Bernoulli’s Equation Part II

Solutions to Exercise Four – Bernoulli’s Equation Part II 1 Water flows through the pipe contraction shown in the figure 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. (Problem and figure 3.30 from Munson et al., Fluid Mechanics text.) 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 ρg 2g The measurement tube at point 1 is facing the flow so it will measure the stagnation pressure. This means that the elevation difference shown in the diagram (h = 0.2 m) times the specific weight of the fluid is the difference between the stagnation pressure at point (1) (p1 + ρV12/2) and the static pressure, p2 at point (2). We thus have the following interpretation of the height difference. γh = ρgh = p1 + ρV12 2 − p2 ⇒ p 2 = p1 + ρV12 2 − ρgh We can substitute this expression for p2 into our Bernoulli equation along with the fact that at the center of the pipe z2 = z1. This gives the Bernoulli equation as ( ) 2 2 (z 2 − z1 ) + p 2 − p1 + V2 − V1 = 0 + 2g ρg p1 + ρV12 2 − ρgh − p1 ρg + (V 2 2 ) − V12 =0 2g We see that the terms in p1 and V12 cancel leaving the following result. V22 −h+ =0 2g V2 = 2 gh ⇒ The flow rate is simply V2A2 = V2πD22/4, with D2 = D. Q = A2V2 = A2 2 gh = π 4 D 2 2 gh = π 4 2(0.2 m ) The flow rate will have units of m3/s when D is in meters. 9.81 m 1.556 m 2 = D s s2 Exercise four solutions 2 Page 2 A Pitot-static tube is used to measure the velocity of helium in a pipe. The temperature and pressure are 40oF and 25 psia. A water manometer connected to the Pitot-static tube indicates a reading of 2.3 in. Determine the helium velocity. Is it reasonable to consider the flow as incompressible? Explain. (Problem 3.34 from Munson et al., Fluid Mechanics text. Figure below from solutions manual.) The flow can be assumed incompressible if the Mach number, Ma = V/c, where c = (kRT)1/2 is the sound speed, is less than 0.3. We can assume that the flow is incompressible, then compute the velocity and Mach number, Ma. If Ma < 0.3 our answer will be correct. 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 velocitt, V, is reduced to zero.. (See the discussion of the Pitotstatic 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. When the pressure difference is measured by a manometer, as shown at the right (the top part of the diagram, linking each side with the flowing fluid was truncated in copying), the measured value of p0 – ps = (γmano – γfluid)h. In this case where the manometer fluid is water and the flowing fluid is helium we can neglect the specific weight of heilum and write p0 – ps = γmanoh so the equation for velocity becomes. V = 2 p0 − p s ρ fluid = 2γ mano h ρ fluid We can compute the density of the fluid, Heliuim, from the ideal gas law using the value of R = 2.446x104 ft•lbf/slug•R for helium found in Table 1.7 in the inside front conver. We can use the pressure of 25 psia directly since it is already an absolute pressure, but we have to convert the temparature of 40oF to a Rankine temperaure by the equation R = oF + 459.67, so we use a temperature of 499.67 R in both the density calculation and the Mach number calculation below. 25 lb f 144 in 2 5.80 x10 − 4 slugs P in 2 ft 2 ρ= = = RT 2.446 x10 4 ft ⋅ lb f ft 3 (499.67 R ) slug ⋅ R We can check the assumption that the specific weight of the helium was neglegible compared to the manometer fluid, water, as follows: γ mano − γ He = γ mano − ρ He g = 62.4 lb f ft 3 2 62.38 lb f 5.80 x10 −4 slugs 32.174 ft lb f ⋅ s − = 3 2 1 slug ⋅ ft ft s ft 3 This gives an error of only 0.03%. Other gases, with higher molecular weights, whould give a larger, but still neglegible, error. Subsituting the density found above and the specific weight of water, the manometer fluid, as 62.4 lbf/ft3 in the equation for velocity gives. Exercise four solutions V = Page 3 2γ mano h ρ fluid 2 62.4 lb f = (2.3 in ) 1 ft 12 in ft = 203 ft/s −4 5.80 x10 slugs ft 3 3 We can now check the compressibliity assumption by computing the Mach number. The value of k for Helium is also found in Table 1.7 on the inside front cover. V Ma = = c V kRT = (1.66) 203 ft s 4 2.446 x10 ft ⋅ lb f (499.67 R )1 slug ⋅ 2ft slug ⋅ R = 0.063 lb f ⋅ s This value is well below the upper limt of Ma = 0.3 for which we said we could consider a gas flow incompressible so we conclude that the helium flow can be considered incompressible. 3 Water flows from a large tank as shown in the figure at the right. Atmospheric pressure is 14.5 psia and the vapor pressure is 1.60 psia. If viscous effects are neglected, at what height, h, will cavitation begin? To avoid cavitation, should the value of D1 be increased or decreased? To avoid cavitation, should the value of D2 be increased or decreased?. Explain. (Problem and figure 3.60 from Munson et al., Fluid Mechanics text.) Cavitation is likely to occur at the thin diameter point where D1 = 1 in. (Point 2 is open to the atmosphere so cavitation will not occur here. To determine if cavitation will occur we have to find the pressure at the point where D1 = 1 in. To do this we have to know the overall flow rate. We can find this by applying Bernoulli’s equation between the top of the large tank (point 0) and the exit (point 2). This gives ( ) p 0 − p 2 V02 − V22 (z 0 − z 2 ) + + =0 2g ρg From the diagram we have z0 – z2 = h, p0 = p2 = 0 since both are open to the atmosphere, and V02 =( A2V2/A0)2 << V22 and can be neglected because the area ratio, A2/A0 is so small. Inserting these values into the Bernoulli equation gives. h+ ( ) 0 − 0 0 − V22 + = 0 ⇒ V2 = 2 gh 2g ρg The flow rate, Q, can then be written as Q = A2V2 = A2 2 gh . Now that we know the flow rate, we can find the pressure at point (1) where the diameter is 1 in. Applying Bernoulli’s equation between this point and the exit (point 2) gives. Exercise four solutions Page 4 ( ) p 1 − p 2 V12 − V22 (z 1 − z 2 ) + + =0 2g ρg Here z1 = z2 and p2 = 0. Also, by continuity, V1A1 = V2A2 so that we can rewrite this equation as follows. (z 1 − z 2 ) + ( ) p 1 − p 2 V12 − V22 p −0 V 2 A 2 / A 2 − V22 + = 0+ 1 + 2 2 1 =0 2g 2g ρg ρg We can use the previous result that V22 = 2gh and the definition of γ = ρg to rewrite the previous result as follows. p 1 V22 A22 / A12 − V22 p 1 A22 / A12 − 1 2 p 1 A22 / A12 − 1 p 2 gh = 1 + A22 / A12 − 1 h = 0 + = + V2 = + 2g 2g 2g ρg γ γ γ ( ) Solving this equation for the gage pressure, p1, and noting that the area ratio is the same as the diameter ratio squared gives. ( ) p1 = γ 1 − D24 / D14 h If we add the atmospheric pressure to both sides of the equation we will obtain the absolute pressure at point 1, which must be geater than the vapor pressure to avoid cavitation.. ( ) p1,abs = p1 + p atm = γ 1 − D24 / D 4 h + p atm > p v Substituting the given data that D1 = 1 in, D2 = 2 in, patm = 14.3 psia, pv = 1.60 psia, and using a specific weight of 62.4 lbf/ft3 for water gives the following numerical result. 4 ft 2 1 psia ⋅ in 2 62.4 lb f   2 in     h + 14.3 psia > 1.60 psia 1 −  lb f 144 in 2 ft 3   1 in     − 6,5 psia h > −12.7 psia ⇒ h < 1.96 ft ft Note than when an inequality is multiplied or divided by a negative number the direction of the inequality is reversed. To determine the effects of D1 and D2 we can solve γ (1 − D24 / D 4 )h + p atm > p v for h. Note that since D2 > D1 we will be dividing by a negative number so we have to change the direction of the inequality. h< p v − p atm p atm − p v = 4 4 γ 1 − D2 / D1 γ D24 / D14 − 1 ( ) ( ) This shows that increasing D2 or decreasing D1 will increase h. The physical reasoning for this is as follows. Neglecting visclous forces, increasing D2 will decrease the overall flow rate and less material will have to flow through D1 for a given h. Thus we can increase h. However, with D2 fixed, the flow rate is fixed, and decreasing D2 increases the flow rate through point 1; thus we would have to reduce h to avoid cavitation.