13_ConvectionCoolant_web

# 13_ConvectionCoolant_web - ENU 4134 – Convection to...

This preview shows pages 1–9. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ENU 4134 – Convection to Coolant – Part I D. Schubring Fall 2011 Learning Objectives I 3-e Apply knowledge of single-phase and two-phase convective heat transfer to convection from fuel rods to coolants, including adjustments to correlations regarded in assembly geometry I 4-a Use equations from nuclear heat transfer (as simple functions of axial coordinate z ) to perform simple, analytical single-channel analysis (SCA) I 4-b Derive finite volume method for use in steady-state SCA I 4-c Implement finite volume method in scripting or compiled language to solve a wide range of LWR SCA problems (Pr. 2) I 4-d Use general SCA code to efficiently perform design/safety analyses for LWR (Pr. 2) I 4-e Integrate knowledge from course to evaluate safety- and design-significance of results from SCA code (Pr. 2) + all cross-cutting technical objectives (#5) General Equation for T co- T m Convective heat transfer is typically understood through the heat transfer coefficient: q 00 co = htc ( T co- T m ) (1) T co- T m = q 00 co htc = q 2 π R co htc (2) Computation of htc can be non-trivial, particularly for the 2-phase case, but this equation is general for any htc . Solution of the nuclear TH (single-channel analysis) problem is greatly aided by the fact that q is usually given. Must consider adjustments to htc due to rod bundle geometry. Example Problem I P = 14.608 MPa I Geometry: P = 12 mm, D =10 mm, R fo = 4.6 mm, R ci = 4.7 mm I htc g = 5000 W m- 2 K- 1 I Equation for htc (heat transfer coefficient) (Weisman 1 φ , Schrock and Grossman 2 φ ) I k c = 20 W m- 1 K- 1 and k f = 3 W m- 1 K- 1 I q = 25 kW/m I ˙ m = 0 . 15 kg/s (per channel) I T 1 = 320 ◦ C , T 2 = T sat = 340 x 2 = 0 . 2 Equations for Solution T co- T m = q 2 π R co htc (3) T ci- T co = q 2 π k c ln R co R ci (4) T fo- T ci = q 2 π R g htc g = q π ( R ci + R fo ) htc g (5) T max- T fo = q 4 π k f (6) This is the end of material in the scope of Exam 2 Single Channel Analysis I Single-phase coolant, constant properties & heat transfer coefficient (analytical solutions) I Single-phase coolant, variable properties & heat transfer coefficient (numerical solution) I Two-phase coolant I Miscellanea I Engineering judgment and single-channel analysis Axial Dependence There are simple(-ish) equations for the unknown temperatures. However, many of the variables are actually functions of z (if not also local T ’s). T co ( z )- T m ( z ) = q ( z ) 2 π R co htc ( z ) (7) T ci ( z )- T co ( z ) = q ( z ) 2 π k c ( z ) ln R co R ci (8) T fo ( z )- T ci ( z ) = q ( z ) π ( R ci + R fo ) htc g ( z ) (9) ffT max ( z )- T fo ( z ) = q ( z ) 4 π k f ( z ) (10) Axial Dependence (2) Typically only T m (- L / 2) & P (- L / 2) – coolant bulk temperature, pressure at inlet – are known. T m ( z ) must be determined from an energy balance. Neglecting boiling for the moment: q ( z ) dz = c p ( z ) ˙ mdT m (11) dT m dz = q ( z ) c p ( z ) ˙ m (12) T m ( z ) = T m , in + Z z- L / 2 q ( z ) c p ( z ) ˙ m dz (13) Axial Dependence (3)...
View Full Document

## This note was uploaded on 10/22/2011 for the course ENU 4134 taught by Professor Schubring during the Fall '11 term at University of Florida.

### Page1 / 54

13_ConvectionCoolant_web - ENU 4134 – Convection to...

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online