Centre Of Pressure Expt

CENTER OF PRESSURE AIM:- To measure the moment due to fluid pressure on a wholly or partially submerged plane surface and compare with theoretical analysis.

DESCRIPTION OF APPARATUS:-

Water is contained in a quadrant of a semi-circular perspex tank assembly which is allowed to swing on a smooth bar. The cylindrical sides of the quadrant have their axes coincident with the centre of rotation of the tank assembly, and therefore the total fluid pressure acting on these surfaces exerts no moment about that centre. The only moment present is that due to the fluid pressure acting on the plane surface. This moment is measured experimentally by applying weights to a weight hanger mounted on the opposite side to the quadrant tank. A second tank, situated on the same side of the assembly as the weight hanger, provides a trimming facility and enables different angles of balance to be achieved. The angular position of the plane and the height of water above it are measured on an angular scale engraved on the tank and a linear scale on the back panel. Included with the apparatus are base leveling feet and a spirit level.

THEORY:Centre of Pressure Centre of pressure may be defined as the point in a plane at which the total fluid thrust can be said to be acting normal to that plane.

The following analysis applies to the condition of a plane surface at various angles when it is wholly or partially submerged in a fluid. Let breadth of quadrant = B and weight per unit volume = w Referring to the diagram above, consider an element at start depth y, width dy. Force on element dF = w( y cos  h) Bdy and the Moment of Force on element about 0 = wB( y cos  h) ydy Therefore, total moment about 0 = M = wB ( y 2 cos  hy)dy Case 1: Plane Fully Submerged Limits R1 and R2 R2

M  wB  ( y 2 cos  hy)dy R1

R2

 cosy 3 hy2  M  wB   2 R  3 1

M

wB cos 3 wB 2 R2  R13  R2  R12 h 3 2









This equation is of the form of y = mx + c. A plot of M against h will yield a straight line graph. Case 2: Plane Partially Submerged Limits R2 and hsecθ Hence:

M  wB

R2

y  (cos 

2

 hy)dy

h sec

R2

 cosy 3 hy 2  M  wB  2  h sec  3 wB cos 3 wBh 2 R2  h 3 sec3   R2  h 2 sec 2  3 2





M

B cosR23

B sec2 h3

M

B cosR23

M

3

3





3







wBR22h wB sec2 h3  2 2

wBR22h wB sec2 h3  2 6

PROCEDURE:1. Once the equipment has been set-up, affix the weight hanger to the hanger support located on the top left of the hopper. The apparatus will now require trimming in order to bring the submerged plane to the vertical (i.e. 0º position). This is achieved by gently pouring water into the trim tank until the desired position is achieved.

The horizontal line on the tank

assembly should be read against the zero line on the back scale. 2. Add a weight to the weight hanger. Pour water, with dye added if necessary, into the quadrant tank until a 0º balance is restored. Note the weight and the height reading of the water (h). 3. Repeat the procedure for the full range of weights. 4. Empty both tanks of water. Again, with the weight hanger alone in position, trim the assembly by gently adding water to the trim tank until a 30º balance is achieved. Add a weight; restore balance to 30º point and record values for h for the full range of weights.

RESULTS AND CALCULATIONS:Radius to lower edge

R2

200 mm

=

0.2

m

Radius to upper edge

R1

100 mm

=

0.1

m

Slant depth

R2-R1

100 mm

=

0.1

m

Width of rectangle

B

75 mm

=

0.075 m

Radius to weight hanger

Rw

200 mm

=

0.2

m

1) θ = 00 Sr.no.

W

Mexpt W  9.81 Rw 10 3 (Nm)

H

Mthe

(mm)

(Nm)

W

Mexpt

H

Mthe

(gm)

W  9.81 Rw 10 3

(mm)

(mm)

(gm)

2) θ =300 Sr.no.

(Nm)

Sample Calculations:-

GRAPHS:1) Mexpt and Mthe against h (θ =00 ) 2) Mexpt and Mthe against h (θ =300) CONCLUSION:-