Fluid Mechanics
FFlluiuid Mechd Mechananiics cs
Chapter 6 Momentum Equation Momentum Equation: Derivation (Read Chapter 6 and Section 7.6).6)
1 MomenMomenttumum EqEquauattionion:: DDererivivaattionion
Ø For the analysis of forces and moments produced by flowing fluids Ø RTT is applied to Newton’ s 2nd law of motion (F=ma) Ø Develop the Eulerian form of the momentum equation
2 MomenMomenttumum EqEquauattionion:: DDererivivaattionion
Ø Principle: The rate of change of momentum of a body is equal to the force acting on the body Ø Newton’ s second law of motion: ΣF = mm a
l where FF and a
l and mass is constant Ø Writing this in terms of momentum for a single particle… …
3 MomenMomenttumum EqEquauattionion:: DDererivivaattionion
Ø Writing this in terms of momentum for a group of particles (like a fluid system)m) …
l where Mom sys id the total momentum of all mass comprising the system
l This is the Lagrangian equation
4 MomenMomenttumum EqEquauattionion:: DDererivivaattionion
Ø Now use RTT to derive the Eulerian eqn.
l where B sys = Mom sys l and b = vector velocity (v)
l Note: velocity must be relative to an inertial reference frame (does not rotate) and can be either stationary or moving at a constant velocity..
5 MomenMomenttumum EqEquauattionion:: DDererivivaattionion
Ø Now use RTT to derive the Eulerian eqn.
Ø Substitute B sys and b… …
6 MomenMomenttumum EqEquauattionion:: DDererivivaattionion
Ø Combining...
Ø Yields the integral form of the momentum equation… …
7 MomenMomenttumum EqEquauattionion:: DDererivivaattionion
Ø Integral form of the momentum equation…
Ø Can be expressed in words as
Ø Note: the momentum equation is a vector equation (i.e. there is direction associated with it)
8 MomenMomenttumum EqEquauattionion:: DDererivivaattionion
Ø If velocity is uniformly distributed across each port, then…
Ø becomes...
l where “o o ” refers to outlet and “i i ” refers to inlet
l Note that ṁṁ v has the same units as F and is the momentum per time
l Applies to any cv including one that is moving, deforming or bothh
9 MomenMomenttumum EqEquauattionion:: DDererivivaattionion
Ø The three components of…
Ø in cartesian coordinates are...
10 FFlluiuid Mechd Mechananiics cs
Chapter 6 Momentum Equation Momentum Equation: Interpretation (Read Chapter 6 and Section 7.6).6)
11 ForForcce e TerTermsms
Ø Terminology
l Force diagramm : shows the forces acting on the matter contained within a control volume. • A force diagram is equivalent to a free b ody diagram at the instant in time when the momentum equation is applied.
l Body force: A force that acts on mass elements within the body is defined (gravity)
l Surface forcerce : as a force that requires physical contact, meaning that surface forces act at the control surface ( ( pAA )
12 ForForcce e TerTermsms
Ø Consider the force diagrams (FD) in (b) and (c) Ø What are the forces acting on the pipe of D diameter and L length?
13 ForForcce e TerTermsms
Ø Figure (b): cv inside the pipe
14 ForForcce e TerTermsms
Ø Figure (c): cv outside the pipe
15 ForForcce e TerTermsms
Ø Choice of cv is dependent on what you want to know. Ø (b) to relate the pressure change between 1 & 2 to wall shear stress Ø (c) to calculate tensile force carried by pipee
16 MomenMomenttumum AAccccumuumulatilationon
Ø The momentum accumulation term is the first term on the right side of = sign
Ø It is the momentum of the material inside the control volume Ø Many problem it equals zero if
l Steady flow through a control volume and
l Stationary structure and
l Constant density
17 MomenMomenttumum DDiiagagraram m
Ø Momentum diagramm : created by sketching a control volume and then drawing a vector to represent the momentum accumulation term and a vector to represent momentum flow at each section where mass crosses the control surface
18 MomenMomenttumum DDiiagagraram m
Ø Consider the moment diagram beloww …
Ø And assume that the velocity is constant. Ø Therefore, the momentum accumulation term = 0
19 MomenMomenttumum DDiiagagraram m
Ø Outlet momentum flow is
Ø Inlet momentum flow iss
20 MomenMomenttumum DDiiagagraram m
Ø Therefore, net outward flow of momentum iss ……
21 IInn cclalassss ExaExammpleple ProbProblelem m
Ø A horizontal circular jet of water at 15 0 C is discharged from a tank and then strikes a vertical plate. Ø The axis of the jet is normal to the plate Ø The cross s ectional area of the jet is 0.015m 2 . Ø What velocity in the jet is needed to produce a force of 2000N on the plate? Ø What pressure at A in the tank is needed to produce this velocity??
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