final_exam_soln - Buongiorno Spring 2007 22.313J 2.59J...

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Buongiorno, Spring 2007 22.313J, 2.59J, 10.536J THERMAL-HYDRAULICS IN POWER TECHNOLOGY Tuesday, May 22 nd , 2007, 9 a.m. – 12 p.m. OPEN BOOK FINAL (solutions) Problem 1 (35%) – Steady-state natural circulation in a steam generation system i) The flow in the loop is due to natural circulation, driven by the density difference between the two-phase riser and the single-phase downcomer. The momentum equation for the loop is: m 2 ( ρ down riser ) gL = φ l 2 o K 2 f A 2 ( 1 ) The fluid in the downcomer is saturated water therefore its density is ρ = ρ f in the riser is: ρ riser = αρ g + (1 ) ρ f ( 2 where is the void fraction. If HEM is used: α = ρ g 1 1 x 3 ) 1 + ρ f x where is the flow quality in the riser. The two-phase multiplier for the form loss in the steam separator is: 2 f φ l o = x ρ ρ 1 ( 4 ) g per the problem assumption. The flow quality x can be found from the energy balance for the heater: Q = xh fg m x = Q & /( h fg m &
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Buongiorno, Spring 2007 2 ( ρ f g ) gL = 1 + Q /( m h ) f 1 K m (6) 1 + 1 Q /( m h fg ) g fg g 2 f A 2 Q /( m h fg ) f which could be solved to find m = m (Q ,A,L,K) . ii) If Q =0 (no steam), one has x =0, α =0, riser = f , and therefore m =0. For Q = m h fg (complete vaporization), one has x =1, =1, riser = g , φ l 2 o = f and From Eq. (1): g 2 g A 2 ( f g ) gL m = K ( 7 ) An increase in heat rate, Q & , increases the density difference between the riser and the downcomer, which would tend to increase the flow. However, an increase in Q & also increases
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This note was uploaded on 11/27/2011 for the course CHEMICAL E 20.410j taught by Professor Rogerd.kamm during the Spring '03 term at MIT.

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final_exam_soln - Buongiorno Spring 2007 22.313J 2.59J...

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