Control Volume Analysis of Momentum and Energy

Solutions to Exercise Seven – Control Volume Analysis of Momentum and Energy 1 A 100-ft-wide river with a flow rate of 2400 ft3/s flows over a rock pile as shown in the figure at the right. Determine the direction of flow and the head loss associated with the flow across the rock pile. (Problem 5.87 and Figure P5.87 copied from Munson et al., Fluid Mechanics text.) We know that the head loss must be positive so we can assume a flow direction and compute the head loss. If the head loss is negative, we have assumed the incorrect direction. Start by assuming that the flow goes from point 1 to point 2 (from right to left) and apply the energy equation to a control volume bounded by the top and bottom of the river and vertical lines at section 1 and section 2. Note that our equation has to have point 1 as the assumed inlet and point 2 as the assumed outlet. p2 ρ + V22 p V2 + z 2 = 1 + 1 + z1 + hs − hL 2g ρ 2g Along the top of the river, there is essentially no difference in pressure between two liquid elevations because the air density is negligible. Thus, we have p1 = p2 = atmospheric pressure. There is no shaft work so hs = 0. We can find the velocities from the stated flow rate and the areas. 2400 ft 3 6 ft Q s = = V1 = A1 (4 ft )(100 ft ) s 2400 ft 3 12 ft Q s = = V2 = A2 (2 ft )(100 ft ) s Rearranging the energy equation to solve for head loss with p2 – p1 = hs = 0 and substituting the velocities just found and the elevations shown in the diagram gives the head loss shown below. 2 2  6 ft   12 ft    −  2 2 V1 − V2 s   s   hL = + 4 ft − 2 ft = 0.32 ft + z1 − z 2 = 2g  32.174 ft  2  s2   Since this head loss is positive our original assumption that the river flows from right to left is correct, and the head loss, hL = 0.32 ft . Exercise seven solutions 2 Page 2 If a ¾ hp motor is required to produce a 24-in stream of air having a velocity of 40 ft/s as shown in the figure at the right, estimate (a) the efficiency of the fan and (b) the thrust of the supporting member on the conduit enclosing the fan. (Problem 5.88 and figure taken from Munson, Fluid Mechanics.) This problem is similar to example 5.24 starting on page 239 of the text. In that example the following equation is used to provide “a reasonable estimate of the efficiency”. Useful effect η= = Work input wshaft − loss net in wshaft = net in hs − hL hs We can find the numerator of this efficiency definition from the energy equation. 1 po ρ + Vo2 p V2 + gz o = i + i + gz i + wshaft − loss 2 2 ρ net in For this fan, po = pi = 0 because both are atmospheric pressure, and zo = zi, so the pressure and elevation terms vanish. In addition we can assume that the air entering the duct from the outer atmosphere has such a low velocity that Vi2 << Vo2, and can be neglected. With these assumptions, we have the following result. 2 wshaft net in Vo2 1  40 ft  800 ft 2 − loss = =   = 2 2 s  s2 The shaft work in this equation is defined as the power input divided by the mass flow rate. If we use the standard density of air = 0.00238 slug/ft3 from Table 1.7 in the inside front cover, we have the following mass flow rate from the continuity equation. 0.00238 slug 40 ft π (24 in ) = m = ρVA = ρV 4 4 s ft 3 πD 2 2 2  ft  0.2991 slug   = s  12 in  We can then compute the shaft work from the input power of ¾ hp and this mass flow rate. wshaft = net in W shaft net in m (0.75 hp ) = 550 ft ⋅ lb f 1 slug ⋅ ft hp ⋅ s lb f ⋅ s 2 1379 ft 2 = 0.2991 slug s2 s We now have all the required information to compute the estimate of the efficiency. η= wshaft − loss net in wshaft net in 800 ft 2 s2 = 1379 ft 2 s2 η = 58.0% Note that we could also use the equation with shaft head and head loss; then we would have to divide the shaft work and loss term by g. The efficiency would have the same result. 1 Exercise seven solutions Page 3 To obtain the force, we use the usual momentum balance equation. N inlets ∂ (mVk )cv N outlets + ∑ ρ oVo AoVk ,o − ∑ ρ iVi AiVk ,i = ∑ Fx ∂t o =1 i =1 Here the problem is steady and we have only one inlet and one outlet so there is a common mass flow rate as found above. The only velocity components are in the direction of flow, which we call the x-direction here. This gives our momentum equation as m (V x ,o − V x ,i ) = ∑ Fx In the previous part we neglected the inlet velocity and we will continue to do so here, giving Vx,I = 0. The pressure at both the inlet and exit are zero, so the only force that we have to consider is the anchoring force, FAx. FAx = m V x ,o 2 0.2991 slug 40 ft 1lb f ⋅ s = s s slug ⋅ ft Fax = 12.0 lbf 3. Air flows past an object in a pipe of 2-m diameter and exits as a free jet as shown in the figure at the right. The velocity and pressure upstream are uniform at 10 m/s and 50 N/m2, respectively. At the pipe exit, the velocity is nonuniform as indicated. The shear stress along the wall is negligible. (a) Determine the head loss associated with a particle as it flows from the uniform velocity upstream of the object to a location in the wake at the exit plane of the pipe and the total head loss for the section shown. (b) Determine the force that the air puts on the object. The wake has a uniform velocity of 4 m/s so we can use the simple head loss equation to look at the effect from the uniform inlet flow of 10 m/s to the uniform wake exit flow of 4 m/s. This head loss equation is po γ + Vo2 p V2 + z o = i + i + z i + hshaft − hL 2g γ 2g net in The elevation of the inlet and outlet are the same so zo = zi and the outlet is open to the atmosphere so that po = 0. There is no shaft work input. With all these considerations our head loss equation becomes. 2  10 m   4 m  50 N   −  2 2 2 pi Vi − Vo s   s   m hL = + = + 9.80665 m 12.0 N γ 2g 2 3 s2 m 2 hL = 8.45 m In order to find the head loss for the entire flow we have to consider the nonuniform velocity profile at the exit. To do this we use the nonuniform flow equation based on the average velocity and the Exercise seven solutions Page 4 kinetic energy coefficient, α, that accounts for the nonuniform velocity profile. (See equation 5.89 on page 243 of the text.) po γ + α oVo2 2g + zo = pi γ + α iVi 2 2g + z i + hshaft − hL net in The kinetic energy coefficient is defined by equation 5.86 on page 243 of the text. That equation can be modified as shown below for constant density and a circular geometry where A = πR2 and dA = rdrdθ. ∫ V ρ (V • n )dA 2 α= Area m V 2 = ρ ∫ V 2 (V • n )dA Area (ρV A)V 2 = 1 ∫ V (V • n )dA = V 3 V A Area 2π R 1 2 3 πR 2 ∫ ∫ V (V • n )rdrdθ 2 0 0 Since the outlet velocity is normal to the exit plane the V•n dot product is simply the exit velocity, V. Since the exit velocity profile does not depend on θ, we can integrate over this variable giving a factor of 2π in the numerator. This gives the following. α= 2π R 1 V 3πR 2 2 ∫ ∫ V (V • n )rdrdθ = 0 0 R R 2π 2 V 2Vrdr = 3 2 ∫ V 3 rdr 3 2 ∫ V πR 0 V R 0 The average velocity is defined in equation 5.7 on page 195. We can perform the same manipulations on that equation that we did on the equation for the kinetic energy coefficient. This gives the following result. V = ∫ ρ (V • n )dA Area ρA ρ = ∫ (V • n )dA Area ρA = 1 A 2π R 2 ∫ ∫ V (V • n )rdrdθ = 0 0 R R 2π 2 Vrdr = 2 ∫ Vrdr 2 ∫ R 0 πR 0 The velocity profile at the exit has V = 4 m/s from r = 0 to r = 0.5 m and V = 12 m/s from r = 0.5 m to r = 1 m. This gives the following result for the average velocity. R 0.5 m 1m 1m   0.5 m 4 m 12 m 2 4m r2 12 m r 2   + rdr + ∫ rdr  =  ∫ 2 s s 2 0.5 m   (1 m )  s 2 0  0 s 0.5 m  2 2 2   ( ) ( ) ( ) − − 2 4 m 0.5 m 0 12 m 1 m 0.5 m 10 m + = 2  s s 2 2 (1 m )  s  2 2 V = 2 ∫ Vrdr = R 0 (1 m )2 Note that we could have found this result from a simple application of the continuity equation. Because the density and area at the exit are the same as those at the inlet, the average velocity at exit must be the same as the average velocity at inlet. We can now apply the velocity profile into the integral for αο. (We know that αi = 1 for the uniform inlet velocity profile.) 3 3 1m R  0.5 m  4 m 3 2 2 s 3  4 m  r 2  12 m  3 α = 3 2 ∫ V rdr =  ∫   rdr + ∫   rdr  =  3 5  V R 0 s   10 m  2   500 m  s  2  0  s  0.5 m     (1 m )  s  3 2 1m  3 3  (0.5 m )2 − 0 + 1728 m3 (1 m )2 − (0.5 m )2  = 1.312  12 m  r  = s 5  64 m +   3 2 2 s3  s  2 0.5 m  500 m  s   0.5 m 0 Exercise seven solutions Page 5 Using this value in the non-uniform head loss equation with the previous observations that z0 = zi, and po = hs = 0 gives the following result. 2  10 m   10 m  50 N 1 − 1.312   2 2 p α V − α oVo s   s  m2 hL = i + i i = +  9.80665 m 12.0 N γ 2g 2 3 s2 m 2 hL = 2.58 m To determine the force that the air exerts on the object we use the momentum equation, whose general form is N inlets ∂ (mVk )cv N outlets + ∑ ρ oVo AoVk ,o − ∑ ρ iVi AiVk ,i = ∑ Fx ∂t o =1 i =1 In this case the flow is steady and we effectively have two outlets: the first (with subscript o1) has a velocity of 4 m/s and the second (subscript o2) hash a velocity of 12 m/s. The inlet and both outlets have a flow in the x direction only so the main velocity, V, is the same as the x-component, Vx for each location. We thus have our momentum balance as ρ o1Vo21 Ao1 + ρ o 2Vo22 Ao 2 − ρ iVi 2 Ai = ∑ Fx The exit pressure is zero and the inlet pressure of 50 N/m2 creates a force in the +x direction. We are asked to find the force that the fluid exerts on the object, so this will be –Rx. There will also be a wall shear stress force, but we are told that this force is negligible. With the additional assumption that the density is the same at inlet and outlet 2, our equation becomes ρ (Vo21 Ao1 + Vo22 Ao 2 − Vi 2 Ai ) = ∑ Fx = pi Ai − R x We will use the standard density of air, ρ = 1.23 kg/m3 from Table 1.8 in the inside front cover. The inlet area is π(1 m)2 = 3.1416 m2. The inner area at the exit, with a velocity of 4 m/s is π(0.5 m)2 = 0.78540 m2. The outer area at the exit, with a velocity of 12 m/s, is π[(1 m)2 π(0.5 m)2] = 2.3562 m2. Using these areas and the data for pressure and velocities gives the reaction force as follows. 50 N 3.1416 m 2 − 2 m 2 2 2  1.23 kg  4 m   10 m   12 m  2 2 2  3.1416 m  2.3562 m −   0.78540 m +   3 m  s   s   s  ( ( ) Rx = pi Ai − ρ Vo21 Ao1 + Vo22 Ao 2 − Vi 2 Ai = ( ) ( ) ) (  1N ⋅ s2   kg ⋅ m ) Rx = 110.7 N We have already made this assumption in solving for the head loss. If the exit pressure is the standard atmospheric pressure of 101325 N/m2 the inlet pressure will be 101375 N/m2 which is only a 0.05% difference in pressure (and hence a 0.05% difference in density). 2