Document QXnYdEk275LR0bRBR9yedX084

102 CHAPTER 5 1953 Guide Rz, and into the water stream through the convection,resistance,, Rx. 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, Rx, hour Fahrenheit degrees per Btu, is the summation of the individual resistances: /t = Ri 4- Rt + Ri- + R* where the resultant parallel resistance Rt is obtained from (9) R, = ~Rt + Rc Provided the individual resistances may be evaluated, the total resistancecan be obtained from this relation. Then the heat transfer for the length of pipe (N, ft) can be established by the relation (/Btitu per ihour)\ = (--1 ----fi) Rt For a unit length of the pipe the heat transfer rate is (H) Btu per (hour) (foot) = - (12) The temperature drop, At, through an individual resistance may then be calculated from the relation: At = R g,, (13) 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 manipulation of the rate Equa tions 1, 2 and 3 to produce expressions of the form: M where q is the heat transfer rate, and At is the potential drop or tempera ture difference through the resistance R. Table 6 lists such solutions for six different conduction systems. Table 2 in Chapter 9 and Table 1 of this chapter indicate the magnitudes of the thermal conductivities, k, to be employed in the expressions of Table 6, after dividing k by 12. The solution applicable to the problem depicted in Fig. 8, for the cal culation of Rz and Rs, is case 2 in Table 6. Thus for a 1 ft length of 2 in. nominal size pipe (I. D. = 2.067 in., 0. D. = 2.375 in.) insulated with 1 in. of material having a, conductivity of 0.025: , 1.188: g'~ 1.033 Rt = v "IT X .. = 8.5 X 10 ` (hr) (F deg) per Btu. X1 , 2.188 B'fl88 * = 2x~X M25"xl = 3`9 (hf) (F d6g) P6r BtU' Heat Transfer 103 The convection resistances to heat transfer from the pipe wall to the cold water, Ri, 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. These resistances are also directly dependent upon the temperature distribution, and for this reason it is necessary first, to guess on the basis of the problem statement, a tempera ture distribution upon which to base the initial calculations. Since the values of hc for heat transfer between water and pipe walls are relatively high in this temperature range, it is logical to assume only a small tempera ture difference between the temperature of the fluid body and the tempera ture of the pipe wall. For the purpose of an initial guess, this temperature difference will be assumed to be 2 deg. On the other hand, hc for heat transfer from air to a body is relatively small, and a higher temperature difference would be expected between these masses. The value initially assumed here will be 20 deg. In summary, the temperature distribution in the system is assumed as follows: Fluid temperature = 34 F. Inner pipe wall temperature = 36 F. Outer insulation surface temperature = 100 F. Ambient air temperature = 120 F. With these assumptions and the problem statement, it is now possible to calculate values for the convective resistances. If it is found in the ultimate solution of the problem that the temperature distribution is different from that assumed, it will then be necessary to repeat the solution procedure. If reference now be made to Table 2, it is found that Case 3 of this table is a system similar to that encountered in the convection between the water and the pipe wall. The equation for this case is the following: where h0 = 13.9(ld-* --* UJ Um = 5 fps (15) D= = 0.1725 ft 12 ho = 139 x 69 x 362 = 494 Btu per (hr) (sq ft) (F deg). 0.703 This heat transfer rate is through the inner surface of the pipe and it is, therefore, this area that determines the resistance R\. and therefore A = vD = 0.542 sq ft per unit length of pipe, l_ = 1 Ri hoA ~ 494 X 0.512 R, = 3.73 X 10-* (hr) (F deg) per Btu.