Q Uo AoTlmUo11h1rk1h2Heat TransferHEAT EXCHANGERSRev. 0Page 37HT-02transfer rate of the steam generator can also be determined by comparing the temperatures onthe primary and secondary sides with the heat transfer characteristics of the steam generatorusing the equation . Condensers are also examples of components found in nuclear facilities where the concept ofLMTD is needed to address certain problems. When the steam enters the condenser, it gives upits latent heat of vaporization to the circulating water and changes phase to a liquid. Becausecondensation is taking place, it is appropriate to term this the latent heat of condensation. Afterthe steam condenses, the saturated liquid will continue to transfer some heat to the circulatingwater system as it continues to fall to the bottom (hotwell) of the condenser. This continuedcooling is called subcooling and is necessary to prevent cavitation in the condensate pumps. The solution to condenser problems is approached in the same manner as those for steamgenerators, as shown in the following example.OverallHeatTransferCoefficientWhen dealing with heat transfer across heat exchanger tubes, an overall heat transfer coefficient,U , must be calculated. Earlier in this module we looked at a method for calculating U for bothoorectangular and cylindrical coordinates. Since the thickness of a condenser tube wall is so smalland the cross-sectional area for heat transfer is relatively constant, we can use Equation 2-11 tocalculate U .oExample:Referring to the convection section of this manual, calculate the heat rate per foot oftube from a condenser under the following conditions. T = 232- F. The outerlmdiameter of the copper condenser tube is 0.75 in. with a wall thickness of 0.1 in. Assumethe inner convective heat transfer coefficient is 2000 Btu/hr-ft -- F, and the thermal2conductivity of copper is 200 Btu/hr-ft-- F. The outer convective heat transfercoefficient is 1500 Btu/hr-ft -- F.2
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