A 2.5-cm-diameter cylindrical refractory crucible for melting lead is to be built for thermocouple calibration. An electrical heater immersed in the metal is shut off at some temperature above the melting point. The fusion-cooling curve is obtained by observing the thermocouple emf as a function of time. Neglecting heat losses through the wall of the crucible, estimate the cooling rate (W) for the molten lead surface (melting point 327.3°C, surface emissivity 0.8) if the crucible depth above the lead surface is (a) 2.5 cm, (b) 17 cm. Assume that the emissivity of the refractory surface is unity and the surroundings are at 21°C, (c) Noting that the crucible would hold about 0.09 kg of lead for which the heat of fusion is 23,260 J/kg, comment on the suitability of the crucible for the

purpose intended.



GIVEN

A cylindrical refractory crucible filled with molten lead

Cylinder diameter (D) = 2.5 cm

Melting point of lead (T1) = 327.2°C = 600.3 K

Surface emissivity of lead (?1) = 0.8

Mass of lead in crucible (m) = 0.09

Heat of fusion of lead (hfg) = 23,260 J/kg

FIND

The cooling rate (q) if the crucible depth above the lead surface (L) is

(a) 2.5 cm = 0.025 m

(b) 17 cm = 0.17 m

(c) Comment on the suitability of the crucible for thermocouple calibration

ASSUMPTIONS

Heat loss through the wall of the crucible is negligible

The emissivity of the refractory surface (crucible wall above the lead) is unity (?2 =1)

The surroundings behave as a blackbody enclosure

The temperature of the refractory surface is uniform at TR

SKETCH



PROPERTIES AND CONSTANTS

the Stephan-Boltzmann constant



for air at the film temperature of



Thermal expansion coefficient



Thermal conductivity



Kinematic viscosity



Prandtl number (Pr) = 0.71


The total cooling rate is the sum of natural convection and radiation



where q12 is the radiative heat transfer between the two surfaces connected by a refractory wall and is



where A1 f12 (Note that ?2 = 1.0 and A2 = A1)



The shape factor by letting a = b = D/s and



The shape factor by letting a = b = D/s and



For Case (a)



For Case (b)



By symmetry

F21 = F12

The shape factors from a given surface must sum to unity





The heat transfer coefficient, hc, can be calculated



Although this is slightly below its lower Rayleigh number range, Equation (5.15) will be used to estimate the Nusselt number



(a)



b



(c) The time required for the lead to solidify at the cooling rate (q) of 3.62 W is given by



The technician would have about 9.6 minutes to do the calibration. This should be enough time to accomplish the task.

Physics & Space Science

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