College of Engineering Summer Session Heat Transfer - ME 372 Dr. Saeed J. Almalowi,

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2.. Fin s efficiency and effectiveness The fin efficiency is defined as the ratio of the heat transfer to the fin to the heat transfer to an ideal fin.

1 2.. Fin s efficiency and effectiveness The fin efficiency is defined as the ratio of the heat transfer to the fin to the heat transfer to an ideal fin. η th = q fin ha fin (T b T ), T f = T, and A fin = 2A c + A tip (Square and Recatngular ).35 A tip = t W Fig Rectangular Fin For cylindrical: A fin = πdl + πd2 From Eq.(.29 ), the heat transfer to the fin is at x =0 yields 4 q fin = ka c 2M (T t T ) + (T T b )e ML e ML e ML.36 Manipulating of Eq. (36) leads to: q fin = (T b T ) hpka c cosh(ml) T t T T b T sinsh(ml).37 Now, the fin efficiency can be derived clearly from Eq.(.35) and Eq.(.37)

2 η th = hpka c cosh(ml) T t T T b T sinsh(ml) ha sf = ka c cosh(ml) hl 2 sinsh(ml) T t T T b T.38 In case of the infinitely long fin, the efficiency can be evaluated as: η th = hpka c P 2 L 2 h 2 = ka c PL 2 h = L M.39 Table 2.2. Soutions for a uniform and non uniform cross-section area fin

2.2. The Finned Surfaces Effectiveness Fig. 2.5.

without fin attached. No-fin heat transfer rate: q nf = ha b (T s T ).40 While the heat transfer from the fin can be evaluated from Table 2. and 2.

3 2.2. The Finned Surfaces Effectiveness Fig Finned Surface The effectiveness of the fin is the dimensionless parameter which can be measured the ratio of the heat transfer from the fin to the heat transfer occupied by the fin without fin attached. No-fin heat transfer rate: q nf = ha b (T s T ).40 While the heat transfer from the fin can be evaluated from Table 2. and 2.2 as: The effectiveness is ε = heat transfer wihout fin hea transfer with fin = q nf q f Bundles of Fins (Finned Surfaces) Fins are often placed on surfaces in order to improve their heat transfer capability. Examples of finned surfaces can be found within nearly every appliance in your house, such evaporator, refrigerator, condenser, etc. Fig2.6. Fins array

4 R fin = ηha s fin R unfinned = h(a sb N fin A Cb ) Where, N fin is nmber of fins. A sb the base surface area in unit area = W L. A cb is the cross section base area in unit area = N fin t L. h is the convective heat transfer in W m 2. K. The surface area of the fin can be evaluated as: A sf = 2(t + L) Hf.44 Fig.2.6 Resistances of the Fins and un-finned area R total = (h(a sb N fin A Cb ) + ηha s fin N fin).45 The total heat transfer can be calculated as: q total = T b T R total.46 The prime surface area = the surface area of the base +the fin surface area-the base cross section area A total = A sb + N fin A sf N fin A cb.47

The overall efficiency is defined as the ratio of the total heat transfer rate from the surface to rate of the heat transfer rate from the entire surface (overall efficiency).

5 The overall efficiency is defined as the ratio of the total heat transfer rate from the surface to rate of the heat transfer rate from the entire surface (overall efficiency). η o = q total A total h(t b T ) The total resistance can be evaluated as: R total = η o A total h EXAMPLE 2.2. FINS ARRAY (TUTORIAL PROBLEM) Fig. illustrates fins array, determine A- Free hand sketch for all resistance (fined and un-fined resistance), B- Heat transfer for fined, C- Heat transfer for un-fined, a= 0.5 cm, L=0a cm, and D= 0.75a cm Fig Pined Fins Array Solution: R Total = [N fin hη fin A sfin + h(a sb N fin A cb ] q total = (T b T ) R Total A sfin = πdl A cb = πd2 4

6 B- q fin = T b T R fin C-q nofinned = T b T R unfinned CHAPTER 3.0: FIFTH LECTURE: ZERO DIMENSIONAL TRANSIENT LUMPED CAPACITANCE In this chapter, the temperature varies with time. The simplest situation and assumption is the temperature does not vary with the position which is called lumped capacitance situation. Exact Solution: q conv + du dt = 0 q conv = ha s (T T ) Ѳ = T T, Ѳ i = T i T du dt = mc p dt = ρvc p dt ha s Ѳ + ρvc p dt = 0 dt + ha s ρvc p Ѳ = 0 The first order homogenous differential equation: Assume the lumped capacitance is = ρvc p ha s

7 dt + Ѳ = 0 This separable ODE, it can be solved as: Ѳ = dt ln ( Ѳ Ѳ i ) = Ѳ = Ѳ i e t T T = (T T i )e Biot Number = hl k t (t t 0 ) Ѳ Ѳ i = e t

Conduction Heat Transfer. Fourier Law of Heat Conduction. Thermal Resistance Networks. Resistances in Series. x=l Q x+ Dx. insulated x+ Dx.

Conduction Heat Transfer. Fourier Law of Heat Conduction. Thermal Resistance Networks. Resistances in Series. x=l Q x+ Dx. insulated x+ Dx. Conduction Heat Transfer Reading Problems 17-1 17-6 17-35, 17-57, 17-68, 17-81, 17-88, 17-110 18-1 18-2 18-14, 18-20, 18-34, 18-52, 18-80, 18-104 Fourier Law of Heat Conduction insulated x+ Dx x=l Q x+