRp AbBa

• We know, p2 = p3

gRgZpgRZp ABbBa )(

gRpp BAba )(

• 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:

p = (m - )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:

= 28.8 x 101325/(8314 x 293) = 1.2 kg/m3

p = (m - )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 proportional to the velocity gradient (velocity/distance).

• Introduce the proportionality constant “viscosity”… we get “Newton’s law of viscosity”

• A fluid obeys this law is Newtonian fluid....(i.e. constant viscosity)• -----otherwise Non-Newtonian fluid

dy

du

dy

du

• 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….

• It can be seen that all the units cancel out; i.e. Re is dimensionless.

Dv

N Re

mskg

mkgsmmN

/

)/)(/)(( 3

Re

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 1Reynolds 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 =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) (/4)D2 m2}

• =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) / {(/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