Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Fluids (continued) Last class: Density: ρ = m V kg 3 m F p = Pressure: A N m 2 ≡ Pa 5 = 1 atm 1.013x10 = Pa 760 = Torr 14.7 lb / in 2 Gauge pressure as a function of depth d below the surface of a fluid: p = ρgd Absolute pressure as a function of depth d below the surface of a fluid on earth: p= po + ρgd (po = local atmospheric pressure) Pascal’s principle: the change in pressure occurring in an incompressible fluid, in a closed container, is transmitted to every portion of the fluid and the walls of the container. (allows for hydraulic lifts) U-tubes (determining the density of an immiscible fluid) Tube cross-sectional area A. The pressure on the two sides at the lowest dashed line must be equal (or the fluid would move). H-h H h p p pleft = p right mug mk g po + = po + A A mu = mk Initially – fluid of known density ρk Add column h of fluid of unknown density ρu ( ρ u > ρ k case) ρ u Ah = ρ k AH H ρu = ρk h If ρ u < ρ k The pressure on the two sides at the lowest dashed line must be equal (or the fluid would move). pleft = p right h-H H h p p mug mk g po + = po + A A mu = mk ρ u Ah = ρ k AH So again, Initially – fluid of known density ρk Add column h of fluid of unknown density ρu H ρu = ρk h But now H < h. Archimedes' principle Consider the pressure on the top and lower faces of the cube of water labeled C, somewhere in the tall stationary column of water. The pressure at the upper face, pt , is due to the mass of the column of water above C. The pressure at the lower face, pl , is due to the mass of the column above C plus the mass of C itself. With mf the mass of the fluid contained in C, we can write that, mf g p=l p t + A pt C pl area A Or since, p = F A Fl Ft m f g = + A A A F=l Ft + m f g Fl is the force downwards due to C and the column of water above it but since the water is stationary there must be an equal and opposite force, due to the surrounding water on the lower face of C upwards. Hence the net forces acting on C are Fl − Ft = Ft + m f g − Ft = m f g Ft C Fl mf g So we’ve found that the water that surrounds C provides a buoyant force Fb upwards given by, Fb = m f g This net force from the surrounding water is actually independent of what material occupies that volume. Hence if we replace the cube of water with a different material having a different mass, m, Newton’s 2nd law gives, Fb − mg = ma m f g − mg = ma C Fb = mfg mg mf − m a= g m Period 2 class see slide at end (ignoring hydrodynamic drag) If the material has a smaller mass than the equal volume of water that it displaced (meaning that it has a lower density than water, e.g. Styrofoam) then the acceleration due to the buoyant force will be positive (i.e. upwards). Fb= mfg If the material has a greater mass than the equal volume of water that it displaced (meaning that it has a higher density than water, e.g. metal) then the acceleration due to the buoyant force will be negative (i.e. downwards), but smaller than its free fall acceleration if only gravity were acting. Styro foam mg Another way to look at this is to to consider the apparent weight of the object. Suppose the object is placed on a spring scale (in the fluid). Then FS + Fb − mg = ma = 0 So the scale reads, = FS mg − Fb Thus its apparent weight is the actual weight minus the buoyant force. C FB FS mg The buoyant force depends (only) on the weight of the fluid displaced, i.e. Fb = mf g This is Archimedes' principle. If the buoyant force exactly balances the weight of the object the apparent weight FS = 0 and the object neither rises nor sinks. The object is then said to be neutrally buoyant. FS = mg − FB = 0 This is what permits a submarine to hover at particular depth. A submarine has internal ballast tanks that are designed to be filled with seawater. To dive a neutrally buoyant submarine pumps water into these chambers making the weight of the hull plus the water taken on greater than the weight of the water displaced by the hull (i.e. greater than the buoyant force). neutral sinks As the desired depth is approached this water is blown out, making the ship neutrally buoyant again. To rise again, still more water is blown out, making the submarine have an apparent weight that is negative (Fb > mg). When the ship gets to the surface, it continues to rise until its total weight equals the weight of the water it displaces. neutral rises mg mg This sets the depth at which an object floats, i.e. an object will sink into a fluid to a level until the weight of the object equals the weight of fluid displaced. Example Floating in the very salt rich (dense) waters of the Dead Sea keeps about 1/3 of your body above the water line. What is the density of the water there? Assume your density to be 1.0 g/cm3. Let your mass be m, FB – mg = 0 but FB = mf g , mf = displaced water so mf g – mg = 0 mf = m ρfVf = ρV mg FB 1/3 2/3 mf = m ρfVf = ρV ρf = unknown density of the seawater ρ = 1.0 g/cm3 (your density) Now the volume of fluid displaced is 2/3 of your volume so, Vf = 2/3V then, ρf 2/3V = ρV ρf 2/3 = ρ ρf = 3/2ρ ρf = 1.5 g/cm3 mg FB 1/3 2/3 HITT challenge question Iron The iron block is thrown off the float into the pool. Does the water level in the pool A) go up B) go down C) stay the same Answer B) The water level actually goes down Fully submerged the block displaces a quantity of water equal to its volume. On the float, the block displaces a quantity of water equal to its mass. That displaced water raises the level in the pool Since iron has greater density than water, it displaces more water when on the float. So the water level is higher with the block on the float. Fluid Dynamics Types of fluid flow: We can visualize the flow of a fluid by injecting jets of smoke or dye into the moving fluid. There are two distinct regimes of flow: laminar flow and turbulent flow. These are both evident in the figure below. laminar flow turbulent flow laminar flow turbulent flow For laminar flow the velocity of the fluid at each point is a constant, i.e. at each point the speed and the direction of the fluid remains the same over time. For turbulent flow this is not true. For the situation above two photographs taken at different times would show that the streamlines on the left and upper right would be unchanged, while the smoke pattern on the lower right would be quite different. laminar flow turbulent flow For laminar flow a volume element of the fluid that begins on a streamline remains on that streamline and streamlines do not cross each other. In the following developments we assume an ideal fluid for which: 1) The flow is laminar 2) The fluid is incompressible 3) The fluid has very low (negligible) viscosity Equation of continuity Consider the flow of an ideal fluid through a uniform pipe. Cross-sectional area = A ∆x If the fluid is flowing with speed v then an element of the fluid goes a distance ∆x in time ∆t so ∆x v= → ∆x = v∆t ∆t The volume of fluid crossing a point along the pipe, in that time is then ∆V = A∆x = Av∆t The volume flow rate RV , which is the volume of fluid passing a point along the pipe in given time can then be written, R = V ∆V = Av ∆t So the volume flow rate equals the cross-sectional area of the pipe times the velocity of the fluid going through it. Given the volume flow rate through a hose, RV , we can determine e.g. the time taken to fill a pool of volume VP , since with t unknown we can write, VP = R V t VP t= RV Consider now a pipe that has a non-uniform cross-section, necking down as shown below A1 A2 v2 v1 The volume of fluid crossing a point along the tube on the left (in the fatter section) in time ∆t must equal the volume of fluid crossing a point on the right (in the thinner section) in the same time, so ∆V(left) = ∆V(right) A1v1∆t = A2v2∆t A1v1 = A2v2 A1v1 = A2v2 A1 A2 v2 v1 This is the equation of continuity. It says that in order to keep the volume flow rate constant, ∆V R= = Av = constant V ∆t the velocity of the fluid in the thinner section must be greater (in proportion to the ratio of the cross-sectional areas), i.e. A1 v2 = v1 A2 Example A stationary lawn sprinkler consists of a pipe feeding a dead ended container with 12 identical holes of 0.13 cm diameter. Water in the 1.9 cm (inner) diameter pipe has a speed of 0.91 m/s. With what speed does the water leave the holes? The pipe has cross-sectional area, Ap = π (dp / 2) 2 The holes have total cross-sectional area, 2 A = (12) π d / 2 ( h ) (12 holes × area/hole) h Eqn. of continuity, dp ) Ap ( 1 1.9 cm = = vh = vp vp 0.91 m / s 2 Ah 12 0.13 cm 12 ( d h ) 2 Ahvh = Apvp vh= 16.2 m/s 2 A1 A2 v2 v1 For the ideal fluid the volume flow rate, R V (the volume of fluid passing a point in unit time) is related to the mass flow rate, R m (the mass of fluid passing a point in unit time) via, R m = ρR V The textbook uses the continuity equation and conservation of energy to derive Bernoulli’s equation. The Bernoulli Equation P1 + 12 ρv12 + ρgy1= P2 + 12 ρv 22 + ρgy 2 For the flow of an ideal fluid this relates the absolute pressure and flow velocity at one elevation to the absolute pressure and flow velocity (in a possibly different diameter pipe) at a different elevation. Examples A1 P1 P2 A2 v2 = 15 m/s v1 Water flows through the pipe and out into the atmosphere at 15 m/s. The diameter of the right side is 3 cm and the left side is 5 cm. a) What volume of water flows into the atmosphere in 10 minutes? 2 0.03 m ∆V =R V ∆t =A 2 v 2 ∆t =π (15 m / s)(10 min)(60 s / min) 2 ∆V = 6.4 m 3 b) What is the speed v1? A1 P1 P2 A2 v2 = 15 m/s v1 By the continuity equation A1v1 = A2v2 2 d2 π(d 2 / 2) A2 = = v1 = v2 v2 v2 2 π(d1 / 2) A1 d1 2 3 v1 = = (15 m / s) 5.4 m / s 5 2 b) What is the pressure P1? A1 P1 P2 A2 v2 = 15 m/s v1 By Bernoulli’s equation (for no change in elevation) P1 + 12 ρv12 = P2 + 12 ρv 22 Since the right side flows into the atmosphere the pressure P2 = 1 atm. Then P1 = P2 + 12 ρv 22 − 12 ρv12 P1 = P2 + 12 ρ(v 22 − v12 ) P1 = 1.01x105 Pa + 12 (1000 kg / m 3 )[(15 m / s) 2 − (5.4 m / s) 2 ) P1 = 1.01x105 Pa + 9.8x10 4 Pa = 1.99x105 Pa P1 1 atm 1.99x10 = Pa 1.97 atm 5 1.01x10 Pa 5 The pressure in the larger pipe is greater than that in the smaller pipe. This is absolute pressure. If we were asked for gauge pressure, Pgauge = Pabsolute – Patm= 1.97 atm – 1 atm = 0.97 atm Example The water intake at a hydroelectric power station has a 1.0 m diameter and the inflow rate is 0.40 m/s. The inlet to the turbine at the generator building has a smaller diameter and the flow rate there is 9.5 m/s. What is the pressure difference between the inlet to the turbine and the intake? This is a straightforward application of Bernoulli’s equation P1 + 12 ρv12 + ρgy1= P2 + 12 ρv 22 + ρgy 2 P2 − P1 =12 ρv12 − 12 ρv 22 + ρgy1 − ρgy 2 ∆P = 1 2 ρ(v12 − v 22 ) + ρg(y1 − y 2 ) Let 2 be at the bottom (y2 = 0) then, ∆P 1 2 (1000 kg / m 3 )[(0.40m / s) 2 − (9.5 m / s) 2 ] + (1000 kg / m 3 )(9.8m / s2 )(180 m) ∆P =−4.5x10 4 Pa + 1.764x106 Pa 1 atm Pa 17 atm ∆P 1.72x10 = 5 1.01x10 Pa 6 mf − m a= g m (ignoring hydrodynamic drag) Period 2 class: The slides for this section are correct. If the mass of the object, m, is less than the mass of the fluid it displaces the acceleration should be upwards and so the acceleration should be positive (which is what this expression gives). Fb= mfg The acceleration only comes out negative (i.e. downwards) if m > mf again what should be expected. Styro foam mg