Fluid Mechanics Lecture Notes PDF

Fluid mechanics Fluid statics 1 Fluid statics is concerned with the balance of forces which stabilise fluids at rest. In the case of a liquid, as the pressure largely changes according to its height, then the depth is taken into account. In the case of relative rest, the fluid...

Fluid mechanics Fluid statics 1 Fluid statics is concerned with the balance of forces which stabilise fluids at rest. In the case of a liquid, as the pressure largely changes according to its height, then the depth is taken into account. In the case of relative rest, the fluid can be considered to be at rest if the fluid movement is observed in terms of coordinates fixed upon the vessel.  Relative rest → the case where the fluid is stable relative to its vessel even when the vessel is rotating at high speed. 2  Pressure When a uniform pressure acts on a flat plate of area (A) and a force (P) pushes the plate, then p = P/A (1) where p is the pressure. P is the pressure force. When the pressure is not uniform, the pressure acting on the minute area (ΔA) is expressed by (2) 3 The unit of pressure is Pascal (Pa), and can be expressed in bars or metres of water column (mH2O). The conversion table of pressure units is given in the following table. 4 1. Absolute pressure and gauge pressure Two methods are used to express the pressure a. One is based on the perfect vacuum, which is known as the absolute pressure. b. The other on the atmospheric pressure, which is known as the gauge pressure. The gauge pressure is given as gauge pressure = absolute pressure - atmospheric pressure (3) A pressure under 1 atmospheric pressure, then the gauge pressure is expressed as a negative pressure. 5 The following figure shows the relation between the gauge pressure and the absolute pressure. 6 2. Characteristics of pressure  The pressure has the following three characteristics. 1) The pressure of a fluid always acts perpendicular to the wall in contact with the fluid. 2) The values of the pressure acting at any point in a fluid at rest are equal regardless of its direction.  Imagine a minute triangular prism of unit width in a fluid at rest as in the figure to the right. 7 Let the pressure acting on the small surfaces dA 1, dA2, and dA be p1, p2 and p. The following equations are obtained from the balance of forces in the horizontal p1 dA1 = p dA sinθ (4) The vertical directions p2 dA2 = p dA cosθ + ½ dA1 dA2 ρ g (5) The weight of the triangle pillar is insignificant, so it is omitted. 8 From geometry, the following equations are obtained dA sinθ = dA1 (6-a) dA cosθ = dA2 (6-b) As a result, the following relation is obtained p 1 = p2 = p (7) Since angle θ can be given any value, values of the pressure acting at one point in a fluid at rest are equal regardless of its direction. 9 3) The fluid pressure applied to a fluid in a closed vessel is transmitted to all parts at the same pressure value as that applied (Pascal’s law).  In the figure to the right, when the small piston of area A1 is acted upon by the force F1, the liquid pressure is p = F1 / A1 (8)  When the large piston of area A2 is acted upon by the force F2, the liquid pressure is p = F2 / A 2 (9)  Thus 10 F1 / A1 = p = F2 /A2 (10) So the device in the previous device can create the large force F2 from the small force F1, this is the principle of the hydraulic press. 3. Pressure of fluid at rest In a fluid at rest the pressure varies according to the depth. Consider a minute column in the fluid as shown in figure To the right. 11 Assume that a. The sectional area is dA. b. The pressure acting upward on the bottom surface is p. c. The pressure acting downward on the upper surface is p+(dp/dz)dz. Then, from the balance of forces acting on the column, the following is obtained p dA - (p + [dp/dz] dz ) dA- ρ g dA dz = 0 (11-a) ∴ dp / dz = -ρ g (11-b) Since ρ is constant for liquid, the following is obtained 12 p = -ρg ∫ dz = -ρ g z + c (12) If the base point is set at z0 below the upper surface of liquid as shown in figure to the right, and p0 is the pressure acting on that surface, where p = p0 when z = z0, then c = p0 + p g z (13) Substituting equation (13 ) into equation (12) gives p = p0 + (z0 - z) ρ g = p0 + ρ g h (14) Thus it is found that the pressure inside a liquid increases in proportion to the depth. 13  In the case of the gas in the atmosphere, the density of gas changes with pressure.  Thus it is not possible to integrate simply as in the case of a liquid.  As the altitude increases, the temperature decreases.  Assuming this temperature change to be polytrophic, then pvn = constant is the defining relationship.  Putting the pressure and density at z = 0 (sea level) as p 0 and ρ0 then (15) 14 Substituting p into equation (11-b) (16) Integrating equation 16 from z = 0 (sea level) (17) The relation between the height and the atmospheric pressure is obtained by integrating equation (17) (18) 15 The density is obtained from equations (15) and (17) (19) When the absolute temperatures at sea level and at the point of height z are T0 and T respectively, the following equation is obtained (20) From equations 18, 19, and 20 (21) 16 From equation (21) (22) In aeronautics, the following values were united p0 = 101.325 kPa, T0 = 288.15 K, ρ0 = 1.225 kg/m3, and the standard atmospheric conditions at sea level The temperature decreases by 0.65 °C every l00 m of height in the troposphere up to approximately 1 km high. The temperature is constant at -50.5°C from 1 km to l0 km high. For the troposphere, substituting the above values for p0, T0, and ρ0 into equation (16), then n = 1.235. 17 4. Measurement of pressure Manometer is a device which measures the fluid pressure by the height of a liquid column. For example, in the case of measuring the pressure of liquid flowing inside a pipe, the pressure (p) can be obtained by measuring the height of liquid (H) coming upwards into a manometer made to stand upright as in the figure to the right. 18 If p0 is the atmospheric pressure and ρ is the density, the following equation is obtained p = p0 + ρ g H (23) When the pressure (p) is large, then equation (23) is inconvenient because (H) is too high. Thus a U-tube manometer (previous figure/b), containing a high- density liquid such as mercury is used. As a result when the density is ρ’, then p + ρ g H = p0 + ρ’ g H’ 19 ∴ p = p0 + ρ’ g H’ - ρ g H (24) In the case of measuring the air pressure (ρ’ >> ρ) so ρgH in equation (24) may be omitted. If the pressure difference between two pipes is measured, where in both pipes a liquid of density (ρ) flows, then a differential manometer as in the figure to the right is used. 20 In the case of the previous figure, where the differential pressure of the liquid is small  Then measurements are made by filling the upper section of the meter with a liquid whose density is less than that of the liquid to be measured, or with a gas. Thus p1 – p2 = (ρ – ρ’) g H (25) In the case where ρ’ is a gas, then p1 – p2 = ρ g H (26) 21 The figure to the right shows the case when the differential pressure is large.  Then a liquid column of a larger density than the measuring fluid is used. Thus p1 – p2 = (ρ’ - ρ) g H’ (27) In the case where p is a gas, p1 – p2 = ρ’ g H’ (28) 22  A U-tube manometer is inconvenient for measuring fluctuating pressure.  Because it is necessary to read both the right and left water levels at the same time to measure the differential pressure.  For measuring the differential pressure, if the sectional area of one tube is made large enough (as shown in the figure).  Then the water column of height (H) could be measured by just reading the liquid surface level in the other tube.  Because the liquid surface instability in the tank can be ignored. 23  To measure a minute pressure, a glass tube inclined at an appropriate angle is used as an inclined manometer.  When the angle of inclination is (α) and the movement of the liquid surface level is (L), then the differential pressure (H)is H = L sinα (29)  If (α) is decreased then the reading of the pressure is magnified. 24  Force acting on the vessel of liquid This section is about the force acting on the whole face of a solid wall subject to water pressure, such as the bank of a dam. 1) Water pressure acting on a bank or a sluice gate The following figure shows the total force due to the water pressure acting on a bank built at an angle (θ) to the water surface. Ignoring the atmospheric pressure, then the pressure acting on the surface is zero. 25 26  The total pressure (dP) acting on a minute area (dA) is ρ g h dA = ρ g y sinθ dA (30)  Thus, the total pressure (P) acting on the under water area of the bank wall (A) is (31)  When the centric of A is G, its y coordinate is y G and the depth to G is hG, the following equation is obtained P = ρ g yG A sinθ = ρ g hG A (32)  The total force (P) equals the product of the pressure at the centric 27 (G) and the underwater area of the bank wall.  For a rectangular sluice gate (following figure), the force (P) on the whole plane of the gate is P = ρ g yG A (33)  The force acting on a minute area dA (a horizontal strip of the gate face) is P = ρ g y dA (34)  The moment of this force around the x axis is P = ρ g y dA y (35)  The total moment on the gate which is called the geometrical moment of inertial (Ix) is (36) 28 29  Locate the action point of (P) at which a single force (P) produces a moment equal to the total sum of the moments around the turning axis (x-axis) of the gate produced by the total water pressure acting on all points of the gate.  When the location of C is yc P yc = ρ g Ix (37)  If IG is the geometrical moment of inertia of area for the axis that is parallel to the x-axis and passes through the centre G, then the following relation is obtained I x = I G + A y G2 (38)  Values of IG for a rectangular plate and for a circular plate are given in the figure to the right. 30 31 Substitute equation (38) into (37) then (39) From equation (39), it is clear that the action point (C) of the total pressure (P) is located deeper than the centred (G) by h2/12yG. The position of (yc) in such a case where the gate is located under the water surface as shown in the figure in slide (32) is given as follows (40) 32 33 2) Force acting on a tear cylinder  In the case of a thin cylinder where the inside pressure is acting outward (following figure).  Consider the cylinder longitudinally half sectioned with diameter (d), length (l), and inside pressure (p).  The force acting on the vertical centre wall balances the force in the x direction acting outward on the cylinder wall.  The force generated by the pressure in the x direction on a curved surface equals the pressure pdl.  The same pressure acts on the projected area of the curved surface. 34 In addition, this force is the force 2Tl (T is the force acting per unit length of wall which tears this cylinder longitudinally in halves along the lines BC and AD) 2Tl=pdl ∴T=pd/2 (41) 35  A body floating in a fluid For a body floating in a fluid, then fluid pressure acts all over the wetted surface.  The resulting pressure acts in a vertical upward direction.  This force is called buoyancy. The buoyancy of air is small compared with the gravitational force of the immersed body, so it is normally ignored. 36  Suppose that a cube is located in a liquid of density (ρ) as shown in the following figure.  The pressure acting on the cube due to the liquid in the horizontal direction is balanced right and left.  For the vertical direction, where the atmospheric pressure is (p0), the force (F1) acting on the upper surface (A) is F1 = (p0 + ρgh1) A (42)  The force (F2) acting on the lower surface is F2 = (p0 + ρgh2) A (43) 37 When the volume of the body in the liquid is (V), the resultant force (F) from the pressure acting on the whole surface of the body is F = F2 - F1 = ρ g (h2 – h1) A = ρ g h A = ρ g V (44) The same applies to the case where a cube is floating (as in the previous figure [b]). From equation (44), the body in the liquid experiences a buoyancy equal to the weight of the liquid displaced by the body, and this is known as Archimedes’ principle. The centre of gravity of the displaced liquid is called centre of buoyancy and is the point of action of the buoyancy force. 38 For a ship of weight (W) floating in the water with an inclination of angle (θ) (following figure). Then the location of the centre (G) does not change with the inclination of the ship. Since the centre of buoyancy (C) moves to the new point (C’), a couple of forces is produced Ws = Fs (45)  39 This couple restores the ship’s position to stability. The forces of the couple (Ws) is called restoring forces. The intersecting point (M) on the vertical line passing through the centre of buoyancy C’ (action line of the buoyancy F) and the centre line of the ship is called the metacentre. GM is called the metacentre height. If M is located higher than G, the restoring force acts to stabilise the ship. If M is located lower than G, the couple of forces acts to increase the roll of the ship and so make the ship unstable. 40  Relative stationary state When a vessel containing a liquid moves in a straight line or rotates.  If there is no relative flow of the liquid while the vessel and liquid moves as a body.  Then it is possible to treat this as the mechanics of a stationary state.  This state is called a relatively stationary state. 41 1) Equi-accelerated straight-line motion Suppose that a vessel filled with liquid is moving in a straight line at constant acceleration on the horizontal level. Consider a minute element of mass (m) on the liquid surface, where its acceleration is (α), the forces acting on (m) are gravity in a vertical downward direction (-mg) and the inertial force in the reverse direction to the direction of acceleration (-mα). 42 No force component normal to the direction of (F). The pressure must be constant on the plane normal to the direction of (F). When (θ) is the angle formed by the free surface and the x direction, the following is obtained tan θ = α / g (46) If (h) is the depth measured in the vertical direction to the free surface, the acceleration in this direction is β=F/m (47) Therefore, p=ρβh (48) 43 2) Rotational motion In the case where a cylindrical vessel filled with liquid is rotating at constant angular velocity (ω). The movement at constant angular velocity is called gyrostatics. Take cylindrical coordinates (r, θ, z). Consider a minute element of mass (m) on the equi-pressure plane. 44 The forces acting on the equi-pressure plane are 1. (-m g) due to the gravitational acceleration (g) in the vertical direction. 2. (-m r ω2) due to the centripetal acceleration (r ω2) in the horizontal direction. If (ϕ) is the angle formed by the free surface and the horizontal direction tan ϕ = (m r ω2) / (m g) = (r ω2) / g (49) Also tan ϕ = dz/dr (50) 45 Therefore dz/dr = (r ω2) / g (51) Putting (c) as a constant of integration z = [(ω r2)/2g] + c (52) If z = h0, at r = 0, c = h0, the following equation is obtained from equation (52) z – h0 = (ω2 r2) / 2 g (53) 46