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CHAPTER 6: PROBLEMS WITH FREEZING AND THAWING

Shortcuts:

CASE 6.01 - Neumann's Freezing Test Case

CASE 6.02 - Stefan Problem

CASE 6.03 - Buried Warm Pipe in Homogeneous Soil without Insulation

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TITLe CASE 6.02 - Stefan Problem
///// Analytic solution given by:
///// Lunardini, V.J. (1981). Heat Transfer in Cold Climates
///// Van Nostrand Reinhold Co., Page 367 - 370
///// Analytic formula ====> interface y proportional to sqrt (t)
************************************************************************
/
PROBLEM TYPE: FREEZING & THAWING AT 0 DEG C
/
********** WHEELER freeze/thaw saturation characteristics
********** Default parameters (n=10, td=1.e-30, sr=0.)
/
GRID NODEs 3 BY 101
COOR Y MINIMUM = -1.0, MAXIMUM 0.
/
SET T = -0.1 DEG Centigrade
/
BOUNdary FOR T, at Y+ VALUE = 0.5 DEG Centigrade
disable flow
/
/PROPERTY OPTION FOR T ARITH
ROCK density 1, porosity = 1.
DENSity of first fluid: 1
DENSity of second fluid: 1
THERmal for soil not required since porosity is unity
FLUID SPECIFIC HEAT: = 0.01
FLUID THERMAL CONDUCTIVITY = 1.00
FLUID SPECIFIC HEAT: = 0.01 FOR SECOND PHASE:
FLUID THERMAL CONDUCTIVITY = 1.00 FOR SECOND PHASE
FLUID ENTHALPY: = 1. (LATENT HEAT)
/
Diagnostic node 2,5 every 100 steps
DEBUG GEOMETRY OFF
/
SELEct window (2,1) to (2,999)
OUTPut for S only in SELEcted window
/
save off
/
SOLVE FOR 0.01 DAYS, DT=0.00001 DAY
OUTPut NOW for SELEcted window
///===> Interface should be at 0.1
/
SOLVE for 0.03, DT=0.00001 fac 1.05, max = 0.0001 !!! total T = 0.04
OUTPut NOW for SELEcted window
///===> Interface should be at 0.2
/
SOLVE FOR 0.05 DAYS, DT=.0001 !!! total T = 0.09
OUTPut NOW for SELEcted window
///===> Interface should be at 0.3
/
SOLVE FOR 0.07 DAYS, DT=.0001!!! total T = 0.16
/OUTPut NOW for SELEcted window
///===> Interface should be at 0.4
/
OUTPut for T and S for SELEcted window
/
/
END
/
/

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************************************************************************
TITLe CASE 6.03 - Buried Warm Pipe in Homogeneous Soil without Insulation
************************************************************************
/
PROBLEM WITH FREEZING/THAWING AT T=0.
/
********** WHEELER freeze/thaw saturation characteristics
********** Default parameters (n=10, td=1.e-30, sr=0.)
/
GRID NODEs 26 BY 19 BY 39
/
COORdinate X NODEs are:
0.00 0.09 0.286 0.429 0.571 0.713 0.90 1.10 1.30 1.75
2.25 3.00 4.00 5.00 7.00 9.00 11.0 13.0 15.0 17.0
19.0 21.0 25.0 30.0 35.0 37.50
/
COORdinate Y MINIMUM = -20 MAX 20
/
COORdinate Z NODEs are:
-40.0 -35.0 -30.0 -25.0 -21.0 -19.0 -17.0 -15.0 -13.0 -11.0
-9.00 -7.00 -5.00 -4.00 -3.10 -2.30 -1.75 -1.30 -1.10 -0.90
-.713 -.571 -.429 -.286 -.143 0.00 0.143 0.286 0.429 0.571
0.713 0.90 1.10 1.30 1.75 2.25 3.00 4.00 4.50
/
/
/ BACKGROUND AREA IS DESIGNATED AS ZONE 1
ZONE 1 IS FROM (1,1,1) to (99,99,99)
/
/ CONCRETE AND INSULATION COMBINED IS ZONE 2
ZONE 2 IS FROM (01,01,25) to (05,99,27)
ZONE 2 IS FROM (01,01,24) to (04,99,28)
ZONE 2 IS FROM (01,01,23) to (03,99,29)
ZONE 2 IS FROM (01,01,22) to (02,99,30)
/
/ WARM PIPE IS ZONE 3
ZONE 3 IS FROM (01,01,25) to (04,99,27)
ZONE 3 IS FROM (01,01,24) to (03,99,28)
ZONE 3 IS FROM (01,01,23) to (02,99,29)
/
/ FROZEN BLOCK IS ZONE 4
ZONE 4 IS FROM (17,09,18) to (18,10,20)
/
RENAME S to 'S - LIQUID WATER FRACTION'
RENAME S2 to 'ICE - SOLID WATER FRACTION'
/
SET P = 0 EVERYWHERE
SET T = -0.10 deg Centigrade EVERYWHERE
SET ICE fraction = 1.0
/
SELEct material 3
SET ICE fraction = 0.0 in SELEcted material
SET T = 80. in SELEcted material
FIXED T SELEct material
/
SELEct material 4
FIXED T SELEct material
SET T = -20. in SELEct material !!!! DEG C
/
BOUNdary FOR P, INDEX = X-, FLUX = 0.
BOUNdary FOR P, INDEX = Y-, FLUX = 0.
BOUNdary FOR P, INDEX = Y+, FLUX = 0.
BOUNdary FOR P, INDEX = Z+, FLUX = 0.02 !!! Infilteration at top
/
BOUNdary FOR T, INDEX = X-, FLUX = 0.
BOUNdary FOR T, INDEX = X+, GRAD = 0.
BOUNdary FOR T, INDEX = Y-, FLUX = 0.
BOUNdary FOR T, INDEX = Y-, FLUX = 0.
BOUNdary FOR T, INDEX = Z-, FLUX = 0.
BOUNdary FOR T, INDEX = Z+, value = 5.
/
****************************************************
***** >>>>>>>>>>>>>> UNITS ARE: kg, m, hour, deg C
****************************************************
/
DENSity of first fluid: 1000
DENSity of second fluid: 1000
VISCosity of first fluid: 1
/
FLUID SPECIFIC HEAT: = 4.18700
FLUID THERMAL CONDUCTIVITY = 2.16
FLUID SPECIFIC HEAT: = 1.46545 FOR SECOND PHASE:
FLUID THERMAL CONDUCTIVITY = 3.00 FOR SECOND PHASE
FLUID ENTHALPY: = 100 (LATENT HEAT)
/
/ Hydrothermal properties for each zone
/
FOR ZONES 1 THROUGH 4
MULTIPHASE FREEZING: A=10, TD=1.E-30, SR=0. ALPHA = 1.
MULTIPHASE: VAN, MUAL N=2.5, ALFA = 1., SM=0.0
ROCK DENSity 1600.
THERMAL CP = 1.00, KT = 5.0
/
FOR ZONES 1 AND 2
ROCK POROsity = 0.31
FOR ZONES 3 AND 4
ROCK POROsity = 0.00
/
FOR zones 1 to 4
HYDRAULIC ss = 0.31, K=3*0.24 m/hr
FOR zone 3
HYDRAULIC ss = 0.01, K=3*1.E-9 m/hr !!! No flow across pipe
/
CONVergence for P, 1.E-4, 1
CONVergence tolerance for Flow, 1.E-3, 1
/
Diagnostic output at (2,10,22) every 10 steps
DEBUG GEOMETRY OFF
SAVE OFF !!!out
!!!in SAVE U,V,W,P,T, and ICE EVERY 100 STEPS on 'F2.ARC'
/
SELEct window from (01,10,11) to (21,10,99) step (2,1,2)
OUTPut in XZ plane for U,V,W,P,T, and ICE for SELEcted window
/
!!!<in> SOLVE FOR 9600 UNITS STEP 0.5 DTFAC=1.02 DTMAX = 5 DAYS
SOLVE FOR 50 days, dt=0.5, fac=1.02, dtmax=5 !!!out
/
END
/
/

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