Document e5N7yw755d1GqvkzOd3KJNXm4

HEATINC VENTILATING AIR CONDITIONING GUIDE 1944 Table 8. Solutions fob Some Steady-State Thermal Conduction Problems'* No. System Expressions for the resistance R entering into the equation: * _________ 9 83 At/R (Btu per hour) w cuxved wall if curvature la amall (waU^thickness less than 0.1 of Inside dia- Surface area. A RkA Radial flow through a right circular cylinder. Long cylinder of lenglfi N' The buried cylinder. ts 1 a k At.lp-t* Long cyfinderTrJ of length. N Radial flow in a hollow sphere. &1- l0*7r 2rkN (See footnote c). log, 2a 2*kN * K for > 3 (See footnote c). cosh-* A 2_______ T 2rkN R- 1 *o 4tA [ The straight fin or rod heated at one *"<1 .9/k Conduction ' cross-section area A R th~ip- tanh m ,l (see footnotes d and e). For ml > 2.3, tanh ml ** 1 m -\/h*i>/kA A = conduction cross-section area. p b perimeter of crossHsection A.T he = unit conductance to the surroundings ' from the fin surface. k = thermal conductivity fin material. A/ = wall temperature--ambient temperature (* +6) k* ^--tanh m / + r ^ j V5 At defined as in Case 5 above. . aThe`dimensions to be employed in these solutions are: length of dimension p,l,r = feet; units of A Btu per hour per square foot per degree Fahrenheit for one foot thickness; units of A, -Btu per hour per square foot per degree Fahrenheit; units-of area;-j4 ** square feet. bThe thermal conductivity, k, in these solutions should be taken at the average material temperature (see Table 5); ; Log' x -- 2.303 logio x. dTtus expression can also be employed as an approximation for tapered fins or of annular fins by employ ing average magnitudes of A and p. Tanh is the hyperbolic tangent. 82 CHAPTER 3. FUNDAMENTALS OF HEAT TRANSFER where the resultant parallel resistance R, is obtained from: J_=JL+_L Rt Rr T 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: ffrc (Btu per hour) = * _ - Kt For a unit length of the pipe the heat transfer rate is: (10) (Btu per hour foot) = (11) The temperature drop, A/, through an individual resistance may then be calculated from the relation: RAi -- rc 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: where q is the heat transfer rate, and Af is the potential drop or- tempera ture difference through the resistance R., Table 8 lists such solutions for six different conduction systems. Table 2 in Chapter 4 and Table 1 of this chapter indicate the magnitudes of the thermal conductivities, k, to be employed in the expressions of Table 8. The solution applicable to the problem depicted in Fig. 4, for the calculation of Rt and Rt, is case 2 in Table 8. Thus for- a 1 ft length of 2 in. nominal size pipe (I. D'. = 2.067 in., O. D. = 2.375 in.) insulated with 1 in. of cork: 1.188 R, = 2* X 1.033 26 X 1 = 8.5 X 10-4 hr degree Fahrenheit per Btu. 2.188 R, = 2k loge 1.188 X 0.025 X 1 3.9 hr degree Fahrenheit per Btu. The convection resistances to heat transfer from the pipe wall to the cold water, R\, and from the air to the surface of the insulating material, .Rc,' are dependent on the flow conditions prevailing at these surfaces, and on the thermal properties of the fluids. The unit conductances for thermal convection, h, Btu per hour per square foot per degree Fahrenheit, have been determined by test for many flow systems. These data may be employed to predict the conductances for similarflow systems. Table 5 summarizes some empirical equations expressing such test results; 83