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LECTURE NO.6 HEAT SOURCE SYSTEMS: ONE-DIMENSIONAL STEADY STATE HEAT CONDUCTION WITH HEAT GENERATION: HEAT FLOW THROUGH SLAB / PLANE WALL HEAT-SOURCE SYSTEMS A number of interesting applications of the principles of heat transfer are concerned with systems in which heat may be generated internally. Nuclear reactors are one example; electrical conductors and chemically reacting systems are others. Fig.6.1 Sketch illustrating one- dimensional conduction problem with heat generation. One-dimensional steady state heat conduction with heat generation: Heat Flow through slab / Plane Wall Consider the plane wall with uniformly distributed heat sources shown in the above Figure. The thickness of the wall in the x direction is 2L, and it is assumed that, the dimensions in the other directions are sufficiently large that the heat flow may be considered as one-dimensional. The heat generated per unit volume is q • , and we assume that the thermal conductivity does not vary with temperature. (This situation might be produced in a practical situation by passing a current through an electrically conducting material) The differential equation which governs the heat flow is
2 d T 2 dx . q + = 0 k 2 d T q =− 2 dx k Integrating twice with respect to x results in dT dx . . q =− x+ c k . q 2 T = − x + c1x + c -------------- (1) 2 2k For the boundary conditions we specify the temperatures on either side of the wall, i.e., T = T w at x = ± L 1 Since the temperature must be the same on each side of the wall, c 1 must be zero. dT = 0, q • = 0⇒c 1 = 0 The temperature at the mid plane (x = 0) is denoted by T o and from Equation (1) At mid plane = x=0 and T= T 0 T o = c 2 The temperature distribution equation (1) becomes 2 T =− x + . q 2k T 0 T . q 2k x 2 −T 0 = − --------------- (2) Assumed T=T w at x= L . q 2k L 2 T w −T 0 = − ----------------- (3) (2) (3) T −T0 ⇒ T −T w 0 2 ⎛ x ⎞ = ⎜ ⎟ ⎝ L ⎠ ------------ (4)
Page 1 and 2: For Class use only ACHARYA N.G. RAN
Page 3 and 4: Simplified Case For Systems With Ne
Page 5 and 6: LECTURE NO.1 INTRODUCTION TO HEAT T
Page 7 and 8: q A k = ( T ) 1 −T 2 -------- (4)
Page 9 and 10: LECTURE NO.2 THE BASIC EQUATION THA
Page 11 and 12: The general three-dimensional heat-
Page 13 and 14: LECTURE NO.3 ONE-DIMENSIONAL STEADY
Page 15 and 16: The cross-sectional area normal to
Page 17 and 18: Total temperature difference, ∆T
Page 19 and 20: LECTURE NO.5 THE OVERALL HEAT-TRANS
Page 21 and 22: Note that the area for convection i
Page 23: T 1 ,T 0 , K, L, r o , r i are cons
Page 27 and 28: LECTURE NO.7 STEADY STATE HEAT COND
Page 29 and 30: Different fin configurations Differ
Page 31 and 32: LECTURE NO.8 EQUATION OF TEMPERATUR
Page 33 and 34: That is, the rate at which energy i
Page 35 and 36: eason, there is no assurance that t
Page 37 and 38: LECTURE NO.9 PRINCIPLES OF UNSTEADY
Page 39 and 40: where α is k / ρc p , thermal dif
Page 41 and 42: calculate the temperature of the ba
Page 43 and 44: dimensionless numbers can be used i
Page 45 and 46: Repeating this procedure for π 2 a
Page 47 and 48: q k ( Tw T* = ) L " − w The Nusse
Page 49 and 50: The mass transfer analog of the Pra
Page 51 and 52: The transition to turbulent flow oc
Page 53 and 54: long-wavelength thermal radiation.
Page 55 and 56: Figure 11.4 Method of constructing
Page 57 and 58: LECTURE NO.12 INTRODUCTION TO CONDE
Page 59 and 60: are plotted against temperature exc
Page 61 and 62: LECTURE NO.13 HEAT EXCHANGERS- GENE
Page 63 and 64: counterflow to the hot shell-side f
Page 65 and 66: to be calculated in the above equat
Page 67 and 68: divided by the natural logarithm of
Page 69 and 70: Figure 13.7 Correction-factor plot
Page 71 and 72: Then, Since q= UA ∆T m , 5 1.895
Page 73 and 74: 3.8 A = = 0.0104 (1000 )(0.366 ) m
Page 75 and 76:
LECTURE NO.15 INTRODUCTION TO MASS
Page 77 and 78:
concentration of water vapor at the
Page 79 and 80:
T is temperature in K, R is 8314.3
Page 81 and 82:
Equating Eq. J J * Az = −D AB dc
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