*Flow MeasurementsIn process industries, Flowing systems require energy, typically provided by -pumps and compressors, The two produce a pressure difference as the driving forceWide variety of methods and instruments for measuring flowDevices based on Bernoulli principle; orifice plate, rotameter, Pitot-static tube, venturi meterWeir and NotchMeters: Turbine meter, Target meter, Thermal flow meter

*Turbine flow meterTurbine flow meter consists of a small in-line turbine placed inside a section of pipe, the rotation speed, which can be transmitted electrically to a recorder, depends on the flow rate.

*Target flow meterTypically consisting of a disk mounted on flexible arm and placed normal to the flow in a pipe. The displacement of the disk, and hence the flow rate, is determined from the output of a strain gauge attached to the arm

*Devices based on Bernoulli PrincipleOrifice plate meterAn orifice plate is a very simple device installed in a straight run of pipe. The orifice plate contains a hole smaller than the pipe diameter. The flow constricts, experiences a pressure drop, and then the differential pressure can be related to a flow.

*It is also important to note that relating differential pressure to flow across an orifice depends on the location of the pressure taps in relation to the orifice. In the Figure below, the pressure taps are designated as P1 and P2. "D" is the diameter of the pipe and "d" is the diameter of the orifice.

*Let apply Bernoullis equation at points 1 and 2 (see orifice plate arrangement above) which have the same elevation (z1=z2)

5.1

Conservation of mass between point 1 and 2 gives continuity equation

u1A1= u2A2 5.2

Elimination of u2 gives in Eq. 5.1 gives

5.3

_1251457901.unknown

*

Solution for u1 yields

5.4

So that the volumetric flow rate Q is :

5.5

Where the coefficient of contraction Cc is approximately 0.63. In most cases you will find the following equation that uses a dimensionless discharge coefficient CD to be most frequently used.

5.6

The values of CD depends on the NRe and are usually presented graphically.

_1251458376.unknown

_1251459086.unknown

_1251458139.unknown

*Pitot tubeThe Pitot tube (named after Henri Pitot in 1732) measures a fluid velocity by converting the kinetic energy of the flow into potential energy. The conversion takes place at the stagnation point, located at the Pitot tube entrance (see the schematic below). The device is based on Bernoulli principle and is used in finding the velocity of a moving craft such as a boat and airplane.

*

Let u1 be the upstream velocity of approaching water. Opposite the Pitot tube, the oncoming water decelerates and in fact comes to rest at the stagnation point at the tip of the tube.

5.7

The first zero on the LHS is taken for z1=0 (datum) and at stagnation point u2=0.

From hydrostatics, the pressure at point 1 is given by

P1=P2 + (gd 5.8

Combining Eqs. 5.7 and 5.8 we get

5.9

_1251461119.unknown

_1251461348.unknown

*Pitot-static tubeFor pipe flow, we make use of pitot-static tube as shown below for measuring velocity at

different radial locations in a pipe. Two tubes are used. The left hand tube simply measures the pressure and the movable right-hand one is essentially a Pitot tube as before. As was before the velocity u1 is given by

*A venturi meter is a tube with a constricted throat that increases velocity and decreases pressure. They are used for measuring the flowrate of compressible and incompressible fluids in pipeline 3. Venturi meter

*

Apply Bernoullis equation and law of mass conservation between between stations 1 and 2:

(5.10)

(5.11)

where

p = average pressure

V = average velocity

A = area of crossection

( = kinetic energy correction factor (( 1)

( = liquid density

gc = Newtons law conversion factor (equals 1 in Metric system and

32.17 ft-lbm/lbf/s2 in British system)

Solve for V2 from eqs. (5.10) and (5.11) to get

_1071570961.unknown

_1071571233.unknown

*

(5.12)

To account for friction between stations 1 and 2, eq. (5.12) is modified to read

(5.13)

where

CV = Venturi coefficient (( 0.98 for 2-8 diameter pipes and 0.99 for larger sizes)

_1071571776.unknown

_1071571884.unknown

_1071571486.unknown

*

Note: For circular pipes, (A2/A1)2 = (D2/D1)4 = (4 where ( = D2/D1. Also, CC = 1.0062 to 1.0328 for D2 = (1/3 to ) D1.

Therefore,

q = volumetric flowrate =

(5.14)

m = mass flowrate = (q

(4.15)

with

C = CV CC

(5.16)

Typically, in a well-designed venturi meter, about 90% of the pressure drop (p1 p2) is recovered.

_1071573593.unknown

*Problem: Water is flowing in a schedule 40 pipe having nominal D1 = 4 inch with a flowrate of 325 gal/min at 60F. The pressure differential in the manometer connected to the two taps of the venturi meter is 50 inch Hg. Calculate (a) throat diameter D2 of the venturi meter to the nearest 1/8 inch and (b) power consumed by the venturi meter (assume 10% pressure loss).

Solution:

q = (325 gal/min) / (60 s/min) / (7.48 gal/ft3) = 0.725 ft3/s

( = 62.37 lbm/ft3 at 60(F

*

= (50 inch) / (12 inch/ft) (32.17 ft/s2)/(32.17 ft-lbm/lbf/s2) (13.6 1) (62.37 lbm/ft3)

= 3275 lbf/ft2

Substitute in eq. (5.14) to get

0.725 =

which on using

yields

_1071575393.unknown

_1071575464.unknown

_1071576547.unknown

_1071575251.unknown

*

which has to be solved (iteratively) for D2. For a first approximation , assume

. Then the above equation gives D2 = 0.127 ft = 1.53 inch and we get ( = 1.53 inch / 4.026 inch = 0.38, and thus

which is very close to 1.

Hence, to the nearest 1/8 inch, D2 = 1.5 inch.

permanent pressure loss ( 10% of (p1 p2) = 327.5 lbf/ft2

Volumetric flowrate q = (325 gal/min) / (7.48 ft3/gal) = 43.4 ft3/min

Hence, power consumed by meter = (43.4 ft3/min)(327.5 lbf/ft2)/(33,000 lbf-ft/min/hp)

= 0.43 hp (= 0.32 kW)

_1071575614.unknown

_1071575790.unknown

*4. Rota MeterA rotameter consists of a tapered tube, typically made of glass, with a float inside that is pushed up by flow and pulled down by gravity. At a higher flow rate more area (between the float and the tube) is needed to accommodate the flow, so the float rises.

*

Mass, energy and momentum balances in the upward direction yield;

Continuity: m =

Bernoulli:

Momentum:

(i ) due to pressure

(ii) convection

(iii) Gravity : (a)mass of fluid x g

Volume of fluid x density x g

(b) mass of float x g

Mg

_1252661057.unknown

_1252661407.unknown

_1252661717.unknown

_1252661271.unknown

_1252648980.unknown

*

Combination of (i) through (iii), we get

Pressure convection gravity

In which ( and (f are the densities of the fluid and the float, respectively. Starting with elimination of terms involving pressure and elevation in the momentum balance using energy balance equation

_1252662319.unknown

_1252662353.unknown

_1252649308.unknown

*

Eliminating terms involving m and u2 using mass balance equation

Simplifying further we get the following equation:

EMBED Equation.3

_1252650267.unknown

_1252662643.unknown

_1252663323.unknown

_1252663392.unknown

_1252662739.unknown

_1252662463.unknown

_1252650166.unknown

*The flow rate is then

*Buoyancy Example Problem # 1Displaced Volume of Water:Vdisp-W = 4/3 x p x R3Vdisp-W = 33.51 ft3Buoyancy Force:FB = gW x Vdisp-wFB = 62.4 x 33.51FB = 2091.024 lbs upSum of the Forces:SFy = 0 = 500 - 2091.024 + TT = 1591.024 lbs down

*FR = g A YC sinqor FR = g A HcYR = (Ixc / YcA) + YcXR = (Ixyc / YcA) + Xcbut for a rectangle or circle: XR = XcFor 90 degree walls:FR = g A Hc

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