CONSERVATION OF ENERGY FOR ACONTINUUM

Chapter 6 CONSERVATION OF ENERGY FOR ACONTINUUM Figure 6.1: 6.1 Conservation of Energ In order to define conservation of energ, we will follow a derivation similar to those in previous chapters, using a differential volume element in Cartesian coordinates. Energ ma take on several forms that are generall associated with mechanical and thermal characteristics such as 137

138 CHAPTER 6. CONSERVATION OF ENERGY FOR A CONTINUUM internal energ due to deformation; pressure, temperature or volume change, etc., kinetic energ due to motion, work due to external forces, tractions or bod forces. Consider the various forms of energ flux associated with the continuum differential volume. Onl the x-component is listed below flow through area z: Internal energ of the mass flux: ρv x Û Û Internal Energ mass Kinetic energ of the mass flux: ρv x KE KE Kinetic Energ mass m 2 s m 2 s Work rate power done b boundar tractions: t i v Heat energ flux b conduction: q x m 2 s Work rate power done b the bod forces: ρg v m 3 s Radiation source or other bod heat source or sink: ρφ m 2 s m 3 s Note: Other heat energ flux terms that could be considered include convection and radiation. Consistent with most thermodnamics treatments of energ flow, we assume that energ leaving a sstem is positive. Summing all of the energ entering and leaving the element x z during the time period t and setting this equal to the change in energ of the sstem differential element gives: = + + { [ ] ρ t+ t { [ ρv x ] { [ ρv ] x [ ] } t ρ x z ] } x+ x [ρv x z t ] } + [ρv z x t { [ ] ] } z ρv z [ρv z+ z z x t + ρg v x z t 6.1 { ti + v x+ x t i v } { tj x z t + v + t j v } z x t { tk + v z+ z t k v } z x t + { } { } q x x q x x+ x z t + q q + z x t + { } q z z q z z+ z x t + ρφ x z t Take the limit as x z 0 and as t 0toobtain:

6.1. CONSERVATION OF ENERGY 139 [ ] ρ ] [ρv x = + [ ti v + + t j v ] [ρv + t k v ] ] [ρv z + + ρg v 6.2 [ qx + q + q ] z + ρφ The partial derivative of product terms like ρv x and ρ ma be expanded using the product rule and the traction terms can be replaced b Cauch stresses to obtain: ρ + ρ = ρv x + + { ρvx + ρv i + j + k ρv + + ρv } Û z + KE σ v+ρg v q + ρφ ρv z 6.3 Note that conservation of mass is given b: [ ρ = ρvx + ρv + ρv ] z Using conservation of mass, the terms indicated ma be struck from equation 6.3. The energ term from the right side of the equation is moved to the left and equation 6.3 is simplified b writing all terms in their vector form to obtain: Conservation of Energ ρ v + ρ = ρg v + ρφ q + σ v 6.4 The above equation is the total energ form of the conservation of energ. Note that the energ equation is a scalar equation. The total energ conservation equation ma be further simplified if conservation of linear momentum is applied. Thus, we start with the conservation of linear momentum equations with the terms rearranged as given follows: ρg = ρ v + ρ v v σ. 6.5 Find the dot product of velocit and each side of the linear momentum equation and equate them to obtain

140 CHAPTER 6. CONSERVATION OF ENERGY FOR A CONTINUUM ρg v = ρ v + ρ v v σ v. 6.6 Some of the terms in the above equation can be identified as kinetic energ. Note that the kinetic energ per unit mass can be written as Consequentl, terms like v v x v v = v x 1 2 v v KE = kinectic energ per unit mass = v v ma be written as 1 2massv v = 1 v v mass 2 v = 1 2 v v = v KE x. Hence, equation 6.6 ma be written as: = KE and similarl ρg v = ρ v + ρ v v v σ v = ρ 1 1 2 v v + ρ v 2 v v σ v 6.7 = ρ KE + ρ v KE σ v Substituting equation 6.7 into the vector form of conservation of energ 6.4, the kinetic energ terms cancel out, and the final result is where ρ v Û + Û = ρφ q + σ v σ v 6.8 ρ v Û + ρ Û is the total rate of internal energ accumulation per unit volume ρφ isthe net energ input rate due to radiation per unit volume q is the net energ input rate due to heat conduction per unit volume σ v isthe net energ input rate due to traction per unit volume σ v is the net input rate of mechanical energ due to traction per unit volume Note that the sum of the last two terms, σ v σ v is the net conversion rate of mechanical energ to thermal energ per unit volume. This term is also equal to tr σ v where tr = trace of a matrix = sum of the diagonal terms see Appendix. The above form of the conservation of energ equation ma also be written in terms of pressure P and the deviatoric or extra stress tensor, S. Asnoted above, the last two terms in 6.8 can be written as σ v σ v = trσ v. The relationship between Cauch and deviatoric stress ma be utilized to write: tr σ v =tr[s PI v] =trs v P v 6.9 Substituting the last result 6.9 into 6.8, one obtains for conservation of energ ρ Û +v Û = ρφ q +trs v P v 6.10

6.1. CONSERVATION OF ENERGY 141 Equations 6.8 and 6.10 provide two equivalent forms of the conservation of energ. Equation 6.10 is more useful for fluid applications since it is written in terms of the average hdrostatic pressure P. Unfortunatel, the two conservation of energ equations [6.8 and 6.10] are not ver useful in their current form since the contain internal energ Û, which is not a directl observable quantit. The internal energ must be related to observable measurable quantities such as heat capacit ĈV or ĈP, pressure P, specific volume V ortemperature T. This requires the use of thermodnamics equations that relate Û and other thermodnamics properties to measurable variables such as P, V and T. Utilizing prior work on thermodnamics ENGR 212, the heat capacit at constant volume ma be defined as: Û Ĉ V = T Internal energ, temperature, pressure and volume are then related as follows: V 6.11 Û Û = Û T T = ĈV + U V T + V T P T V V P 6.12 and the gradient of internal energ is given b Û = ĈV T + T P P T V 6.13 V Utilizing 6.11-6.13, the final form of conservation of energ for a constant volume process is given b: [ ] T P ρĉ + v T = ρφ q T v +trs v 6.14 T V Note that T P T V v is the conversion of mechanical energ to thermal energ due to compression of the material. If the material is incompressible, this term becomes zero because when ρ = constant, v =0from conservation of mass. Also, for incompressible materials, Ĉ V = ĈP = Ĉ. Consequentl, for incompressible continua, conservation of energ becomes: [ ] T ρĉ + v T = ρφ q +trs v 6.15 where tr S v represents the conversion of mechanical energ to thermal energ due to viscous friction rate of dissipation. For the static case v = 0, the conservation of energ equation becomes: ρĉ T = ρφ q 6.16 The last result, equation 6.16, represents the most common cases of conservation of thermal energ in solids, which will be considered further in the following chapter. A note on the derivation of equations 6.8 and 6.10

142 CHAPTER 6. CONSERVATION OF ENERGY FOR A CONTINUUM For students interested in some of the vector and matrix operations leading to the term σ v σ v =trσ v, the derivation is provided below: [ ] σ v = σ xx σ x σ xz σ x σ σ z v x v σ zx σ z σ zz v z = [ ] σ xxv x + σ x v + σ xz v z σ x v x + σ v + σ z v z σ zx v x + σ z v + σ zz v z 1 = σ xxv x + σ x v + σ xz v z + σ xv x + σ v + σ z v z + σ zxv x + σ z v + σ zz v z σ v = [ ] σ xx σ x σ xz σ x σ σ z v x v σ zx σ z σ zz v z σxx 2 = + σ x + σ zx σx v x + + σ + σ z σxz v + + σ z + σ zz v z Now subtract 2 from 1 to get: σ v σ v = σ xx + σ v x + σ v z xz + σ x + σ v + σ z v z + σ zx The two tensors σ and v are dotted together, and all terms are shown below. σ xx σ x σ xz σ x σ σ z v v v σ zx σ z σ zz = σ xx σ x σ zx + σ x v + σ v + σ z v v z + σ xz vz + σ z vz + σ zz vz v z v z σ xx σ x σ zx + σ x v + σ v + σ z v + σ xz vz + σ z vz + σ zz vz + σ v z + σ v z zz σ xx σ x σ zx + σ x v + σ v + σ z v + σ xz vz + σ z vz + σ zz vz Notice that the trace is simpl the sum of the diagonal terms of the resulting matrix, as shown below. trσ v = σ xx + σ v x + σ v z xz + σ x + σ v + σ v z z + σ zx + σ v z + σ v z zz Now observe that σ v σ v = trσ v. 6.2 Heat Transfer in a Solid Heat flux is a vector quantit since heat flows in all directions: q = q x i + q j + q z k 6.17 Heat flux is defined as the amount of heat described b the vector q which flows through a given area that is described b its unit normal n. As with all flux terms, the onl component of

6.2. HEAT TRANSFER IN A SOLID 143 q q q n n q x q z x z Figure 6.2: Heat Flux through surface n heat flux that passes through an area with unit normal n is that component that is in the direction of n. This requirement is easil written in terms of the dot product q with n, which will give the component of q in the direction of n. q n = q n 6.18 Suppose q =5i s-m 2 a Find the heat flux through plane i. Example 6-1 b Find the heat flux through plane j. c Find the heat flux through plane n n = cos π 4 i + sin π 4 j. a q i =5i i =5 s-m 2 b q j =5i j =0 s-m 2 c q n =5i cos π 4 i + sin π 4 j = 5 2 2 s-m 2 Solution Heat Transfer Recall from the discussion above equation 6.16 that for an incompressible bod in static equilibrium v = 0 the conservation of energ simplifies to: ρĉ T = q + ρφ 6.19 Unfortunatel, the above equation cannot be solved b itself because there are seven unknowns: T, q 3 components, Φ; ρ and C. Assuming that ρ remains constant incompressible bod one can independentl measure its value. The same can be done for C. The heat source term Φ can normall be specified if the phsical nature of the source is known. This leaves one equation with four unknowns, T, q x, q, and q z. The previousl derived conservation equations for mass and momentum

144 CHAPTER 6. CONSERVATION OF ENERGY FOR A CONTINUUM can not help us, because the involve more unknowns than equations. In fact, if we consider all conservation laws we have eight equations 1 COM+3 COLM + 3 COAM + 1 COE = 8 and 17 unknowns 1 ρ +3 v +9 σ +1 T +3 q = 17. This is a tpical circumstance that often occurs in nature: conservation laws are in and of themselves insufficient to predict the response of continua except in special cases. This is due to the fact that conservation laws sa nothing regarding material behavior. Thus, additional equations must be specified regarding the material makeup. Those equations generall are based on experimental observation. For the heat transfer case, Fourier made the observation in 1822 that for a broad range of solid materials the heat flux is proportional to the negative of the temperature gradient: Fourier s Law of Heat Conduction: q = k T or q x = k T, q = k T, q z = k T 6.20 q x k=slope=thermal conductivit dt dx Figure 6.3: Plot of 1-D Heat Flux vs. Temperature Gradient Fourier s Law The parameter k is called the thermal conductivit, which is a measured material propert that is generall either constant or a weak function of temperature. The negative sign in equation 6.20 is a convenience that arises due to the second law of thermodnamics, wherein it can be shown that k is never negative, due to the fact that heat alwas flows from hot to cold. Substituting 6.20 into 6.19 results in ρĉ T = k T +ρφ 6.21 If k = constant, 6.21 reduces to Conservation of Energ in a Solid with Constant Thermal Conductivit ρĉ T 2 = k 2 T + ρφ =k 2 + 2 2 + 2 2 T + ρφ 6.22 It can be seen that the above represents a single partial differential equation that can be used to solve for the unknown primar variable, T = T x,, z, t. As with an differential equation, the complete solution will require knowledge of the boundar conditions for the problem. That is, the temperature T or its spatial gradient must be specified or known on the phsical boundar of the problem. Applications and the solution of equation 6.22 together with appropriate boundar conditions will be considered the next chapter. The secondar variables q x, q, q z can subsequentl be obtained using 6.20.

6.2. HEAT TRANSFER IN A SOLID 145 Example 6-2 A 1-m long aluminum plate is insulated so that heat flows onl in the x direction. The temperature on the left boundar is 100 C and on the right boundar 0 C. The heat flux is measured to be 24700 under stead state conditions. m 2 s insulated q = 0 Aluminum Material 100 0 0 1 x insulated q = 0 q Aluminum =24,700 T=Tx 2 m s Figure 6.4: T = T x q Aluminum = 24700 m 2 s a Plot the temperature with respect to position. b Write the equation for T x. c For the given q value, solve for the thermal conductivit, k, in this situation. a Solution 100 0 0 1 x Figure 6.5: b The equation for the temperature is simpl a straight line: T x = 100x + 100 C. c Use Fourier s Law, equation 6.20, to obtain the temperature gradients and make use of the specified heat flux to obtain:

146 CHAPTER 6. CONSERVATION OF ENERGY FOR A CONTINUUM 24700 m 2 s [ T q = k T = k = k 100 k = 247 m-s- C C m i + T j + T ] k

6.2. HEAT TRANSFER IN A SOLID 147 Deep Thought I bet if we could capture the energ used in tapping one s pencil incessantl we could power the entire state of North Dakota.

148 CHAPTER 6. CONSERVATION OF ENERGY FOR A CONTINUUM 6.3 Questions 6.1 In our own words, describe the conservation of energ. 6.2 Which wa does heat flux? 6.3 What are the SI dimensions for the convection coefficient, h? 6.4 What tpe of equation is Fourier s Law of heat conduction and explain its phsical significance? 6.5 Wh is there a negative sign in Fourier s Law of heat conduction? 6.4 Problems 6.6 Suppose q =8 [ m s] j 2 a Find the heat flux through plane i. b Find the heat flux through plane j. c Find the heat flux through plane n where n = cos 30i + sin 30j. 6.7 Suppose a heat flux vector is given b: q =2x 2 + i +z 2 +3 2 j +5k Determine the heat flux across the following surfaces at the point given b r = i + j + k 6.8 a x =1 b z = 2 c x 2 + 2 + z 2 =3 2m insulated q = 0 Aluminum Material 150 C 20 C 0m 3m x q Aluminum = 24,700 m s T=Tx 2 Problem 6.8 T = T x q Aluminum = 24700 m-s a Plot the temperature with respect to position. b Write the equation for T x. c For the given q value, solve for the thermal conductivit, k, inthis situation.

6.4. PROBLEMS 149 6.9 A slab of thickness, t. Inside the slab, heat is being generated at a rate of ax 2 BTU hr-ft 3, where: a is a constant. One side of the slab is insulated. The other side of the slab is maintained at a constant temperature, T 0. a Draw a sketch of the problem including a useful coordinate sstem. b Determine the governing differential equation to be used and list an assumptions made. c List all available boundar conditions. d Determine T x and qx e Plot T x vs. x and qx vs. x f What additional information would be needed to determine the heat transfer rate for the non-insulated side of the slab 6.10 For Parts 1-3, suppose q =12i +6j +9k m 2 -s a Find the heat flux through plane k. b Find the heat flux through n if n = cos 30 i + sin 30 j. c Find the heat flux through n if n =3i +5k. For d g, use the following diagram: 10 ft 30 Copper Insulation q x = 0 0ft 150 5ft x Problem 6.10 d Of what is the temperature a function? Explain. e Plot the temperature with respect to position. f Write the equation for the temperature. g If q Cu = 39800 m 2 -s, solve for the thermal conductivit, k, inthis situation. 6.11 A heat flux vector, q, isgiven b: q =2x 2 + 2 i +5xj 2z 3 k A surface is defined b: f x,, z =x 2 + 2 + z 2 =3 Apoint, r, isgiven b: r = i + j + k a Evaluate the heat flux vector at point r b Determine the component vector of q fluxing across the surface at point r c Complete a sketch which includes the surface and the results of a and b 6.12 In each case, a through c, a specific surface and heat flux are given. Find their respective heat transfer rates and draw a rough sketch, showing the surface indicated and the heat flux vector at one point on the surface.

150 CHAPTER 6. CONSERVATION OF ENERGY FOR A CONTINUUM a q = q0 R e r. The surface is a clindrical surface of radius R and length L with the z-axis as the clinder s axis of smmetr. b q =4i +6j k. The surface is a plane normal to the x-axis and of width a in the z-direction and height b in the -direction. c q = 4i + 6j k. The surface is a plane parallel to the z-axis which makes equal angles with the x and axes and which is of height b in the -direction and width a.