BOOKCOMP, Inc. — John Wiley & Sons / Page 20 / 2nd Proofs / Heat Transfer Handbook / Bejan 20 BASIC CONCEPTS 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [20], (20) Lines: 940 to 1001 ——— 0.99356pt PgVar ——— Normal Page PgEnds: T E X [20], (20) F = m d ˆ V dt = d(m ˆ V) dt (1.64) where mV is the momentum. Equation (1.64) is the statement of the conservation of momentum principle. Note that the conservation of momentum principle is stated in terms of the proper- ties of particles and not in terms of the properties of a field. To derive the momentum theorem, a region in a fluid confined by the control surface S 1 , shown in Fig. 1.7, is employed. The surface S 1 contains a definite and fixed number of particles at time t 1 . At time t 2 , these particles will have moved to a region bounded by the control surface S 2 , which is shown as a dashed curve to distinguish it from S 1 . The control surfaces S 1 and S 2 enclose three separate and distinct regions, desig- nated by a,b, and c. Let the momentum in the three regions be P a , P b , and P c , re- spectively. At time t 1 the particles within surface S 1 will possess momentum P a +P b1 . At time t 2 these particles will have momentum, P b2 + P c because they have moved into the region enclosed by surface S 2 . Hence the momentum change during the time interval t 2 − t 1 may be described by (P b2 + P c ) − (P b1 + P a ) = (P b2 − P b1 ) + (P c − P a ) and the time rate of change of momentum will be lim t 2 →t 1 P b 2 − P b1 t 2 − t 1 + P c − P a t 2 − t 1 (1.65) As t 2 approaches t 1 as a limit, the control surface S 2 will coincide with S 1 . The first term in eq. (1.65) is therefore the time rate of change of momentum of the fluid contained within region 1,R 1 , contained within S 1 . This may be written as the integral over R 1 . Because the mass of fluid contained in R 1 is R 1 ρ dR 1 the time rate of change of momentum of the fluid contained within region 1 will be ∂ ∂t R 1 ρ ˆ V dR 1 The second term in eq. (1.65) is the momentum efflux through the control surface S 1 . If the flux in the outward direction is taken as positive, this efflux can be expressed by the integral S 1 ρ ˆ V ˆ V n dS 1 where V n is the component of velocity normal to S 1 . BOOKCOMP, Inc. — John Wiley & Sons / Page 21 / 2nd Proofs / Heat Transfer Handbook / Bejan MOMENTUM AND THE MOMENTUM THEOREM 21 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [21], (21) Lines: 1001 to 1042 ——— 0.42567pt PgVar ——— Normal Page PgEnds: T E X [21], (21) S 1 S 2 c b a Figure 1.7 Regions bounded by control surfaces used for the development of the momentum theorem. The conservation of momentum principle then becomes F = m d ˆ V dt = ∂ ∂t R 1 ρ ˆ V dR 1 + S 1 ρ ˆ V ˆ V n dS 1 (1.66) or, by rearrangement of terms, ∂ ∂t R 1 ρ ˆ V x dR 1 = F x − S 1 ρ ˆ V x ˆ V n dS 1 (1.67a) ∂ ∂t R 1 ρ ˆ V y dR 1 = F y − S 1 ρ ˆ V y ˆ V n dS 1 (1.67b) ∂ ∂t R 1 ρ ˆ V z dR 1 = F z − S 1 ρ ˆ V x ˆ V n dS 1 (1.67c) in the three rectangular coordinate directions. The foregoing development leads to the statement of the momentum theorem: The time rate of increase of momentum of a fluid within a fixed control volume R will be equal to the rate at which momentum flows into R through its confining surface S, plus the net force acting on the fluid within R. When the flow is incompressible, the viscosity is constant, and the flow is laminar, the Navier–Stokes equations result. In Cartesian coordinates, with F x , F y , and F z taken as the components of the body force per unit volume, the Navier–Stokes equations are ρ ∂ ˆ V x ∂t + ˆ V x ∂ ˆ V x ∂x + ˆ V y ∂ ˆ V x ∂y + ˆ V z ∂ ˆ V z ∂z BOOKCOMP, Inc. — John Wiley & Sons / Page 22 / 2nd Proofs / Heat Transfer Handbook / Bejan 22 BASIC CONCEPTS 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [22], (22) Lines: 1042 to 1077 ——— 1.68011pt PgVar ——— Long Page * PgEnds: Eject [22], (22) =− ∂P ∂x + µ ∂ 2 ˆ V x ∂x 2 + ∂ 2 ˆ V x ∂y 2 + ∂ 2 ˆ V x ∂z 2 + F x (1.68a) ρ ∂ ˆ V y ∂t + ˆ V x ∂ ˆ V y ∂x + ˆ V y ∂ ˆ V y ∂y + ˆ V z ∂ ˆ V y ∂z =− ∂P ∂y + µ ∂ 2 ˆ V y ∂x 2 + ∂ 2 ˆ V y ∂y 2 + ∂ 2 ˆ V y ∂z 2 + F y (1.68b) ρ ∂ ˆ V z ∂t + ˆ V z ∂ ˆ V x ∂x + ˆ V y ∂ ˆ V z ∂y + ˆ V z ∂ ˆ V z ∂z =− ∂P ∂z + µ ∂ 2 ˆ V z ∂x 2 + ∂ 2 ˆ V z ∂y 2 + ∂ 2 ˆ V z ∂z 2 + F z (1.68c) In cylindrical coordinates with F r , F θ , and F z taken as the components of the body force per unit volume, the Navier–Stokes equations are ρ ∂ ˆ V r ∂t + ˆ V r ∂ ˆ V r ∂r + ˆ V θ r ∂ ˆ V r ∂θ − ˆ V 2 θ r + ˆ V z ∂ ˆ V r ∂z =− ∂P ∂r + µ ∂ 2 ˆ V r ∂r 2 + 1 r ∂ ˆ V r ∂r − ˆ V r r 2 + 1 r 2 ∂ 2 ˆ V r ∂θ 2 − 2 r 2 ˆ V θ ∂θ + ∂ 2 ˆ V r ∂z 2 + F r (1.69a) ρ ∂ ˆ V θ ∂t + ˆ V r ∂ ˆ V θ ∂r + ˆ V θ r ∂ ˆ V θ ∂θ − ˆ V r ˆ V θ r + ˆ V z ∂ ˆ V θ ∂z =− 1 r ∂P ∂θ + µ ∂ 2 ˆ V θ ∂r 2 + 1 r ∂ ˆ V θ ∂r − ˆ V θ r 2 + 1 r 2 ∂ 2 ˆ V θ ∂θ 2 + 2 r 2 ∂ ˆ V r ∂θ + ∂ 2 ˆ V θ ∂z 2 + F θ (1.69b) ρ ∂ ˆ V z ∂t + ˆ V r ∂ ˆ V z ∂r + ˆ V θ r ∂ ˆ V z ∂θ + ˆ V z ∂ ˆ V z ∂z =− ∂P dz + µ ∂ 2 ˆ V z ∂r 2 + 1 r ∂ ˆ V z ∂r + 1 r 2 ∂ 2 ˆ V z ∂θ 2 + ∂ 2 ˆ V z ∂z 2 + F z (1.69c) Finally, in spherical coordinates with F r , F θ , and F φ taken as the components of the body force per unit volume and with D DT = ∂ dt + ˆ V r ∂ ∂r + ˆ V φ r ∂ ∂φ + ˆ V θ r sin φ ∂ ∂θ ∇ 2 = 1 r 2 ∂ ∂r r 2 ∂ ∂r + 1 r 2 sin φ ∂ ∂φ sin φ ∂ ∂φ + 1 r 2 sin 2 φ ∂ 2 ∂θ 2 BOOKCOMP, Inc. — John Wiley & Sons / Page 23 / 2nd Proofs / Heat Transfer Handbook / Bejan CONSERVATION OF ENERGY 23 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [23], (23) Lines: 1077 to 1154 ——— 2.01216pt PgVar ——— Long Page * PgEnds: Eject [23], (23) the Navier–Stokes equations are ρ D ˆ V r Dt − ˆ V 2 φ + ˆ V 2 θ r =− ∂P ∂r + µ ∇ 2 ˆ V r − 2 ˆ V r r 2 − 2 r 2 ∂ ˆ V φ ∂φ − 2 ˆ V θ cot φ r 2 − 2 r 2 sin φ ∂ ˆ V θ ∂θ + F r (1.70a) ρ D ˆ V φ Dt − ˆ V r ˆ V φ r − ˆ V 2 θ cot φ r =− 1 r ∂P ∂φ + µ ∇ 2 ˆ V φ + 2 r 2 ∂ ˆ V r ∂φ − ˆ V φ r 2 sin φ − 2 cos φ r 2 sin 2 φ ∂ ˆ V θ ∂θ + F φ (1.70b) ρ D ˆ V θ Dt + ˆ V θ ˆ V r r + ˆ V φ ˆ V θ cot φ r =− ˆ V θ r sin φ ∂P ∂θ + µ ∇ 2 ˆ V θ − ˆ V θ r 2 sin 2 φ + 2 r 2 sin φ ∂ ˆ V r ∂θ + 2 cos φ r 2 sin 2 φ ∂ ˆ V θ ∂θ + F θ (1.70c) 1.5 CONSERVATION OF ENERGY In Fig. 1.8, an imaginary two-dimensional control volume of finite size ∆x ∆y with flow velocity ˆ V = e x ˆ V x +e y ˆ V y , heat flux q = e x q x +e y q y , specific internal energy u, and rate of internal heat generation q , the first law of thermodynamics requires that rate of energy accumulation within the control volume = net transfer of energy by fluid flow + net heat transfer by conduction × rate of internal heat generation − net work transfer from the control volume to the environment (1.71) Four of the five terms indicated do not involve work transfer from the control volume to the environment. • The rate of energy accumulated in the control volume is ∆x ∆y ∂ ∂t (ρu) (1.72a) BOOKCOMP, Inc. — John Wiley & Sons / Page 24 / 2nd Proofs / Heat Transfer Handbook / Bejan 24 BASIC CONCEPTS 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [24], (24) Lines: 1154 to 1178 ——— 2.91211pt PgVar ——— Long Page * PgEnds: Eject [24], (24) ∂ ∂ ()ρe t xy⌬⌬ y x q ∂q ∂y yxЉϩ Љ ⌬⌬ y y qyЉ x ⌬ qx yٞ⌬⌬ qxЉ y ⌬ q ∂q ∂x xyЉϩ Љ ⌬ x x ⌬ ( ( ) ) ρ ρ ˆˆ Vu ∂ ∂y Vu y x y y ϩ⌬⌬() ρ ρ ˆˆ Vu ∂ ∂y Vu x y x x ϩ⌬⌬() ρ ˆ V uy x ρ ˆ V ux y Figure 1.8 First law of thermodynamics applied to an imaginary control volume in two- dimensional flow. • The net transfer of energy by fluid flow is − (∆x ∆y) ∂ ∂x ρ ˆ V x u + ∂ ∂y ρ ˆ V y u (1.72b) • The net heat transfer by conduction is − (∆x ∆y) ∂q x ∂x + ∂q y ∂y (1.72c) BOOKCOMP, Inc. — John Wiley & Sons / Page 25 / 2nd Proofs / Heat Transfer Handbook / Bejan CONSERVATION OF ENERGY 25 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [25], (25) Lines: 1178 to 1196 ——— 3.754pt PgVar ——— Long Page PgEnds: T E X [25], (25) • The rate of internal heat generation is (∆x ∆y)q (1.72d) The origin of the term involving the net work transferred from the control volume to the environment is shown in Fig. 1.9, where the normal and tangential stresses are sketched. For example, the work done per unit time by the normal stress σ x on the left ρρ ˆˆ ˆˆ VV ∂ ∂y VV y x xy xy ϩ ()⌬⌬ ∂ ∂t ()ρ ⌬⌬yx ˆ V x ρ ρ 2 2 ∂ ∂x xϩ () ⌬⌬y ˆ ˆ V V x x ϩ xy xy ∂ ∂y yx ⌬ ⌬ xy ⌬x x y ⌬ ρ 2 y⌬ ˆ V x ˆˆ VV xy x⌬ ϩ x x ∂ ∂x xy⌬⌬ Xx y⌬⌬ ( ( ) ) y x Figure 1.9 Force balance in the x direction of an imaginary control volume in two-dimensional flow. BOOKCOMP, Inc. — John Wiley & Sons / Page 26 / 2nd Proofs / Heat Transfer Handbook / Bejan 26 BASIC CONCEPTS 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [26], (26) Lines: 1196 to 1263 ——— 3.16624pt PgVar ——— Long Page * PgEnds: Eject [26], (26) side of the ∆x ∆y element is negative and equal to the force acting on the boundary σ y , multiplied by the boundary displacement per unit time ˆ V x . This yields − ˆ V x σ x ∆ y . Similarly, the work transfer associated with normal stresses acting on the right side of the element is positive and equal to σ x + ∂σ x ∂x ∆x ˆ V x + ∂ ˆ V x ∂x ∆x ∆y The net work transfer rate due to these two contributions is σ x ∂ ˆ V x ∂x + ˆ V x ∂σ x ∂x (∆x ∆y) Three more work transfer rates can be calculated in the same way by examining the effect of the remaining three stresses, τ xy in the x direction and σ y and τ yx in the y direction. This gives (∆x ∆y) σ x ∂ ˆ V x ∂x − τ xy ∂ ˆ V x ∂y + σ y ∂ ˆ V y ∂y − τ yx ∂ ˆ V y ∂x + (∆x ∆y) ˆ V x ∂σ x ∂x − ˆ V x ∂τ xy ∂y + ˆ V y ∂σ y ∂y − ˆ V x ∂τ yx ∂x where the eight terms have been separated into two groups. It can be shown that the second group reduces to − ρ D Dt ˆ V 2 x + ˆ V 2 y 2 which represents the change of kinetic energy of the fluid in the control volume. This change may be considered negligible relative to the internal energy change, ∂(ρu)/∂t, so that the work transfer becomes (∆x ∆y) σ x ∂ ˆ V x ∂x − τ xy ∂ ˆ V x ∂y + σ y ∂ ˆ V y ∂y − τ yx ∂ ˆ V y ∂x (1.72e) The stresses σ x and τ xy can be related to the flow field via the constitutive relations given by Rohsenow and Choi (1961): σ x = P − 2µ ∂ ˆ V x ∂x + 2 3 µ ∂ ˆ V x ∂x + ∂ ˆ V y ∂y (1.73a) and τ xy = µ ∂ ˆ V x ∂y + ∂ ˆ V y ∂x (1.73b) BOOKCOMP, Inc. — John Wiley & Sons / Page 27 / 2nd Proofs / Heat Transfer Handbook / Bejan CONSERVATION OF ENERGY 27 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [27], (27) Lines: 1263 to 1328 ——— 0.00224pt PgVar ——— Long Page * PgEnds: Eject [27], (27) The bookkeeping required by eq. (1.71) dictates the assembly of eqs. (1.72), using eqs. (1.73), into ρ Du Dt + u Dp Dt + ρ∇· ˆ V =−∇·q + q − P∇· ˆ V + µΦ (1.74) where µ is the dynamic viscosity of the fluid and Φ is the viscous dissipation function, which is detailed subsequently in rectangular, cylindrical, and spherical coordinates. However, eq. (1.53) shows that the term in parentheses on the left-hand side of eq. (1.74) is equal to zero, so that eq. (1.74) reduces to ρ Du Dt =−∇·q + q − P∇· ˆ V + µΦ (1.75) In the special case where the flow can be modeled as incompressible and two- dimensional, the viscous dissipation function reduces to Φ = 2 ∂ ˆ V x ∂x 2 + ∂ ˆ V y ∂y 2 + ∂ ˆ V x ∂y + ∂ ˆ V y ∂x 2 (1.76) To express eq. (1.74) in terms of enthalpy, the definition h = u + Pv = u + P ρ is invoked. Hence, Dh Dt = Du Dt + 1 ρ DP Dt − P ρ 2 Dρ Dt (1.77) Moreover, the heat fluxes q x and q y can be expressed in terms of local temperature gradients through use of Fourier’s law: q =−k ∇T (1.78) Thus, the combination of eqs. (1.74), (1.77), and (1.78) results in ρ Dh Dt =∇·(k ∇T)+ q + DP Dt + µΦ − P ρ Dρ Dt + ρ ∇· ˆ V (1.79) Here, too, eq. (1.53) points out that the terms in parentheses in eq. (1.79) are equal to zero, so that eq. (1.75) reduces to ρ Dh Dt =∇·(k ∇T)+ q + DP Dt + µΦ (1.80) Bejan (1995) points out that the change in specific enthalpy for a single-phase fluid is given by BOOKCOMP, Inc. — John Wiley & Sons / Page 28 / 2nd Proofs / Heat Transfer Handbook / Bejan 28 BASIC CONCEPTS 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [28], (28) Lines: 1328 to 1411 ——— 7.50035pt PgVar ——— Long Page PgEnds: T E X [28], (28) dh = Tds+ vdP = Tds+ dP ρ (1.81) where T is the absolute temperature and ds is the specific entrophy change: ds = ds dT P dT + ds dP T dP (1.82) The last of the Maxwell relations given by Bejan (1997) ∂s ∂P T =− ∂(1/ρ) ∂T P = 1 ρ 2 ∂ρ ∂T P =− β ρ (1.83) where β is the volumetric coefficient of thermal expansion, β =− 1 ρ ∂ρ ∂T P (1.84) With c p taken as the specific heat at constant pressures, it can be shown that ds dT P = c p T (1.85) and eqs. (1.81) through (1.85) can be combined, dh = c p dT + 1 ρ (1 − βT)dP (1.86) so that the left-hand side of eq. (1.80) can be written as ρ Dh DT = ρc p DT Dt + (1 − βT) DP Dt (1.87) Thus, the temperature formulation of the first law of thermodynamics is ρc p DT Dt =∇·(k ∇T)+ q + βT DT Dt + µΦ (1.88) with the special forms for the ideal gas where β = 1/T, ρc p DT Dt =∇·(k ∇T)+ q + DP Dt + µΦ (1.89a) and for an incompressible fluid where β = 0, ρc p DT Dt =∇·(k ∇T)+ q + µΦ (1.89b) Most convection problems concern an even simpler model where the fluid has constant thermal conductivity k, neglible viscous dissipation Φ, zero internal heat BOOKCOMP, Inc. — John Wiley & Sons / Page 29 / 2nd Proofs / Heat Transfer Handbook / Bejan CONSERVATION OF ENERGY 29 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [29], (29) Lines: 1411 to 1462 ——— 9.43025pt PgVar ——— Long Page * PgEnds: Eject [29], (29) generation (q = 0), and a negligible compressibility effect, βT(DP/Dt)≈ 0. The energy equation for this model is simply ρc p DT Dt = k ∇ 2 T (1.90) or, in the rectangular coordinate system, ρc p ∂T ∂t + ˆ V x ∂T ∂x + ˆ V y ∂T ∂y + ˆ V z ∂T ∂z = k ∂ 2 T ∂x 2 + ∂ 2 T ∂y 2 + ∂ 2 T ∂z 2 (1.91) in the cylindrical coordinate system, ρc p ∂T ∂t + ˆ V r ∂T ∂r + ˆ V θ r ∂T ∂θ + ˆ V z ∂T ∂z = k 1 r ∂ ∂r r ∂T ∂r + 1 r 2 ∂ 2 T ∂θ 2 + ∂ 2 T ∂z 2 (1.92) and in the spherical coordinate system, ρc p ∂T ∂t + ˆ V r ∂T ∂r + ˆ V φ r ∂T ∂φ + ˆ V θ r sin φ ∂T ∂θ = k 1 r 2 ∂ ∂r r 2 ∂T ∂r + 1 r 2 sin φ ∂ ∂φ sin φ ∂T ∂φ + 1 sin 2 φ ∂ 2 T ∂θ 2 (1.93) If the fluid can be modeled as incompressible then, as in eq. (1.89b), the specific heat at constant pressure c p is replaced by c. And when dealing with extremely viscous flows, the model is improved by taking into account the internal heating due to viscous dissipation, ρc p DT Dt = k ∇ 2 T + µΦ (1.94) In the rectangular coordinate system, the viscous dissipation can be expressed as Φ = 2 ∂ ˆ V x ∂x 2 + ∂ ˆ V y ∂y 2 + ∂ ˆ V z ∂z 2 + ∂ ˆ V x ∂y + ∂ ˆ V y ∂x 2 + ∂ ˆ V y ∂z + ∂ ˆ V z ∂y 2 + ∂ ˆ V z ∂x + ∂ ˆ V x ∂z 2 − 2 3 ∂ ˆ V x ∂x + ∂ ˆ V y ∂y + ∂ ˆ V z ∂z 2 (1.95) . Proofs / Heat Transfer Handbook / Bejan 24 BASIC CONCEPTS 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [ 24] , ( 24) Lines:. Proofs / Heat Transfer Handbook / Bejan 22 BASIC CONCEPTS 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [22], (22) Lines:. Proofs / Heat Transfer Handbook / Bejan CONSERVATION OF ENERGY 23 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [23],