Experiment 5 Center of Pressure

November 23, 2017 | Author: sakura9999 | Category: Pressure, Applied And Interdisciplinary Physics, Mechanics, Physics, Physics & Mathematics
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Faculty of Engineering DEPARTMENT OF MECHANICAL ENGINEERING ME 316 Thermo Fluid Lab Experiment 5- Center of Pressure on a Plane Surface

AIMS The objective of this experiment is to determine the hydrostatic force and the centre of pressure on a rectangular surface when it is partially or fully submerged in water. APPARATUS

Figure 1: Hydrostatic Pressure Apparatus BASIC THEORETICAL BACKGROUND Hydrostatic force acting on the rectangular face:

P  ghC A

(1)

Center of pressure (point of application):

y D  yC 

I yC A

(2) where yc: position of the centroid and yd: position of center of pressure

1

 Partial immersion

L

m

Water surface y

yD

h

a

yC

d

b hC  y C  P

Hence

y ; area: A = by 2

(3)

1 gby 2 2

(4)

by 3 y D  yC 

12  y 6 by 2

(5)

2

y y    h   a  d    y d  y c    a  d   2 3   

And

(6)

Moment M of pressure force (P) about knife-edge axis is given by:

y y y   1  M  Ph  P a  d    y d  y c   gby 2  a  d    2 2 6   2  M

And then

y 1  gby 2  a  d   2 3 

(7)

(8)

At equilibrium, we must have: M =Ph = mgL ; where m = mass added to balance pan

mL 

Thereby: And also:

y 1  by 2  a  d   2 3 

(9)

h= mgL/P with P given by Eq. (4)

 

Eq. (6) h   a  d 

(10)

y  can be considered as the theoretical value of h and Eq. (10) the experimental value of h. 3

2

 Complete immersion

L

h

a

Water surface yD

m

yC

y

d

b

Hence

d hC  y C  y  ; A = bd 2

(11)

d  P  g  y  bd 2 

(12)

y D  yC 

And

bd 3

d2 12  bd  y  d / 2 12 y  d / 2

(13)

 y d2     h   a  d    y d  y c    a  d  2 12 y  d / 2    

(14)

Moment M of P about knife-edge axis is given by:

 d d  d d2     M  P a    y d  y c   gbd  y   a   2 2  2 12 y  d / 2    

(15)

At equilibrium M=Ph =mgL; thereby,

 d  d2   mL  bd  y   a  d  2  12 y  d / 2   And also:

h= mgL/P with P given by Eq. (12)



Eq. (14) h   a  d 



(16)

(17)

 d2  gives the theoretical value of h and Eq. (17) the experimental value of h. 12 y  d / 2  3

PROCEDURE 1. Measure the dimensions a, b, and d, and the distance L from the knife – edge axis to the balance pan axis. 2. Level the tank, using the adjustable feet in conjunction with the spirit level and make sure to close the drain valve. 3. Start with the tank empty. Move the counter-balance weight until the balance arm is horizontal. 4. Add a small mass (typically 50g) to the weight hanger. 5. 6. 7. 8.

Slowly add water to the tank allowing time for the water levels to stabilize. Add water until the hydrostatic pressure causes the balance arm to recover the horizontal position. Fine adjustment of the water level may be achieved by over – filling and slowly draining, using the drain cock. Note the water level on the scale.

9. Repeat the procedure under section (4) for different masses : 5 masses for water levels y > d (complete immersion) and 5 masses for y < d (partial immersion) 10. Repeat readings for reducing masses on the balance pan. 11. All record data can be arranged as shown in table 1 and 2. 12. Make sure to convert all measurements to SI units in order to obtain the right values of P and mgL (in Newton). Convert the experimental value of h to cm. REPORT 

For y< d (partial immersion)

1/ Plot P against y (depth) and comment on the graph. From (8)

ba  d  b m b  y  ba  d  b ; With (theoretical values) A  and B   a  d    y  A  By   2L 6L 3 2L 6L y 2 2L 

2/ Plot m *  

m against y with the measured values and fit the curve with a straight line (experimental curve). y2

For y> d (complete immersion)

3/ Plot P against y (depth) and comment on the graph.

 

From Eq. (16) mL  bd  y 



 d  d2 d  ; if we introduce the variable y *   y    a  d  2  12 y  d / 2  2 

 bd 3   1  m bd  d d 2  bd  d   bd 3  1  bd  d            a    a    A  B where A  a  and B       y*  L  2 12 y *  L  2   12 L  y *  L  2 y*    12 L 

4/ Plot m * 

m 1 against * with the measured values and fit the curve with a straight line (experimental curve). * y y

CONCLUSIONS Comment on the variation of P and h with y. Give reasons for the discrepancies, if any, between the measured and theoretical values of h. Compare the theoretical and experimental values of A and B. Give reasons for the discrepancies, if any, between the measured and theoretical values. 4

Table 1. Between brackets: given values; confirm with your own measurements. a+d (cm)

b( cm)

d(cm)

L (cm)

……….(200mm)

………..(75 mm)

………….(100 mm)

………….(275 mm)

Table. 2 : Notation: CP= Center of Pressure Patital immersion ; y < d

P

1 gby 2 2

m* 

Complete immersion ; y > d

y  htheor.   a  d   3 

d  P  g  y  bd 2 

m y2

m* 

m y*

  d2  htheor.   a  d  12 y  d / 2  

d  y*   y   2 

Increasing masses Case

m (g)

y (cm)

Theoretical CP position (cm)

P (N)

Experimental CP position (cm)

m*

1 y*

hexp= mgL/P

Patital immersion yd

Decreasing masses Case

m (g)

y (cm)

Theoretical CP position

(N)

Experimental CP position

m*

1 y*

Complete immersion y>d

Patital immersion y
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