Thermal Conductivity Nomographs for Wood D. S. DAVIS
Michigan Alkali Co., Wyandotte, Mioh.
F
ROM exhaustive experimental data MacLeanl developed two expressions for calculating the thermal conductivity of wood : K = 0.0833S (1.39-I-0.028M ) 0.0138 0.0833 Sn'(1.39
[email protected] 1 S, (0.27 - 0.009 M> where K = thermal conductivity, (B. t. u.) (ft.)/(hr.) (sq. ft.) (" F.) S = sp. gr. at current moisture content and weight when oven dry So = sp. gr. based on volume when green and weight when oven dry M = moisture, % ~
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+
The use of Figure 1, which presents a nomographic solution of Equation 1, is illustrated as follows: What is the thermal conductivity of wood when the specific gravity, based on the current moisture content and the weight when oven dry, is 0.41 and the moisture content is 10 per cent? Connect 0.41 on the S scale with 10 on the M scale and read the thermal conductivity as 0.071 (B. t. u.)(ft.)/(hr.)(sq. ft.)(" F.) on the K scale. The use of Figure 2, which presents a nomographic solution of Equation 2, is illustrated as follows: What is the thermal conductivity of wood when the specific gravity, based on the volume when green and the weight when oven dry, is
+
+
FIGURE1
+ 0.16
0.14
042 0.037
1
0/0
\
\
\
\
\
006 O ' O I
.-1 OVO
\
0.30 and the moisture content is 20 per cent? Connect 20 on the M scale with 0.30 on the S, scale and read the thermal conductivity as 0.064 (B. t. u.)(ft.)/(hr.)(sq. f t . ) ( O F.) on the K scale.
0.15 0.17
FIGURE 2
1
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Heating, Piping. Air Conditioning, 13, 380 (1941).
