1. 2.4 MANOMETERS
Manometers are devices that employ liquid columns for determining differences in pressure.
Figure 2.6a: the most elementary manometer – piezometer
It measures the pressure in a liquid when it is above zero gage
Glass tube is mounted vertically so that it is connected to the space within the container
Liquid rises in the tube until equilibrium is reached
The pressure is then given by the vertical distance h from the meniscus (liquid surface) to the point where the pressure is to be measured, expressed in units of length of the liquid in the container.
Piezometer would not work for negative gage pressures, because air would flow into the container through the tube

2. Figure 2.6 Simple manometers.

3. Figure 2.6b: for small negative or positive gage pressures in a liquid
With this arrangement the meniscus may come to rest below A, as shown. Since the pressure at the meniscus is zero gage and since pressure decreases with elevation,
    units of length H2O
Figure 2.6c: for greater negative or positive gage pressures (a second liquid of greater relative density employed)
It must be immiscible in the first fluid, which may now be a gas
If the relative density of the fluid at A is S1 (based on water) and the relative density of the manometer liquid is S2, the equation for pressure at A
hA - the unknown pressure, expressed in length units of water,
h1, h2 - in length units

4. A general procedure in working all manometer problems :
Start at one end (or any meniscus if the circuit is continuous) and write the pressure there in an appropriate unit (say pascals) or in an appropriate symbol if it is unknown.
Add to this the change in pressure, in the same unit, from one meniscus to the next (plus if the next meniscus is lower, minus if higher). (For pascals this is the product of the difference in elevation in metres and the unit gravity force of the fluid in newtons per cubic metre.)
Continue until the other end of the gage (or the starting meniscus) is reached and equate the expression to the pressure at that point, known or unknown.
The expression will contain one unknown for a simple manometer or will give a difference in pressures for the differential manometer. In equation form,

5. Figure 2.7 Differential manometers

6. A differential manometer (Fig. 2
A differential manometer (Fig. 2.7) determines the difference in pressures at two points A and B when the actual pressure at any point in the system cannot be determined
Application of the procedure outlined above to Fig. 2.7a produces
For Fig. 2.7b:
If the pressures at A and B are expressed in length of the water column, the above results can be written, for Fig. 2.7a,
For Fig 2.7b:

7. Example 2.4  In Fig. 2.7a the liquids at A and B are water and the manometer liquid is oil. S = 0.80; h1 = 300 mm; h2 = 200 mm; and h3 = 600 mm.
(a) Determine pA - pB, in pacals.
(b) If pB = 50kPa and the barometer reading is 730 mm Hg, find the pressure at A, in meters of water absolute.
(a)
(b)
(a)

8. Micromanometers
For determining very small differences in pressure or determining large pressure differences precisely – several types of manometers
One type very accurately measures the differences in elevation of two menisci of a manometer.
By means of small telescopes with horizontal cross hairs mounted along the tubes on a rack which is raised and lowered by a pinion and slow motion screw so that the cross hairs can be set accurately, the difference in elevation of menisci (the gage difference) can be read with verniers.

9. Figure 2.8 Micromanometer using two gage liquids

10. Fig. 2.8: two gage liquids, immiscible in each other and in the fluid to be measured  a large gage difference R can be produced for a small pressure difference.
The heavier gage liquid fills the lower U tube up to 0-0; then the lighter gage liquid is added to both sides, filling the larger reservoirs up to 1-1.
The gas or liquid in the system fills the space above 1-1. When the pressure at C is slightly greater than at D, the menisci move as indicated in Fig. 2.8.
The volume of liquid displaced in each reservoir equals the displacement in the U tube 
Manometer equation
γ1, γ2 and γ3 are the unit gravity force

11. Example 2. 5 In the micromanometer of Fig 2
Example In the micromanometer of Fig 2.8 the pressure difference is wanted, in pascals, when air is in the system, S2 = 1.0, S3 = 1.10, a/A = 0.01, R = 5 mm, t = 20oC, and the barometer reads 760 mm Hg.
The term γ1(a/A) may be neglected. Substituting into Eq. (2.4.1) gives #

12. Figure 2.9
Inclined manometer
The inclined manometer: frequently used for measuring small differences in gas pressures.
Adjusted to read zero, by moving the inclined scale, when A and B are open. Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube, it affords greater accuracy in reading the scale.
Surface tension causes a capillary rise in small tubes. If a U tube is used with a meniscus in each leg, the surface-tension effects cancel.

13. 2.5 FORCES ON PLANE AREAS
In the preceding sections variations oF pressure throughout a fluid have been considered.
The distributed forces resulting from the action of fluid on a finite area may be conveniently replaced by a resultant force, insofar as external reactions to the force system are concerned.
In this section the magnitude of resultant force and its line of action (pressure center) are determined by integration, by formula, and by use of the concept of the pressure prism.

14. Horizontal Surfaces
A plane surface in a horizontal position in a fluid at rest is subjected to a constant pressure.
The magnitude of the force acting on one side of the surface is
The elemental forces pdA acting on A are all parallel and in the same sense  a scalar summation of all such elements yields the magnitude of the resultant force. Its direction is normal to the surface and toward the surface if p is positive.
Fig. 2.10: arbitrary xy axes - to find the line of action of the resultant, i.e., the point in the area where the moment of the distributed force about any axis through the point is zero,
Then, since the moment of the resultant must equal the moment of the distributed force system about any axis, say the y axis,
x’ – the distance from the y axis to the resultant

15. Figure 2.10 Notation for determining the line
of action of a force

16. Momentum (1) First moment
The moment of an area A about the y axis
The moment about a parallel axis, for example, x = k, the moment
Centroidal axis
Volume center
Mass center: center of gravity of a body

17. Figure A.1 Notation for first and second moments

18. (2) Second moment
The second moment of an area A (the moment of inertia of the area)
The moment about a parallel axis, for example, x = k, the moment
Figure A.2 Moments of inertia of simple areas about centroidal axes

19. The product of inertia Ixy of an area
- the product of inertia about centroidal axes parallel to the xy axes.

20. Inclined Surfaces
Fig. 2.11: a plane surface is indicated by its trace A'B‘;it is inclined θo from the horizontal. x axis: intersection of the plane of the area and the free surface. y axis: taken in the plane of the area, with origin O in the free surface. The xy plane portrays the arbitrary inclined area. The magnitude, direction, and line of action of the resultant force due to the liquid, acting on one side of the area, are sought.
For δA:
Since all such elemental forces are parallel, the integral over the area yields the magnitude of force F, acting on one side of the area,
Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid
The presence of a free surface is unnecessary

21. Figure 2.11 Notation for force of liquid on one side of a plane inclined area.

22. Center of Pressure
Fig. 2.11: the line of action of the resultant force has its piercing point in the surface at a point called the pressure center, with coordinates (xp, yp). Center of pressure of an inclined surface is not at the centroid. To find the pressure center, the moments of the resultant xpF, ypF are equated to the moment of the distributed forces about the y axis and x axis, respectively 
- may be evaluated conveniently through graphical integration, for simple areas they may be transformed into general formulas:

23. When either of the centroidal axes is an axis of symmetry for the surface, vanishes and the pressure center lies on x = x- . Since may be either positive or negative, the pressure center may lie on either side of the line x = x-. To determine yp by formula, with Eqs. (2.5.2) and (2.5.6)
In the parallel-axis theorem for moments of inertia
in which IG is the second moment or the area about its horizontal centroidal axis. If IG is eliminated from Eq. (2.5.9)

24. Figure 2.12 Triangular gate
Example 2.6 The triangular gate CDE (Fig. 2.12) is hinged along CD and is opened by a normal force P applied at E. It holds oil, relative density 0.80, above it and is open to the atmosphere on its lower side Neglecting the weight of the gate, find (a) the magnitude of force exerted on the gate by integration and by Eq. (2.5.2); (b) the location of pressure center; (c) the force P needed to open the gate.
Figure Triangular gate

25. (a) By integration with reference to Fig. 2.12,
When y = 4, x = 0, and when y = 6.5, x = 3, with x varying linearly with y; thus
in which the coordinates have been substituted to find x in terms of y. Solving for a and b gives
Similarly, y = 6.5, x = 3; y = 9, x = 0; and x = 6/5(9 - y). Hence,
Integrating and substituting for γsinθ leads to
By Eq. (2.5.2)

26. (b) With the axes as shown,
In Eq. (2.5.8)
I-xyis zero owing to symmetry about the centroidal axis parallel to the x axis; hence
In Eq. (2.5.11),
i.e., the pressure center is 0.16 m below the centroid, measured in the plane of the area.
(c) When moments about CD are taken and the action of the oil is replaced by the resultant,

27. The Pressure Prism
Pressure prism: another approach to determine the resultant force and line of action of the force on a plane surface - prismatic volume with its base the given surface area and with altitude at any point of the base given by p = γh. h is the vertical distance to the free surface, Fig (An imaginary free surface may be used to define h if no real free surface exists.) (in the figure, γh may be laid off to any convenient scale such that its trace is OM)
The force acting on an elemental area δA is
(2.5.12)
- an element of volume of the pressure prism. After integrating, F = ϑ
From Eqs. (2.5.5) and (2.5.6),
(2.5.13)
 xp, yp are distances to the centroid of the pressure prism  the line of action of the resultant passes through the centroid of the pressure prism

28. Figure Pressure prism

29. Effects of Atmospheric Pressure on Forces on Plane Areas
In the discussion of pressure forces the pressure datum was not mentioned : p = γh  the datum taken was gage pressure zero, or the local atmospheric pressure
When the opposite side of the surface is open to the atmosphere, a force is exerted on it by the atmosphere equal to the product of the atmospheric pressure P0 and the area, or P0A , based on absolute zero as datum. On the liquid side the force is
The effect P0A of the atmosphere acts equally on both sides and in no way contributes to the resultant force or its location
So long as the same pressure datum is selected for all sides of a free body, the resultant and moment can be determined by constructing a free surface at pressure zero on this datum and using the above methods

30. Example 2.8 An application of pressure forces on plane areas is given in the design of a gravity dam. The maximum and minimum compressive stresses in the base of the dam are computed from the forces which act on the dam. Figure 2.15 shows a cross section through a concrete dam where the unit gravity force of concrete has been taken as 2.5γ and γ is the unit gravity force of water. A 1 m section of dam is considered as a free body; the forces are due to the concrete, the water, the foundation pressure, and the hydrostatic uplift. Determining amount of hydrostatic uplift is beyond the scope of this treatment. but it will be assumed to be one-half the hydrostatic head at the upstream edge, decreasing linearly to zero at the downstream edge of the dam. Enough friction or shear stress must be developed at the base of the dam to balance the thrust due to the water that is Rx = 5000γ. The resultant upward force on the base equals the gravity force of the dam less the hydrostatic uplift Ry = 6750γ γ γ = 7625γ N. The position of Ry is such that the free body is in equilibrium. For moments around O,

31. Figure 2.15 Concrete gravity dam

32. It is customary to assume that the foundation pressure varies linearly over the base of the dam, i.e., that the pressure prism is a trapezoid with a volume equal to Ry; thus
in which σmax, σmin are the maximum and minimum compressive stresses in pascals. The centroid of the pressure prism is at the point where x = 44.8 m. By taking moments about 0 to express the position of the centroid in terms of σmax and σmin,
Simplifying gives
When the resultant falls within the middle third of the base of the dam, σmin will always be a compressive stress. Owing to the poor tensile properties of concrete, good design requires the resultant to fall within the middle third of the base.