Total energy in volume

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1 General Heat Transfer Equations (Set #3) ChE 1B Fundamental Energy Postulate time rate of change of internal +kinetic energy = rate of heat transfer + surface work transfer (viscous & other deformations) + work done y ody shaft work + Electromagnetic energy or other sources if we take an aritrary volume of material let U internal energy per unit mass 1 ρv kinetic energy per unit volume internal + kinetic 1 V t U v dv = m () ρ + ρ Total energy in volume 3-1

2 ChE 1B Then the time rate of change of this quantity is: D Vm () t ρ 1 + ρ U v dv for an oserver moving with the volume element. Hence, it is called a material or total derivative qn heat transfer rate = m () A t da This is no more than a definition of q. From fluid mechanics, the force df acting on the surface is df =t n da t n is the stress vector. Imagine the area moving in a direction λ y an amount Δ in time Δt. The work done on this surface is just the force times the distance: t n da λδ and the rate of work is just t da λ n Δ Δt 3 -

3 ChE 1B In the limit as λ Δ Δ t, Δt The velocity vector and then λ Δ = v, Δt surface work on element da = t n v and the total work done on volume is Am () t tn vda similarly, work = ρ dv V m () q v t g is for gravity ody force et Φ = energy source/unit volume, and comining D 1 ρu + ρv dv = A () () n () ρ m t qn da + Am t t vda + Vm t gv dv + Vm() t Φ dv We can re-write the gravity term as D ρg v dv = ρφ dv Vm( t) Vm( t) in which φ is the potential energy/mass and D ρe+ 1 ρv +ρφ dv = qn da + t v da + ΦdV V ( t) A () () n () m m t Am t Vm t or ( n ) T v da T is the stress tensor T=ρ I +τ U + KE + PE = Q + W which is just our old way of writing the conservation of energy. imiting Cases Steady-state, v =, = qn da + ΦdV Am Vm Now A m, V m are fixed in space 3-3

4 ChE 1B Rememer the divergence theorem from calculus (or physics)? qn da = q dv so A V [An aside: v= ( v) = ( ) Φ dv nt T ] A V dv v q - For this to e true over an aritrary volume, the integrand must e identically ( q ) Φ= or in various forms q q x y qz Rectangular: = + + Φ y z Cylindrical: 1 1 qθ qz = ( rqr r ) + + Φ r r θ z qθ = r + θ + Φ r r rsinθ θ rsinθ θ Spherical: ( rq) ( sinθ q) In the general case, we have D 1 V ρ U+ v + φ dv ( ( T v) ) V q = Φ dv Reynold's transport theorem states D 1 D 1 V ρ U+ v + φ dv V ρ U+ v + φ dv = D 1 V ρ U+ v + φ + ( q) ( T v) Φ dv = hence inside the integral is also zero to account for aritrary volumes. D 1 ρ U+ v + φ = ( q) + ( v) +Φ T Or D 1 ρ U+ v = ( q) + ( v) + ( ρ v) +Φ T g We can expand the sustantial derivative 1 1 D 1 ρ U+ v U+ v ρ U+ v t + = v for v = we get U ρ This is the transient energy alance in the = q+ Φ t asence of convection. 3-4

5 ChE 1B Rememer ˆ i ˆ = + j+ kˆ y z Reorganizing using T = pi + τ DU ρ ( ) τ : v = q p v + These equations can e further simplified if we use the equations of motion: Dv ρ = ρg + T If we take the dot product with v, we get the mechanical energy equation: Dv ρv = ρ g v+ v T ( ) We can recognize the first term as: Dv D 1 D 1 ρ v = ρ = ρ v v v and rewrite the last term as: ( v T) v:t= v ( T ) If we sustitute these results into the mechanical energy equation Dv 1 v ρ = ρg v + ( v T) v:t If this is sutracted-from the total energy equation; we get the thermal energy equation: DU ρ = + q v:t and T = pi + τ and 3-5

6 ChE 1B DU ρ = q p v + v : τ To convert this equation into something useful we need a more readily accessile quantity than U, hopefully T To get there, rememer H=U+PV U=H PV=H P/ρ so ρ DH D ( P / ) p ρ ρ = q v+ v: τ DH 1 Dp p Dρ p ρ = q+ ρ ( ρ ) v + τ ρ ρ ρ v: Dp p Dρ = q+ ρ ( ) + τ ρ v v: y continuity ρ DH Dp = q + + v: τ and DH H DT H Dp = + T p P T DH 1 = 1 Dp ρ T ( T β ) 1 ρ H β = = C p ρ T p T p and q = k T leads to: DT Dp ρ Cp = ( k T) + Tβ + v: τ For the general thermal energy equation. 3-6

7 A second general conversion to T goes through constant V and T processes: Recall from Thermodynamics: U U du = dv + dt V V T p = p + T dv + CvdT T v V is the specific volume v ChE 1B Rememer, also the continuity equation DV D 1 1 Dρ ρ = ρ = = ( v) ρ ρ Dρ continuity equation says = ρ ( v) A few manipulations gives DT p ρ Cv = q T v τ:v T ( ) ( ) ( ) call μ Φ (usually omitted in practical cases) ook at different cases p P 1. Ideal gas = T v T DT ρ Cv = ( q) p( v ) (neglect μφ ) v For incompressile fluids and solids du = CvdT CpdT and T ρ C p = q +Φ t Assuming Fourier's aw (q = k T, constant k) T k = T +Φ t ρ Cp k α thermal diffusivity ρ C p For solids, with no Φ T T T T ρ Cp = k + + t y z k length α thermal diffusivity units ρ C time p and T = α T t 3-7

8 ChE 1B α is analogous to kinematic viscosity v ρ unit same mass diffusivity D unit same for solids v = usually and T ρ Cp = k T t The groupings all refer to a diffusion-type constant: k α is analogous to D and to ρ C Thermal Diffusivity p Mass Diffusivity Momentum Diffusivity v ρ In general, T = D T t is known as the diffusion equation. 3-8

9 Examples of general alances Uniform heat generation in sla dt Φ + = for k constant, dx k Solution to this is: d dt Φ = dx dx k dt Φ d = dx dx k dt Φ = B x dx k Φ T = A+ Bx x k If T = x=, T = x= we get x Φ T = T ( T T1 ) + ( x) x a quadratic profile. k ChE 1B We know if there is no source term, we should get ack a linear profile. Check if Φ, we get ack linear profile!!!! Maximum temperature now where will the temperature e greatest? dt T T Φ = = + dx k ( x) 1 ( T ) k T 1 x max = Φ k( T T1) We can say for this to work (otherwise x = ) Φ Φ k( T ) T1 Tmax = T + k Φ Assume copper plate (k = Btu/hr ft F) 1 inch thick 6 3 Φ= 4 x 1 Btu/hr ft T = 1 F T1 = 7 F = 49.6 F Tmax 3-9

10 ChE 1B Order of Magnitude Analysis Depending on the detail necessary for an engineering answer to a complex prolem, we might want to determine how hard we want to work on that prolem. This involves 1) First estimating average values of quantities ) Comparing magnitudes of these quantities. 1 Define ( f ( x )), order of magnitude of f ( x) f dx This is just the average of the asolute value of that function over 1 ( g ( x, y) ) g( x, y) dx dy Of course, to do this efore knowing g relies on a proper physical intuition How aout derivatives? f 1 df dx dx Assume that f ( x ) is monotonic, then this result is pretty simple f 1 df 1 dx = f f dx If we have a reasonale f ( x ), then ( ( ) ( )) 1 ( f ( x) ) = f ( x) dx= F( ξ) ξ which is just the mean value theorem and ( ) = ( ( ξ )) = ( ) ( ) = ( ( ξ )) = ( ) f f f f f f and 3-1

11 ChE 1B ( ) ( ) ( ) f = f = f, hence f ( f )/ in a similar fashion, f ( f )/ let s get more concrete T T1 we assumed earlier that if >>, then qy = k Can we justify this quantitatively? T T Start with + = y T y = 1 T = y = 1 ( ) T = T T T sin 3 π x= 1 ( )( ) T = T + T T x= 1 Order of magnitude stuff is always easier with dimensionless equations: θ ( T T ) = X = / T T 1 x Y = y/ β = / 3-11

12 θ θ * + = X Y θ Y = θ = Y = 1 θ = sin 3 π Y / x= θ = Y x= β ChE 1B and ( ) 1 θ θ = = β β ( θ ) θ 1 = = = y 1 ( θ ) ( ) and for the case β 1 we might say θ θ, y ut we also know that θ θ from y * A etter way is as follows: η = Y θ θ = dη C + y η = 1 τ = Y η = Y θ θ = dη dτ C Y C + + τ= η= 1 We can now apply the B.C. s to get 3-1

13 when Y =, θ = C = Y = 1, θ = 1 ChE 1B τ = 1 η= τ θ C = 1+ d η d τ and 1 τ= η= τ= 1 η= τ τ= Y η= T θ θ θ = Y 1+ dη dτ dη dτ x τ= η= τ= η= 1 1 θ = Y + Y + β β and whenever β 1 this simplifies to θ = Y or in terms of parameters ( )( ) T = T + T T y 1 / And rememer: ( AB) = ( A) ( B) ( A+ B) ( A) + ( B) 3-13

14 ChE 1B Fins Revisited We can start with the general alance at steady state T T T + + = y z If the fin is very long in y direction ( ) then T T z + = and our primary ojective is to determine the heat flux from the wall due to the fin. Assume top and ottom of fin are identical, then T z = z and h,k independent of position so that T k = h( T Ta ) for top, ottom z = ± z T k = hend T T z ( ) x and for the end of fin and that wall is of constant temperature x = We now have a messy P.D.E. in x,z. et s make the further assumption that and that most of the variation in T is along x direction so let's average temperature over z, and determine the variation in x 3-14

15 T T dz + dz = z T T Tdz z z= z z= + = T T = ; = and z z= z z= Applying oundary conditions k h( T Ta ) ChE 1B 1 Define d h dx k = Tdz= T. Then T + T T z = ( a ) Now we simply need to guess what T is? z= From our analysis of the Biot numer, h k If k is ig and h is small T T, or the fin is uniform in z= temperature, Bi << 1. k is small and h is large T = Ta z= or the fin surface is the same temperature as surroundings Bi >>

16 ChE 1B et s look at a case in which T = T as second case leads to no heat flux z= d T h + ( T Ta ) = dx k and we can non-dimensionalize temperature as T Ta = θ and we know intuitively T Ta <θ < 1 in the fin and d θ h θ = dx k T k = hend ( T Ta x= T x= in terms of θ θ k = hend x= θ = 1@ x = We can also non-dimensionalize distance η = x/ η = x η 1 and d θ ( η ) d h = a k θ d θ h θ dη k = h = γ and k d θ γθ= d ( η ) and the oundary conditions are θ h = η = 1 η k θ = η =

17 ChE 1B Often, y making our equations dimensionless, we can isolate important parameters, and estimate what is dominant. Mathematically, we can ound a variale and the formalities are nicer. Solutions are also not too tough θ = C sinh mη+ C cosh mη η = θ = 1 C = η = 1 θ = h θ, h = m η k k θ nc mη + C mη = = Cm mη+ m mη [ 1sinh cosh ] cosh sinh η ( msinh mη+ n cosh mη) = C [ m cosh m+ nsinh m] msinh m+ n cosh m C1 = m cosh m+ nsinh m msinh m+ n cosh m θ = sinh mη+ cosh mη m cosh m+ nsinh m which can e manipulated into cosh m( 1 η ) + n sinh m( 1 η m ) θ = cosh m+ n sinh m m When is this solution valid we said for small h/large k Can we e more specific? maximum is T ΔT = Δ T = ( T Ta ) z Δ T = T T T k = h T T ( a ) z z= T T = ( z= a) z= z= k h T T ΔT k = h T T ΔT h T T k z = a z = a 1 z = z = and we want this to e small so ΔT h or k 3-17

18 ChE 1B This grouping is commonly known as the Biot numer and is the ratio of convective/conductive heat transfer. For Bi 1, conduction is fast and temperature on fin are near uniform for Bi 1, the oject takes on the amient temperature. If n = = and m cosh m( 1 η ) θ cosh m 1/ 1/ Biend Bi. then Bi, Bi 1 and the rate of heat flow through the fin is d T qx x= = k dx k = ( T T ) d d θ η a η = ( Bi ) ( a ) 1/ 1/ = N tanh N / k T T 3-18

Consider a volume Ω enclosing a mass M and bounded by a surface δω. d dt. q n ds. The Work done by the body on the surroundings is.

The Energy Balance Consider a volume Ω enclosing a mass M and bounded by a surface δω. δω At a point x, the density is ρ, the local velocity is v, and the local Energy density is U. U v The rate of change