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Unformatted text preview: EGN 3353C  Fluid Mechanics Spring Semester 2011 Supplementary Handout: ControlVolume Solutions 1. Assumptions: • Steady flow (no indication given that unsteady effects present) • Uniform flow at nozzle exit (no indication given that nonuniformity present) Schematic: The control volume, indicated by the dashed line, is chosen as shown below. Note that we choose the control volume to be on the outside of the rocket to avoid having to specify the surface forces on the inside walls. Note also that the control surface cuts through the test stand because we need to determine the force acting on it. p ∞ p e u e T Analysis: (a) The linear momentum equation is X ~ F = d dt Z CV ρ ~ V d ∀ + Z CS ~ V ρ ~ V · ~ndA. (1) We only need to consider the xcomponent of the momentum equation because the thrust force acts in the xcoordinate direction. Because the flow is assumed to be steady, the first term is zero and the linear momentum equation reduces to X F x = Z CS uρ ~ V · ~ndA. (2) The thrust force produced by the rocket acts in the negative xdirection, so the force exerted by the test stand on the control volume is in the positive xdirection. Because the flow is supersonic at the nozzle exit and hence compressible, the pressure at the exit is not atmospheric. The force due to the gage pressure acting at the nozzle exit is thus ( p e p ∞ ) A e in the negative xdirection, where A e is the area of the nozzle exit. Therefore, Eq. (2) becomes T ( p e p ∞ ) A e = Z CS uρ ~ V · ~ndA. (3) 1 For uniform flow at the nozzle exit, the surface integral simplifies as follows Z CS uρ ~ V · ~ndA = u e ρ e ( u e ~ i ) · (+ ~ i ) Z e dA = u e ρ e u e A e = ˙ mu e , (4) where ˙ m = ρ e u e A e is the mass flow through the nozzle. Combining with Eq. (3), we obtain the desired expression for the thrust produced by the rocket, T = ˙ mu e + ( p e p ∞ ) A e . (5) The two terms are called momentum and pressure thrust. Note that the pressure thrust can be positive or negative. (b) With the given diameter at the nozzle exit, the area is A e = π 4 d 2 e = π 4 (3 . 8 m) 2 = 11 . 34 m 2 . (6) Inserting the area and the given values into Eq. (5) gives T = 4200 kg s × 3200 m s + ( 6 · 10 4 1 . · 10 5 ) N m 2 × ( 11 . 34 m 2 ) = 13 . · 10 6 N ....
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This note was uploaded on 09/05/2011 for the course EGN 3353C taught by Professor Lear during the Spring '07 term at University of Florida.
 Spring '07
 Lear
 Fluid Mechanics

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