Example

Using the conductance heat transfer equation, calculate the amount of heat lost through 2 m2 (21.528 ft2) of a cold storage wall that is composed of three layers of material. The wall consists of 4-in. (0.1016-m) thick concrete block (xc) on the outside, 6-in. (0.1524-m) thick polystyrene insulation (xp), and 2-in. (0.0508-m) thick fir wood (xf ) on the inside. The inside wall of the cold room is - 40°F (- 40°C) and the outside wall is 68°F 20°C). From Table

2, the thermal conductivities are kc = 0.76 W • m/m2 • K = 5.27 (BtuXin.)/ft2Xhj(°F) kp = 0.038 W • m/m2 • K = 0.263 (BtuXin.MftiXhJCF) kf = 0.11 W ■ m/m2 • K = 0.76 (Btu)(in.)/(ft?)(h)(°F)

Using equation 12 where the area is constant (Fig. 1), t, - t„

In SI units:

Table 2. Thermal Conductivity of Selected Foods and Processing Facility Materials

Thermal conductivity, ka

Table 2. Thermal Conductivity of Selected Foods and Processing Facility Materials

Thermal conductivity, ka

°C

(W)(m) (m4)(K)

(BtuXin.) (ft2)(h)(°F)

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