A study has been made to determine the thermal conductivity of a thin specimen of fabrics placed between a standard sample and a slab under a predetermined temperature by heating the slab stepwise. Following results have been obtained :
(1) Assuming that an initial temperature beeing zero degree, the portion of x=-l is maintained at a temperature ct during 0<t<t
1 and is maintained at a temperature V during t
1≤t, for the composite solids -l<x<∞ of which the portion of -l<x<0 is the test specimen less than about 5 mm in thick, and the portion of x>0 is a standard sample, it has been proved that the surface temperature of the standard sample v
2(0, t) is approximately given by :
v2(0, t)/V=1-β
-2/t
1/t{e
β2erfcβ-exp[β
2(1-t
1/t)]}·erfc(β√1-t
1/t)+2/√πβ(1-√1-t
1/t)
β=K
1/K
2√k
2t/l
Where, k
1 : thermal diffusivity of a specimen
K
1 : thermal conductivity of a specimen
k
2 : thermal diffusivity of a standard sample
K
2 : thermal conductivity of a standard sample
c, V : constant
t
1 : the time required to reach the temperature V
Using these equations and borosilicate glass as the standard sample, 0.038 kcal/m·hr·°C has been obtained for K
1 of polyester taffeta fabrics of four to six layeres.
(2) Assuming k
1=0.284 × 10
-3m
2/h for the same fabrics of 30-40 layers, K
1=0.040 kcal/m·hr · °C has been obtained by the following equations :
v
2(0, t)/V=2/1+σ(a
1t
1/t+a
2)
a
1=∞Σn=0θ
npn(1/pn+2pn)(erfcpn-erfcpn/√1-t
1/t)-2/√π(e
-pn2-√1-t
1/texp(-pn
2/1-t
1/t))
a
2=∞Σn=0θerfcpn/√1-t
1/t
σ=K
2/K
1√k
1/k
2, θ=σ-1/σ+1, pn=(2n+1)p, p=l/2√k
1t
(3) Assuming that an initial temperature beeing zero degree, a surface temperature is maintained at a temperature ct during 0<t<t
1 and then the portions at x=0 and x=2l are maintained at a temperature V
1 and V
2 respectively during t
1≤ t, the central plane temperature v(l, t) of the specimen 0<x<2l is represented by:
v(l, t)/V
1+V
2=a
1/t
1/t+a
2where, θ=-1.
By using this equation, 0.284 ×10
-3m
2/h has been obtained for k
1 of polyester taffeta fabrics.
View full abstract