5.6. DIFFERENTIAL EQUATION OF ENERGY CHANGE

5.6A. Introduction

In Sections 3.6 and 3.7 we derived a differential equation of continuity and a differential equation of momentum transfer for a pure fluid. These equations were derived because overall mass, energy, and momentum balances made on a finite volume in the earlier parts of Chapter 2 did not tell us what goes on inside a control volume. In the overall balances performed, a new balance was made for each new system studied. However, it is often easier to start with the differential equations of continuity and momentum transfer in general form and then simplify the equations by discarding unneeded terms for each specific problem.

In Chapter 4 on steady-state heat transfer and Chapter 5 on unsteady-state heat transfer, new overall energy balances were made on a finite control volume for each new situation. To progress further in our study of heat or energy transfer in flow and nonflow systems, we must use a differential volume to investigate in greater detail what goes on inside this volume. The balance will be made on a single phase and the boundary conditions at the phase boundary will be used for integration.

In the next section we derive a general differential equation of energy change: the conservation-of-energy equation. Then this equation is modified for certain special cases that occur frequently. Finally, applications of the uses of these equations are given. Cases for both steady-state and unsteady-state energy transfer are studied using this conservation-of-energy equation, which is perfectly general and holds for steady- or unsteady-state conditions.

5.6B. Derivation of Differential Equation of Energy Change

As in the derivation of the differential equation of momentum transfer, we write a balance on an element of volume of size Δ×, Δy, Δz which is stationary. We then write the law of conservation of energy, which is really the first law of thermodynamics for the fluid in this volume element at any time. The following is the same as Eq. (2.7-7) for a control volume given in Section 2.7.

Equation 5.6-1


As in momentum transfer, the transfer of energy into and out of the volume element is by convection and molecular transport or conduction. There are two kinds of energy being transferred. The first is internal energy U in J/kg (btu/lbm) or any other set of units. This is the energy associated with random translational and internal motions of the molecules plus molecular interactions. The second is kinetic energy ρν2/2, which is the energy associated with the bulk fluid motion, where ν is the local fluid velocity, m/s (ft/s). Hence, the total energy per unit volume is (ρU + ρν2/2). The rate of accumulation of energy in the volume element in m3 (ft3) is then

Equation 5.6-2


The total energy entering by convection in the x direction at x minus that leaving at x + Δx is

Equation 5.6-3


Similar equations can be written for the y and z directions using velocities νy and νz, respectively.

The net rate of energy entering the element by conduction in the x direction is

Equation 5.6-4


Similar equations can be written for the y and z directions, where qx, qy, and qz are the components of the heat flux vector q, which is in W/m2 (btu/s · ft2) or any other convenient set of units.

The net work done by the system on its surroundings is the sum of the following three parts for the x direction. For the net work done against the gravitational force,

Equation 5.6-5


where gx is gravitational force. The net work done against the static pressure p is

Equation 5.6-6


where p is N/m2 (lbf/ft2) or any other convenient set of units. For the net work against the viscous forces,

Equation 5.6-7


In Section 3.7 these viscous forces are discussed in more detail.

Writing equations similar to (5.6-3)–(5.6-7) in all three directions; substituting these equations and Eq. (5.6-2) into (5.6-1); dividing by Δx, Δy, and Δz; and letting Δx, Δy, and Δz approach zero, we obtain

Equation 5.6-8


For further details of this derivation, see (B2).

Equation (5.6-8) is the final equation of energy change relative to a stationary point. However, it is not in a convenient form. We first combine Eq. (5.6-8) with the equation of continuity, Eq. (3.6-23), with the equation of motion, Eq. (3.7-13), and express the internal energy in terms of fluid temperature T and heat capacity. Then writing the resultant equation for a Newtonian fluid with constant thermal conductivity k, we obtain

Equation 5.6-9


This equation utilizes Fourier's second law in three directions, where

Equation 5.6-10


The viscous-dissipation term μφ is generally negligible except where extremely large velocity gradients exist. It will be omitted in the discussions to follow. Equation (5.6-9) is the equation of energy change for a Newtonian fluid with constant k in terms of the fluid temperature T.

5.6C. Special Cases of the Equation of Energy Change

The following special forms of Eq. (5.6-9) for a Newtonian fluid with constant thermal conductivity are commonly encountered. First, Eq. (5.6-9) will be written in rectangular coordinates without the μφ term:

Equation 5.6-11


1. Fluid at constant pressure

The equations below can be used for constant-density fluids as well as for constant pressure.

Equation 5.6-12


In rectangular coordinates,

Equation 5.6-13


In cylindrical coordinates,

Equation 5.6-14


In spherical coordinates,

Equation 5.6-15


For definitions of cylindrical and spherical coordinates, see Section 3.6. If the velocity ν is zero, DT/Dt becomes ∂T/∂t.

2. Fluid at constant density

Equation 5.6-16


Note that this is identical to Eq. (5.6-12) for constant pressure.

3. Solid

Here we consider ρ is constant and ν = 0.

Equation 5.6-17


This is often referred to as Fourier's second law of heat conduction. This also holds for a fluid with zero velocity at constant pressure.

4. Heat generation

If there is heat generation in the fluid by electrical or chemical means, then can be added to the right side of Eq. (5.6-17).

Equation 5.6-18


where is the rate of heat generation in W/m3 (btu/h · ft3) or other suitable units. Viscous dissipation is also a heat source, but its inclusion greatly complicates problem solving because the equations for energy and motion are then coupled.

5. Other coordinate systems

Fourier's second law of unsteady-state heat conduction can be written as follows.

For rectangular coordinates,

Equation 5.6-19


where α = k/pcp, thermal diffusivity in m2/s (ft2/h).

For cylindrical coordinates,

Equation 5.6-20


For spherical coordinates,

Equation 5.6-21


5.6D. Uses of Equation of Energy Change

In Section 3.8 we used the differential equations of continuity and motion to set up fluid-flow problems. We did this by discarding the terms that are zero or near zero and using the remaining equations to solve for the velocity and pressure distributions. This was done instead of making new mass and momentum balances for each new situation. In a similar manner, to solve problems of heat transfer, the differential equations of continuity, motion, and energy will be used, with the unneeded terms being discarded. Several examples will be given to illustrate the general methods used.

EXAMPLE 5.6-1. Temperature Profile with Heat Generation

A solid cylinder in which heat generation is occurring uniformly as W/m3 is insulated on the ends. The temperature of the surface of the cylinder is held constant at Tw K. The radius of the cylinder is r = R m. Heat flows only in the radial direction. Derive the equation for the temperature profile at steady state if the solid has a constant thermal conductivity.

Solution: Equation (5.6-20) will be used for cylindrical coordinates. The term /ρcp for generation will be added to the right side, giving

Equation 5.6-22


For steady state ∂T/∂t = 0. Also, for conduction only in the radial direction, ∂2T/∂z2 = 0 and ∂2T/∂θ2 = 0. This gives the following differential equation:

Equation 5.6-23


This can be rewritten as

Equation 5.6-24


Note that Eq. (5.6-24) can be rewritten as follows:

Equation 5.6-25


Integrating Eq. (5.6-25) once,

Equation 5.6-26


where K1 is a constant. Integrating again,

Equation 5.6-27


where K2 is a constant. The boundary conditions are when r = 0, dT/dr = 0 (by symmetry), and when r = R, T = Tw. The final equation is

Equation 5.6-28


This is the same as Eq. (4.3-29), which was obtained by another method.


EXAMPLE 5.6-2. Laminar Flow and Heat Transfer Using Equation of Energy Change

Using the differential equation of energy change, derive the partial differential equation and boundary conditions needed for the case of laminar flow of a constant-density fluid in a horizontal tube which is being heated. The fluid is flowing at a constant velocity νz. At the wall of the pipe where the radius r = r0, the heat flux is constant at q0. The process is at steady state and it is assumed at z = 0 at the inlet that the velocity profile is established. Constant physical properties will be assumed.

Solution: From Example 3.8-3, the equation of continuity gives ∂νz/∂z = 0. Solution of the equation of motion for steady state using cylindrical coordinates gives the parabolic velocity profile:

Equation 5.6-29


Since the fluid has a constant density, Eq. (5.6-14) in cylindrical coordinates will be used for the equation of energy change. For this case νr = 0 and νθ = 0. Since this will be symmetrical, ∂T/∂θ and ∂2T/∂θ2 will be zero. For steady state, ∂T/∂t = 0. Hence, Eq. (5.6-14) reduces to

Equation 5.6-30


Usually conduction in the z direction (∂2T/∂z2 term) is small compared to the convective term νzT/∂z and can be dropped. Finally, substituting Eq. (5.6-29) into (5.6-30), we obtain

Equation 5.6-31


The boundary conditions are


For details on the actual solution of this equation, see Siegel et al. (S2).


..................Content has been hidden....................

You can't read the all page of ebook, please click here login for view all page.
Reset
44.213.80.203