Chemical Engineering 693R Reactor Design and Analysis Lecture 4 Reactor Flow and Pump Sizing
Spiritual Thought 2
Rod Analysis with non-constant q 3 Now q = qq zz = qqq mmmmmm sin ππzz Steady state Know mm, T b,in Develop expressions for: T max (z) T fuel (z) T clad (z) T bulk (z) LL q mm T b,in
Bulk Flow 4 Energy increase in Flow: qq (zz) = mm ddd dddd If no phase change, dh = C p dt b mm ddcc pptt bb dddd = qq zz = qqq mmmmmm sin ππππ LL Solve for T b T b (z) = qqq mmmmmm mmcc pp LL ππ 1 cos ππππ LL
Solution 5 T b becomes new boundary condition Solve clad, gap, and fuel regions based on this Final solution of T max based upon new boundary condition This problem is reserved for HW #4
Flow Analysis 6 Size pumps Evaluate Erosion Flow rates for heat transfer Core sizing/dimensions Vibration analysis Core Orificing Valve/component sizing
Flux 7 Flux = quantity per unit area per unit time Heat flux : J/m 2 *s Mass flux: kg/m 2 *s Neutron flux: neutrons/m 2 *s Momentum flux: kg*m/s*m 2 *s = kg/ms 2 ρv = kg/m 3 * m/s = kg/m 2 *s = mass flux Then the flux of any quantity per unit mass (q) is.ρqv q = h J/m 2 *s Heat flux q=1 kg/m 2 *s Mass flux q=v kg/s*m 2 *s = kg/m*s 2 Momentum flux -ρ*q*v. n*a is the rate of quantity through surface A
Lagrangian/Eulerian 8 Lagrangian Eulerian Motion of system of fixed mass CONSERVATION LAWS Fluid elements move around and deform Some fixed control volume CONVENIENT FOR ENGINEERING Don t care about fluid elements Want pressure and velocity fields at a point. Pressure on a wing Drag on a car Not the pressure of a chunk of fluid as it moves along
Lagrangian Solution 9 Differential Analysis Solution at any point in space Produce vector/scalar fields Pressure Velocity Enthalpy CFD Common in Licensing, beyond class scope For concept reactors, use Eulerian approach
Eulerian Solution 10 Utilize Reynolds Transport Theorem: ddbb ssssss = dd dddd dddd ρρρρ dddd + ρρρρ vv nn dddd CCVV CCSS Recall that any property can be B sys Mass Continuity Equation Momentum Force Vector Balance Energy Mechanical Energy Balance
Governing Equation 11 0 = dd dddd CCVV ρρ dddd + CCSS ρρ vv nn dddd FF = dd dddd CCVV ρρmm vvdddd + CCSS ρρmm vv vv nn dddd. Combined, these give integral view of fluid dynamics, useful in reactor design
Mechanical Energy Equation 12 WW ss FF = mm PP ρρ + vv2 2 + gg zz Can be viewed as multiple components of energy balance for control volume: Gravity Stagnation (acceleration) Pressure difference Friction Shaft work
Non-circular flow regions 13 Non-circular flow Hydraulic Diameter, DD ee = 4AA PP ww, f lam = shape specific f turb ~ round tube f d P s DD ee DD ee = 4AA PP ww = 4ss2 4ss = s = 4AA = 4 PP ππ 4 DD PP ww ππππ f lam =57/Re f = CC ffff RRRR nn n = 1, laminar, n = 0.18, turbulent
Tube Bundle Friction 14 Lam - CC ffff = aa + bb 1 PP DD 1 + bb 2 PP 1 2 DD For edge and corners, P/D replaced by W/D Turb - RRRR DDDD = 10 4 ff ff cccc 1.045 + 0.071 PP DD 1 - RRRR DDDD = 10 5 ff ff cccc 1.036 + 0.054 PP DD 1 PP OR CC fftt = aa + bb 1 1 + bb PP DD 2 1 2 DD Coefficients found in Tables 9.5 and 9.6 of Nuclear Systems I Add contributions from spacer grids and mixing vanes
Entrance Region Effects 15 Velocity profile develops here zz DD ee = 0.05RRRR (laminar) zz DD ee = 25 40 (turbulent) Characterized by higher f (flow acceleration) Exists for most flow disruptions Sensors far from orifices, obstructions, etc.
Flow Networks 16 Pipe Networks composed of single pipes Pipes is series Constant VV, additive ΔP Type I find ΔP Type II find VV Pipes in parallel Constant ΔP, additive VV Type I Type II Type III Find D
Sample Problem 17 Calculate the pressure drop across a bare-rod core assembly (no spacer grids or mixing vanes) for a closed stainless steel assembly with 264 tubes in square configuration, P/D = 1.13, D = 0.34 in, L = 10 ft, mm = 100 kg/s water.
Single Phase Heat Transfer (I) 18 qq zz = mm ddd = mm ddcc pptt = qqqq(zz)ππππ dddd dddd Need a boundary condition: 1. Constant T wall 2. Constant q Case 1: TT bb zz = TT bb,iiii + qqqqππππππ mmcc pp q ` T b (z) mm T b,in
Single Phase Heat Transfer (II) 19 T wall Case 2, Constant T wall (BC is T(0) = T b,in ) mm ddcc pptt dddd = h TT wwaaaaaa TT bb ππππ T b (z) mm TT bb zz = CC 1 ee hππππ mmccpp zz + TT wwwwwwww T b,in TT bb zz = TT wwwwwwww T b,in 1ee hππππ mmccpp zz + TT wwwwwwww
How to find h? 20 Pick correct flow correlation. 5 questions:l Fuid? (Pr~1, Pr>1, Pr<<1) Flow Regime? Geometry of channel? BC? q or T w? Fully Developed Flow vs. Entry Flow? Nu + correlations can be found in books, online, textbook, etc.
Pump Curve Schematic 21
Pump Operation Curves 22 Piping system requires a given V and a given H. H loss is friction and minor losses, etc. Pump has a corresponding V and H. These must match, forming the operating point. This may not be the best efficiency. Select a pump so that the best efficiency point (BEP) occurs at the operating point. Generally oversize the pump a bit higher flow for given H req or Higher H avail for given flow Add a valve after pump raises H req to match H avail for given flow Somewhat wasteful, but offers control. Also may increase efficieny. (But higher efficieny may not compensate for extra work wasted in the valve (see example 14.2)