coriolli's coponent

7/29/2019 Coriolli's Coponent

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TECHNICAL MANUAL

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CORIOLLI'S COMPONENT OF ACCELERATION APPARATUS

If a point is moving along a line, with the line having rotational motion,

the absolute acceleration of the point, is vector sum of -

I) Absolute acceleration of coincident point over the link relative to fixed

center.

ii) Acceleration of point under consideration relative to coincident point

andiii) The third component, called coriolli's component of acceleration.

THEORY-

Consider the motion of slider 'B' on the crank OA. Let OA rotate with

constant angular velocity of rad /sec, and slider B have a radial

outwards velocity V m/ sec relative to crank center O.

REF.FIG. A

In the velocity diagram, Oa represents tangential velocity of slider at

crank position OA, and ab represents radial velocity of slider, at same

crank position. Oa ' is the tangential velocity of slider at crank position

OA' and a'b' represents radial velocity of slider at same crank position.

Hence, bb' represents the resultant change of velocity of slider. This

velocity has two component b' T and bT in tangential and radial

directions respectively.

Now, Tangential component, b'T= bs + sT

= V sin d + [ (r + dr) - r]

= V d + dr ------------------------------- (I )


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Therefore Rate of change of tangential velocity

d dr

= V ------- + ---------

dt dt

= V + V

= 2 V ----------------------------------( ii )

Equation (ii) represents coriolli's component of acceleration. This

acceleration is made up of two components, one due to increase in

radius and other from change in the direction of crank.

Hydraulic Analogy -

REF.FIG. B

Consider a short column of fluid of length dr at radius r from axis of

rotation of the tube. Then, if velocity of fluid relative to tube is V and

angular velocity of tube is then coriolli's component of acceleration is 2

V in a direction perpendicular to rotation of tube. The torque dT

applied by the tube to produce this acceleration is then

dW

dT = ------------- . 2 V r

g

Where, dW is weight of short column of fluid.

If W be the specific weight of fluid and a is cross sectional area of

tube, then ,

dW = W a dr

W a dr

dT = ----------- 2 V r

gand total torque applied to column of length l,

l W

T = 2 ---------- V . a. r. dr

0 g


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WT = ---------- w V . a. l 2

g

Hence, coriollis component of acceleration,

^ ^ 2 g TC A = ------------- --------------------------- ( iii )

W a l 2

THE APPARATUS -

The apparatus uses hydraulic analogy to determine coriolli's

component of acceleration. The apparatus consists of two brass tubes

connected to a central rotor distributor. The distributor is rotated by a

variable speed d. c. motor. Water is supplied to a distributor by a pump

through water meter. When tubes are rotating with water flowing through

tubes, with various measurements provided, coriolli's component can be

determined experimentally and theoretically.

SPECIFICATIONS -

1) Pipes - 6 mm I. D, 300 mm. Effective length - 2Nos.

2) Drive motor 0.5 HP. 750 rpm d. c. series motor, swinging field type

with speed control.3) Torque arm -Radius 0.095 m with 5 Kg. Capacity spring balance.

4) Pump - 0.5 H.P. capacity 25 x 25 mm connection single ph.

5) Water meter - 250 to 2500 liters per hour capacity.

6) Housing for rotating pipes, this also acts as water reservoir, with

two sides of Perspex sheets.

EXPERIMENTAL PROCEDURE -

Fill up sufficient water in the tank. Rotate the coupling to ensure free

rotation. Check the nut bolts for tightening. Start the motor and set the

speed as required, e.g. say 200 rpm. Measure the torque required for

free rotation of tubes at that speed. (for measuring the torque, pointer

over the torque arm must coincide with the stationary pointer, before

taking spring balance reading.)

Now start the pump and adjust the flow rate with the help of bypass

valve, so that water does not overflow through central glass tube and


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also pipes run full of water. Now adjust the speed to previous value and

measure the torque. Note down water flow rate.

Repeat the procedure at different speeds.

OBSERVATIONS-

Sr.No.

Speed N Spring balance( Kg )

Water flow

( rpm ) Initial Final ( LPH )

CALCULATIONS -

1) Bore dia. Of tubes = 8 mm

Flow area of tube = 0.05 x 10 -3 m2

Total flow area = 0.0001 m2

2) Flow rate = Q = L P H / 3.6 x 106 m3 / Sec.

Velocity of water through the tubes,

QV = -------- m / s

a

2 N T

3) Power P = ___________

60

P = V I Watts (nm/s )

60 x P

T = ________

2 N

4) Now theoretically,

CA = 2 V

= 2 V [ 2 N / 60 ]


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Practically,

^ ^ 2 g TCA = ------------------ m / Sec2

W . a . l 2

Where, W = Specific weight of water

= 1000 Kg / m3

l = Effective length of tube

= 0.285 m.

a = Flow area of tube

= 0.0001 m

2

and T = Torque in Kg-m.

PRECAUTIONS -

1) While filling the water, see that there is sufficient gap between the

tubes and water surface.

2) Remove the glass tube with its water supply nozzle after the

experiment is over. Store glass tube safety.

3) Drain the water after the experiment is over.

4) Operate all the switches and controls gently.

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- : SAMPLE CALCULATION : -

(CORIOLLIS COMPONENT OF ACCELERATION)

OBSERVATIONS: -

Sr.No.

Speed(N )

Spring balanceReading

(Kg)

Water flow

( LPH )

( RPM ) Initial Final

1 100 0.225 0.525 6002 150 0.250 0.750 675

3 200 0.275 1.000 725

4 250 0.350 1.500 9505 300 0.375 1.875 975

CALCULATIONS - for reading No. - 3

1) Bore dia. Of tubes = 8 mm

Flow area of tube = 0.05 x 10 -3 m2

Total flow area = 0.0001 m2

L P H2) Flow rate , Q = ---------------

3.6 x 106

725

Q = ---------------3.6 x 106

Q = 2.01 x 10-4 m3 / Sec.

Velocity of water through the tubes,

QV = -------- m / s

a2.01 x 10-4

V = ----------------1 x 10-4

V = 2.01 m / s


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2 N T

3) Power P = ___________

60

P = V I Watts (nm/s )

60 x P

T = ________ =

2 N

4) Now theoretically ,

CA = 2 V

2 x x N= 2 x V x --------------

60

2 x x 200

= 2 x 2.01 x -------------------60

= 84.19 m / Sec2 -------------------- ( 1 )

Practically ,

^ ^ 2 g TCA = ------------------

W . a. l 2

2 x 9.81 x 0.0435= -----------------------------------

1000 x 1 x 10-4 x (0.329 ) 2

= 78.85 m / Sec2 ---------------------- ( 2 )

compare the values of equation ( 1 ) & ( 2 )

RESULT TABLE: -

SR.NO.

Discharge -LPH

Q = ---------3.6 x

106

VelocityQ

v = --------A

( m /

Torque

T = 60 x P 2 N

C. A. Theo.

= 2 v

( m/sec2 )

C . A. act.2 g T

= ---------W.A.l2

( m/sec2 )


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( m3/sec. ) sec )

1 1.670 x 10-4 1.67 0.018 34.98 32.63

2 1.875 x 10-

4

1.875 0.030 58.90 54.38

3 2.010 x 10-

4

2.01 0.0435 84.19 78.85

4 2.640 x 10-4 2.64 0.069 138.23 125.075 2.710 x 10-4 2.71 0.090 170.27 163.14

NOTE - This calculation is only for reference purpose.

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coriolli's coponent

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