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7 Pages

HW3Prob5

Course: NE 402, Fall 2010
School: N.C. State
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Word Count: 1133

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PWR A has dimensions and operating conditions given below. Core Height Core Mass Flux Number of Fuel Rods Rod Diameter Rod Pitch Core Inlet Loss Coefficient Core Exit Loss Coefficient Grid Loss Coefficient Number of Grids Pressure Core Inlet Temperature Maximum Channel Heat Flux Extrapolation Distance You may assume an axial heat flux profile of the form (z + ) q ( z ) = q0 sin He 144 2.62 x 106 56,876...

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PWR A has dimensions and operating conditions given below. Core Height Core Mass Flux Number of Fuel Rods Rod Diameter Rod Pitch Core Inlet Loss Coefficient Core Exit Loss Coefficient Grid Loss Coefficient Number of Grids Pressure Core Inlet Temperature Maximum Channel Heat Flux Extrapolation Distance You may assume an axial heat flux profile of the form (z + ) q ( z ) = q0 sin He 144 2.62 x 106 56,876 0.374 0.496 1.5 1.5 0.5 8 2250 560 265,700 0.866 inches lbm/hr-ft2 inches inches psia F Btu/hr-ft2 ft Determine the acceleration, friction, forms, elevation and total pressure loss. Compare to that which would be obtained from a simple Bernoullis Equation approach where the density is approximated by the average channel density ( exit + inlet ) / 2 and the density is evaluated at the average coolant temperature. Assume the same treatment holds for viscosity also. Solution Integrate the conservative form of the steady-state Momentum Equation over the channel length 11 P P g ( vvAx ) = z w w g sin gc Ax z Ax c (1) or 1 G2 P f G 2 = + z De 2 gc gc z j G2 g sin K j (z z j ) 2 gc gc (2) where we have assumed a uniform flow area and a vertical flow channel. For convenience the momentum equation has been written in terms of the mass flux. 1 gc 2 1 G2 dz = z 2 1 P dz z 2 1 f G2 dz + De 2 gc 2 1 j G2 K j (z z j ) dz 2 gc G2 j 2 j gc 2 1 g dz gc (3) 1 G2 1 = (P2 P ) 1 g c 2 1 i fi Li Gi2 Dei 2 i g c j Kj g (H 2 H1 ) gc (4) Rearranging gives for the core pressure drop Single Phase 22 1 Pcore = 1 fH core G 2 G2 1 + + g c 2 1 De 2 g c j Kj G2 g + H core 2 j gc gc (5) where location 1 is the core inlet, location 2 is the core exit and and are appropriate average fluid densities in the core. The core pressure drop then consists of four components: 1) The acceleration pressure drop Pacc = G2 gc 1 1 2 1 2) The frictional pressure drop Pfriction = fHcore G 2 De 2 gc 3) The local or forms losses Plocal = j Kj G2 2 j gc 4) The elevation or hydrostatic losses Pelev = g H core gc Acceleration Pressure Drop To determine the acceleration drop, requires the fluid density at the core inlet and exit. The inlet density is given 3 directly in terms of the core inlet temperature, i.e. 1 = (560 F ) = 45.905 lbm / ft . To determine the fluid density at the core exit requires the core exit temperature. This is obtained from the energy balance h2 = h1 + 1 GAx H q( z )Ddz = h1 + 0 q0 DH e GAx ( H + ) cos cos He He (6) The flow area for an arbitrary flow channel is given by Ax = S D 4 2 2 = 0. 496 ( )( 0.374 ) / 4 2 2 = 0.136 in 2 For the given data, the core exit enthalpy is Single Phase 22 2 h2 = 562.4 + (265,700)(0 .374 12)(13.732) 0.866 (12.866) cos cos 6 (2.62 10 )(0.136 / 144) 13.732 13.732 = 652.5 Btu /lbm This value of enthalpy corresponds to a core exit temperature of Texit = 623.6 F . The core exit density can then be obtained from steam tables, or other fluid property tables giving 2 = 40. 21 lbm/ ft . The acceleration pressure drop is then 3 Pacc = = G2 gc 1 1 2 1 1 1 40.21 45.905 (2.62 106 ) 2 (4.17 108 ) = 50.8 lbf / ft 2 = 0.353 lbf /in 2 Frictional Pressure Drop The general form of the Frictional Drop is Pfriction = H 0 f G2 fH core G 2 dz = De 2 g c De 2 g c The friction factor f is a function of the Reynolds number Re = GDe where the viscosity is in general a function of temperature. The equivalent diameter is De = 4 Ax Pw = 4[S D /4] 2 2 D which for the data given in this problem gives De = 4( 0.136 ) = 0. 463 in ( 0.374) A reasonable approximation for the friction factor in rod bundles is the friction factor in smooth tubes from the correlation f = 0.184 Re 0.2 If we assume that over the temperature range of interest, the density and viscosity can be obtained from the simple polynomial fits (T = ) 34.7441 + 0.3519 T 3.7098 104 T 2 Single Phase 22 3 (T ) = 0.1325 + 7.7571 104 T 1.0714 106 T 2 then for a given temperature distribution the frictional drop can be integrated over the channel height. The temperature distribution in the channel is given by T ( z ) = T (0) + 1 GAxC p z q( z)Ddz = T (0) + 0 q0 DH e GAxC p ( z + ) cos cos H He e where we take the specific heat to be Cp = Q h2 h1 652.5 562.4 = 1.417 = = GAx (Texit Tinlet ) Texit Tinlet 623.6 560 The integration is performed numerically, giving Pfriction = H 0 f G2 dz = 797.66 lbf/ft 2 = 5.54 lbf/in 2 De 2 g c If we evaluate the frictional drop by Pfriction = fH core G 2 De 2 g c where the average fluid properties are taken to be the average between the inlet and exit values, then = = giving for the Reynolds number (0) + ( H ) 2 = 0.231 + 0.200 = 0.215 2 45.905 + 40.21 = 43.06 2 ( 0) + ( H ) 2 = Re = (2.62 106 )(0.463 / 12) = 470,178.3 0.215 and the friction factor f = 0.0135 such that the frictional pressure drop is Pfriction = = fH core G 2 De 2 g c (0.0135)(144) (2.62 106 ) 2 (0.463) (2)(43.06)(4.17 108 ) = 802.56 lbf / ft 2 = 5.57 lbf /in 2 If we assume the average fluid properties can be evaluated at the average fluid temperature, then = (T ) = (591.8) = 0.216 Single Phase 22 4 = (T ) = (591.8) = 43.56 giving for the Reynolds number Re = (2.62 106 )(0.463 / 12) = 468,001.5 0.216 and the friction factor f = 0.0135 such that the frictional pressure drop is Pfriction = = fH core G 2 De 2 g c (0.0135)(144) (2.62 106 ) 2 (0.463) (2)(43.56)(4.17 108 ) = 793.3 lbf / ft 2 = 5.51 lbf /in 2 Local Losses The local or forms losses are due to the grid spacers as well as the core inlet and exit losses. The local loss coefficients and their locations are zj Kj 0 1 1/3 2 2/3 4 5 1/3 6 2/3 8 9 1/3 10 2/3 12 For Plocal = 1.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 1.5 j Kj G2 2 j gc where j = (T ( z j )) then Plocal = 1333 lbf/ft 2 = 9.256 lbf/in 2 If we assume the density can be taken as a constant average value, then for Single Phase 22 5 Plocal = j Kj G2 2 j gc j Kj G2 2 gc = (8 0.5 + 2 1.5) (2.62 106 ) 2 (2)(43.06)(4.17 108 ) = 1338 lbf /ft 2 = 9.292 lbf /in 2 where density has been taken as the average between the inlet and exit values and Plocal = j Kj G2 2 j gc j Kj G2 2 gc = (8 0.5 + 2 1.5) (2.62 106 ) 2 (2)(43.56)(4.17 108 ) = 1322.6 lbf /ft 2 = 9.185 lbf /in 2 if the density is taken at the average fluid temperature. Elevation Pressure Drop The elevation pressure drop is in general given by Pelev = H ( z) 0 g dz = gc H (T ( z )) 0 g dz gc which for the temperature given here yields Pelev = H ( z) 0 g dz = 520.99 lbf/ft 2 = 3.618 lbf/in 2 gc For Pelev = H ( z) 0 g g dz = H gc gc then if the average density is approximated as the average between the inlet and exit densities Pelev = g H core gc (43.06)(1)(144 / 12) = 516.72 lbf /ft 2 = 3.59 lbf /in 2 If the average density is approximated as the density at the average channel temperature, then Single Phase 22 6 Pelev = g H core gc (43.56)(1)(144 / 12) = 522.72 lbf /ft 2 = 3.63 lbf /in 2 The total pressure drop is then Pcore = Pacc + Pfriction + Plocal + Pelev = 0.353 + 5.54 + 9.256 + 3.618 = 18.767 lbf /in 2 if the fluid properties (density and viscosity) are allowed to vary continuously along the channel. If we assume Bernoullis Equation is valid, then the acceleration pressure drop is zero. If we approximate the fluid properties as the average between the inlet and exit values, then the total pressure drop is Pcore = Pacc + Pfriction + Plocal + Pelev = 0.0 + 5.57 + 9.292 + 3.59 = 18.452 lbf /in 2 and if we approximate the fluid properties as those taken at the average channel temperature, then the total pressure drop is Pcore = Pacc + Pfriction + Plocal + Pelev = 0.0 + 5.51 + 9.185 + 3.63 = 18.325 lbf /in 2 Single Phase 22 7
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