Page 255 - Handbook of Energy Engineering Calculations
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absorbs  heat  from  the  NaK  coolant,  leaving  the  evaporator  at  607°F
               (319.4°C).
                  Draw  a  line  vertically  downward  from  point  3  until  the  liquid  enthalpy
               curve  is  intersected,  point  4.  Point  4  represents  the  boiler  feedwater  inlet
               temperature,  or  442°F  (227.8°C),  based  on  the  valid  assumption  that  the

               feedwater leaving the condenser hot well is in the saturated state.


               5. Compute the reactor coolant flow rate
               Sodium  enters  the  reactor  at  600°F  (315.6°C)  and  leaves  at  1000°F
               (537.8°C),  Fig.  1.  Thus,  the  temperature  rise  of  the  Na  during  passage

               through  the  reactor  is  1000  −  600  =  400°F  (222.2°C).  Also,  the  average
               specific heat of Na is 0.306 Btu/(lb · °F) [1.28 kJ/(kg · °C)], found from a
               tabulation of Na properties in an engineering handbook.
                  Compute the Na flow from f = 3413 kw/Δtc, where f = Na flow rate, lb/h;

               kw  =  reactor  heat  rating,  kW;  Δt  =  Na  temperature  rise  during  passage
               through  the  reactor,  °F; c  =  specific  heat  of  the  Na  coolant,  Btu/(lb  ·  °F).
               Substituting  gives  us  f  =  3413(400,000)/[400(0.306)]  =  11,130,000  lb/h
               (1402.4 kg/s).



               6. Compute the boiler heating liquid flow rate
               Use the same relation as in step 5, substituting the temperature change and
               specific heat of NaK. Since the NaK enters the boiler at 950°F (510.0°C) and
               leaves  at  550°F  (287.8°C),  its  temperature  change  is  950  −  550  =  400°F
               (222.2°C). Also, the specific heat of NaK is 0.251 Btu/(lb · °F) [1.05 kJ/(kg ·

               °C)], as found from NaK properties tabulated in an engineering handbook. So
               f = 3413(400,000)/[400(0.251)] = 13,600,000 lb/h (1713.6 kg/s).


               7. Compute the plant thermal efficiency
               The net station output kw = gross output of turbine, kW, minus the total plant

               auxiliary demand, kW = 155,000 − 12,000 = 143,000 kW. Then overall plant
               thermal  efficiency  =  net  station  output,  kW/reactor  heat  output,  kW  =
               143,000/400,000 = 0.357, or 35.7 percent.


               Related Calculations. This analysis is valid for a cycle in which the reactor
               coolant does not do work in the turbine. In general, designers prefer to avoid

               using the reactor coolant in the turbine. Although the thermodynamic aspects
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