prac_final_solns

# prac_final_solns - 22.313 THERMAL-HYDRAULICS IN NUCLEAR...

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22.313 THERMAL-HYDRAULICS IN NUCLEAR POWER TECHNOLOGY Tuesday, May 17 th , 2005, 9 a.m. – 12 p.m. OPEN BOOK FINAL (solutions) 3 HOURS Problem 1 (30%) – Hydraulic analysis of the PWR primary system at cold zero-power conditions i) The momentum equation for the loop is: 2 2 sg core pump A 2 m ) K K ( P dt dm A L l ρ + Δ = ( 1 ) where m is the mass flow rate, L=40 m is the total length of the loop, A=1.65 m 2 is the flow area, K core =7 and K sg =4 are the form loss coefficients for the core and steam generator, respectively. The acceleration and friction terms were neglected in Equation 1, as per the problem statement. Moreover the gravity term is zero because the fluid is isothermal. At steady-state 0 dt dm = and Equation 1 can be easily solved for the steady-state mass flow rate, m ss : ) K K ( P A 2 m sg core pump 2 ss + Δ ρ = l 9,960 kg/s (2) ii) Equation 1 can be re-written as follows: 2 2 ss sg core m m dt dm ) K K ( L A 2 = + ρ l ( 3 ) Equation 3 can be integrated to find m(t) during start-up. Separating the variables, making use of the hint in the problem statement, and setting the initial condition m(0)=0, one gets: τ τ + = / t / t ss e 1 e 1 m ) t ( m ( 4 ) where the time constant, τ , is defined as follows: pump sg core ss sg core P ) K K ( 2 L m ) K K ( L A Δ + ρ = + ρ = τ l l 0.6 s (5)

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Equation 4 is plotted in Figure 1. The time it takes to reach 50% of the steady-state value can be calculated by setting m=0.5 m ss in Equation 4, and solving for t. 0 . 6 6 s ( 6 ) ) 3 ln( 50 τ = τ Figure 1. PWR primary system mass flow rate during cold zero-power start-up. iii) Equation 5 indicates that the time constant is proportional to the loop length and inversely
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prac_final_solns - 22.313 THERMAL-HYDRAULICS IN NUCLEAR...

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