Introduction to Basic Fluid Mechanics and Hydraulics

Fluid flow is classified by how density, the motion of particles and the flow properties behave in space and time.

Compressible versus incompressible flow

• Compressible flow: density changes from point to point, i.e. $\rho \neq$ constant. Examples: flow of gases at high speed, air in a compressor or nozzle, flow of steam in a turbine.
• Incompressible flow: density is (practically) constant, $\rho =$ constant. Liquids are almost always treated as incompressible. Examples: water flow in a penstock, river flow, oil in a pipe; air at low speed (Mach number below about 0.3).

Laminar versus turbulent flow

• Laminar flow: particles move in smooth, parallel layers (laminae) that slide over each other without mixing. Viscous forces dominate. In pipes $Re < 2000$. Examples: flow of thick oil in a thin tube, blood flow in capillaries, groundwater flow.
• Turbulent flow: particles move in a random, zig-zag way with eddies and strong mixing. Inertia forces dominate. In pipes $Re > 4000$. Examples: flow in rivers and canals, water in a penstock, smoke rising far from a chimney.
• Reynolds number: $Re = \dfrac{\rho V D}{\mu} = \dfrac{VD}{\nu}$; between 2000 and 4000 the flow is transitional.

Steady versus unsteady flow

• Steady flow: flow properties (velocity, pressure, density) at a point do not change with time:

$$\frac{\partial V}{\partial t} = 0,\quad \frac{\partial p}{\partial t} = 0,\quad \frac{\partial \rho}{\partial t} = 0$$

Example: flow through a pipe from a large reservoir at constant level; constant discharge in a canal.

• Unsteady flow: properties at a point change with time, $\dfrac{\partial V}{\partial t} \neq 0$. Examples: flood flow in a river, flow during valve closure (water hammer) in a penstock, emptying of a tank.

These are the basic physical properties used to describe a fluid.

• Specific weight (weight density), $w$ or $\gamma$: weight per unit volume, $\gamma = \rho g$. Unit N/m³. Water: 9810 N/m³.
• Specific gravity, $S$: ratio of the density (or specific weight) of a fluid to that of a standard fluid (water at 4 °C for liquids, air for gases). $S = \rho/\rho_w$; no unit. Mercury: 13.6.
• Specific mass (mass density), $\rho$: mass per unit volume, $\rho = m/V$. Unit kg/m³. Water: 1000 kg/m³.
• Specific volume, $v$: volume per unit mass, $v = 1/\rho$. Unit m³/kg. Mostly used for gases.
• Viscosity: the property by which a fluid resists the relative motion (shear) between its layers. By Newton's law, $\tau = \mu \dfrac{du}{dy}$, where $\mu$ is the dynamic viscosity (N·s/m² or Pa·s). Kinematic viscosity $\nu = \mu/\rho$ (m²/s).
• Surface tension, $\sigma$: the tensile force per unit length acting on the free surface of a liquid, caused by cohesion between liquid molecules. Unit N/m. Water–air at 20 °C: about 0.073 N/m.
• Capillarity: rise or fall of a liquid in a thin tube due to surface tension, adhesion and cohesion. Rise $h = \dfrac{4\sigma \cos\theta}{\rho g d}$. Water rises (adhesion > cohesion); mercury falls.
• Compressibility: the reciprocal of bulk modulus; it measures the fractional change in volume per unit change of pressure, $\beta = \dfrac{1}{K} = -\dfrac{dV/V}{dp}$. Unit m²/N. Liquids have very low compressibility, so they are treated as incompressible.

Total pressure is the resultant force exerted by a static fluid on a surface in contact with it; it acts normal to the surface. Centre of pressure is the point on the surface through which this resultant force acts.

Setup

Consider a plane surface of any shape, area $A$, immersed vertically in a liquid of specific weight $\rho g$. Let $\bar h$ be the depth of its centroid G and $h^*$ the depth of the centre of pressure P, both measured from the free surface.

  free surface
 ---------------------------------
      |          |
      | h-bar    | h*
      v          |
  +---G------+   |
  |  [strip] |<- dh at depth h
  |       P  |   v
  +----------+

Total pressure

Take a thin horizontal strip of width $b$ and thickness $dh$ at depth $h$. Pressure on it is $p = \rho g h$, so

$$dF = \rho g h\, (b\, dh) = \rho g h\, dA$$ $$F = \int \rho g h\, dA = \rho g \int h\, dA = \rho g A \bar h$$

since $\int h\, dA = A\bar h$ (first moment of area about the free surface).

Centre of pressure

Take moments about the free surface line. The moment of $dF$ is $dF\cdot h = \rho g h^2 dA$. Sum of moments equals the moment of the resultant:

$$F\, h^* = \rho g \int h^2 dA = \rho g\, I_o$$

where $I_o$ is the second moment of area about the free surface. By the parallel axis theorem $I_o = I_G + A\bar h^2$. So

$$\begin{aligned} h^* &= \frac{\rho g (I_G + A\bar h^2)}{\rho g A \bar h} \\ &= \bar h + \frac{I_G}{A\bar h} \end{aligned}$$

Results

• $F = \rho g A \bar h$ (independent of the shape; depends only on area and centroid depth).
• $h^* = \bar h + \dfrac{I_G}{A\bar h}$. Because $I_G/(A\bar h)$ is always positive, the centre of pressure lies below the centroid; the gap reduces as depth increases.
• Example: rectangle $b \times d$ with top at the surface: $\bar h = d/2$, $I_G = bd^3/12$, so $h^* = 2d/3$.

In a moving liquid the energy per unit weight is called head (unit m). There are three main heads.

Types of heads

1. Pressure head, $p/\rho g$: energy due to pressure; the height of liquid column the pressure can support.
2. Velocity (kinetic) head, $V^2/2g$: energy due to motion; the height a particle would rise if its velocity were turned upwards.
3. Potential (datum) head, $z$: energy due to position above an arbitrary datum.

• Piezometric head $= p/\rho g + z$ (level shown by a piezometer, defines the HGL).
• Total head $H = p/\rho g + V^2/2g + z$ (defines the energy line). In real flows a loss of head $h_L$ due to friction and fittings also appears.

Bernoulli's equation

For steady, ideal, incompressible flow along a streamline, the total head is constant:

$$\frac{p}{\rho g} + \frac{V^2}{2g} + z = \text{constant}$$

or between sections 1 and 2:

$$\frac{p_1}{\rho g} + \frac{V_1^2}{2g} + z_1 = \frac{p_2}{\rho g} + \frac{V_2^2}{2g} + z_2$$

Assumptions

1. The fluid is ideal (non-viscous), so there is no friction loss.
2. The flow is steady.
3. The flow is incompressible ($\rho$ constant).
4. The flow is irrotational and the equation is applied along a streamline.
5. Only gravity and pressure forces act; no energy is added or removed (no pump or turbine) and no heat transfer.
6. Velocity is uniform over the section (one-dimensional flow).

A most economical (best hydraulic) section of an open channel is the section which, for a given area, bed slope and roughness, carries the maximum discharge. Since $Q = A C\sqrt{RS}$ (Chezy) or $Q = \frac{1}{n}AR^{2/3}S^{1/2}$ (Manning), $Q$ is maximum when $R = A/P$ is maximum, i.e. when the wetted perimeter $P$ is minimum for the given area. Such a section also needs the least lining and excavation.

Proof for a rectangular section

   |<------- b ------->|
   |                   |  ^
   |      water        |  y
   |___________________|  v

Let width $= b$, depth $= y$.

$$A = by \quad \Rightarrow \quad b = \frac{A}{y}$$ $$P = b + 2y = \frac{A}{y} + 2y$$

For given $A$, $P$ is minimum when $dP/dy = 0$:

$$\frac{dP}{dy} = -\frac{A}{y^2} + 2 = 0 \;\Rightarrow\; A = 2y^2$$

Since $A = by$: $by = 2y^2 \Rightarrow b = 2y$. (Check: $d^2P/dy^2 = 2A/y^3 > 0$, so it is a minimum.)

Hydraulic mean depth:

$$\begin{aligned} R &= \frac{A}{P} = \frac{by}{b + 2y} \\ &= \frac{2y\cdot y}{2y + 2y} = \frac{2y^2}{4y} = \frac{y}{2} \end{aligned}$$

Result

For the most economical rectangular channel:

• width is twice the depth, $b = 2y$;
• hydraulic mean depth is half the depth, $R = y/2$.

Example: if $y = 1.5$ m, then $b = 3$ m and $R = 0.75$ m.

Surface tension

Surface tension ($\sigma$) is the tensile force per unit length acting along the free surface of a liquid, so that the surface behaves like a stretched membrane. Unit: N/m.

• Cause: a molecule inside the liquid is pulled equally in all directions by cohesion, but a molecule at the surface is pulled only inwards. This creates a net inward pull and a "skin" at the surface.
• Examples: water drops are spherical, small insects walk on water, a needle can float.
• Pressure inside a droplet: $p = \dfrac{4\sigma}{d}$; inside a soap bubble (two surfaces): $p = \dfrac{8\sigma}{d}$; inside a liquid jet: $p = \dfrac{2\sigma}{d}$.
• $\sigma$ decreases with increase in temperature. For water at 20 °C, $\sigma \approx 0.073$ N/m.

Capillarity

Capillarity is the rise or fall of a liquid in a small-diameter tube dipped in it, caused by surface tension together with adhesion and cohesion.

• If adhesion > cohesion (water and glass), the liquid wets the tube, the meniscus is concave and the liquid rises.
• If cohesion > adhesion (mercury and glass), the meniscus is convex and the liquid falls (capillary depression).

Equating the upward surface tension force to the weight of the raised column:

$$\sigma \pi d \cos\theta = \rho g \frac{\pi d^2}{4} h \;\Rightarrow\; h = \frac{4\sigma\cos\theta}{\rho g d}$$

where $\theta$ is the contact angle ($\approx 0^\circ$ for water–glass, $\approx 130^\circ$ for mercury–glass). The rise is inversely proportional to tube diameter, so manometer and piezometer tubes are made wider than about 6 mm to keep capillary error small.

Specific gravity and specific weight

• Specific weight ($\gamma$ or $w$): weight per unit volume of a fluid, $\gamma = \rho g$ (N/m³). For water $\gamma = 9810$ N/m³.
• Specific gravity ($S$): ratio of the specific weight (or density) of a fluid to that of a standard fluid (water at 4 °C for liquids). $S = \gamma/\gamma_w$; it has no unit. For mercury $S = 13.6$.

Newtonian and non-Newtonian fluids

A Newtonian fluid obeys Newton's law of viscosity: shear stress is directly proportional to velocity gradient, with a constant viscosity:

$$\tau = \mu \frac{du}{dy}$$

A non-Newtonian fluid does not obey this linear law; its apparent viscosity changes with the rate of shear, in general $\tau = A\left(\frac{du}{dy}\right)^n + B$.

 tau ^          B
     |         /        P
     |        /     _.-'
     |       /   .-'      N
     |      /  .'      .-'
     |     / .'     .-'
  ty +----'.'    .-'       D
     |   .'   .-'      _.-'
     |  /  .-'    _.--'
     | / .-' _.--'
     |/.-_.-'
     +--------------------> du/dy
 N = Newtonian (straight line from 0)
 P = pseudo-plastic (slope decreases)
 D = dilatant (slope increases)
 B = Bingham plastic (flows only after
     yield stress ty, then linear)

Types of non-Newtonian fluids: pseudo-plastic ($n<1$, e.g. paint, blood), dilatant ($n>1$, e.g. rice starch, quicksand), Bingham plastic (needs a yield stress before flowing, e.g. toothpaste, sewage sludge).

Newton's law of viscosity: the shear stress on a fluid layer is directly proportional to the rate of shear strain (velocity gradient) normal to the layer:

$$\tau = \mu \frac{du}{dy}$$

where $\mu$ is the coefficient of dynamic viscosity (N·s/m²). Fluids that obey this law are Newtonian fluids.

Importance of viscosity in fluid motion

1. Resistance to flow and head loss: viscosity causes shear between layers and friction at the walls. This produces head loss in pipes, penstocks and canals ($h_f = \frac{fLV^2}{2gD}$). In hydropower, friction loss in the penstock reduces the net head and power.
2. Type of flow: viscosity appears in the Reynolds number $Re = \rho VD/\mu$, which decides whether flow is laminar or turbulent.
3. No-slip condition and velocity profile: because of viscosity, fluid touching a wall has zero velocity, giving the parabolic (laminar) or flatter (turbulent) velocity profile and the boundary layer.
4. Drag and power needs: viscous drag on bodies and the power needed to pump fluids depend on viscosity.
5. Lubrication: a thin oil film separates moving surfaces (bearings of turbines and generators), reducing wear; correct viscosity is essential.
6. Energy dissipation: viscosity converts kinetic energy into heat, for example in hydraulic jumps below spillways.
7. Temperature effect: viscosity of liquids falls with rise in temperature (weaker cohesion) while that of gases rises (more molecular momentum exchange); this must be considered in design.

Use Newton's law of viscosity, $\tau = \mu\, du/dy$, with a linear velocity distribution between the plates.

Given: gap $dy = 5$ mm $= 0.005$ m, $F = 2$ N, $u = 0.8$ m/s, $A = 0.09$ m², $S = 0.9$.

Dynamic viscosity

$$\begin{aligned} \tau &= \frac{F}{A} = \frac{2}{0.09} = 22.22\ \text{N/m}^2 \\ \frac{du}{dy} &= \frac{0.8 - 0}{0.005} = 160\ \text{s}^{-1} \\ \mu &= \frac{\tau}{du/dy} = \frac{22.22}{160} = 0.1389\ \text{N·s/m}^2 \end{aligned}$$

Kinematic viscosity

$$\begin{aligned} \rho &= 0.9 \times 1000 = 900\ \text{kg/m}^3 \\ \nu &= \frac{\mu}{\rho} = \frac{0.1389}{900} = 1.543 \times 10^{-4}\ \text{m}^2/\text{s} \end{aligned}$$

Answer: $\mu = 0.1389$ N·s/m² (= 1.389 poise), $\nu = 1.543 \times 10^{-4}$ m²/s (= 1.543 stokes).

The plate slides at uniform velocity, so the component of its weight along the plane is balanced by the viscous shear force of the oil film.

Given: $A = 0.8 \times 0.8 = 0.64$ m², $W = 300$ N, $\theta = 30^\circ$, $u = 0.3$ m/s, $t = 1.5$ mm $= 0.0015$ m.

          plate (W = 300 N)
         _/_/
        /  /  ---> slides down at 0.3 m/s
       /  / oil film 1.5 mm
      /__/________
     /  30 deg

Shear force and shear stress

$$\begin{aligned} F &= W \sin 30^\circ = 300 \times 0.5 = 150\ \text{N} \\ \tau &= \frac{F}{A} = \frac{150}{0.64} = 234.375\ \text{N/m}^2 \end{aligned}$$

Velocity gradient (linear profile)

$$\frac{du}{dy} = \frac{0.3}{0.0015} = 200\ \text{s}^{-1}$$

Viscosity

$$\mu = \frac{\tau}{du/dy} = \frac{234.375}{200} = 1.172\ \text{N·s/m}^2$$

Answer: dynamic viscosity of the oil $\mu = 1.172$ N·s/m² (= 11.72 poise).

Bulk modulus is the ratio of the increase in pressure to the resulting volumetric strain:

$$K = \frac{dp}{-\,dV/V}$$

Given: $V_1 = 3600$ cm³ at $p_1 = 4$ MN/m², $V_2 = 3200$ cm³ at $p_2 = 8$ MN/m².

$$\begin{aligned} dp &= 8 - 4 = 4\ \text{MN/m}^2 \\ dV &= 3200 - 3600 = -400\ \text{cm}^3 \\ -\frac{dV}{V} &= \frac{400}{3600} = 0.1111 \\ K &= \frac{4}{0.1111} = 36\ \text{MN/m}^2 \end{aligned}$$

(The initial volume 3600 cm³ is used as the reference volume.)

Answer: $K = 36$ MN/m² $= 3.6 \times 10^7$ N/m².

Pascal's law: the pressure at a point in a fluid at rest is the same in all directions, i.e. $p_x = p_y = p_z$.

Proof

Consider a small wedge-shaped fluid element at rest, of unit width perpendicular to the paper, with sides $dx$ (horizontal), $dy$ (vertical) and inclined face $ds$ at angle $\theta$ to the vertical.

          |\
          | \  ds  (pressure p_z
   p_x -> |  \       normal to it)
       dy |   \
          |    \
          |_____\
            dx
            ^ p_y

Pressures: $p_x$ on the vertical face, $p_y$ on the bottom face, $p_z$ on the inclined face. Geometry: $dy = ds\cos\theta$, $dx = ds\sin\theta$ (with $\theta$ the angle between the inclined face and the vertical).

Since the fluid is at rest, there are no shear forces; only pressure forces and weight act.

Horizontal equilibrium:

$$p_x (dy \cdot 1) - p_z (ds \cdot 1)\cos\theta = 0$$

Using $ds\cos\theta = dy$: $p_x\,dy = p_z\,dy \Rightarrow p_x = p_z$.

Vertical equilibrium:

$$p_y (dx\cdot 1) - p_z (ds\cdot 1)\sin\theta - \rho g \frac{dx\, dy}{2}\cdot 1 = 0$$

Using $ds\sin\theta = dx$:

$$p_y\,dx - p_z\,dx - \rho g \frac{dx\,dy}{2} = 0$$

As the element shrinks to a point ($dy \to 0$), the weight term (second-order small) vanishes, so $p_y = p_z$.

Therefore

$$p_x = p_y = p_z$$

Since $\theta$ was arbitrary, the pressure at a point is the same in every direction.

Applications

Hydraulic press, hydraulic jack and brakes, where a small force on a small piston gives a large force on a large piston ($F_2 = F_1 A_2/A_1$).

Use the manometer principle: pressures at the same level in the same continuous liquid are equal. Take the datum at the lower mercury level, which lies $x$ below Z–Z.

Given: $p_M = 105$ kN/m², $S_1 = 1.594$ (CCl₄); $p_N = 175$ kN/m², $S_2 = 0.8$ (oil); $S_{Hg} = 13.6$; $g = 9.81$ m/s².

Since $p_N > p_M$, mercury is pushed down in the limb under N and up in the limb under M. So Z–Z (higher mercury level) is in M's limb and the lower level is in N's limb.

• Height of M above Z–Z $= 2.5 + 1.5 = 4.0$ m.
• Oil column in N's limb $= 1.5 + x$.

Pressure balance at the datum (lower mercury level)

Left limb (M side):

$$p_M + \rho_1 g (4.0) + \rho_{Hg} g\, x$$

Right limb (N side):

$$p_N + \rho_2 g (1.5 + x)$$

Equating (in kN/m², $\rho g$ for water $= 9.81$ kN/m³):

$$\begin{aligned} 105 + 1.594(9.81)(4.0) + 13.6(9.81)x &= 175 + 0.8(9.81)(1.5 + x) \\ 105 + 62.549 + 133.416x &= 175 + 11.772 + 7.848x \\ 125.568\,x &= 186.772 - 167.549 = 19.223 \\ x &= 0.153\ \text{m} \end{aligned}$$

Answer: the difference between the mercury levels is $x \approx 0.153$ m $= 153$ mm (mercury higher in the limb below pipe M).

Write the pressure balance at the datum X–X (mercury surface in the left limb), moving through each fluid column.

Assumed reading of the figure: the air–water surface on the A side is 200 mm above X–X. On the B side, mercury stands 100 mm above X–X, and B is 550 mm above X–X, so the oil column is $550 - 100 = 450$ mm.

Given: $p_A = 60$ kN/m² (air; its weight is neglected), water $S_1 = 1$, oil $S_2 = 0.9$, mercury $S = 13.6$, $\rho_w g = 9.81$ kN/m³.

   A (air)               B (oil)
   |~~~~~| 200 mm water    |  450 mm oil
   |     |                 |------ Hg top (+100)
 X-|-----|------------ X   | 100 mm Hg
   |__ mercury ____________|

Pressure at X–X, left limb

$$p_L = p_A + \rho_w g(0.20) = 60 + 9.81(0.20) = 60 + 1.962 = 61.962\ \text{kN/m}^2$$

Pressure at X–X, right limb

$$p_R = p_B + 0.9(9.81)(0.45) + 13.6(9.81)(0.10) = p_B + 3.973 + 13.342$$

Equate

$$\begin{aligned} 61.962 &= p_B + 17.315 \\ p_B &= 44.647\ \text{kN/m}^2 \end{aligned}$$

Answer: $p_B \approx 44.65$ kN/m² (gauge), i.e. about 15.35 kN/m² less than at A. (If the figure's heights are placed differently, use the same column-by-column balance with those heights.)

Use the manometer rule: start at A, add $\rho g h$ when going down a column, subtract when going up, and end at B.

Assumed reading of the figure: A (oil) is 12 cm above the oil–mercury surface in the left limb. Mercury in the right limb stands 4 cm higher than in the left limb. B (water) is 3 cm above the right mercury surface.

Given: oil $S = 0.9$, mercury $S = 13.6$, water $S = 1$; $\rho_w g = 9810$ N/m³.

  A (oil)
  |                         B (water)
  | 12 cm oil               | 3 cm water
  |              Hg  -------+ (right Hg level)
  |              | 4 cm
  +-- Hg level --+
  |_____ mercury ______|

Manometer equation

$$p_A + \rho_{oil} g(0.12) - \rho_{Hg} g(0.04) - \rho_w g(0.03) = p_B$$ $$\begin{aligned} p_A - p_B &= 9810\,[13.6(0.04) + 1(0.03) - 0.9(0.12)] \\ &= 9810\,[0.544 + 0.030 - 0.108] \\ &= 9810 \times 0.466 = 4571\ \text{N/m}^2 \end{aligned}$$

Answer: $p_A - p_B \approx 4.57$ kN/m² (pressure at A is higher). If the column heights are arranged differently in the figure, the same step-by-step balance applies with those heights.

Equate the pressures on the horizontal plane D–D (the lower oil level / top of the manometric liquid in the left limb), since below it the same manometric liquid is continuous.

Given: oil $S_1 = 0.875$, manometric liquid $S = 1.25$, water $S = 1$. Centre of A is 1.5 m above D–D. Manometric liquid in the right limb stands 2.25 m above D–D (level E–F). The centre of B is taken at level E–F, as no other height is given.

   A (oil) o                    o B (water) -- E-F
           |                    |
           | 1.5 m oil          | 2.25 m of
           |                    | liquid S=1.25
   D ------+---- S = 1.25 ------+------ D

Pressure at D–D, left limb

$$p_A + 0.875 \times 9.81 \times 1.5 = p_A + 12.876\ \text{kN/m}^2$$

Pressure at D–D, right limb

$$p_B + 1.25 \times 9.81 \times 2.25 = p_B + 27.591\ \text{kN/m}^2$$

Equate

$$\begin{aligned} p_A + 12.876 &= p_B + 27.591 \\ p_A - p_B &= 27.591 - 12.876 = 14.715\ \text{kN/m}^2 \end{aligned}$$

Answer: $p_A - p_B = 14.715$ kN/m² (≈ 14.72 kPa); pipe A (oil) is at the higher pressure.

Go from A to B through each column: add $\rho g h$ going down, subtract going up.

Assumed reading of the figure (heights above the bottom datum): centre of A = 0.90 m. In the left mercury U-tube, the mercury surface is at 0.40 m in A's limb and 0.60 m in the other limb. Oil fills the inverted U-tube, down to the right U-tube's mercury surface, taken at 0.40 m. The right U-tube's mercury in B's limb is at 0.50 m. B is 0.80 m above that, i.e. at 1.30 m.

Given: $S_{oil} = 0.8$, $S_{Hg} = 13.6$, water $S = 1$, $\rho_w g = 9.81$ kN/m³.

Manometer equation

$$p_A + 4.905 - 26.683 + 1.570 - 13.342 - 7.848 = p_B$$ $$p_A - p_B = 41.398\ \text{kN/m}^2$$

Answer: $p_A - p_B \approx 41.4$ kN/m² (A at higher pressure), for the heights read as above. The method is the same for any other placement of the marked heights.

In an inclined limb only the vertical height of a column produces pressure: $h = L\sin\theta$.

Assumed reading of the figure: liquid of $S = 0.8$ stands 550 mm vertically above the mercury surface in A's limb, and water stands 300 mm vertically above the mercury surface in B's limb. The mercury surface lies 130 mm along the 45° limb on the A side and 230 mm along the 60° limb on the B side, measured from the common bottom of the U-tube.

Vertical heights of the mercury surfaces

$$\begin{aligned} h_A &= 0.130\sin 45^\circ = 0.0919\ \text{m} \\ h_B &= 0.230\sin 60^\circ = 0.1992\ \text{m} \\ \Delta h &= 0.1992 - 0.0919 = 0.1073\ \text{m} \end{aligned}$$

So mercury on the B side stands 0.1073 m higher.

Pressure balance at the level of the lower (A-side) mercury surface

$$p_A + \rho_{0.8}g(0.55) = p_B + \rho_w g(0.30) + \rho_{Hg}g(0.1073)$$ $$\begin{aligned} p_A - p_B &= 9.81\,[0.30 + 13.6(0.1073) - 0.8(0.55)] \\ &= 2.943 + 14.310 - 4.316 \\ &= 12.94\ \text{kN/m}^2 \end{aligned}$$

Answer: $p_A - p_B \approx 12.94$ kN/m² (container A at the higher pressure), for the reading above. Use the same $L\sin\theta$ method if the figure places the lengths differently.

The pressure difference between two points in a pipe (or between two pipes) is measured with a differential manometer: a U-tube with a heavy manometric liquid (usually mercury) whose two limbs are connected to the two points.

U-tube differential manometer (points at different levels)

     A o                     
       |  pipe liquid S1      o B
       |                      |  pipe liquid S2
       | h_A                  | h_B
       |       h              |
   X --+-----.  .-------------+
       |  Hg |  | Hg surface
        \____|__|____/
          mercury (S_g)

Let A be $h_A$ above the datum X–X (lower mercury surface), B be $h_B$ above the higher mercury surface, and $h$ the mercury difference.

Pressure at X–X in the left limb $= p_A + \rho_1 g h_A$. Pressure at X–X in the right limb $= p_B + \rho_2 g h_B + \rho_g g h$.

$$p_A - p_B = \rho_g g h + \rho_2 g h_B - \rho_1 g h_A$$

Both points at the same level, same liquid

If A and B are in the same horizontal pipe carrying liquid of density $\rho$:

$$p_A - p_B = g h(\rho_g - \rho)$$

For water and mercury: $p_A - p_B = 12.6\,\rho_w g h$.

Inverted U-tube manometer

For small pressure differences a light liquid (air or oil) is used at the top of an inverted U-tube:

$$p_A - p_B = \rho_1 g h_1 - \rho_2 g h_2 - \rho_s g h$$

Steps of measurement

1. Connect each limb to the pipe tapping, filling the leads with the pipe liquid (no air bubbles).
2. Let the manometer settle and read the level difference $h$.
3. Choose a datum at the lower interface and equate pressures in both limbs.

Other devices: Bourdon gauge pair or electronic differential pressure transducers for large differences.

Use of a manometer

A manometer measures pressure at a point (simple manometer) or the difference of pressure between two points (differential manometer) by balancing it against a column of liquid. Used for pipes, venturimeters, orifice meters and laboratory flow measurement; it is simple, cheap and accurate.

Inverted U-tube differential manometer

It is an inverted U-tube whose top part contains a light fluid (air or a light oil) lighter than the liquid in the pipes. It is used to measure small pressure differences, where a heavy liquid like mercury would give too small a reading.

        .---------------.
        | light liquid  |  (air or oil, rho_s)
  X ----+------.        |
        |      |  h     |
        |      '--------+---- right interface
   h_1  | liquid        |
        | rho_1         | h_2   liquid rho_2
        o A             |
                        o B

Let X–X be the left (higher) interface, A be $h_1$ below it, the right interface be $h$ lower than X–X, and B be $h_2$ below the right interface.

Working principle

Pressure at level X–X in the left limb:

$$p_A - \rho_1 g h_1$$

Pressure at level X–X in the right limb (go up $h_2$ of pipe liquid from B, then up $h$ of light liquid):

$$p_B - \rho_2 g h_2 - \rho_s g h$$

Equating:

$$p_A - p_B = \rho_1 g h_1 - \rho_2 g h_2 - \rho_s g h$$

If air is used at the top, $\rho_s g h \approx 0$.

Example: water in both pipes, oil $S = 0.8$ at top, $h_1 = 0.30$ m, $h_2 = 0.20$ m, $h = 0.12$ m: $p_A - p_B = 9.81(0.30 - 0.20 - 0.8\times0.12) = 0.039$ kN/m², a very small difference that an inverted manometer can read well.

Total pressure is the resultant force exerted by a static fluid on a surface, acting normal to it. Centre of pressure is the point on the surface where this resultant force acts.

Derivation for a vertical plane surface

Let a plane surface of area $A$ be immersed vertically in a liquid of density $\rho$. Let $\bar h$ = depth of centroid G and $h^*$ = depth of centre of pressure P below the free surface.

 free surface ===========================
        |                |
        | h-bar          | h*
  +-----G------+         |
  | ---strip-- | dh at h |
  |        P   |  <------'
  +------------+

Total pressure: for a horizontal strip of area $dA$ at depth $h$, $dF = \rho g h\, dA$.

$$F = \rho g \int h\, dA = \rho g A \bar h$$

Centre of pressure: moment of all strip forces about the free surface equals the moment of $F$:

$$F h^* = \int \rho g h^2\, dA = \rho g I_o$$

By the parallel axis theorem, $I_o = I_G + A\bar h^2$, where $I_G$ is the second moment of area about the horizontal axis through G. Therefore

$$\begin{aligned} h^* &= \frac{\rho g (I_G + A\bar h^2)}{\rho g A\bar h} \\ &= \bar h + \frac{I_G}{A\bar h} \end{aligned}$$

Proof that P is below G

• $I_G$ is a second moment of area, always positive.
• $A$ and $\bar h$ are positive for a submerged surface.

So $\dfrac{I_G}{A\bar h} > 0$ and hence $h^* > \bar h$: the centre of pressure always lies below the centroid.

Physical reason: pressure increases linearly with depth, so the lower part of the surface carries more force than the upper part, pulling the resultant below the centroid.

Note: as the surface goes deeper ($\bar h$ large), $I_G/(A\bar h) \to 0$ and P approaches G. Example: rectangle $b \times d$ with top edge at the surface: $h^* = d/2 + \frac{bd^3/12}{bd \cdot d/2} = \frac{2d}{3}$, which is below $\bar h = d/2$.

On a curved surface the pressure acts normal to each element, so the elemental forces have different directions. The total force is found by resolving it into horizontal and vertical components.

   free surface ==================
        :  water above   |
        :  (volume ABCD) |  F_V = weight of
        :                |  this water
        D________________A
                         \    <- curved surface AB
                          \
                           B
     F_H = force on vertical projection of AB

Horizontal component

$$F_H = \rho g A_v \bar h_v$$

• $A_v$ = area of the vertical projection of the curved surface, $\bar h_v$ = depth of its centroid.
• $F_H$ acts through the centre of pressure of this projected area, at depth $\bar h_v + \dfrac{I_G}{A_v \bar h_v}$.

Vertical component

$$F_V = \rho g V$$

• $V$ = volume of liquid (real or imaginary) lying vertically above the curved surface up to the free surface.
• $F_V$ acts through the centroid of that volume.
• If the liquid is above the surface, $F_V$ acts downward. If the liquid is below the surface, $F_V$ acts upward and equals the weight of the imaginary liquid above it.

Resultant

$$F = \sqrt{F_H^2 + F_V^2}, \qquad \tan\theta = \frac{F_V}{F_H}$$

For a circular (cylindrical) surface, every elemental force is normal to the surface and passes through the centre, so the resultant also passes through the centre of curvature.

Applications: radial (Tainter) gates, sector gates and drum gates on spillways, curved dam faces and pipe bends.

The plate is not vertical because its depth range (4.5 − 2.5 = 2 m) is less than its diameter, so it is inclined. Use the formulas for an inclined plane surface.

Given: $d = 2.2$ m, $S = 1.2$ so $\rho = 1200$ kg/m³, depths of top and bottom points $= 2.5$ m and $4.5$ m.

  free surface =====================
         :  2.5 m
         :     * top point
         :    / \
         :   /   \   plate, d = 2.2 m
         :  /  G  \   inclined at theta
         :   \   /
         :    \ /
         :     * bottom point (4.5 m)

Inclination and centroid depth

$$\begin{aligned} \sin\theta &= \frac{4.5 - 2.5}{2.2} = 0.9091 \\ \bar h &= \frac{2.5 + 4.5}{2} = 3.5\ \text{m} \end{aligned}$$

Area and second moment of area

$$\begin{aligned} A &= \frac{\pi}{4}(2.2)^2 = 3.8013\ \text{m}^2 \\ I_G &= \frac{\pi d^4}{64} = \frac{\pi (2.2)^4}{64} = 1.1499\ \text{m}^4 \end{aligned}$$

Total pressure

$$\begin{aligned} F &= \rho g A \bar h = 1200 \times 9.81 \times 3.8013 \times 3.5 \\ &= 156\,622\ \text{N} = 156.62\ \text{kN} \end{aligned}$$

Centre of pressure

$$\begin{aligned} h^* &= \bar h + \frac{I_G \sin^2\theta}{A \bar h} \\ &= 3.5 + \frac{1.1499 \times 0.8264}{3.8013 \times 3.5} \\ &= 3.5 + 0.0714 = 3.571\ \text{m} \end{aligned}$$

Answer: total pressure $F \approx 156.62$ kN; centre of pressure at a depth of $3.571$ m below the free surface (on the plate's vertical diameter, below the centroid).

The plate is not vertical because its depth range (3 − 1 = 2 m) is less than its diameter, so it is inclined. Use the formulas for an inclined plane surface.

Given: $d = 2.5$ m, $S = 0.9$ so $\rho = 900$ kg/m³, depths of top and bottom points $= 1$ m and $3$ m.

  free surface =====================
         :  1 m
         :     * top point
         :    / \
         :   /   \   plate, d = 2.5 m
         :  /  G  \   inclined at theta
         :   \   /
         :    \ /
         :     * bottom point (3 m)

Inclination and centroid depth

$$\begin{aligned} \sin\theta &= \frac{3 - 1}{2.5} = 0.8000 \\ \bar h &= \frac{1 + 3}{2} = 2\ \text{m} \end{aligned}$$

Area and second moment of area

$$\begin{aligned} A &= \frac{\pi}{4}(2.5)^2 = 4.9087\ \text{m}^2 \\ I_G &= \frac{\pi d^4}{64} = \frac{\pi (2.5)^4}{64} = 1.9175\ \text{m}^4 \end{aligned}$$

Total pressure

$$\begin{aligned} F &= \rho g A \bar h = 900 \times 9.81 \times 4.9087 \times 2 \\ &= 86\,679\ \text{N} = 86.68\ \text{kN} \end{aligned}$$

Centre of pressure

$$\begin{aligned} h^* &= \bar h + \frac{I_G \sin^2\theta}{A \bar h} \\ &= 2 + \frac{1.9175 \times 0.6400}{4.9087 \times 2} \\ &= 2 + 0.1250 = 2.125\ \text{m} \end{aligned}$$

Answer: total pressure $F \approx 86.68$ kN; centre of pressure at a depth of $2.125$ m below the free surface (on the plate's vertical diameter, below the centroid).

The plate is not vertical because its depth range (3 − 1 = 2 m) is less than its diameter, so it is inclined. Use the formulas for an inclined plane surface.

Given: $d = 2.6$ m, water, $\rho = 1000$ kg/m³, depths of top and bottom points $= 1$ m and $3$ m.

  free surface =====================
         :  1 m
         :     * top point
         :    / \
         :   /   \   plate, d = 2.6 m
         :  /  G  \   inclined at theta
         :   \   /
         :    \ /
         :     * bottom point (3 m)

Inclination and centroid depth

$$\begin{aligned} \sin\theta &= \frac{3 - 1}{2.6} = 0.7692 \\ \bar h &= \frac{1 + 3}{2} = 2\ \text{m} \end{aligned}$$

Area and second moment of area

$$\begin{aligned} A &= \frac{\pi}{4}(2.6)^2 = 5.3093\ \text{m}^2 \\ I_G &= \frac{\pi d^4}{64} = \frac{\pi (2.6)^4}{64} = 2.2432\ \text{m}^4 \end{aligned}$$

Total pressure

$$\begin{aligned} F &= \rho g A \bar h = 1000 \times 9.81 \times 5.3093 \times 2 \\ &= 104\,168\ \text{N} = 104.17\ \text{kN} \end{aligned}$$

Centre of pressure

$$\begin{aligned} h^* &= \bar h + \frac{I_G \sin^2\theta}{A \bar h} \\ &= 2 + \frac{2.2432 \times 0.5917}{5.3093 \times 2} \\ &= 2 + 0.1250 = 2.125\ \text{m} \end{aligned}$$

Answer: total pressure $F \approx 104.17$ kN; centre of pressure at a depth of $2.125$ m below the free surface (on the plate's vertical diameter, below the centroid).

For a vertical plane surface: $F = \rho g A \bar h$ and $h^* = \bar h + \dfrac{I_G}{A\bar h}$.

Given: height $d = 2$ m, width $b = 3$ m, top edge 1 m below the surface; water, $\rho = 1000$ kg/m³.

 free surface ===================
           | 1 m
       +---------+  -
       |    G    |  2 m   (b = 3 m)
       |    P    |
       +---------+  -

Data

$$\begin{aligned} A &= 3 \times 2 = 6\ \text{m}^2 \\ \bar h &= 1 + \frac{2}{2} = 2\ \text{m} \\ I_G &= \frac{b d^3}{12} = \frac{3 \times 2^3}{12} = 2\ \text{m}^4 \end{aligned}$$

Total pressure

$$F = 1000 \times 9.81 \times 6 \times 2 = 117\,720\ \text{N} = 117.72\ \text{kN}$$

Centre of pressure

$$\begin{aligned} h^* &= \bar h + \frac{I_G}{A\bar h} = 2 + \frac{2}{6 \times 2} \\ &= 2 + 0.1667 = 2.167\ \text{m} \end{aligned}$$

Answer: $F = 117.72$ kN; centre of pressure is 2.167 m below the free surface (0.167 m below the centroid, on the vertical centre line).

The plate is vertical with the 3 m side horizontal, so width $b = 3$ m and depth $d = 5$ m. Use $F = \rho g A\bar h$ and $h^* = \bar h + \dfrac{I_G}{A\bar h}$.

$$A = 3 \times 5 = 15\ \text{m}^2, \qquad I_G = \frac{b d^3}{12} = \frac{3 \times 5^3}{12} = 31.25\ \text{m}^4$$

(i) Top edge flush with the water surface

$$\begin{aligned} \bar h &= \frac{5}{2} = 2.5\ \text{m} \\ F &= 1000 \times 9.81 \times 15 \times 2.5 = 367\,875\ \text{N} = 367.88\ \text{kN} \\ h^* &= 2.5 + \frac{31.25}{15 \times 2.5} = 2.5 + 0.8333 = 3.333\ \text{m} \end{aligned}$$

(Check: $h^* = \frac{2}{3}d = \frac{2}{3}\times 5 = 3.333$ m.)

(ii) Top edge 2 m below the water surface

$$\begin{aligned} \bar h &= 2 + 2.5 = 4.5\ \text{m} \\ F &= 1000 \times 9.81 \times 15 \times 4.5 = 662\,175\ \text{N} = 662.18\ \text{kN} \\ h^* &= 4.5 + \frac{31.25}{15 \times 4.5} = 4.5 + 0.4630 = 4.963\ \text{m} \end{aligned}$$

Answer: (i) $F = 367.88$ kN at 3.333 m depth; (ii) $F = 662.18$ kN at 4.963 m depth. The deeper the plate, the closer the centre of pressure comes to the centroid.

For an inclined plane surface: $F = \rho g A\bar h$ and $h^* = \bar h + \dfrac{I_G \sin^2\theta}{A\bar h}$, where $\theta$ is the angle with the horizontal (free surface).

Given: width $b = 0.6$ m, length along the slope $d = 1.2$ m, $\theta = 40^\circ$, top edge 0.70 m below the surface; water.

 free surface ===========================
          : 0.70 m
          +.                 
            \  .   plate 1.2 m long,
             \    .  0.6 m wide
              \ G    .
               \    P   .  40 deg to
                +----------- horizontal

Centroid depth

$$\bar h = 0.70 + \frac{1.2}{2}\sin 40^\circ = 0.70 + 0.6 \times 0.6428 = 1.0857\ \text{m}$$

Area and second moment

$$\begin{aligned} A &= 0.6 \times 1.2 = 0.72\ \text{m}^2 \\ I_G &= \frac{b d^3}{12} = \frac{0.6 \times 1.2^3}{12} = 0.0864\ \text{m}^4 \end{aligned}$$

Total pressure force

$$F = 1000 \times 9.81 \times 0.72 \times 1.0857 = 7668\ \text{N} = 7.67\ \text{kN}$$

Centre of pressure

$$\begin{aligned} h^* &= \bar h + \frac{I_G \sin^2\theta}{A\bar h} \\ &= 1.0857 + \frac{0.0864 \times 0.4132}{0.72 \times 1.0857} \\ &= 1.0857 + 0.0457 = 1.131\ \text{m} \end{aligned}$$

Measured along the plate from the top edge: $\dfrac{1.131 - 0.70}{\sin 40^\circ} = 0.671$ m (the centroid is at 0.600 m).

Answer: total pressure force $\approx 7.67$ kN; centre of pressure at a vertical depth of 1.131 m below the free surface, i.e. 0.671 m down the plate from its top edge, on the centre line.

The plate is vertical and its axis of symmetry (the altitude) is horizontal, at 6 m depth. So the 4 m base is vertical, and the centroid lies on the axis at depth 6 m.

 free surface ==========================
                 : 6 m
        |\       :
   4 m  |  \     v
  base  |----G-----> axis (horizontal)
 (vert.)|  /   6 m altitude
        |/

Given: base $b = 4$ m, altitude $h = 6$ m, water.

Area and centroid

$$A = \frac{1}{2}\times 4 \times 6 = 12\ \text{m}^2, \qquad \bar h = 6\ \text{m}$$

The centroid lies on the axis, at $h/3 = 2$ m from the base (4 m from the apex).

Total pressure

$$F = \rho g A\bar h = 1000 \times 9.81 \times 12 \times 6 = 706\,320\ \text{N} = 706.32\ \text{kN}$$

Second moment about the horizontal centroidal axis (the axis of symmetry)

The triangle is two right triangles, each of base 6 m (along the axis) and height 2 m, on either side of the axis:

$$I_G = 2 \times \frac{6 \times 2^3}{12} = \frac{h b^3}{48} = \frac{6 \times 4^3}{48} = 8\ \text{m}^4$$

Centre of pressure (depth)

$$\begin{aligned} h^* &= \bar h + \frac{I_G}{A\bar h} = 6 + \frac{8}{12 \times 6} \\ &= 6 + 0.111 = 6.111\ \text{m} \end{aligned}$$

Horizontally, the plate is symmetric about the axis (product of inertia about the centroidal axes is zero), so the centre of pressure lies at the same horizontal position as the centroid: 2 m from the base.

Answer: $F = 706.32$ kN, acting at 6.111 m depth (0.111 m below the axis), 2 m from the vertical base (4 m from the apex).