V 2 Q 1 A t Q 2 (3) The two inlet volume flows are Q 1 A 1 V 1 1 4 ( 1 1 2 ft) 2 (3 ft/s) 0.016 ft 3 /s Q 2 A 2 V 2 1 4 ( 1 3 2 ft) 2 (2 ft/s) 0.098 ft 3 /s Then, from Eq. (3), d d h t 0.057 ft/s Ans . (b) Suggestion: Repeat this problem with the top of the tank open. An illustration of a mass balance with a deforming control volume has already been given in Example 3.2. The control-volume mass relations, Eq. (3.20) or (3.21), are fundamental to all fluid- flow analyses. They involve only velocity and density. Vector directions are of no con- sequence except to determine the normal velocity at the surface and hence whether the flow is in or out . Although your specific analysis may concern forces or moments or energy, you must always make sure that mass is balanced as part of the analysis; oth- erwise the results will be unrealistic and probably rotten. We shall see in the examples which follow how mass conservation is constantly checked in performing an analysis of other fluid properties. In Newton ’ s law, Eq. (3.2), the property being differentiated is the linear momentum m V . Therefore our dummy variable is B m V and d B / dm V , and application of the Reynolds transport theorem gives the linear-momentum relation for a deformable control volume d d t ( m V ) syst F d d t CV V d CS V ( V r n ) dA (3.35) The following points concerning this relation should be strongly emphasized: 1. The term V is the fluid velocity relative to an inertial (nonaccelerating) coordi- nate system; otherwise Newton ’ s law must be modified to include noninertial relative-acceleration terms (see the end of this section). 2. The term F is the vector sum of all forces acting on the control-volume mate- rial considered as a free body; i.e., it includes surface forces on all fluids and (0.016 0.098) ft 3 /s 2 ft 2 146 Chapter 3 Integral Relations for a Control Volume
One-Dimensional Momentum Flux Net Pressure Force on a Closed Control Surface solids cut by the control surface plus all body forces (gravity and electromag- netic) acting on the masses within the control volume. 3. The entire equation is a vector relation; both the integrals are vectors due to the term V in the integrands. The equation thus has three components. If we want only, say, the x component, the equation reduces to F x d d t CV u d CS u ( V r n ) dA (3.36) and similarly, F y and F z would involve v and w , respectively. Failure to ac- count for the vector nature of the linear-momentum relation (3.35) is probably the greatest source of student error in control-volume analyses. For a fixed control volume, the relative velocity V r V , and F d d t CV V d CS V ( V n ) dA (3.37) Again we stress that this is a vector relation and that V must be an inertial-frame ve- locity. Most of the momentum analyses in this text are concerned with Eq. (3.37). By analogy with the term mass flow used in Eq. (3.28), the surface integral in Eq.
- Fall '16
- Prof. A. Dasgupta
- Fluid Dynamics, Fig