Question

Liver paste for sandwiches. Chubby liver sausages (assume cylindrical, 5 cm across, 10 cm long), originally at 20 C, are to be processed in an autoclave kept at 115 C, and we want every part of the sausage to reach 105 C. Estimate the lowest temperature of the sausage after 3 h in the autoclave. Data: For food products containing a water fraction x, we have the following estimates: Cp= 4184x + 800(1-x) , J=kg K k = 0.56x + 0.25(1- x) , W=m K Liver sausage has about the same water content as canned dog food or 73 %; its density is about 1,050 kg/m3, and in the autoclave, h=7.6W/m2 K.

Answer 1

Given:
Initial Temperature of Sausage (Ti)            =			20	°C
Radius of sausage r'		=	5/2         =	2.5 cm    =	0.025 m
Height of sausage (h)		=	10 cm    =	0.1 m
Temperature of Autoclave (Ta)                   =			115	°C
Time given (t)                                              =			3 hrs       =	3*3600 sec
Moisture content (x)                                    =			73%        =	0.73
Cp = 4184x + 800(1-x)		=	3270.32	J/kgK
Density of sausage		=	1050	kg/m3
Heat transfer coefficient (h)                        =			7.6	W/m2K
According to unsteady state heat conduction,
	(T-Ta)/(Ti-Ta) = exp(-hAt/mCp)
where,
T= req. temperature
A= Area of sausage=		2*π*r*h + 2*π*r*r =	0.019635
m= mass of sausage=		Volume*Density
		π*r*r*h*density
	=	0.206	kg
Substitute all values in above equation
We get,
	(T-115)/(20-115) = 0.091599
	T=115-8.70195
	T= 106.298 °C
Lowest Temperature of sausage after 3 hrs = 106.298 °C