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Fluids in motion

Publisher Summary

This chapter focuses on the fluids in motion. Bernoulli’s equation is the principle of the conservation of energy applied to fluids in motion. The two significant differences between this application of the principle of the conservation of energy and the application of the principle to solids in motion are: this application is to a steady process mass (or weight) flowing per second has to be considered instead of a given fixed mass or weight and a third form of energy, that is, pressure energy, must be considered. The corresponding form of energy in dealing with solids, strain energy, is only occasionally met with. The force exerted by a jet of water on a plate is, from Newton’s third law of motion, equal and opposite to the force exerted by the plate on the water. From Newton’s second law this is equal to the rate of change of momentum of water.

Bernoulli’s equation

1. This is the principle of the conservation of energy applied to fluids in motion.

$\frac{p}{w} + \frac{v^{2}}{2 g} + Z = constant$ $\frac{p}{w} + \frac{v^{2}}{2 g} + Z = constant$ (1)

(1)

$or \frac{p}{w} + \frac{v^{2}}{g} + Z g = constant$ $or \frac{p}{w} + \frac{v^{2}}{g} + Z g = constant$ (2)

(2)

All of the quantities on the left of each equation apply to a specified fixed point in the moving fluid:

p = pressure (gauge pressure unless otherwise specified)

w = specific weight (weight per unit volume)

v = velocity

g = acceleration due to gravity

Z = height above some specified datum

ρ = density

The two significant differences between this application of the principle of the conservation of energy and the application of the principle to solids in motion are:

(i) this application is to a steady process mass (or weight) flowing per second has to be considered instead of a given fixed mass or weight;

(ii) a third form of energy, that is, pressure energy, must be considered; the corresponding form of energy in dealing with solids, strain energy, is only occasionally met with.

Each of the terms in equation (1) represents energy per unit weight of fluid. The basic unit of each term is the metre,

$i .e . \frac{Nm}{N} or \frac{J}{N} .$ $i .e . \frac{Nm}{N} or \frac{J}{N} .$

The basic unit for pressure isthe same as for stress. N/m² or pascal (Pa). Each term in the equation is called a head.

$\begin{array}{l} \frac{p}{w} is the pressure head (\frac{{N/m}^{2}}{{N/m}^{3}} = m) \\ \frac{υ^{2}}{2 g} is the velocity or kinetic head (\frac{{m/s}^{2}}{{m/s}^{2}} = m) \\ z is the potential head (m) . \end{array}$ $\begin{array}{l} \frac{p}{w} is the pressure head (\frac{{N/m}^{2}}{{N/m}^{3}} = m) \\ \frac{υ^{2}}{2 g} is the velocity or kinetic head (\frac{{m/s}^{2}}{{m/s}^{2}} = m) \\ z is the potential head (m) . \end{array}$

The sum of the three heads is called total head (H).

Equation (2) gives energies per unit mass.

In practice, as with solids, some energy is lost, converted into heat. The estimation of this loss is an important aspect of the mechanics of fluids at a higher level.

If a pipe is filled by moving liquid the volumetric rate of flow $\dot{V}$ $\dot{V}$ , that is, the volume passing per second, must be the same at every section: V = A₁v₂ = A₂v₂ where A₁ and v₁ are the cross-sectional area and velocity at one selected section and A₂ and v₂ are area and velocity at a second section. The equation A₁v₁ = A₂v₂ is called the equation of continuity.

The basic unit of volumetric rate of flow $\dot{V}$ $\dot{V}$ is m³/s which is a large unit. The smaller unit, litre per second, is often preferred.

${1 litre = 1000 cm}^{3} {=1 m}^{3} {x l 0}^{-3}$ ${1 litre = 1000 cm}^{3} {=1 m}^{3} {x l 0}^{-3}$

(For very low rates of flow litres per minute units may be preferred.)

For example, let the rate of flow of water through a pipe of 32 mm diameter be 2.8 litres/s. The total head at a point where the pressure is 28.4 kPa with reference to a datum 1.84 m below is determined as follows:

$Velocity of flow υ = \frac{V}{A} = \frac{2 {.8}^{\times} 10^{-3} m^{3/s}}{\frac{π}{4} {(32)}^{2} \times 10^{-6} m^{2}} = 3 .4815 m/s .$ $Velocity of flow υ = \frac{V}{A} = \frac{2 {.8}^{\times} 10^{-3} m^{3/s}}{\frac{π}{4} {(32)}^{2} \times 10^{-6} m^{2}} = 3 .4815 m/s .$

$The velocity head, \frac{υ^{2}}{2 g} = \frac{{(3 .4815)}^{2}}{(2)(9 .81)} =0 .6178 m .$ $The velocity head, \frac{υ^{2}}{2 g} = \frac{{(3 .4815)}^{2}}{(2)(9 .81)} =0 .6178 m .$

The specific weight

$ω = ρ {g=(1000kg/m}^{3})(9 {.81 m/s}^{2}) = 9 {.81 kN/m}^{3} .$ $ω = ρ {g=(1000kg/m}^{3})(9 {.81 m/s}^{2}) = 9 {.81 kN/m}^{3} .$

$The pressure head, \frac{p}{w} = \frac{28 .4 kPa}{9 {.81 kN/m}^{2}} =2 .895 m .$ $The pressure head, \frac{p}{w} = \frac{28 .4 kPa}{9 {.81 kN/m}^{2}} =2 .895 m .$

The potential head Z = 1.84 m.

$\begin{array}{l} The pressure head H= \frac{p}{w} + \frac{υ^{2}}{2 g} + z = 2 .895 +0 .61784+1 .84=5353 m . \end{array}$ $\begin{array}{l} The pressure head H= \frac{p}{w} + \frac{υ^{2}}{2 g} + z = 2 .895 +0 .61784+1 .84=5353 m . \end{array}$

Flow through orifices

2. Water issuing from a tank as a horizontal jet, as shown in Figure 48.1, has a velocity head only, if the datum is taken at the level of orifice. Water which will eventually form the jet starts at the top of the tank with a potential head only, h. Equating initial potential and final velocity heads,

$\begin{array}{l} \frac{υ^{2}}{2 g} + h \end{array}$ $\begin{array}{l} \frac{υ^{2}}{2 g} + h \end{array}$

In practice, some energy loss occurs. The ratio

$\frac{actual velocity of jet}{theoretical velocity of jet}$ $\frac{actual velocity of jet}{theoretical velocity of jet}$

is called the coefficient of velocity of the orifice (C_c). The actual velocity can be obtained from accurate observation of co-ordinates x and y of points on the jet trajectory.

The diameter of the jet is also found in practice to be less than the diameter of the orifice. The ratio:

$\frac{cross-sectional area of jet}{area of orifice}$ $\frac{cross-sectional area of jet}{area of orifice}$

is called the coefficient of contraction of the orifice (C_c). The ratio

$\frac{actual rate of discharge}{theoretical rate of discharge}$ $\frac{actual rate of discharge}{theoretical rate of discharge}$

is called the coefficient of discharge of the orifice (C_d). But

$\begin{array}{l} \frac{actual rate of discharge}{theoretical rate of discharge} = \frac{{actual velocity of jet}^{x} c .s .a . of jet}{{theoretical velocity of jet}^{x} area of orifice} \\ i .e {. C}_{d} {=C}_{v} {xC}_{c} . \end{array}$ $\begin{array}{l} \frac{actual rate of discharge}{theoretical rate of discharge} = \frac{{actual velocity of jet}^{x} c .s .a . of jet}{{theoretical velocity of jet}^{x} area of orifice} \\ i .e {. C}_{d} {=C}_{v} {xC}_{c} . \end{array}$

For example, an orifice in the bottom of a water tank has a diameter of 12.5 mm. Assuming coefficients of contraction and velocity of 0.64 and 0.96 respectively, the depth of water required in the tank to give a rate of discharge through the orifice of 0.25 litres per second is determined as follows:

The theoretical rate of discharge from the tank,

$\begin{array}{l} \dot{V} = A υ = (\frac{π d^{2}}{4}) \sqrt (2 g h) \\ = \frac{π {(0 .0125)}^{2}}{4} \sqrt [(2) (9.81) h] \\ = 0 .00054357 \sqrt h m^{3} /s \\ =0 .54357 \sqrt h litres/s, where A is in metres . \end{array}$ $\begin{array}{l} \dot{V} = A υ = (\frac{π d^{2}}{4}) \sqrt (2 g h) \\ = \frac{π {(0 .0125)}^{2}}{4} \sqrt [(2) (9.81) h] \\ = 0 .00054357 \sqrt h m^{3} /s \\ =0 .54357 \sqrt h litres/s, where A is in metres . \end{array}$

Coefficient of discharge C_d = C_v × C_c = 0.96 × 0.64 = 0.6144.

$C_{d} = \frac{actual rate of discharge}{theoretical rate of discharge},$ $C_{d} = \frac{actual rate of discharge}{theoretical rate of discharge},$

thus (actual rate of discharge) = (C_d) (theoretical rate of discharge)

$\begin{array}{l} = (0 .6144) (0 .54357 √h) \\ = 0 .33397 √h litres/s . \end{array}$ $\begin{array}{l} = (0 .6144) (0 .54357 √h) \\ = 0 .33397 √h litres/s . \end{array}$

Hence, 0.25 = 0.33397 $\sqrt{h}$ $\sqrt{h}$ , from which, depth of water

$h = {(\frac{0 .25}{0.33397})}^{2} = 0 .5604 m=560 .4 mm .$ $h = {(\frac{0 .25}{0.33397})}^{2} = 0 .5604 m=560 .4 mm .$

Impact of a jet

3. The force exerted by a jet of water on a plate is, from Newton’s third law of motion, equal and opposite to the force exerted by the plate on the water. From Newton’s second law this is equal to the rate of change of momentum of water.

$Rate of change of momentum = \frac{mass x change of velocity}{time}$ $Rate of change of momentum = \frac{mass x change of velocity}{time}$

In dealing with solids this is interpreted as

$mass = \frac{change of velocity}{time} = mass \times acceleration$ $mass = \frac{change of velocity}{time} = mass \times acceleration$

In dealing with the continuous process of fluid flow it must be interpreted as

$\frac{mass}{time} \times change of velocity$ $\frac{mass}{time} \times change of velocity$

= mass rate of flow × change of velocity (Mv).

In the case of the jet striking a flat plate at right angles (see Figure 48.2) the final velocity in the original direction is zero, so that v is the change of velocity in this direction. Also, if d is the diameter of the jet:

$\begin{array}{l} M = \dot{V} p = A υ ρ \\ = \frac{π d^{2}}{4} υ ρ \end{array}$ $\begin{array}{l} M = \dot{V} p = A υ ρ \\ = \frac{π d^{2}}{4} υ ρ \end{array}$

Force on plate,

$F = \dot{M} v = \frac{π d^{2}}{4} v^{2} ρ .$ $F = \dot{M} v = \frac{π d^{2}}{4} v^{2} ρ .$

The force will be in newtons if the jet diameter is in metres, the jet velocity in metres per second and density in kilograms per metre cubed (mass rate of flow $\dot{M}$ $\dot{M}$ in kg/s).

For example, let a jet of water with a diameter of 12.5 mm and a velocity of 40 m/s strike a stationary flat plate at right angles.

Mass rate of flow

$\begin{array}{l} M = A υ ρ = \frac{π d^{2}}{4} υ ρ = \frac{π {(0 .0125)}^{2}}{4} (40) (1000) \\ = 4 .9087 kg/s . \end{array}$ $\begin{array}{l} M = A υ ρ = \frac{π d^{2}}{4} υ ρ = \frac{π {(0 .0125)}^{2}}{4} (40) (1000) \\ = 4 .9087 kg/s . \end{array}$

Force on plate F = $\dot{M}$ $\dot{M}$ v = (4.9087 kg/s)(40 m/s) = 196.3 kg m/s²= 196.3 N.

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Table of Contents for Chapter 48: Fluids in motion

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Publisher Summary

Bernoulli’s equation

Flow through orifices

Impact of a jet

Table of Contents for
Chapter 48: Fluids in motion