Fluid Mechanics Notes

February 26 Homework Solutions 3.23 A differential pressure gage attached to a Pitot-static tube (See Video 3.4) is calibrated to give speed rather than the difference between stagnation and static pressures. The calibration is done so that the speed indicated on the gage is the actual fluid speed if the fluid flowing past the Pitot-static tube is air at standard sea level conditions. Assume the same device is used in water and the gage indicates a speed of 200 knots. Determine the water speed. A Pitot-static tube measures the difference between the static pressure, ps, where the fluid has a velocity, V, and the stagnation pressure, p0 = ps + ρV2/2, which occurs when the original velocity, V, is reduced to zero.. (See the discussion of the Pitot-static tube in the text and lecture presentations for more information.) We can obtain the velocity, V, from the difference in static and stagnation pressures by solving the equation p0 = ps + ρV2/2 for V. In the problem given here, the gage has been calibrated to translate the pressure difference directly into speed when the fluid is air. When we have a speed reading of 200 knots in any fluid the Pitot-static tube thinks that it is in air so the pressure difference measured by the Pitot-static tube can be found from the following formula: p0 − p s = ρ airVair2 2 = ρ air (200 knots )2 2 When we put the Pitot-static tube in another fluid and measure this velocity we are actually measuring the pressure diference that the Pitot-static tube would show in air at that velocity. The other fluid, water in this problem, with a different density, will give the same Pitot-static tube pressure difference if it has the following velocity relationship. p0 − p s = 2 ρ waterVwater ρ (200 knots )2 = air 2 2 So the velocity of water is given by the following formula where we use the tabulated values for the standard densities of air and water. 2 = Vwater ρ air (200 knots )2 ρ water ⇒ Vwater = (200 knots ) ρ air 0.00238 slugs ft 3 = (200 knots ) ρ water 1.94 slugs ft 3 Vwater = 7.01 knots 3.24 When an airplane is flying 200 mph at 5000-ft altitude in a standard atmosphere, the air velocity at a certain point on the wing is 273 mph relative to the airplane. What suction pressure is developed on the wing at that point? What is the pressure at the leading edge (a stagnation point) of the wing? The diagram above (taken from the solutions manual for the text) shows the flow along two streamlines, both of which start far from the airplane where the velocity, relative to the airplane is 200 mph and the gage pressure can be taken as zero. At an elevation of 5000 ft the density in the standard atmosphere (see Table C-1 in Appendix C on page 764) is 0.002048 slugs/ft3. The Bernoulli equation can be written in the following form gz + p ρ + V2 = C = Constant along a streamline 2 Although there may be slight elevation differences in the wing, they will have a negligible impact on the force-momentum balance because of the low specific weight of air. Consequently we can write that P/ρ + V2/2 = C along a streamline. Before starting the calculations we can convert the velocities given in the problem from mph to ft/s using the conversion factor that 30 mph = 44 ft/s. This gives the velocity far from the airplane as 293 ft/s and the velocity over the airplane wing as 400 ft/s. We first consider the streamline that starts far from the airplane and passes over the wing at a point where the velocity is 273 mph = 400 ft/s. The Bernoulli equation with no effect of elevation gives us p1 ρ + V12 p3 V32 = + ρ 2 2 ⇒ p3 = p1 + ρ V12 − V32 2 Substituting data for the velocities and the definition of the reference gage pressure as p1 = 0 gives 2 2 V12 − V32 1 0.002048 slugs  293 ft   400 ft   p3 = p1 + ρ = 0+     − 2 2 ft 3  s   s   p3 = –70.6 lbf/ft2 Applying the Bernoulli equation along a stagnation streamline to point 2, where V2= 0 gives p1 V12 p 2 V22 + = + 2 2 ρ ρ ⇒ p 2 = p1 + ρ V12 − 0 1 0.002048 slugs  293 ft  = 0+   2 2  s  ft 3 2 p2 = 88.0 lbf/ft2 Both p3 and p2 are gage pressures relative to a pressure far from the airplane at the same elevation as the airplane. 3.25 Water flows steadily downward through the pipe shown in the figure at the right. Viscous effects are negligible, and the pressure gage indicates the pressure is zero at point (1). Determine the pressure and flow rate at point (2). To solve this problem we have to apply the Bernoulli equation (and the continuity equation) two times. First we apply it to points one and the exit, which we will label as point 3. Since the exit is a free jet we know that it has zero gage pressure. We also know the gage pressure at point 1 is zero. Thus, we can use these pressure data to find the flow rate in the pipe. Once, we know the flow rate, we can apply the Bernoulli equation and the continuity equation between points (1) and (2) to get the desired pressure at point (2). Applying the Bernoulli equation between point (1) and the exit (3) gives. gz1 + p V2 p1 V12 + = gz 3 + 3 + 3 2 2 ρ ρ (3) The two velocities are related by the continuity equation: V1A1 = V3A3. This equation can be used to eliminate V1. Doing this and substituting z2 – z3= 3 ft as shown in the diagram, p3 = 0 because it is open to the atmosphere, and p1 = 0 as given in the problem statement gives the following result. g ( z1 − z 3 ) + p1 − p3 V12 − V32 0 − 0 V32 A32 / A12 − V32 + = g (3 ft ) + + =0 2 2 ρ ρ We can solve the resulting equation for V3, using the fact that the area ratio is the diameter ratio squared in the intermediate step below. V3 = 2 g (3 ft ) 1− A32 / A12 = 2 g (3 ft ) 1− D34 / D14 2 = 32.174 ft s2 (3 ft )  0.1 ft   1 −   0.12 ft  4 = 19.3 ft s We can now compute the flow rate as Q = V3A3 From the continuity equation we know that his flow rate will be the same throughout the pipe and gives the desired result for Q2. 0.152 ft 3 19.3 ft π 2 (0.1 ft ) = Q = Q1 = Q2 = Q3 = V3 A3 = s s 4 To get the pressure at point (2) we can apply the Bernoulli equation between points (1) and (2). Since the flow area is the same at these points we must have V1 = V2; we know that p1 = 0 and, from the diagram, we see that z1 – z2 = –2 ft. This gives the following result for the Bernoulli equation. p1 − p 2 V12 − V22 0 − p2 (z1 − z 2 ) + + = −2 ft + 1 +0=0 ⇒ ρg 2g γ p 2 = (− 2 ft )γ Using the specific weight of water as g = 62.4 lbf/ft3 gives the desired pressure: p2 = [(62.4 lbf)/ft3]·(-2 ft), so p2=–125 lbf/ft2 . 3.32 Water flows through the pipe contraction shown in Figure P3.32 at the right. For the given 0.2-m difference in manometer level, determine the flow rate as a function of the diameter of the small pipe, D. Apply the Bernoulli equation for incompressible, inviscid flows, shown below, between two points (1) and (2) along a streamline in the center of the pipe. (z 2 − z1 ) + (1) ( (2) ) p 2 − p1 V22 − V12 + =0 2g ρg The diagram shows that the measurement tube at point 2 is facing the flow; this means that it will measure the stagnation pressure. The Piezometer tube at point 1 is measuring the static pressure at that point. The measured difference h (= 0.2 m) times the specific weight of the fluid is the difference between the stagnation pressure at point (2) (p2 + ρV22/2) and the static pressure, p1 at point (1). This gives the following interpretation of the height difference. γh = ρgh = p 2 + ρV22 − p1 ⇒ 2 p1 = p 2 + ρV22 − ρgh 2 We can substitute this expression for p1 into our Bernoulli equation along with the fact that at the center of the pipe z2 = z1. This gives the Bernoulli equation as (z 2 − z1 ) + ( p 2 − p1 V22 − V12 + ρg 2g )   ρV 2 p 2 −  p 2 + 2 − ρgh    V 2 −V 2 2  + 2 1 = 0+ =0 ρg 2g ( ) We see that the terms in p2 and V22 cancel leaving the following result. h− V12 =0 2g V1 = 2 gh ⇒ The flow rate is simply V1A1 = V1πD12/4. 0.01556 m3 9.81 m π 2 π 2 2 gh = D1 2 gh = (0.1 m ) 2(0.2 m ) 2 = s 4 4 s Q = A1V1 = A1 From the equation for Q, we see that the flow rate does not depend on the smaller diameter D. 3.33 The speed of an airplane through the air is obtained by use of a Pitot-static tube that measures the difference between the stagnation and static pressures. (See Video 3.4.) Rather than indicating this pressure difference (psi or N/m2) directly, the indicator is calibrated in speed (mph or knots). This calibration is done using the density of standard sea level air. Thus the air speed displayed (termed the indicated air speed) is the actual air speed only at standard sea level conditions. If the aircraft is flying at an altitude of 20,000 ft and the indicated air speed is 220 knots, what is the actual air speed? A Pitot-static tube measures the difference between the static pressure, ps, where the fluid has a velocity, V, and the stagnation pressure, p0 = ps + ρV2/2, which occurs when the original velocity, V, is reduced to zero.. (See the discussion of the Pitot-static tube in the text and lecture presentations for more information.) We can obtain the velocity, V, from the difference in static and stagnation pressures by solving the equation p0 = ps + ρV2/2 for V. In the problem given here, the Pitot-static has been calibrated to translate the pressure difference directly into speed when the fluid is air at standard sea-level conditions. When we have a speed reading of 220 knots at an altitude of 20,000 ft., the pressure difference in the Pitot-static tube is pressure difference it would see if it were travelling at that speed in sea level air. That pressure difference is given by the following equation: p0 − p s = 2 ρ sea levelVindicated 2 = ρ sea level (220 knots )2 2 At 20,000 ft, that same pressure difference is generated at a different density and speed. The pressure difference can then be written for both the actual and indicated speeds as follows. p0 − p s = 2 ρ 20,000 ft Vactual 2 = 2 ρ sea levelVindicated 2 = ρ sea level (220 knots )2 2 So the actual velocity at velocity at 20,000 ft is given by the following formula. ρ sea level Vactual = Vindicated ρ 20,000 ft We can find the air density at sea level and an elevation of 20,000 ft for the standard atmosphere from Table C-1 in Appendix C on page 764: ρsea level = 0.00238 slugs/ft3 and ρ20,000 ftl = 0.001267 slugs/ft3. So, the actual air speed for an indicated air speed of 220 knots is found as follows. Vactual = Vindicated ρ sea level ρ 20,000 ft = (220 knots ) 0.00238 slugs ft 3 0.001267 slugs ft 3 Vactual = 302 knots 3.39 Water is siphoned from the tank shown in Figure P3.39. The water barometer indicates a reading of 30.2 ft. Determine the maximum value of h allowed without cavitation occurring. Note that the pressure of the vapor in the closed end of the barometer equals the vapor pressure. In order to avoid cavitation the pressure at the top of the siphon tube must be greater than the vapor pressure. We have to get a relationship for the (1) pressure at this point. (2) For this problem we will work with absolute pressures because the vapor pressure is an absolute pressure. The pressure, p1, is the (absolute) atmospheric pressure, patm. This is also the pressure at point (3). From the barometer equation we have the following value for the atmospheric pressure. (3) (hbaro p atm = γ water hbaro + pv = 30.2 m ) We can use the Bernoulli equation for incompressible, inviscid flows, applied between the water level in the tank, point (1), and the exit, point (3), to obtain an equation for the velocity in the siphon. g ( z1 − z 3 ) + ( ) p1 − p3 V12 − V32 + =0 ρ 2 We are not given the vapor pressure or any temperature data from which we can determine the vapor pressure. In applying the Bernoulli equation between the liquid level at point (1) and the exit at point (3), we have p1 = p3 = patm, and z1 – z3 = h. Because the tank is large we assume that V12 << V22 and can be neglected. This gives the follow results for the Bernoulli equation. g ( z1 − z 3 ) + ( ) ( ) 0 − V32 p1 − p3 V12 − V32 p −p + = gh + atm atm + =0 2 2 ρ ρ This gives the simple relationship between the velocity V3 and the height h. V3 = 2 gh The velocity at point (2), the top of the siphon, is related to V3 by the continuity equation. In the last step below we use the fact that the area ratio for a circular pipe is the diameter ratio squared. V2 A2 = V3 A3 = A3 2 gh ⇒ V2 = A3 2 gh D32 2 gh = A2 D22 We can now apply the Bernoulli equation between the liquid surface in the tank, point (1) and the top of the siphon, point (3) to determine the pressure there. Here we choose to write the equation using the “head” terms that have dimensions of length. (z1 − z 2 ) + ( ) p1 − p 2 V12 − V22 + =0 γ 2g We can solve this equation for p2 using (i) the previous assumption that V1 is negligible, (ii) the equation derived above for V2 in terms of the height, h, and (iii) the height difference, z2 – z1 = 6 ft from the diagram. p 2 = γ ( z1 − z 2 ) + p1 + γ (V 2 1 ) 2 ghD34  − V22 γ  = γ ( z1 − z 2 ) + p atm + 0− 2g 2 g  D24  Simplifying this equation and using the result of the barometer equation that p atm = γhbaro + pv we can obtain the following inequality expressing the fact that the pressure at point (2) must be greater than the vapor pressure to avoid cavitation. p 2 = γ ( z1 − z 2 ) + γhbaro + p v − γhD34 D24 > pv The pv terms cancel and we have the following inequalities for the height, h. The second is obtained by dividing the first by –γ, which changed the direction of the inequality − γD34 D24 h > γ ( z 2 − z1 ) − γhbaro D34 ⇒ D24 h < hbaro − ( z 2 − z1 ) We can now solve for the height, h and enter the data to obtain the maximum value for h. h< D24 D34 [hbaro 4  3 in   (30.2 ft − 6 ft ) = 3.14 ft − ( z 2 − z1 )] =   5 in  So there restriction to avoid cavitation is that h < 3.14 ft .