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Static (Pg) and Total Pressure (P,)

A fluid is flowing through a pipe. The static pressure is defined as the pressure on the wall by a static fluid. For fluid flow situation as shown in Fig 2.5, we can measure the static pressure by just putting a vertical tube in the wall of the pipe. If we like to measure the total pressure we have to fit a tube (pi tot tube) as shown in Fig. 2.5 (right side tube). P dis the indicated dynamic pressure which is the head difference between the two tubes.

SOLVED EXAMPLES

Example 2.1. A hydraulic press has a ram of 40 cm diameter and plunger of 10 cm diameter.Find the force required at the plunger to lift a load of 30 KN. The stroke length of plunger is 40 cm and it executes 120 strokes per minute. Calculate the distance through which the load is raised per minute and power required in driving the plunger in this case.

Solution. p1 = f / a , [ a = cross –section area of plunger]

P2 = W /A , [ A = cross – section area of ram]

P1 = p2

or               f / a = W / A or W = F.A /a

or         f = W. a/ A = 30 x 103 x  = 1875 N .

(b) Stroke volume (plunger side) = 0.4 x π / 4 (0.1)2 m3

Stroke volume (ram side) = S r x π / 4 (0.4)2 m3

S r x π / 4 (0.4)2 = 0.4 x π /4 (0.1)2

S r = 0.4 (0.1)2 / (0.4)2 = 0.025 m

Distance raised on ram side

= 0.025 x 120m/ minute = 3m/minute Ans.

(c) Power required = 30 x 103 x 3 Nm/minute

= 9 x 104 Nm / minute = 9 / 60 x 104 watt

= 1.5 KW

 

Example 2.2. In Fig. 2.7 find the pressure at A and also at B.

 Solution.

Pressure at A,             PA = P oil x g x z1,

0.8 x 1000 x 9.81 x 2

= 15696 N I m2 Ans.

Pressure at B = PA + P water x g x z2,

= 15696 + 1000 x 9.81 X 4

= 54936 N / m2 Ans.

 

Example 2.3. The diameters of a small piston and a large piston are 2 cm and 8 cm respectively. A force of 100 N is applied on the small piston of the hydraulic jack. Calculate the load lifted by the large piston when (i) the pistons are at same level (Fig. 2.8) (ii) small piston is 50 cm above the large piston (Fig. 2.9).

Solution. (i)    

                   

                    (ii) PAA = 0.5 x 1000 x 9.81 +

= (4905 + 318310) N/m2

= 323215 N/m2

W2 = PAA x A

= 323215 x π / 4 (0.08)2 N.

= 1624.6 N = force on large piston.

 

Example 2.4. The pressure at a point in liquid is 4 N/cm2. Find the corresponding height when the liquid is (i) water (ii) oil of sp. gravity 0.8.

Solution. For water

z  p / pg = 4 x 104 / 1000 x 9.81 m

= 4.08 m

For oil

z  = p / p oil g = 4 x 104 / 0.8 x 1000 x 9.81 = 5.1 m

Example 2.5. An oil of specific gravity 0.8 is contained in a vessel. At a point the height of oil is 50m. Determine the corresponding water head.

Solution                                p = 0.8 x 1000 x 9.81 x 50 N/m2

Z water = p / p water x g = 0.8 x 1000 x9.81 x 50 / 1000 x 9.81 m

=40 m