# AA210A Fundamentals of Compressible Flow. Chapter 5 -The conservation equations

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1 AA210A Fundamentals of Compressible Flow Chapter 5 -The conservation equations 1

2 5.1 Leibniz rule for differentiation of integrals Differentiation under the integral sign. According to the fundamental theorem of calculus if then Similarly if then 2

3 Suppose the function depends on two variables where the limits of integration are constant. The derivative of the integral with respect to time is But suppose the limits of the integral depend on time. 3

4 From the chain rule. In this case the derivative of the integral with respect to time is Time rate of change due to movement of the boundaries. 4

5 In three dimensions Leibniz rule describes the time rate of change of the integral of some function of space and time, F, contained inside a control volume V. 5

6 Consider a fluid with the velocity field defined at every point. Let the velocity of each surface element coincide with the fluid velocity. This is called a Lagrangian control volume. Use Gauss s theorem to convert the surface integral to a volume integral. Reynolds transport theorem 6

7 5.2 Conservation of mass The Reynolds transport theorem applied to the density is Since there are no sources of mass contained in the control volume and the choice of control volume is arbitrary the kernel of the integral must be zero. This is the general procedure that we will use to derive the differential form of the equations of motion. 7

8 Incompressible flow Expand the continuity equation. If the density is constant then the continuity equation reduces to 8

9 5.3 Conservation of momentum The stress tensor in a fluid is composed of two parts; an isotropic part due to the pressure and a symmetric part due to viscous friction. where We deal only with Newtonian fluids for which the stress is linearly related to the rate-of-strain. where 9

10 Notice that viscous forces contribute to the normal stresses through the non-zero diagonal terms in the stress tensor. Sum the diagonal terms to generate the mean normal stress The bulk viscosity that appears here is often assumed to be zero. This is the so-called Stokes hypothesis. In general the bulk viscosity is not zero except for monatomic gases but the Stokes hypothesis is often invoked anyway. 10

11 The rate of change of the total amount of momentum inside the control volume is determined by the external forces that act on the control volume surface. Use the Reynolds transport theorem to replace the left-handside and Gauss s theorem to replace the surface integrals. Since there are no sources of momentum inside the control volume and the choice of control volume is arbitrary, the kernel must be zero. 11

12 5.4 Conservation of energy The rate of change of the total energy inside the control volume is determined by the rate at which the external forces do work on the control volume plus the rate of heat transfer across the control volume surface. In a linear heat conducting medium Again, use the Reynolds transport theorem to replace the lefthand-side and Gauss s theorem to replace the surface integrals. 12

13 We make the usual argument. Since there are no sources of energy inside the control volume and the choice of control volume is arbitrary, the kernel must be zero. Stagnation enthalpy 13

14 Typical gas transport properties at 300K and one atmosphere. 14

15 5.5 Summary - differential equations of motion 15

16 5.6 Integral form of the equations of motion Recall the Leibniz rule Integral equations on an Eulerian control volume If the surface of the control volume is fixed in space, ie, the velocity of the surface is zero then This is called an Eulerian control volume. 16

17 The integral form of the continuity equation on an Eulerian control volume is derived as follows. Let F= density Use the differential equation for continuity to replace the partial derivative inside the integral on the right-hand-side Use the Gauss theorem to convert the volume integral to a surface integral. The integral form of the continuity equation is: 17

18 The integral form of the conservation equations on an Eulerian control volume is 18

19 5.6.2 Mixed Eulerian-Lagrangian control volumes The integral form of the continuity equation on a Mixed Eulerian-Lagrangian control volume is derived as follows. Let F in Liebniz rule be the fluid density. Use the differential equation for continuity to replace the partial derivative inside the first integral on the right-handside and use the Gauss theorem to convert the volume integral to a surface integral. The integral form of the continuity equation on a Mixed Eulerian-Lagrangian control volume is 19

20 The integral equations of motion on a general control volume where the surface velocity is not the same as the fluid velocity are derived in a similar way. The most general integral form of the conservation equations is Remember U A is the velocity of the control volume surface. 20

21 5.7 Applications of control volume analysis Example 1 - Solid body at rest, steady flow Integral form of mass conservation Integral form of momentum conservation 21

22 Momentum fluxes in the streamwise and normal directions are equal to the lift and drag forces exerted by the flow on the body. 22

23 5.7.2 Example 2 -Channel flow with heat addition Mass conservation Energy conservation 23

24 To a good approximation the energy balance becomes Most of the conductive heat transfer is through the wall. The energy balance reduces to When the vector multiplication is carried out the energy balance becomes The heat addition (or removal) per unit mass flow is equal to the change in stagnation enthalpy of the flow. 24

25 5.7.3 Example 3 - A Rotating fan in a stationary flow The control volume is attached to and moves with the fan surface. The integrated mass fluxes are zero. Momentum fluxes are equal to the surface forces on the fan 25

26 The vector force by the flow on the fan is The integrated energy fluxes are equal to the work done by the flow on the fan. The flow and fan velocity on the fan surface are the same due to the no-slip condition. If the flow is adiabatic and work by viscous normal stresses is neglected the energy equation becomes. The work per unit mass flow is equal to the change in stagnation enthalpy of the flow. 26

27 5.7.4 Example 3 - Combined heat transfer and work In a general situation with heat transfer and work 27

28 Example - Drag of a 2-D airfoil 28

29 29

30 30

31 5.8 Stagnation enthalpy, temperature and pressure Stagnation enthalpy of a fluid element Using the energy equation Along with the continuity equation and the identity One can derive the transport equation for the stagnation enthalpy. 31

32 5.8.2 Blowdown from a pressure vessel revisited Assume the divergence term is small then Neglect the kinetic energy and assume the temperature and pressure are constant over the interior of the sphere. Then Which is the Gibbs equation for an isentropic process. If the gas is calorically perfect 32

33 5.8.3 Stagnation enthalpy and temperature in steady flow For flow in an adiabatic streamtube Since the mass flow at any point in the tube is the same then the stagnation enthalpy per unit mass is also the same and we would expect Neglecting viscous work, the stagnation enthalpy is conserved along an adiabatic path. 33

34 Stagnation or total temperature If the heat capacity is constant in the range of temperatures between T and T t C p T t + const = CpT +const+(1/2)u i U i The enthalpy constant is the same on both sides and can be cancelled. In terms of the Mach number 34

35 5.8.4 Frames of reference Stagnation temperature in fixed and moving frames Transformation of kinetic energy Stagnation temperature 35

36 5.8.5 Stagnation pressure Gibbs equation along an isentropic path Integrate the pressure term 36

37 If an element of fluid is brought to rest isentropically If the heat capacity is constant Changes in the stagnation state are related by the Gibbs equation. 37

38 5.8.6 Transforming the stagnation pressure between fixed and moving frames Divide out the static pressure and temperature 2 38

39 5.9 Problems 39

40 40

41 41

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