- Email: [email protected]
Heat transfer for slab-on-grade floor with stepped ground
Energy Convers, Mgmt Vol. 39, No, 7, pp. 691-701, 1998
~
Pergamon PII: S0196-8904(97)00045-9
© 1998 Elsevier Science Ltd. All rights reserved Printed in Great Britain 0196-8904/98 $19.00 + 0.00
HEAT TRANSFER FOR SLAB-ON-GRADE FLOOR WITH STEPPED G R O U N D
SANGHO CHOI and MONCEF KRARTI Joint Center for Energy Management, CEAE Department, University of Colorado, Boulder, CO 80309-0428, U.S.A.
(Received 15 August 1996)
Abstraet--A general solution is developed to calculate foundation heat transfer for a slab-on-grade floor with lifted ground. The Interzone Temperature Profile Estimation (ITPE) technique is applied to determine the steady state variation of temperature within the ground. A water table at constant temperature is assumed to exist at a given depth below the soil surface. The derived solution addresses all the common horizontal insulation configurations for slab-on-grade floor foundations. The effect of various insulation configurations and the height of the lifted ground on foundation heat loss are discussed in detail. © 1998 Elsevier Science Ltd. Slab-on-grade floor
Isotherms
Stepped ground
Total slab heat loss
NOMENCLATURE a ~
Distance from origin to left side of stepped ground, m
b= Distance from origin to boundary where soil temperature is undisturbed, m C =
Distance from origin to stepped ground surface, m
d= Distance from origin to soil ground surface, m f l = Distance from origin to left side of foundation, m f 4 = Distance from origin to right side of foundation, m f2,3 = Distance from origin to start and end point of insulation, m
Hf 1-3 = Ratio of U-factor of slab portion to soil thermal conductivity, m -~ Ratio of U-factor of soil surface portion to soil thermal conductivity, m -~ Soil thermal conductivity, W/m°C Foundation heat loss, W/m2 Qlou A . , B . , C . , D . , E . = Fourier coefficients T= temperature, K T~= building air temperature, K T,= outside air, K Tw= water table temperature, K T~= soil surface temperature, K x,y= space coordinates, m Hfs =
Greek symbols • , /L ?, 6-- Coefficients defined in equations (5), (6), (7), (8) and (9) #., v., Z., ~. = eigenvalues, m -~
Subscripts f = floor I = zone (I) II -- zone (II) III -- zone (III)
1.
INTRODUCTION
It is well k n o w n t h a t the g r o u n d s u r f a c e t o p o g r a p h y affects the e n e r g y p e r f o r m a n c e o f buildings. I n p a r t i c u l a r , the t e r r a i n t o p o g r a p h y c a u s e s c h a n g e s in t h e a i r f l o w a r o u n d b u i l d i n g s w h i c h influences b u i l d i n g H V A C e q u i p m e n t o p e r a t i o n , i n f i l t r a t i o n a n d e x f i l t r a t i o n , a n d e v e n t r a n s m i s s i o n h e a t t r a n s f e r t h r o u g h b u i l d i n g e n v e l o p e s . H o w e v e r , t h e effect o f t h e t o p o g r a p h y o n t h e f o u n d a t i o n h e a t t r a n s f e r is n o t well u n d e r s t o o d . M o s t o f t h e w o r k in f o u n d a t i o n h e a t t r a n s f e r a s s u m e s t h a t the g r o u n d s u r f a c e is leveled o r b e r m e d . 691
692
CHOI and KRARTI: HEAT TRANSFER FOR SLAB-ON-GRADE FLOOR
In some hilly and mountainous locations, the building foundations are constructed over or near the edge of a lifted ground or even a cliff. In cold regions, it is a common practice to use a lifted floor system (typically with compacted granular fill) with in-ground insulation to dissipate heat losses from the structures in order to maintain the permafrost level at the desired level [2]. The vast majority of existing heat transfer calculations for slab-on-grade floor assume that the ground surface is flat. No work has been done to investigate the temperature variation along the slab surface and within the ground when the topography of the ground surface is not regular. In this paper, a solution for a slab-on-grade floor with a stepped ground is presented. A water table at a finite depth of the soil surface is assumed. The ITPE technique [3, 4] is used to solve the steady state heat conduction equation within the soil medium. The first section of the paper gives the general solution for a slab-on-grade floor with stepped ground surface. In particular, the formal expression for soil temperature and heat loss are derived. The second section of the paper deals with parametric analysis to determine the effects on the foundation heat loss of the stepped ground with various insulation levels, depths, and distances between the stepped ground and the slab.
2.
FORMULATION
OF
THE
PROBLEM
The slab-on-grade floor problem with an uneven soil surface is modeled as shown in Fig. 1. The ground is assumed homogeneous with constant thermal properties. Along the bounding surfaces x = 0 and b, the soil temperature is assumed to be undisturbed and is simply considered to be a linear function of depth. Under these conditions, the steady state temperature distribution T(x,y) beneath the slab-on-grade floor with the stepped ground, is a solution of the following equation: ~2T ~2T ~x---~ + ~ = 0
(1)
Y
SoH
Surface .....................................
d
................................... "'wl ITa T ................. ~ ...................... / Son Surface s C Of 4
(n)
(m) Tw
o,o
X
Water Table
iiii~ 7 " ~
A Fig. 1. Slab-on-grade floor with the stepped ground.
CHOI and KRARTI: HEAT TRANSFER FOR SLAB-ON-GRADE FLOOR
693
with the boundary conditions: T=T~
@ y=d,O<~x<~a
T=T~
@ y=c,a<~x<~b
@y=0
T=Tw
Hfs × d
T = Ts(y)=~T~ @ x = O
with Ts = Ta Hfs × d + 1
T = T,(y) =
with T~ = Ta Hfs x c + l
Ts @ x = a
Hfs x c
Note that, without loss of generality, the water table temperature, Tw, can be set to zero (i.e. Tw = 0). This assumption will be made for the remainder of this paper. Using the ITPE technique [3, 4], the ground medium is divided into three zones as shown in Fig. 1. Also, the temperature profiles at the surfaces are defined below as functions of x and y, respectively. T(x,d)= A(x) T(x,c) = C(x)
0
T(a,y) = B ( y ) O < x < a, T(a,y) = D(x) T(x,c) = E ( y ) a < x < b,
c
Using the separation of variables technique [1], the solution in zone (I) is
T~(x,y)
+
=
~_e~.
sinhg.(y-c) sinh #. (d - c)
sinh/~.(d - y) sin ~. x sinh #. (d - c)
2•lC"
+-7-=-=_
+
~ a .-l
2 ~ A. sin/z, x
B. sin X. (Y - c .- t
)sinh Z . x sinh X. a
' 7", (d - e ( - 1)") sin ;(.(y - c) sinh z.(a - X) Z. sinh Z. a
. "-d
(2)
In zone (II), the solution is given by _2 ~ Tl~(x,y) =
+2
en-I
a..t
sinh/~, y
C. sin/~, x sinh #. c
sinh q/. x D. sin ~O. y sinh q,. a
2 o~ T, ( - 1) ~ sinh ~ . ( a - x) -c ~. ~ ~. sin~0.y sinhqJ, a
(3)
694
CHOI and KRARTI: HEAT TRANSFER FOR SLAB-ON-GRADE FLOOR
The expression for the temperature in zone (III) is given by ~ E. sin v.(x - a) sinh v. y sinh v.c
2
Tm(x,y) - b - a = t
sinh ~h.(b - x) + 2 E D . sin 0 . y c =, sinh~k.(b a) 2 ~ T~(- 1)" sinh ~.(x - a) -c=~ qJ. sin ~O.y sinh ~b.(b a)
(4)
with nn
nn
~°=--a' V " - b - a '
nn
Z" = d - c '
i
A. =
nn
O"
c
a
A(x)sin # . x d x
B. = fdB( y)sin Z.( Y -- c)dy
i j
C. =
a
C(x)sin p. xdx
cDc
D. =
D(y)sin O.ydy
0
~a b
E(x)sin v.(x - a)dx
E. =
The Fourier coefficients A.-E. can be determined from the required continuity of the heat flux along the surfaces and the third-kind boundary conditions along the various surfaces. First, the third-kind boundary condition at y = d and 0 ~< x ~< a gives the condition +~ A
+E
+~
2c,
.SA.+
n=l
n=l
with
A
Id-~c ~ Ts{c(-1)'-d} t.
--
n=
1
pp
A~(tanh#Ad-c))
~P
2.[
7.A __
Ab
2
A~ ) [ t a n h / # ( d - c ) )
Z.(-- 1)"(-- 1)p tanh #p(d - e) tanh #p(d - c)
(5)
CHOI and KRARTI: HEAT TRANSFER FOR SLAB-ON-GRADE FLOOR
695
where, A ~ = HfsTa(1 -
cos
ltpf
ktpf 1 -
1) + H f 1T(cos
cos/6f2)
+ H f 2 T i ( c o s #p f 2 - cos ktpf 3)
+ H f 3 Ti(cos # p f 3 - cos # ~ f 4) + HfsTa(cos #pf 4 - ( - 1) p) A b =/~. cos p . f 1 sin p p f l(Hfs - H f 1) +/~, cos # , f 2 sin # . f 2 ( H f I - H f 2 ) + # ° cos p , f 3 sin # p f 3 ( H f 2 - H f 3) + #, cos # . f 4 sin # p f 4(Hfs - H f 3) A ¢ = pp cos p p f 1 sin # . f l(Hfs -- H f 1) + pp cos p p f 2 sin # . f 2 ( H f 1 - H f 2 ) +ktp cos p p f 3 sin # , f 3 ( H f 2 - H f 3) + I~p cos #p f 4 sin # . f 4(Hfs - H f 3) The third-kind b o u n d a r y condition at surface x = a and c ~< y < d gives the condition Bp=a~+
+oc
+(x
Efl~,A.+
ET~,B.+
.=1
+oc
E6B..pC.
.=1
(6)
n=l
with
STs{c-d(-1)P}+ B ' ~ ( t a n h z e a )
s ~P = [
Zp
~-pJ\
dsinzpa
2
]
tanh zpa
/7,,~ = a kl.(- 1)"(- 1) p X2 + kl: 2
{
/~
Bb
]tanh#p(d-
7.,5 - d - c 2(Z.-Tzp) + 2(z, - z.)J"
=
c)
z,
2 i~. tanh ;ca -a..(--, zJ +
where, B a = H w l T a ( c o s Zp(wl - c) - ( - 1) p) + Hw2Ta(1 - cos Xp(wl - c)) B b = sin((Xp + X,)(c - wl))(hw2 - h w l ) B ~ = sin((zp - Z.)(wl - c))(hw2 - h w l ) The heat flux continuity at surface y = c and 0 ~< x ~< a gives the condition +~
c, =
+~ C
n=l
(7)
Z ~.. c D. n=l
with
S2 ~7" #"
~ = ~ ~ : ~ + t " . :~ "p " ~'"
2
~ T,(c-d(-l')) .=,
fl~ =
b.c = -c
#& ~S
tanh#p(d-c)tanh#xec
#J + Z.2J[#p(tanh # p ( d - c) + tanh #p c)
/6 tanh # e ( d - c)tanh #e c sinh ~tp(d - c) #p(tanh/~.(d - c) + tanh/zp c) 2 d-c
?.c=
d
Z. P e ( - 1) p tanh #e(d - c)tanh #e c #~+Z. 2 #p(tanhpp(d-c)+tanhgpc)
2 q/. # p ( - 1)P(- 1)" tanh gp(d - c)tanh #e c p~ + $.2 pp(tanh gp(d - c) + tanh #p c)
}
696
CHOI and KRARTI:
HEAT TRANSFER
FOR SLAB-ON-GRADE
FLOOR
The heat flux continuity at surface y --- d and a ~< x ~< b gives the condition + oc
+
Op=~D+ 2~.,~Cnnt- 2"7.,pE. D
n=l
(8)
n=l
with ~'.T s ( - 1)Pc T s ( - 1) p "(~. tanh~kp(b--a)tanhO, a e~ = - [ d sinh ~,, a + d sinh Op(b - a ) J [~,p(tanh Op(b - a) + tanh O, a ) J 7~ - 2 p. ~bp(- 1 ) " ( - 1) p ~" tanh Op(b - a)tanh ~bp a "( a 0 2 + p~ [0p (ta--~ ~ ---~ ta--~p a)J
7Dp _
2
tanh ~bp(b - a)tanh Op a O~ + v2, [ O p ( t ~ - h O - - - ~ - - ~ a ~ - ~ p a ) J
v. 0 p ( - 1) p ~"
b-a
"(
The third-kind b o u n d a r y condition at surface y = c and a ~< x ~< b gives the condition +oc
=
E
]~,,p D.
(9)
n=l
with
0~p =
--
.=t
Vp2 + ~2.
fie =c
+
Vp
v, + tanh VpC
2~._Vp(--1)"/'. tanhvpc ) vp2+~k2 ~ v ~ + t a n h v p c
The Fourier coefficients Ap-Ep are determined by truncating the sums in equations (4)-(9) to N terms. By doing so, a system o f 5N equations with 5N u n k n o w n s (A~, A2 . . . . . A~E~, E2. . . . , EN) is obtained. This system can be easily solved using standard methods (e.g. G a u s s - J o r d a n elimination). 3.
TEMPERATURE
DISTRIBUTION
Figure 2 shows the soil temperature field beneath a slab-on-grade floor of width ( f 4 - f 1) = 10 m. The interior air temperature above the slab is T~ = 20°C, while the soil surface temperature is Ts = 15°C. A water table at a depth d = 7 m below the soil surface and c = 5 m below the stepped ground surface has a temperature Tw = 10°C. Five cases with various step depths, insulation levels, and distances between the slab and the step are considered in Fig. 2(a)-Fig. 2(f) as described below. (a) a = 31.0m, Fig. 2(a). (b) a = 3 1 . 0 m Fig. 2(b). (c) a = 3 1 . 0 m Fig. 2(c). (d) a = 4 1 . 0 m Fig. 2(d). (e) a = 31.0 m K / W [R-10 (f) a = 3 1 . 0 m K / W [R-10
b = 50.0m, c = 7.0 m, d = 7.0 m, f 1 = 20.0 m, f 4 = 30.0 m, no insulation, b=50.0m,
c -- 5.0 m, d = 7.0 m, f 1 = 20.0 m, f 4 = 30.0 m, no insulation,
b=50.0m,
c = 2.0 m, d = 7.0 m, f 1 = 20.0 m, f 4 = 30.0 m, no insulation,
b=60.0m,
c=5.0m,
b = 50.0 m, in English] b = 50.0 m, in English]
c = 5.0 m, insulation c = 5.0m, insulation
d = 7.0 m, f 1 = 20.0 m, f 4 = 30.0 m, no insulation, d = 7.0 m , f 1 = 20.0 re, f 4 = 30.0 m, uniform R-1.76 m 2 on slab-on-grade floor, Fig. 2(e). d = 7.0 m , f 1 = 20.0 re, f 4 = 30.0 m, uniform R-1.76 m 2 on step wall, Fig. 2 ( f ) .
(a)
Isothermsfor Slab-on-Grade Floor 15 "C
7
20 "C
E e 0 4 E
i5
1 10 "C 0
10
20
30
40
50
Distance from Origin, m
(b)
Isotherms for Slab-on-Grade Floor 15 °C
E 6 " O~ .rO E
20°C
~ ) ~
14 °C
5
15°C
13"C~
14 °C
4 3
12~~______~
¢0 2 E3 1
tl
12°C
°C 10 °C
0
10
20
30
40
50
Distance from Origin, m
(c)
Isotherms for Slab-on-Grade Floor 15 °C 14 °C
E .=2
.== O E
8 r-
E3
20 "C .
.
~
~
13 °C 4
12~
3
~.
15 °C
2 I 10 "C
0 0
10
20
30
40
50
Distance from Origin, m Fig. 2. (a-c). Soil temperature isotherms for uninsulated slab and even soil surface level, a - - 3 1 . 0 m, b = 50.0 m, c = 7.0 m, d = 7.0 m, f l = 20.0 re, f 4 -- 30.0. (b) Soil temperature isotherms for uninsulated slab and 2 m deep stepped soil surface a = 3 1 . 0 m , b = 50.0m, c = 5.0m, d = 7 . 0 m , f l = 2 0 . 0 m , f 4 = 30.0. (c) Soil temperature isotherms for uninsulated slab and 5 m deep stepped soil surface, a = 3 1 . 0 m , b = 5 0 . 0 m , c = 2 . 0 m , d = 7 . 0 m , f l =20.0m, f 4 = 3 0 . 0 m . 697
(d)
Isotherms for Slab-on-Grade Floor
15 *C
20 °C
14 *C
E 6 r- 5 -tO
E
13 *C 4
12 *C
8 e2
~ 1 1 *C
121 1
10 *C
0
10
20
30
40
50
60
DistancefromOngin, rn
(e)
Isotherms for Slab-on-Grade Floor 20 °C
15 °C
E
6
15 °c
e- 5 ,¢.-
13*C
O 4 E
8
3 *c
12 *C
3
¢-
m 2
O
11 *C
1 10 *c 10
20
30
40
50
D i s t a n c e f r o m Origin, rn
Isotherms for Slab-on-Grade Floor
(f) 15 *C
6 E e- 5
20 *C
*
C
15 *C
~
"r-
0 4 E
g 3 8 ~2
12 *C
t,-
D
1 10 *C 0
10
20
30
40
50
D i s t a n c e f r o m Origin, m Fig. 2. (d-f). (d) Soil temperature isotherms for uninsulated slab and 5 m deep stepped soil surface, a = 41.0 m, b = 60.0 m, c = 5.0 m, d = 7.0 m, f l = 20.0 m, f 4 -- 30.0 m. (e) Soil temperature isotherms for insulated slab and 2 m deep stepped soil surface, a = 31.0m, b = 50.0m, c = 5 . 0 m , d = 7.0m, f l = 20.0 m , f 4 = 30.0. m, Uniform R-10. ( f ) Soil temperature isotherms for uninsulated slab and 2 m deep steemp soil surface with vertical insulation, a = 31.0m, b - - 5 0 . 0 m , c = 5.0m, d = 7.0m, f l = 20.0 m, f 4 = 30.0. m, Uniform R-10. 698
CHOI and KRARTI:
HEAT TRANSFER FOR SLAB-ON-GRADE FLOOR
699
Figure 2(a) illustrates the soil temperature isotherms of the base case which has no stepped ground (i.e. leveled soil surface). Meanwhile, Fig. 2(b)-2(e) shows the effect on soil temperature distribution of uneven soil level. In particular, Fig. 2(a)-2(c) shows the effect of the depth of the stepped ground on soil temperature distribution. The uneven soil surface results in a slight change of the soil temperature along the slab-on-grade floor surface even when the depth of the stepped soil surface is significant, as shown in Fig. 2(c). At the side where the soil surface is even, the soil temperature distribution has not changed significantly. However, at the side where the stepped ground exists, a significant change in the soil temperature occurs. For instance, the isotherm line of 14°C is moved from the 5.7 m level in the case of Fig. 2(a) to the 4 m level in the case of Fig. 2(b) at the side of the stepped ground surface. When the unevenness of the soil surface is significant with a depth of 5 m, as illustrated in Fig. 2(c), the double point of the isotherm 15°C comes into contact with the vertical surface of the stepped ground. The concept of the double point was introduced in Krarti et al. [3, 4]. The double point marks the intersection of two isotherms that have the same temperature. In the case of Fig. 2(a), two double points can be located along the horizontal soil surface near the slab edges. For the slab configuration presented in Fig. 2, the double point is the intersection of the isotherm line 15°C and the soil surface with temperature set at 15°C. One of the properties of the double point is that it provides information on how heat is exchanged between the various bounding surfaces. For instance, and in the case of Fig. 2(a), the part of the soil surface located between the slab edge and the double points gains heat from the slab. However, the remainder part of the soil surface loses heat to the cooler water table surface. Figure 2(b) and (c) indicates that the double point moves along the vertical soil surface when the depth of the stepped ground increases. Figure 2(d) shows that the temperature of the ground beneath the floor slightly increases with increasing distance between the slab and the stepped ground. Indeed, when comparing the temperature field of Fig. 2(b) with that of Fig. 2(d), it is apparent that when the distance between the floor and the stepped ground increases from 1 to 10 m, the isotherm line of 15°C slightly moves away from the right edge of the floor, resulting in a slight increase in the soil temperature just below the slab edge. This slight soil temperature increase can be attributed to the increased mass of the ground around the right edge of the slab. Figure 2(e) shows that an increase in the R-value of the uniform insulation along the floor significantly decreases the temperature of the ground beneath the slab. Indeed, the increase foundation R-value decreases the thermal interactions between the slab, the cold water table and the soil surface. Finally, Fig. 2(f) indicates that placing a thermal insulation along the vertical surface of the stepped ground has very little effect on the soil temperature beneath the foundation. 4.
SLAB
HEAT
LOSSES
The total slab heat loss, Qtossis obtained by integrating the temperature gradient along the slab surface [f 2, f 4]. 2 ~ tanh #,(d A, - c) {cos # , f 1 - cos # , f 4 } Qtoss= a,=,
+ 2 ~. C, {cos # , f 4 - cos # , f 1} a , = ~sinh #,(d - c)
nE1~=sinhB"( - 1)"Z,a {c°shz"f4-c°shz"fl}
+ ~---Z-~2
2
~ Ts(d(-1)"-e) {coshz,(a-fl)-coshz,(a-f4)}
+ d---Z-cc, = ~ d
X,
sinh Z, a
(10)
700
CHOI and KRARTI:
HEAT TRANSFER FOR SLAB-ON-GRADE FLOOR
Parametric Analysis Results: Heat Loss 454035-
O
30. 2S
r,, U)
2O
O ._1
~~Step Insulation Ste~Depth
lS
T
10 S 0 0.00
0.50
1. ~
| .50
2.00
2.50
3.IX)
3.50
4.00
4.50
S.~
Length or Distance, rn Fig. 3. Parametric analysis results: Slab heat loss changes.
Figure 3 shows the effect on the foundation heat loss of varying the step depth, slab or step insulation length and distance between the slab edge and the step of the ground surface. For the base case, the slab width is 10 m with an interior temperature of T~ = 20°C. The water table is at a depth of 7 m with a temperature Tw = 10°C. The soil surface temperature is set at 7", = 15°C and has a 2-m deep step located 1 m from the slab. R-10 thermal insulation is placed along the slab or the vertical surface of the stepped ground. The air ambient temperature above the soil surface is assumed to be T, = 15°C. The results shown in Fig. 3 are normalized and are presented as a percentage of the base case total foundation heat loss. As clearly shown in Fig. 3, the slab total heat loss is significantly reduced with increasing floor insulation length. For a partially insulated slab with R-1.76 m 2 K/W [R-10 in English] insulation placed at 1 m strip around the slab perimeter, the foundation total heat loss is reduced by 17%. For a uniformly insulated slab (i.e. the entire slab surface is insulated using R-1.76 m 2 K/W [R-10 in English] insulation), the foundation heat loss is reduced by 41% compared to the base case of an uninsulated slab. Figure 3 indicates that, except for the slab insulation, the other parameters have little effect on the total slab heat loss. Indeed, the foundation heat loss is reduced only by 5% when the step depth reaches 5 m, by 2% when the distance between the slab and the step is 10 m, and 7% when the 2 m long vertical surface of the step is covered by R-1.76 m 2 K/W [R-10 in English] insulation. These results confirm that the total foundation heat loss is mainly affected by the conditions of the surfaces in direct contact with the foundation. It is interesting to note that there is an optimum length for the vertical insulation as shown in Fig. 3. The slab heat loss reduction is slightly reduced with increasing R-1.76 m' K/W JR-10 in English] insulation lengths along the vertical surface of the step. However, when the insulation length reaches a certain level, 1.6 m in the slab configuration of Fig. 3, the foundation heat loss starts to increase. This result is due to the temperature difference between the ambient air (15°C) and the water table (10°C). Indeed, by insulating the entire vertical surface of the step, the thermal interaction between the slab and the ambient air is almost eliminated. Meanwhile, the thermal interaction between the slab and the colder water table is enhanced, resulting in an increase in foundation heat loss.
CHOI and KRARTI: HEAT TRANSFER FOR SLAB-ON-GRADE FLOOR 5.
SUMMARY
AND
701
CONCLUSIONS
A general solution of the soil temperature distribution is developed for slab-on-grade floors with uneven soil surface levels. A detailed analysis is provided to determine the effect on soil temperature distribution and on slab heat loss of various parameters, such as slab insulation length, vertical insulation depth along the stepped ground surface, distance between the slab and the stepped ground and depth of the stepped ground. It is found that the soil temperature field and the total foundation heat loss is not significantly affected by the existence of a stepped ground. In addition, it is found that the total slab heat loss can be more effectively reduced by placing insulation directly along the floor surface rather than along the step wall. REFERENCES
1. 2. 3. 4.
Delsante, A. E., Stockes, A. N. and Walsh, P. J., Int. J. Heat Mass Transfer, 1982, 26, 121. Johnston, G. H. (Ed.), Permafrost Engineering Design and Calculation. Wiley, Toronto, 1981. Krarti, M., Claridge, D. E. and Kreider, J. F., Int. J. Heat Mass Transfer, 1988, 31, 1885. Krarti, M., Int. J. Heat Mass Transfer, 1989, 32, 961.
~
Pergamon PII: S0196-8904(97)00045-9
© 1998 Elsevier Science Ltd. All rights reserved Printed in Great Britain 0196-8904/98 $19.00 + 0.00
HEAT TRANSFER FOR SLAB-ON-GRADE FLOOR WITH STEPPED G R O U N D
SANGHO CHOI and MONCEF KRARTI Joint Center for Energy Management, CEAE Department, University of Colorado, Boulder, CO 80309-0428, U.S.A.
(Received 15 August 1996)
Abstraet--A general solution is developed to calculate foundation heat transfer for a slab-on-grade floor with lifted ground. The Interzone Temperature Profile Estimation (ITPE) technique is applied to determine the steady state variation of temperature within the ground. A water table at constant temperature is assumed to exist at a given depth below the soil surface. The derived solution addresses all the common horizontal insulation configurations for slab-on-grade floor foundations. The effect of various insulation configurations and the height of the lifted ground on foundation heat loss are discussed in detail. © 1998 Elsevier Science Ltd. Slab-on-grade floor
Isotherms
Stepped ground
Total slab heat loss
NOMENCLATURE a ~
Distance from origin to left side of stepped ground, m
b= Distance from origin to boundary where soil temperature is undisturbed, m C =
Distance from origin to stepped ground surface, m
d= Distance from origin to soil ground surface, m f l = Distance from origin to left side of foundation, m f 4 = Distance from origin to right side of foundation, m f2,3 = Distance from origin to start and end point of insulation, m
Hf 1-3 = Ratio of U-factor of slab portion to soil thermal conductivity, m -~ Ratio of U-factor of soil surface portion to soil thermal conductivity, m -~ Soil thermal conductivity, W/m°C Foundation heat loss, W/m2 Qlou A . , B . , C . , D . , E . = Fourier coefficients T= temperature, K T~= building air temperature, K T,= outside air, K Tw= water table temperature, K T~= soil surface temperature, K x,y= space coordinates, m Hfs =
Greek symbols • , /L ?, 6-- Coefficients defined in equations (5), (6), (7), (8) and (9) #., v., Z., ~. = eigenvalues, m -~
Subscripts f = floor I = zone (I) II -- zone (II) III -- zone (III)
1.
INTRODUCTION
It is well k n o w n t h a t the g r o u n d s u r f a c e t o p o g r a p h y affects the e n e r g y p e r f o r m a n c e o f buildings. I n p a r t i c u l a r , the t e r r a i n t o p o g r a p h y c a u s e s c h a n g e s in t h e a i r f l o w a r o u n d b u i l d i n g s w h i c h influences b u i l d i n g H V A C e q u i p m e n t o p e r a t i o n , i n f i l t r a t i o n a n d e x f i l t r a t i o n , a n d e v e n t r a n s m i s s i o n h e a t t r a n s f e r t h r o u g h b u i l d i n g e n v e l o p e s . H o w e v e r , t h e effect o f t h e t o p o g r a p h y o n t h e f o u n d a t i o n h e a t t r a n s f e r is n o t well u n d e r s t o o d . M o s t o f t h e w o r k in f o u n d a t i o n h e a t t r a n s f e r a s s u m e s t h a t the g r o u n d s u r f a c e is leveled o r b e r m e d . 691
692
CHOI and KRARTI: HEAT TRANSFER FOR SLAB-ON-GRADE FLOOR
In some hilly and mountainous locations, the building foundations are constructed over or near the edge of a lifted ground or even a cliff. In cold regions, it is a common practice to use a lifted floor system (typically with compacted granular fill) with in-ground insulation to dissipate heat losses from the structures in order to maintain the permafrost level at the desired level [2]. The vast majority of existing heat transfer calculations for slab-on-grade floor assume that the ground surface is flat. No work has been done to investigate the temperature variation along the slab surface and within the ground when the topography of the ground surface is not regular. In this paper, a solution for a slab-on-grade floor with a stepped ground is presented. A water table at a finite depth of the soil surface is assumed. The ITPE technique [3, 4] is used to solve the steady state heat conduction equation within the soil medium. The first section of the paper gives the general solution for a slab-on-grade floor with stepped ground surface. In particular, the formal expression for soil temperature and heat loss are derived. The second section of the paper deals with parametric analysis to determine the effects on the foundation heat loss of the stepped ground with various insulation levels, depths, and distances between the stepped ground and the slab.
2.
FORMULATION
OF
THE
PROBLEM
The slab-on-grade floor problem with an uneven soil surface is modeled as shown in Fig. 1. The ground is assumed homogeneous with constant thermal properties. Along the bounding surfaces x = 0 and b, the soil temperature is assumed to be undisturbed and is simply considered to be a linear function of depth. Under these conditions, the steady state temperature distribution T(x,y) beneath the slab-on-grade floor with the stepped ground, is a solution of the following equation: ~2T ~2T ~x---~ + ~ = 0
(1)
Y
SoH
Surface .....................................
d
................................... "'wl ITa T ................. ~ ...................... / Son Surface s C Of 4
(n)
(m) Tw
o,o
X
Water Table
iiii~ 7 " ~
A Fig. 1. Slab-on-grade floor with the stepped ground.
CHOI and KRARTI: HEAT TRANSFER FOR SLAB-ON-GRADE FLOOR
693
with the boundary conditions: T=T~
@ y=d,O<~x<~a
T=T~
@ y=c,a<~x<~b
@y=0
T=Tw
Hfs × d
T = Ts(y)=~T~ @ x = O
with Ts = Ta Hfs × d + 1
T = T,(y) =
with T~ = Ta Hfs x c + l
Ts @ x = a
Hfs x c
Note that, without loss of generality, the water table temperature, Tw, can be set to zero (i.e. Tw = 0). This assumption will be made for the remainder of this paper. Using the ITPE technique [3, 4], the ground medium is divided into three zones as shown in Fig. 1. Also, the temperature profiles at the surfaces are defined below as functions of x and y, respectively. T(x,d)= A(x) T(x,c) = C(x)
0
T(a,y) = B ( y ) O < x < a, T(a,y) = D(x) T(x,c) = E ( y ) a < x < b,
c
Using the separation of variables technique [1], the solution in zone (I) is
T~(x,y)
+
=
~_e~.
sinhg.(y-c) sinh #. (d - c)
sinh/~.(d - y) sin ~. x sinh #. (d - c)
2•lC"
+-7-=-=_
+
~ a .-l
2 ~ A. sin/z, x
B. sin X. (Y - c .- t
)sinh Z . x sinh X. a
' 7", (d - e ( - 1)") sin ;(.(y - c) sinh z.(a - X) Z. sinh Z. a
. "-d
(2)
In zone (II), the solution is given by _2 ~ Tl~(x,y) =
+2
en-I
a..t
sinh/~, y
C. sin/~, x sinh #. c
sinh q/. x D. sin ~O. y sinh q,. a
2 o~ T, ( - 1) ~ sinh ~ . ( a - x) -c ~. ~ ~. sin~0.y sinhqJ, a
(3)
694
CHOI and KRARTI: HEAT TRANSFER FOR SLAB-ON-GRADE FLOOR
The expression for the temperature in zone (III) is given by ~ E. sin v.(x - a) sinh v. y sinh v.c
2
Tm(x,y) - b - a = t
sinh ~h.(b - x) + 2 E D . sin 0 . y c =, sinh~k.(b a) 2 ~ T~(- 1)" sinh ~.(x - a) -c=~ qJ. sin ~O.y sinh ~b.(b a)
(4)
with nn
nn
~°=--a' V " - b - a '
nn
Z" = d - c '
i
A. =
nn
O"
c
a
A(x)sin # . x d x
B. = fdB( y)sin Z.( Y -- c)dy
i j
C. =
a
C(x)sin p. xdx
cDc
D. =
D(y)sin O.ydy
0
~a b
E(x)sin v.(x - a)dx
E. =
The Fourier coefficients A.-E. can be determined from the required continuity of the heat flux along the surfaces and the third-kind boundary conditions along the various surfaces. First, the third-kind boundary condition at y = d and 0 ~< x ~< a gives the condition +~ A
+E
+~
2c,
.SA.+
n=l
n=l
with
A
Id-~c ~ Ts{c(-1)'-d} t.
--
n=
1
pp
A~(tanh#Ad-c))
~P
2.[
7.A __
Ab
2
A~ ) [ t a n h / # ( d - c ) )
Z.(-- 1)"(-- 1)p tanh #p(d - e) tanh #p(d - c)
(5)
CHOI and KRARTI: HEAT TRANSFER FOR SLAB-ON-GRADE FLOOR
695
where, A ~ = HfsTa(1 -
cos
ltpf
ktpf 1 -
1) + H f 1T(cos
cos/6f2)
+ H f 2 T i ( c o s #p f 2 - cos ktpf 3)
+ H f 3 Ti(cos # p f 3 - cos # ~ f 4) + HfsTa(cos #pf 4 - ( - 1) p) A b =/~. cos p . f 1 sin p p f l(Hfs - H f 1) +/~, cos # , f 2 sin # . f 2 ( H f I - H f 2 ) + # ° cos p , f 3 sin # p f 3 ( H f 2 - H f 3) + #, cos # . f 4 sin # p f 4(Hfs - H f 3) A ¢ = pp cos p p f 1 sin # . f l(Hfs -- H f 1) + pp cos p p f 2 sin # . f 2 ( H f 1 - H f 2 ) +ktp cos p p f 3 sin # , f 3 ( H f 2 - H f 3) + I~p cos #p f 4 sin # . f 4(Hfs - H f 3) The third-kind b o u n d a r y condition at surface x = a and c ~< y < d gives the condition Bp=a~+
+oc
+(x
Efl~,A.+
ET~,B.+
.=1
+oc
E6B..pC.
.=1
(6)
n=l
with
STs{c-d(-1)P}+ B ' ~ ( t a n h z e a )
s ~P = [
Zp
~-pJ\
dsinzpa
2
]
tanh zpa
/7,,~ = a kl.(- 1)"(- 1) p X2 + kl: 2
{
/~
Bb
]tanh#p(d-
7.,5 - d - c 2(Z.-Tzp) + 2(z, - z.)J"
=
c)
z,
2 i~. tanh ;ca -a..(--, zJ +
where, B a = H w l T a ( c o s Zp(wl - c) - ( - 1) p) + Hw2Ta(1 - cos Xp(wl - c)) B b = sin((Xp + X,)(c - wl))(hw2 - h w l ) B ~ = sin((zp - Z.)(wl - c))(hw2 - h w l ) The heat flux continuity at surface y = c and 0 ~< x ~< a gives the condition +~
c, =
+~ C
n=l
(7)
Z ~.. c D. n=l
with
S2 ~7" #"
~ = ~ ~ : ~ + t " . :~ "p " ~'"
2
~ T,(c-d(-l')) .=,
fl~ =
b.c = -c
#& ~S
tanh#p(d-c)tanh#xec
#J + Z.2J[#p(tanh # p ( d - c) + tanh #p c)
/6 tanh # e ( d - c)tanh #e c sinh ~tp(d - c) #p(tanh/~.(d - c) + tanh/zp c) 2 d-c
?.c=
d
Z. P e ( - 1) p tanh #e(d - c)tanh #e c #~+Z. 2 #p(tanhpp(d-c)+tanhgpc)
2 q/. # p ( - 1)P(- 1)" tanh gp(d - c)tanh #e c p~ + $.2 pp(tanh gp(d - c) + tanh #p c)
}
696
CHOI and KRARTI:
HEAT TRANSFER
FOR SLAB-ON-GRADE
FLOOR
The heat flux continuity at surface y --- d and a ~< x ~< b gives the condition + oc
+
Op=~D+ 2~.,~Cnnt- 2"7.,pE. D
n=l
(8)
n=l
with ~'.T s ( - 1)Pc T s ( - 1) p "(~. tanh~kp(b--a)tanhO, a e~ = - [ d sinh ~,, a + d sinh Op(b - a ) J [~,p(tanh Op(b - a) + tanh O, a ) J 7~ - 2 p. ~bp(- 1 ) " ( - 1) p ~" tanh Op(b - a)tanh ~bp a "( a 0 2 + p~ [0p (ta--~ ~ ---~ ta--~p a)J
7Dp _
2
tanh ~bp(b - a)tanh Op a O~ + v2, [ O p ( t ~ - h O - - - ~ - - ~ a ~ - ~ p a ) J
v. 0 p ( - 1) p ~"
b-a
"(
The third-kind b o u n d a r y condition at surface y = c and a ~< x ~< b gives the condition +oc
=
E
]~,,p D.
(9)
n=l
with
0~p =
--
.=t
Vp2 + ~2.
fie =c
+
Vp
v, + tanh VpC
2~._Vp(--1)"/'. tanhvpc ) vp2+~k2 ~ v ~ + t a n h v p c
The Fourier coefficients Ap-Ep are determined by truncating the sums in equations (4)-(9) to N terms. By doing so, a system o f 5N equations with 5N u n k n o w n s (A~, A2 . . . . . A~E~, E2. . . . , EN) is obtained. This system can be easily solved using standard methods (e.g. G a u s s - J o r d a n elimination). 3.
TEMPERATURE
DISTRIBUTION
Figure 2 shows the soil temperature field beneath a slab-on-grade floor of width ( f 4 - f 1) = 10 m. The interior air temperature above the slab is T~ = 20°C, while the soil surface temperature is Ts = 15°C. A water table at a depth d = 7 m below the soil surface and c = 5 m below the stepped ground surface has a temperature Tw = 10°C. Five cases with various step depths, insulation levels, and distances between the slab and the step are considered in Fig. 2(a)-Fig. 2(f) as described below. (a) a = 31.0m, Fig. 2(a). (b) a = 3 1 . 0 m Fig. 2(b). (c) a = 3 1 . 0 m Fig. 2(c). (d) a = 4 1 . 0 m Fig. 2(d). (e) a = 31.0 m K / W [R-10 (f) a = 3 1 . 0 m K / W [R-10
b = 50.0m, c = 7.0 m, d = 7.0 m, f 1 = 20.0 m, f 4 = 30.0 m, no insulation, b=50.0m,
c -- 5.0 m, d = 7.0 m, f 1 = 20.0 m, f 4 = 30.0 m, no insulation,
b=50.0m,
c = 2.0 m, d = 7.0 m, f 1 = 20.0 m, f 4 = 30.0 m, no insulation,
b=60.0m,
c=5.0m,
b = 50.0 m, in English] b = 50.0 m, in English]
c = 5.0 m, insulation c = 5.0m, insulation
d = 7.0 m, f 1 = 20.0 m, f 4 = 30.0 m, no insulation, d = 7.0 m , f 1 = 20.0 re, f 4 = 30.0 m, uniform R-1.76 m 2 on slab-on-grade floor, Fig. 2(e). d = 7.0 m , f 1 = 20.0 re, f 4 = 30.0 m, uniform R-1.76 m 2 on step wall, Fig. 2 ( f ) .
(a)
Isothermsfor Slab-on-Grade Floor 15 "C
7
20 "C
E e 0 4 E
i5
1 10 "C 0
10
20
30
40
50
Distance from Origin, m
(b)
Isotherms for Slab-on-Grade Floor 15 °C
E 6 " O~ .rO E
20°C
~ ) ~
14 °C
5
15°C
13"C~
14 °C
4 3
12~~______~
¢0 2 E3 1
tl
12°C
°C 10 °C
0
10
20
30
40
50
Distance from Origin, m
(c)
Isotherms for Slab-on-Grade Floor 15 °C 14 °C
E .=2
.== O E
8 r-
E3
20 "C .
.
~
~
13 °C 4
12~
3
~.
15 °C
2 I 10 "C
0 0
10
20
30
40
50
Distance from Origin, m Fig. 2. (a-c). Soil temperature isotherms for uninsulated slab and even soil surface level, a - - 3 1 . 0 m, b = 50.0 m, c = 7.0 m, d = 7.0 m, f l = 20.0 re, f 4 -- 30.0. (b) Soil temperature isotherms for uninsulated slab and 2 m deep stepped soil surface a = 3 1 . 0 m , b = 50.0m, c = 5.0m, d = 7 . 0 m , f l = 2 0 . 0 m , f 4 = 30.0. (c) Soil temperature isotherms for uninsulated slab and 5 m deep stepped soil surface, a = 3 1 . 0 m , b = 5 0 . 0 m , c = 2 . 0 m , d = 7 . 0 m , f l =20.0m, f 4 = 3 0 . 0 m . 697
(d)
Isotherms for Slab-on-Grade Floor
15 *C
20 °C
14 *C
E 6 r- 5 -tO
E
13 *C 4
12 *C
8 e2
~ 1 1 *C
121 1
10 *C
0
10
20
30
40
50
60
DistancefromOngin, rn
(e)
Isotherms for Slab-on-Grade Floor 20 °C
15 °C
E
6
15 °c
e- 5 ,¢.-
13*C
O 4 E
8
3 *c
12 *C
3
¢-
m 2
O
11 *C
1 10 *c 10
20
30
40
50
D i s t a n c e f r o m Origin, rn
Isotherms for Slab-on-Grade Floor
(f) 15 *C
6 E e- 5
20 *C
*
C
15 *C
~
"r-
0 4 E
g 3 8 ~2
12 *C
t,-
D
1 10 *C 0
10
20
30
40
50
D i s t a n c e f r o m Origin, m Fig. 2. (d-f). (d) Soil temperature isotherms for uninsulated slab and 5 m deep stepped soil surface, a = 41.0 m, b = 60.0 m, c = 5.0 m, d = 7.0 m, f l = 20.0 m, f 4 -- 30.0 m. (e) Soil temperature isotherms for insulated slab and 2 m deep stepped soil surface, a = 31.0m, b = 50.0m, c = 5 . 0 m , d = 7.0m, f l = 20.0 m , f 4 = 30.0. m, Uniform R-10. ( f ) Soil temperature isotherms for uninsulated slab and 2 m deep steemp soil surface with vertical insulation, a = 31.0m, b - - 5 0 . 0 m , c = 5.0m, d = 7.0m, f l = 20.0 m, f 4 = 30.0. m, Uniform R-10. 698
CHOI and KRARTI:
HEAT TRANSFER FOR SLAB-ON-GRADE FLOOR
699
Figure 2(a) illustrates the soil temperature isotherms of the base case which has no stepped ground (i.e. leveled soil surface). Meanwhile, Fig. 2(b)-2(e) shows the effect on soil temperature distribution of uneven soil level. In particular, Fig. 2(a)-2(c) shows the effect of the depth of the stepped ground on soil temperature distribution. The uneven soil surface results in a slight change of the soil temperature along the slab-on-grade floor surface even when the depth of the stepped soil surface is significant, as shown in Fig. 2(c). At the side where the soil surface is even, the soil temperature distribution has not changed significantly. However, at the side where the stepped ground exists, a significant change in the soil temperature occurs. For instance, the isotherm line of 14°C is moved from the 5.7 m level in the case of Fig. 2(a) to the 4 m level in the case of Fig. 2(b) at the side of the stepped ground surface. When the unevenness of the soil surface is significant with a depth of 5 m, as illustrated in Fig. 2(c), the double point of the isotherm 15°C comes into contact with the vertical surface of the stepped ground. The concept of the double point was introduced in Krarti et al. [3, 4]. The double point marks the intersection of two isotherms that have the same temperature. In the case of Fig. 2(a), two double points can be located along the horizontal soil surface near the slab edges. For the slab configuration presented in Fig. 2, the double point is the intersection of the isotherm line 15°C and the soil surface with temperature set at 15°C. One of the properties of the double point is that it provides information on how heat is exchanged between the various bounding surfaces. For instance, and in the case of Fig. 2(a), the part of the soil surface located between the slab edge and the double points gains heat from the slab. However, the remainder part of the soil surface loses heat to the cooler water table surface. Figure 2(b) and (c) indicates that the double point moves along the vertical soil surface when the depth of the stepped ground increases. Figure 2(d) shows that the temperature of the ground beneath the floor slightly increases with increasing distance between the slab and the stepped ground. Indeed, when comparing the temperature field of Fig. 2(b) with that of Fig. 2(d), it is apparent that when the distance between the floor and the stepped ground increases from 1 to 10 m, the isotherm line of 15°C slightly moves away from the right edge of the floor, resulting in a slight increase in the soil temperature just below the slab edge. This slight soil temperature increase can be attributed to the increased mass of the ground around the right edge of the slab. Figure 2(e) shows that an increase in the R-value of the uniform insulation along the floor significantly decreases the temperature of the ground beneath the slab. Indeed, the increase foundation R-value decreases the thermal interactions between the slab, the cold water table and the soil surface. Finally, Fig. 2(f) indicates that placing a thermal insulation along the vertical surface of the stepped ground has very little effect on the soil temperature beneath the foundation. 4.
SLAB
HEAT
LOSSES
The total slab heat loss, Qtossis obtained by integrating the temperature gradient along the slab surface [f 2, f 4]. 2 ~ tanh #,(d A, - c) {cos # , f 1 - cos # , f 4 } Qtoss= a,=,
+ 2 ~. C, {cos # , f 4 - cos # , f 1} a , = ~sinh #,(d - c)
nE1~=sinhB"( - 1)"Z,a {c°shz"f4-c°shz"fl}
+ ~---Z-~2
2
~ Ts(d(-1)"-e) {coshz,(a-fl)-coshz,(a-f4)}
+ d---Z-cc, = ~ d
X,
sinh Z, a
(10)
700
CHOI and KRARTI:
HEAT TRANSFER FOR SLAB-ON-GRADE FLOOR
Parametric Analysis Results: Heat Loss 454035-
O
30. 2S
r,, U)
2O
O ._1
~~Step Insulation Ste~Depth
lS
T
10 S 0 0.00
0.50
1. ~
| .50
2.00
2.50
3.IX)
3.50
4.00
4.50
S.~
Length or Distance, rn Fig. 3. Parametric analysis results: Slab heat loss changes.
Figure 3 shows the effect on the foundation heat loss of varying the step depth, slab or step insulation length and distance between the slab edge and the step of the ground surface. For the base case, the slab width is 10 m with an interior temperature of T~ = 20°C. The water table is at a depth of 7 m with a temperature Tw = 10°C. The soil surface temperature is set at 7", = 15°C and has a 2-m deep step located 1 m from the slab. R-10 thermal insulation is placed along the slab or the vertical surface of the stepped ground. The air ambient temperature above the soil surface is assumed to be T, = 15°C. The results shown in Fig. 3 are normalized and are presented as a percentage of the base case total foundation heat loss. As clearly shown in Fig. 3, the slab total heat loss is significantly reduced with increasing floor insulation length. For a partially insulated slab with R-1.76 m 2 K/W [R-10 in English] insulation placed at 1 m strip around the slab perimeter, the foundation total heat loss is reduced by 17%. For a uniformly insulated slab (i.e. the entire slab surface is insulated using R-1.76 m 2 K/W [R-10 in English] insulation), the foundation heat loss is reduced by 41% compared to the base case of an uninsulated slab. Figure 3 indicates that, except for the slab insulation, the other parameters have little effect on the total slab heat loss. Indeed, the foundation heat loss is reduced only by 5% when the step depth reaches 5 m, by 2% when the distance between the slab and the step is 10 m, and 7% when the 2 m long vertical surface of the step is covered by R-1.76 m 2 K/W [R-10 in English] insulation. These results confirm that the total foundation heat loss is mainly affected by the conditions of the surfaces in direct contact with the foundation. It is interesting to note that there is an optimum length for the vertical insulation as shown in Fig. 3. The slab heat loss reduction is slightly reduced with increasing R-1.76 m' K/W JR-10 in English] insulation lengths along the vertical surface of the step. However, when the insulation length reaches a certain level, 1.6 m in the slab configuration of Fig. 3, the foundation heat loss starts to increase. This result is due to the temperature difference between the ambient air (15°C) and the water table (10°C). Indeed, by insulating the entire vertical surface of the step, the thermal interaction between the slab and the ambient air is almost eliminated. Meanwhile, the thermal interaction between the slab and the colder water table is enhanced, resulting in an increase in foundation heat loss.
CHOI and KRARTI: HEAT TRANSFER FOR SLAB-ON-GRADE FLOOR 5.
SUMMARY
AND
701
CONCLUSIONS
A general solution of the soil temperature distribution is developed for slab-on-grade floors with uneven soil surface levels. A detailed analysis is provided to determine the effect on soil temperature distribution and on slab heat loss of various parameters, such as slab insulation length, vertical insulation depth along the stepped ground surface, distance between the slab and the stepped ground and depth of the stepped ground. It is found that the soil temperature field and the total foundation heat loss is not significantly affected by the existence of a stepped ground. In addition, it is found that the total slab heat loss can be more effectively reduced by placing insulation directly along the floor surface rather than along the step wall. REFERENCES
1. 2. 3. 4.
Delsante, A. E., Stockes, A. N. and Walsh, P. J., Int. J. Heat Mass Transfer, 1982, 26, 121. Johnston, G. H. (Ed.), Permafrost Engineering Design and Calculation. Wiley, Toronto, 1981. Krarti, M., Claridge, D. E. and Kreider, J. F., Int. J. Heat Mass Transfer, 1988, 31, 1885. Krarti, M., Int. J. Heat Mass Transfer, 1989, 32, 961.