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Numericals · Q37

Q.Find the temperature difference between the two sides of a steel plate 4 cm thick, when heat is transmitted through the plate at the rate of 400 kcal per minute per square metre at steady state. Thermal conductivity of steel is 0.026 kcal/m s K.

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Given heat-flow rate 400 kcal min−1m−2400\text{ kcal min}^{-1}\text{m}^{-2}, converted to per-second: 400/60=6.667 kcal s−1m−2400/60 = 6.667\text{ kcal s}^{-1}\text{m}^{-2}; thickness x=4 cm=0.04 mx=4\text{ cm}=0.04\text{ m}; ksteel=0.026 kcal m−1s−1K−1k_{steel}=0.026\text{ kcal m}^{-1}\text{s}^{-1}\text{K}^{-1}. From QAt=kΔTx\dfrac{Q}{At} = \dfrac{k\Delta T}{x} (Eq. 7.36), rearranged: $\Delta T = \dfrac{Q}{At}\times\dfrac{x}{k} = 6.667\times\dfrac{0.04}{ …

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