Fluid Statics

(a) Pascal's law and pressure terms

Pascal's law: the pressure intensity at a point in a fluid at rest is the same in all directions, i.e. $p_x = p_y = p_z$. Any pressure applied to a confined fluid is transmitted equally in all directions.

• Atmospheric pressure: pressure due to the weight of the air above; standard value $101.325\ \text{kPa}$ at sea level.
• Absolute pressure: measured from absolute zero (perfect vacuum).
• Gauge pressure: measured above local atmospheric pressure; $p_{gauge} = p_{abs} - p_{atm}$.
• Vacuum pressure: pressure below atmospheric; $p_{vac} = p_{atm} - p_{abs}$.

 p_abs ^
       |   ______ point A (above atm)
       |  |  gauge pressure
 p_atm |--|-------------------------
       |  |  vacuum pressure
       |  |______ point B (below atm)
       |
 0     +--- absolute zero ----------

$$p_{abs} = p_{atm} + p_{gauge}$$

(b) Derivation of hydrostatic equation

Consider a vertical cylinder of fluid of cross-section $dA$ and height $dz$ in equilibrium. Take $z$ upward. Pressure at the bottom is $p$ and at the top is $p + dp$.

Forces in the vertical direction:

• Upward force on bottom: $p\,dA$
• Downward force on top: $(p + dp)\,dA$
• Weight: $\rho g\, dA\, dz$ (downward)

For equilibrium, $\sum F_z = 0$:

$$\begin{aligned} p\,dA - (p + dp)\,dA - \rho g\,dA\,dz &= 0 \\ \frac{dp}{dz} &= -\rho g = -\gamma \end{aligned}$$

The negative sign shows that pressure decreases as $z$ increases. For horizontal directions there is no net force, so $\partial p/\partial x = \partial p/\partial y = 0$: pressure is constant on a horizontal plane.

For an incompressible fluid ($\rho$ constant), integrating between two points 1 and 2:

$$p_2 - p_1 = -\rho g (z_2 - z_1) \quad \text{or} \quad \frac{p}{\rho g} + z = \text{constant}$$

At depth $h$ below a free surface where the pressure is $p_0$:

$$p = p_0 + \rho g h$$

Here $p/\rho g$ is the pressure head. The sum of pressure head and elevation head is constant throughout a static fluid.

Setup

Let the mercury surface in the A limb be at depth $a$ below the centre of A. This is the datum plane. The mercury in the B limb stands $h = 0.3\ \text{m}$ higher, and B is $z = 0.5\ \text{m}$ above A.

                   o B
   o A             |
   |               | water
   | water      ===+=== Hg, B limb
   |               |   ^ h
 ==+== Hg, A limb  |   v
   |  datum -------|---
   +---- mercury --+

Pressure balance at the datum

Pressure at the datum, from the A side:

$$p_A + \rho_w g a$$

From the B side, the water column above the mercury is $(a + z - h)$ and the mercury column is $h$:

$$p_B + \rho_w g (a + z - h) + \rho_m g h$$

Equating and simplifying:

$$\begin{aligned} p_A - p_B &= \rho_w g (z - h) + \rho_m g h \\ &= \rho_w g z + (\rho_m - \rho_w) g h \end{aligned}$$

Substitution

$$\begin{aligned} p_A - p_B &= 1000 \times 9.81 \times 0.5 + (13600 - 1000) \times 9.81 \times 0.3 \\ &= 4905 + 37081.8 \\ &= 41\,986.8\ \text{Pa} \approx 41.99\ \text{kPa} \end{aligned}$$

Pressure in A

$$p_A = 50 + 41.99 = 91.99\ \text{kPa (gauge)}$$

Answer: $p_A - p_B = 41.99\ \text{kPa}$; $p_A = 91.99\ \text{kPa}$ (gauge)

(a) Vertical rectangular gate

Given: $b = 2\ \text{m}$, $H = 3\ \text{m}$, top edge at 2 m depth.

Depth of centroid: $\bar h = 2 + 1.5 = 3.5\ \text{m}$. Area $A = 2 \times 3 = 6\ \text{m}^2$.

$$F = \rho g \bar h A = 1000 \times 9.81 \times 3.5 \times 6 = 206\,010\ \text{N} = 206.0\ \text{kN}$$

Centre of pressure:

$$\begin{aligned} I_G &= \frac{bH^3}{12} = \frac{2 \times 27}{12} = 4.5\ \text{m}^4 \\ h^* &= \bar h + \frac{I_G}{A\bar h} = 3.5 + \frac{4.5}{6 \times 3.5} = 3.714\ \text{m} \end{aligned}$$

So the centre of pressure is $3.714 - 2 = 1.714\ \text{m}$ below the hinge.

Taking moments about the hinge, with force $P$ at the bottom edge (3 m below the hinge):

$$\begin{aligned} P \times 3 &= F \times 1.714 \\ P &= \frac{206.01 \times 1.714}{3} = 117.7\ \text{kN} \end{aligned}$$

(b) Quadrant gate

Horizontal component equals the force on the vertical projection ($2\ \text{m} \times 3\ \text{m}$), whose centroid is at 1 m depth:

$$F_H = \rho g \bar h A_{proj} = 1000 \times 9.81 \times 1 \times 6 = 58\,860\ \text{N} = 58.86\ \text{kN}$$

Vertical component equals the weight of water above the curved surface (quarter cylinder of water):

$$F_V = \rho g \left(\frac{\pi R^2}{4} \times w\right) = 1000 \times 9.81 \times (3.1416 \times 3) = 92\,457\ \text{N} = 92.46\ \text{kN}$$

Resultant:

$$\begin{aligned} F_R &= \sqrt{F_H^2 + F_V^2} = \sqrt{58.86^2 + 92.46^2} = 109.6\ \text{kN} \\ \theta &= \tan^{-1}\frac{F_V}{F_H} = \tan^{-1}\frac{92.46}{58.86} = 57.5^\circ \text{ below the horizontal} \end{aligned}$$

Because all pressure forces on a circular arc act normal to the surface, the resultant passes through the centre O.

Answer: (a) $F = 206\ \text{kN}$, $h^* = 3.714\ \text{m}$, $P = 117.7\ \text{kN}$; (b) $F_H = 58.86\ \text{kN}$, $F_V = 92.46\ \text{kN}$, $F_R = 109.6\ \text{kN}$ at $57.5^\circ$ to the horizontal

Archimedes' principle and definitions

Archimedes' principle: a body immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced. $F_B = \rho g V_{disp}$.

• Centre of buoyancy (B): centroid of the displaced volume of fluid; the line of action of $F_B$.
• Metacentre (M): the point where the line of action of buoyancy, for a small angle of heel, cuts the original vertical centre line of the body.
• Metacentric height (GM): distance between the centre of gravity G and the metacentre M.

Expression for metacentric height

When a floating body heels through a small angle $\theta$, the shape of the displaced volume changes: a wedge emerges on one side and an equal wedge immerses on the other. B shifts to a new position $B'$.

        M
        |\
        | \
   G ---|--\---
        | B  B'
   ~~~~~~~~~~~~~~~~ water line

The horizontal shift of the centre of buoyancy is found from the moment of the wedge buoyancy about the centre line. For small $\theta$, the moment of the wedge volumes equals $I\theta$, where $I$ is the second moment of area of the waterline plane about the longitudinal axis of tilt. Then

$$V \cdot BB' = I\theta \quad \text{and} \quad BB' = BM \cdot \theta$$

so

$$BM = \frac{I}{V}$$

where $V$ is the volume of fluid displaced. Hence

$$GM = BM - BG = \frac{I}{V} - BG$$

$BG$ is positive when G is above B, and negative when G is below B. The metacentric height is measured positive when M is above G.

Conditions of equilibrium

For a stable floating body, the weight and buoyancy then form a righting couple $W \cdot GM \sin\theta$ that restores the body to its original position. For an unstable body the couple is an overturning couple.

For a submerged body, B and G do not change the shape of displaced volume, so there is no metacentre in the same sense; stability needs B above G.