3 deen 113 neglect gravity solution a in problems

• Homework Help
• 11
• 100% (13) 13 out of 13 people found this document helpful

This preview shows page 8 - 11 out of 11 pages.

3. Deen 11.3 (neglect gravity) Solution a) In problems with one inlet and one outlet, the macroscopic momentum balance in the form of Eq. (11.3-9) can be used. If the velocity fields are flat (i.e., replacing the velocity with the average velocity), then a 1 = a 2 = 1 . Neglect the effect of gravity for this rapid flow through a nozzle, so the force F * that the enclosed fluid exerts on the solid part of the control volume (the nozzle, in this case) is F * = p 1 + ρ v 1 2 ( ) n 1 A 1 p 2 + ρ v 2 2 ( ) n 2 A 2 . But n 1 = e x , n 2 = e x , A 1 = π D 1 2 4 ,A 2 = π D 2 2 4 , v 1 = U, p 2 = p 0 and v 2 A 2 = v 1 A 1 so the net force on the solid nozzle from the fluid inside it is F x * = p 1 A 1 p 0 A 2 + ρ U 2 A 1 1 A 1 A 2 . The net force on the nozzle, including the surrounding air at pressure p 0 , is obtained by replacing absolute pressures by gauge pressures (see Example 11.3-1 in Deen). Replacing p 1 by p 1 -p 0 and p 2 by 0 gives F x 0 = p 1 p 0 ( ) A 1 + ρ U 2 A 1 1 A 1 A 2 . If gravity is negligible, then the y-component force is zero. b) The analysis is the same as part (a), except that now
n 2 = e x cos θ − e y sin θ . Consequently there are both x- and y- components to the force. For the x-direction these turn out to be F x * = p 1 A 1 p 0 A 2 cos θ + ρ U 2 A 1 1 A 1 A 2 cos θ and F x 0 = p 1 p 0 ( ) A 1 + ρ U 2 A 1 1 A 1 A 2 cos θ . For the y-direction they are F y * = p 0 + ρ U 2 A 1 A 2 2 A 2 sin θ and F y 0 = ρ U 2 A 1 2 A 2 sin θ . 4. Deen 11.6 (a,b) a) The total flow rate across the plane at the exit of the jet must equal the total flow rate out, so that with plug flow v an π 4 D 2 − λ 2 D 2 ( ) + v jet π 4 λ 2 D 2 = v out π 4 D 2 or v an 1 − λ 2 ( ) + v jet λ 2 = v out . In terms of flow rates, an equivalent relation is Q jet + Q an = Q out . b) The momentum transfer between the jet and the liquid in the annulus occurs via viscous friction, so it is unlikely that the mechanical energy balance will yield a useful relation between the pressures. However, if the shear force on the wall is negligible, the momentum balance can be used, d dt ρ v dV V a (t) + ρ v v w ( ) n dA A a (t) = ρ g dV V a (t) + n T dA A a (t) .
We use a control volume bounded by planes “1” and “2” at the annulus/jet exit and the tube outlet, respectively, in Fig. P11.6, and following the wall of the pipe. The problem is steady and the control volume is fixed. Take the z component of the momentum balance, where e z is the direction of flow, and neglect viscous stresses. Then we find that ρ v z 2 dA A an ρ v z 2 dA A jet + ρ v z 2 dA A out = p jet dA A jet + A an p 0 dA A out . Here A out =A jet +A an . We assume the velocity profiles are flat, and replace the velocities with their average values (flow rate divided by area). Then we have −ρ Q jet 2 A jet − ρ Q an 2 A an + ρ Q jet + Q an ( ) 2 A out = p jet p 0 ( ) A out . The pressure rise between the jet and the outlet is therefore p 0 p jet = ρ Q jet 2 A jet A out + ρ Q an 2 A an A out − ρ Q jet + Q an ( ) 2 A out 2 . Then p 0 p jet = ρ A jet A out v jet 2 + ρ A an A out v an 2 − ρ λ 2 v jet + 1 − λ 2 ( ) v an ( ) 2 and p 0 p jet = ρλ 2 v jet 2 + ρ 1 − λ 2 ( ) v an 2 − ρ λ 2 v jet + 1 − λ 2 ( ) v an ( ) 2 . Multiplying out the squared term, p 0 p jet ρ = λ 2