ENGINEERING OF NUCLEAR REACTORS. Tuesday, October 9 th, 2014, 1:00 2:30 p.m.

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1 .31 ENGINEERING OF NUCLEAR REACTORS Tuesday, October 9 th, 014, 1:00 :30 p.m. OEN BOOK QUIZ 1 (solutions) roblem 1 (50%) Loss o condensate pump transient in a LWR condenser i) Consider the seaater in the condenser as the control volume or this irst analysis. conservation o energy or steady-state yields the olloing equation: The 0 Q m h m h Q m h h ) m c ( T T ) = 137 W i o ( o i o i here m, T i, T o, and c, are, respectively, the seaater mass lo rate, inlet and let temperatures, and speciic heat, all given in the problem statement. Kinetic and gravitational terms ere neglected; seaater as treated as an incompressible luid; seaater lo as assumed to be isobaric, per the problem statement. Note that the same result could have been obtained by applying the conservation o energy to the steam side o the condenser. ii) No consider the steam side o the condenser as the control volume. During the transient the amount o energy entering the control volume is constant (i.e. the mass lo rate and enthalpy o the incoming et steam is constant); hoever, the heat transer rate removed by the seaater is loer (i.e. 90% o its steady-state value), and the condensate is no longer pumped aay by the condensate pump. As a result a net accumulation o mass and energy is expected on the steam side o the condenser, hich thus results in a higher pressure and temperature. iii) Again the steam side o the condenser is the control volume or the analysis. Conservation o mass d dt Where (1) m i m i = 1,31 kg/s is the mass lo rate o the incoming et steam, given in the problem statement. Integrating beteen initial time (t 1=0) and inal time (t =30 s), e get: () 1 mit Note that the initial mass in the control volume, 1, is easily ound rom the initial masses o condensate and steam, hich in turn are ound rom the initial volumes o condensate and steam given in the problem statement, and the speciic volumes o saturated ater and vapor given in the property table: 1 1 g1 V 1 T / v ) Vg1 / vg ( 1) = 86,150 kg 1

2 Here T 1 = 5C is the initial temperature o the steam and condensate. Thereore, Eq. () can be used to get = 13,094 kg. The conservation o energy equation is: de dt 0. 9 Q m i h i (3) here 0.9 Q is 90% o the heat transer rate to the seaater calculated in art i, and h i = 1,840 kj/kg is the enthalpy o the incoming et steam. Integrating Eq. (3) beteen the initial and inal times, e get: (4) u 1u1 ( 0.9Q m i hi ) t Expanding the LHS o Eq. (4), e get: (5) [ u ( T ) xu g ( T )] 1[ u ) x1u g )] ( 0.9Q mihi ) t here T and x are the inal temperature and steam quality in the condenser, respectively. Note that the initial steam quality is ound rom the initial masses in the condenser: x1 g1 / 1 = We can also rite the olloing volume equation or the inal conditions: V v ( T ) x v ( )] (6) [ g T here the total volume o the condenser is V V 1 Vg1 = 864 m 3. Equations (5) and (6) are to implicit and coupled equations in the unknon T and x, thus the problem can be solved by iterations. The results are T = 37. 4C (corresponding to a pressure = 6.56 ka) and x = Thereore the temperature and pressure increase, as expected, and the steam quality also increases. Finally the mass o condensate 1 ) increases ( x ( = 13,060 kg vs 1= 86,133 kg), hich is expected because the condensate pump has stopped orking.

3 roblem (50%) Thermal parameters in the core o a helium-cooled ast reactor i) The local peaking actor is deined as the ratio o the imum pin poer to the average pin poer in the hot uel assembly: loc q q 16 ( )/ ii) Let us irst calculate the axial peaking actor, deined as the ratio o the imum linear poer to the average linear poer in the hot pin: ax q q 1 L 1 L / L q / q( z) dz L L / L / L q L e z q cos( ) dz L sin Le Le 1.09 The average linear poer in the hot pin is q = 16 kw/ 1. m 13.3 kw/m. Then the imum linear poer in the hot pin is q q 16.1 kw/m. And inally the imum heat lux is: q q 570 kw/m d co ax here d co = 9 mm is the pin diameter. iii) The mass lo rate can be ound rom the inlet coolant velocity, density and lo area as: m in Vin A 110 kg/s here V in = 38 m/s, A = 0.75 m and the coolant at the inlet density is given by the equation o state: in in 3.86 kg/m 3 RTin here T in = 63 K (350C) and in = 5.0 a. iv) The coolant velocity at the let o the core is easily ound as ollos: V m/( A) 6.9 m/s 3

4 here RT.3 kg/m 3 and T = 973 K (700C) and = 4.7 a. v) Conservation o energy applied to the coolant in the core yields: vin v 0 Q m ( hin gzin) m ( h gz) v vin Q m [ cp( T T in) g( z zin)] 00 W here e have used the constitutive relation or enthalpy o an ideal gas h h c T T ) ; and c p c v in p( in R =5193 J/kg-K. Note that in this case the kinetic and gravitational terms actually can be calculated (since e have the velocities and elevations), and turn to be completely negligible ith respect to the enthalpy term, as usual. 4

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