By John H Lienhard

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Extra info for A heat transfer textbook

Example text

4 A thermocouple bead is largely solder, 1 mm in diameter. It is initially at room temperature and is suddenly placed in a 200◦ C gas ﬂow. 18 kJ/kg·K, respectively. Evaluate the response of the thermocouple. 3 Solution. 116 s = ρcV = Therefore, eqn. 116 ◦ C (20 − 200)◦ C This result is plotted in Fig. 12, where we see that, for all practical purposes, this thermocouple catches up with the gas stream in less than 5 s. Indeed, it should be apparent that any such system will come within 95% of the signal in three time constants.

At what rate (in mm/hr) would the lake evaporate away if all of this energy went to evaporating water? Discuss as many other Chapter 1: Introduction 44 ways you can think of that this energy can be distributed (hfg for water is 2,257,000 J/kg). Do you suppose much of the 650 W/m2 goes to evaporation? 15 W/m·K, and a = 10−4 ◦ C−2 . We are concerned with thermal behavior in the extreme case in which T = 100◦ C in the cup and 0◦ C outside. Plot T against position in the cup wall and ﬁnd the heat loss, q.

Without the bar, h denotes the “local” value of the heat transfer coefﬁcient at a point on the surface. The units of h and h are W/m2 K or J/s·m2·K. 18) It turns out that Newton oversimpliﬁed the process of convection when he made his conjecture. Heat convection is complicated and h can depend on the temperature diﬀerence Tbody − T∞ ≡ ∆T . In Chapter 6 we ﬁnd that h really is independent of ∆T in situations in which ﬂuid is forced past a body and ∆T is not too large. This is called forced convection.