Buoyancy-induced Flow: Natural Convection in a Unconfined Space If we examine the flow induced by heat transfer from a single vertical flat plat, we observe that the flow resembles that of a boundary layer. The appropriate description begins with the conservation laws. Mass v y y + v = 0 Momentum v ρ v y y + v v = µ 2 v y 2 + 2 v 2 p ρg Energy v y T y + v T = α 2 T y 2 + 2 T 2 The velocity field is such that the induced flow is viewed as a perturbation of the steady flow. v εv y1,v 0 (y) + εv y 1 Now if we examine the perturbation expansion of v applied to the Navier- Stokes equation, the leading term of the perturbation expansion is the solution to 0 = = dp d + ρg That is, in the eroth approximation, there is no flow in the -direction as a result of the temperature field... the pressure field is given by the hydrostatic law... the pressure field is balanced by the gravitational force. ChE 333 - Lecture 16 1
Velocity Field The velocity field is governed by. v ρ v y y + v v = µ 2 v y 2 + 2 v 2 p ρg ρβg T T The conservation equations are clearly coupled. The solution mirrors the results obtained in Boundary Layer theory analysis. The boundary conditions are at y = 0 v = v y = 0 T = T 0 at y = v = 0 T = T 1 at = - v = v y = 0 T = T 1 The solution is not simple, but involves combination of variables solutions as in Boundary Layer theory. The problem is best solved by putting the equations in dimensionless form. Y = Hαµ B 1/4 ; V y = Bα3 µh 1/ 4 ; V = BHα µ 1/2 B = ρg T ChE 333 - Lecture 16 2
Dimensionless Equations Without belaboring the procedure in detail, the results in dimensionless form are. The differential equations + ϕ ζ = 0 Pr 1 ϕ + ϕ ϕ ζ = 2 ϕ 2 + ε 2 2 ϕ ζ 2 + θ + ϕ ζ = 2 θ 2 + ε2 2 θ ζ 2 For sufficiently small ε, the equations are simpler. It simply means that changes in the ζ are less important than chasnges in the η direction. + ϕ ζ = 0 + ϕ ζ = 2 θ 2 Pr 1 ϕ + ϕ ϕ ζ = 2 ϕ 2 The boundary conditions are at η = 0 φ e = φ y = 0 θ = 1 at η = φ = 0 θ = 0 at ζ = - φ = φ y = 0 θ = 0 1 Heat Transfer Correlations + θ ChE 333 - Lecture 16 3
We can use the solution or at least the form of the solution of the equations, as we note the definition of the heat flux per unit length. q' = 0 H k T y η =0 d = k TH Y 0 1 η =0 dζ It is apparent that the definite integral is a function only of Pr since θ = θ η, ζ, Pr It follows that since the definite intergral is a constant C = C(Pr), that we can express the flux q as where the Grashof number here is q' = Ck T Gr Pr 1 /4 Gr = ρ2 βgh 3 T µ 2 It is easy to recall the definition of a heat transfer coefficient. q' H = h T and we obtain for our model Nu = C Gr Pr 1/4 = CRa 1 /4 ChE 333 - Lecture 16 4
An Empirical Correlation Chu and Churchill developed a more useful quasi-empirical model Nu 1/ 2 = Nu 1/2 + Ra L 300 1 + 9 /16 16/9 0.5 Pr 1/6 Geometry L Nu 0 Vertical Surface L 0.68 Vertical cylinder L 0.68 Horiontal cylinder πd 0.36π Sphere πd/2 π Inclined disk 9D/11 0.56 ChE 333 - Lecture 16 5
An Example Heat Loss from a Horiontal Pipe An uninsulated pipe, 5 cm in outside diam., runs horiontally through a laboratory maintained at 30 C.. Air enter the pipe at 80 C at a rate of 300 kg/hr. The pipe is 20 meters.. The down stream pressure in the pipe is 10 5 Pa gauge. What is the heat loss from the pipe and what is the exit temperatrure of the air? Assume the principal resistance to heat transfer is on the outside of the pipe. We assume as well that he external surface is 10 C lower than the inside of the pipe. (Can we get a better estimate? How? The physical parameters are ρ = 1.0 kg/m 3 ; µ = 2 x 10-5 kg/m-sec ; Pr = 0.7 ; C p = 1.005 kj/kg- K so that we evaluate the Rayleigh number as Ra = GrPr = 350000 From the Chu-Churchill equation, we obtain Nu = 10, so that h = k D Nu = 5.8 watts m 2 C In order to calculate the heat loss we need to know the T T lm = T h T c lm We can make a first approximation that T lm = T h T c lm Th T c max This allows us to get a quick estimate of the heat loss and the exit temperature from the Heat Exchanger Design Equation. w air C p T h h in T out = ha T lm ChE 333 - Lecture 16 6
T h h in T out = ha T lm w air C p = 5.18 3.14 40 300 / 3600 1005 = 8.3 C That means that T ~ 36 C so (T in - T out ) = 7.5 C ChE 333 - Lecture 16 7