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I. Analytical Solutions in Conduction Heat Transfer

C. Lumped formulation of first law of thermodynamics

"A general law is said to be lumped if its terms are independent of space and to be distributed if its terms depend on space." Arpaci (1966)

dEcv V2 V2 = Qcv + Wnonflow + mi h + + gz - me h + + gz dt 2 2 in i out e

(I.C.1)

1 where E = U + mV 2 + mgz 2

dmcv = mi - me (kg / s) dt in out

lesson 4

m = VA =

VA (kg / s) v

(I.C.2)

I. Analytical Solutions in Conduction Heat Transfer

C. Lumped formulation of first law of thermodynamics (cont.)

If we consider an incompressible substance at constant pressure and neglect viscous forces, shaft work, KE and PE, (I.C.1) becomes

c

dT = mi c i (Ti - Tref ) - me ce (Te - Tref ) + q j A j + u' ' ' dt i e j

(I.C.3)

where · qj is heat flux normal to area Aj at the control volume boundary · TR is a reference temperature

lesson 4

1

I. Analytical Solutions in Conduction Heat Transfer

D. Inductive versus deductive approach

1. Analysis of triangular straight fin. We want to find T(x) and heat loss from fin of width w. The drawing is not to scale. It is important to realize that L >> b.

A' = b

x L

Fluid temperature = T h

qx = -k

dT dx

b

qC = hC (T - T )

h L

x

Base temperature = Tb

lesson 4

I. Analytical Solutions in Conduction Heat Transfer

1. Analysis of triangular straight fin (cont.) Assume no T gradients in y-direction. Assume perimeter of fin is P = 2 dx (fin approximation). Assume k = constant. Energy balance per unit width of fin gives

kb d dT x - 2h(T - T ) = 0 L dx dx

Boundary conditions: at x = 0, T(0) = finite at x = L, T(L) = Tb

lesson 4

(I.D.1)

(I.D.2) (I.D.3)

2

I. Analytical Solutions in Conduction Heat Transfer

2. Solution The solution is a modified Bessel function of the first kind of order zero.

I 2c x T - T 2hL where c 2 = = 0 Tb - T I0 2c L kb

The heat rate (W) is

( (

) )

(I.D.4)

Q c L I1 2c L = where A = bw kA (Tb - T ) I0 2c L L

(

(

)

)

(I.D.5)

lesson 4

I. Analytical Solutions in Conduction Heat Transfer

3. Deductive approach with (I.B.18) 0

T ~ c + V T = (kT ) + u' ' ' t

I.B.18 does not fit our geometry nor does it accommodate a mixed formulation.

lesson 4

3

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