Thermodynamics, Heat Transfer, And Fluid Flow. V.2. Heat Transfer (776131), страница 7
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0Page 33HT-02HEAT EXCHANGERSHeat TransferNon-Regenerative Heat ExchangerApplications of heat exchangers may be classified as either regenerative or non-regenerative. Thenon-regenerative application is the most frequent and involves two separate fluids. One fluidcools or heats the other with no interconnection between the two fluids. Heat that is removedfrom the hotter fluid is usually rejected to the environment or some other heat sink (Figure 11).Figure 11Non-Regenerative Heat ExchangerRegenerative Heat ExchangerA regenerative heat exchanger typically uses the fluid from a different area of the same systemfor both the hot and cold fluids.
An example of both regenerative and non-regenerative heatexchangers working in conjunction is commonly found in the purification system of a reactorfacility. The primary coolant to be purified is drawn out of the primary system, passed througha regenerative heat exchanger, non-regenerative heat exchanger, demineralizer, back through theregenerative heat exchanger, and returned to the primary system (Figure 12).In the regenerative heat exchanger, the water returning to the primary system is pre-heated bythe water entering the purification system.
This accomplishes two objectives. The first is tominimize the thermal stress in the primary system piping due to the cold temperature of thepurified coolant being returned to the primary system.HT-02Page 34Rev. 0Heat TransferHEAT EXCHANGERSThe second is to reduce the temperature of the water entering the purification system prior toreaching the non-regenerative heat exchanger, allowing use of a smaller heat exchanger toachieve the desired temperature for purification.
The primary advantage of a regenerative heatexchanger application is conservation of system energy (that is, less loss of system energy dueto the cooling of the fluid).Figure 12 Regenerative Heat ExchangerCooling TowersThe typical function of a cooling tower is to cool the water of a steam power plant by air thatis brought into direct contact with the water. The water is mixed with vapor that diffuses fromthe condensate into the air.
The formation of the vapor requires a considerable removal ofinternal energy from the water; the internal energy becomes "latent heat" of the vapor. Heat andmass exchange are coupled in this process, which is a steady-state process like the heat exchangein the ordinary heat exchanger.Wooden cooling towers are sometimes employed in nuclear facilities and in factories of variousindustries. They generally consists of large chambers loosely filled with trays or similar woodenelements of construction. The water to be cooled is pumped to the top of the tower where it isdistributed by spray or wooden troughs. It then falls through the tower, splashing down fromdeck to deck.
A part of it evaporates into the air that passes through the tower. The enthalpyneeded for the evaporation is taken from the water and transferred to the air, which is heatedwhile the water cools. The air flow is either horizontal due to wind currents (cross flow) orvertically upward in counter-flow to the falling water.
The counter-flow is caused by theRev. 0Page 35HT-02HEAT EXCHANGERSHeat Transferchimney effect of the warm humid air in the tower or by fans at the bottom (forced draft) or atthe top (induced flow) of the tower. Mechanical draft towers are more economical to constructand smaller in size than natural-convection towers of the same cooling capacity.Log Mean Temperature Difference Application To Heat ExchangersIn order to solve certain heat exchanger problems, a log mean temperature difference (LMTDor )Tlm) must be evaluated before the heat removal from the heat exchanger is determined. Thefollowing example demonstrates such a calculation.Example:A liquid-to-liquid counterflow heat exchanger is used as part of an auxiliary system ata nuclear facility.
The heat exchanger is used to heat a cold fluid from 120EF to 310EF.Assuming that the hot fluid enters at 500EF and leaves at 400EF, calculate the LMTDfor the exchanger.Solution:)T2 '400EF& 120EF ' 280EF)T 1 '500EF& 310EF ' 190EF)Tlm '()T2ln& )T 1 ))T2)T1' (280EF & 190EF)ln280EF190EF' 232EFThe solution to the heat exchanger problem may be simple enough to be represented by astraight-forward overall balance or may be so detailed as to require integral calculus.
A steamgenerator, for example, can be analyzed by an overall energy balance from the feedwater inletto the steam outlet in which the amount of heat transferred can be expressed simply as0 ' mQ0 )h , where m0 is the mass flow rate of the secondary coolant and )h is the change inenthalpy of the fluid. The same steam generator can also be analyzed by an energy balance on0 ' mthe primary flow stream with the equation Q0 cp )T , where m0 , cp , and )T are the massflow rate, specific heat capacity, and temperature change of the primary coolant. The heatHT-02Page 36Rev.
0Heat TransferHEAT EXCHANGERStransfer 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 generator0 ' U A )T .using the equation QoolmCondensers 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.Overall Heat Transfer CoefficientWhen dealing with heat transfer across heat exchanger tubes, an overall heat transfer coefficient,Uo, must be calculated. Earlier in this module we looked at a method for calculating Uo for bothrectangular 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 Uo.Uo'1h1%1)rk% 1h2Example:Referring to the convection section of this manual, calculate the heat rate per foot oftube from a condenser under the following conditions.
)Tlm = 232EF. The outerdiameter 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-ft2-EF, and the thermalconductivity of copper is 200 Btu/hr-ft-EF. The outer convective heat transfercoefficient is 1500 Btu/hr-ft2-EF.Rev. 0Page 37HT-02HEAT EXCHANGERSHeat TransferSolution:Uo''1h11)rk%% 1h2112000' 827.6% 0.1 in2001 ft12 in%11500Btuhr&ft 2&EF0Q' U o Ao )Tlm0QL'Uo Ao)TlmL' U o 2B r )Tlm' 827.6Btuhr&ft 2&EF(2B) (0.375 in)1 ft12 in(232EF)' 37,700 Btuhr&ftHT-02Page 38Rev.
0Heat TransferHEAT EXCHANGERSSummaryThe important information in this chapter is summarized below.Heat Exchangers SummaryHeat exchangers remove heat from a high-temperature fluid byconvection and conduction.Counter-flow heat exchangers typically remove more heat thanparallel flow heat exchangers.Parallel flow heat exchangers have a large temperature difference atthe inlet and a small temperature difference at the outlet.Counter-flow heat exchangers have an even temperature differenceacross the heat transfer length.Regenerative heat exchangers improve system efficiency byreturning energy to the system.
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