Experiment to determine the centre of pressure on a partially submerged plane surface

OBJECTIVE OF EXPERIMENT:


To determine the centre of pressure on a partially submerged plane surface


EQUIPMENT SET – UP:


Hydraulics Bench F1 – 10 and Hydrostatic Pressure Apparatus F1 – 12.

SUMMARY OF THEORY
   
 Partial immersion:
    
mgL = [(ρg/2)](by²)[(a+d)-y/3]
    
m/y² = [(ρb)/(2L)][a+d]- [(ρb)/2L](y/3)




PROCEDURE:

1. Place the quadrant on the two dowel pins and using the clamping screw, fasten to the balance arm. Measure a, L, depth d and width b, of the quadrant end face. With the Perspex tank on the bench, position the balance arm of the knife edges    (pivot). Hang the balance pan from the end of the balance arm. Connect a length of hose from the drain cock to the sump and a length from the bench feed to the triangular aperture on the top of the Perspex tank. Level the tank using the adjustable feet and spirit level. Move the counter balance weight until the balance arm is horizontal.

2. Close the drain cock and admit water until the level reaches the bottom edge of the quadrant. Place a weight on the balance pan, slowly adding water into the tank until the balance arm is horizontal. Record the water level on the quadrant and the weight on the balance pan.

3. Fine adjustment of the water level can be achieved by overfilling and slowly draining, using the stop cock.

4. Repeat the above for each increment of weight until the water level reaches the top of the quadrant end face. Then remove each increment of weight noting weights and water levels until the weights have been removed.

RESULTS AND CALCULATIONS:

FILLING TANK		DRAINING TANK		AVERAGE
Weight m (g)	Height of water y (mm)	Weight m (g)	Height of water y (mm)	m (g)	y (mm)	y²	m/y²
50	43	200	44	50	43.5	1892.25	0.026
100	64	150	63	100	63.5	4032.25	0.025
150	79	100	78	150	78.5	6162.25	0.024
200	93	50	93	200	93	8649.00	0.023

Therefore, graph m/y² against as shown:


From the graph


P1 = x : 30
        y : 0.0265


P2 = x : 50
        y : 0.025


My = (y2 – y1) / (x2 – x1)


      = (0.025 – 0.0265) / (50 -80)


      = 0.000033


C   = 0.029




CONCLUSION

From this experiment, we can conclude that center of pressure can be determine by the graph parameters.