Document 2jNox19vmN4jBx77JKZ4m2MZN

108________________________ CHAPTER 5________________ ________ 1946 Guide The same resistance analysis may be applied to complicated steady- ' state conduction problems. Table 8 indicates the solutions in six common cases of steady-state conduction. A complete analysis by the resistance method is well illustrated by considering the heat transfer from the air outside to the cold water inside of an insulated pipe. The temperature gradients and the nature of the resistance analysis are indicated by the two sketches of Fig. 4. Since air is sensibly transparent to radiation, there will be some heat transfer by both radiation and convection to the outer insulation surface. The mechanisms act in parallel on the air side. The total current, by radiation and convection then passes through the insulating layer and the pipe wall by thermal conduction, and thence by convection into main cold water streams. Radiation is not significant on the water side as liquids are sensibly opaque to radiation, although water transmits energy ' in the visible region. The contact resistance between the insulation and . the pipe wall is presumed to be equal to zero. Referring to Fig. 4, the thermal current for a given length N of pipe, 2rc, Btu per hour, may be thought of as flowing through the parallel resistances Rr and 2?c, associated with the insulation surface radiation and convection transfer. Then the flow is through the resistance offered, to thermal conduction by the insulation, R3, through the pipe wall resistance, Rt, and into the water stream through the convection resistance, Ri. Note the analogy to the direct current electrical circuit problem. A temperature (potential) drop is required to overcome these resistances to the flow of thermal current. The total resistance to heat transfer, Rt, hour degrees Fahrenheit per Btu, is the summation of the individual resistances: Rt = Ri + Rt + R> + K. (9) where the resultant parallel resistance Rt is obtained from: -L = _L + _L R, Rr ^ Rc Provided the individual resistances may be evaluated, the total resistance can be obtained from this relation. Then the heat transfer current for the length of pipe (N, ft) can be established by the relation: 3tc (Btu per hour) -- R,--tFor a unit length of the pipe the heat transfer rate is: (Btu per hour foot) = r^xA^ (10) (U) ' The temperature drop, Ai, through an individual resistance may then be calculated from the relation: At -- R Qrc where R is the resistance in question. . The problem is now reduced to one of evaluating the individual resist ances of the system. This entails suitable integration of the rate Equa tions 1, 2 and 3 to produce expressions of the form: Fundamentals of Heat Transfer 109 Table 8. Solutions for Some Steady-State Thermal Conduction Problems8^ Expressions for the resistance R entering into' . the equation: .., 9 = At/R (Btu per hour) Flat wall or curved wall if curvature is small (wall thickness less than 0.1 of inside dia meter). L kA Surface area A Radial flow through a right circular cylinder. >^7 2rkN (See footnote c). The buried cylinder. T k .At:tp-ts Long cyiinderjTrJ of length. N Radial flow in a hollow sphere. jSI- i 2o. loge-^- . . a cosh-1 -- R " 2tkN :R~ 2ikN for S- > 3 r (See footnote cc). The straight fin or rod heated at one end. Conduction II, J-q jk_____ cross-section area, A tt 1 ambient Finned surface of area HB. h^p tanh m L. (see footnotes d and ) For ml > 2.3, tanh m L 1 m = y/ksp/kA A -- conduction cross-section area. p a. perimeter of cross-section A. ht a unit conductance to the surroundings from the fin surface. .. . k thermal conductivity fin material. At wall temperature--ambient temperature K / 2 (* + *) \ ha 1---tanh nt + rj HB A/ defined as in Case 5 above. The dimensions to be employed in these solutions are: length of dimension p, L.r - feet; units of ft = Btu per (hour) (square foot) (degree Fahrenheit for one foot thickness); units of ft, Btu per (hour) (square foot) (degree Fahrenheit); units of area, A = square feet. bThe thermal conductivity, ft, in these solutions should be taken at the average material temperature (see Table 5). eLogs x - 2.303 iogtt x. ,. dThis expression can also be employed as an approximation for tapered fin$ or of annular fins by employ ing average magnitudes of A and p. Tanh is the hyperbolic tangent. ' *