A frozen-food company freezes its spinach by first compressing it into large slabs and then exposing the slab of spinach to a low-temperature cooling medium. The large slab of compressed spinach is initially at a uniform temperature of 21°C; it must be reduced to an average temperature over the entire slab of –34°C. The temperature at any part of the slab, however, must never drop below –51°C. The cooling medium which passes across both sides of the slab is at a constant temperature of –90°C. The following data may be used for the spinach: density = 80 kg/m3; thermal conductivity = 0.87 W/(m K); specific heat = 2100 J/(kg K). Present a detailed analysis outlining a method estimate the maximum thickness of the slab of spinach that can be safely cooled in 60 min.

Question

GIVEN

• Large slabs of spinach are exposed to a low-temperature cooling medium

• Initial uniform temperature (To) = 21°C

• Average temperature must be reduced to –34°C

• The temperature at any part must never drop below –51°C

• Cooling medium temperature (T?) = –90°C

• Density of spinach (?) = 80 kg/m3

• Thermal conductivity (k) = 0.87 W/(m K)

• Specific heat (c) = 2100 J/(kg K)

FIND

• Present a detailed analysis outlining a method to estimate the maximum thickness of the slab of spinach that can be safely cooled in 60 min

ASSUMPTIONS

• One dimensional conduction through the slab

• Constant and uniform thermal

properties

• The average temperature within the slab is equal to the average of the center and surface temperatures

SKETCH

Answer 1

The average temperature for a slab with constant thermal conductivity is arithmetic average of center and end temperature. Thus, for a final average temperature in the slab of –34°C, and a final surface temperature of –51°C, the final center temperature must be



can be used to find the Biot number for the spinach slab



can be used to find the Fourier number



using above information, Fo = 0.5



Solving for L



The thickness of the slab of spinach that can be cooled in 60 minutes is 2L = 0.38 m = 38 cm.

The heat transfer coefficient needed to achieve this cooling can be calculated from the Biot number



The approximate solution for slab of infinite length is given by.



Dividing equn (i) by (ii) we have





For infinite slab with Bi=1.67 we have

for x= 0, we have



Solving for L