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.
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