By Lienhard J.H.

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What does this experiment show you about the transmittance of water to infrared wavelengths? 4 A look ahead What we have done up to this point has been no more than to reveal the tip of the iceberg. The basic mechanisms of heat transfer have been explained and some quantitative relations have been presented. However, this information will barely get you started when you are faced with a real heat transfer problem. Three tasks, in particular, must be completed to solve actual problems: • The heat diﬀusion equation must be solved subject to appropriate boundary conditions if the problem involves heat conduction of any complexity.

The heat convected away from the thermocouple by the air must exactly balance that radiated to it by the hot walls if the system is in steady state. 16 another. 6704 × 10−8 (100 + 273)4 − (Ttc + 273)4 W/m2 since T for radiation must be in kelvin. 4◦ C. We have seen that non-black bodies absorb less radiation than black bodies, which are perfect absorbers. Likewise, non-black bodies emit less radiation than black bodies, which also happen to be perfect emitters. 33) where 0 < ε ≤ 1. 34) where the transfer factor, F1–2 , depends on the emittances of both bodies as well as the geometrical “view”.

6 are suddenly replaced with adiabatic walls, determine the ﬁnal equilibrium temperature of the slab. (b) What is the entropy change for the slab for this process? (c) Does your answer satisfy the Second Law of Thermodynamics in this instance? Explain. 65 Btu/lb·◦ F. 81 J/K]. 5 cm in diameter has a uniform temperature of 40◦ C. The sphere is suspended in a slow-moving air stream at 0◦ C. The air stream produces a convection heat transfer coeﬃcient of 15 W/m2 K. Radiation can be neglected. , Bi 1).