admin@mazurekgravity.in

A fluid food has a viscosity of 5 cp and density of 1009 kg/m³. It is pasteurised in a continuous system that involves heating at 82.2°C, holding in a pipe (diameter = 0.034 m; crossectional area = 9.079×10+ m²) and cooling. For this heating process a SV of 12 for Staphylococcus aureus is desired. The D822 value is equal to 0.0063 min. Calculate the length (in m) of holding tube for a flow rate of 1.666×10+ m³/s.

Question-AnswerCategory: Chemical EngineeringA fluid food has a viscosity of 5 cp and density of 1009 kg/m³. It is pasteurised in a continuous system that involves heating at 82.2°C, holding in a pipe (diameter = 0.034 m; crossectional area = 9.079×10+ m²) and cooling. For this heating process a SV of 12 for Staphylococcus aureus is desired. The D822 value is equal to 0.0063 min. Calculate the length (in m) of holding tube for a flow rate of 1.666×10+ m³/s.
Mazurek Gravity Staff asked 7 months ago

A fluid food has a viscosity of 5 cp and density of 1009 kg/m³. It is pasteurised in a continuous system that involves heating at 82.2°C, holding in a pipe (diameter = 0.034 m; crossectional area = 9.079×10+ m²) and cooling. For this heating process a SV of 12 for Staphylococcus aureus is desired. The D822 value is equal to 0.0063 min. Calculate the length (in m) of holding tube for a flow rate of 1.666×10+ m³/s.

1 Answers
Mazurek Gravity Staff answered 7 months ago

The average velocity is calculated based on flow rate and cross-sectional area after that Reynolds number which decides whether the flow is laminar or turbulent which further decides the maximum velocity. The thermal reduction time has been computed from D and F. Finally the product of Umax and F gives the required tube length.
M= 5cp =5X10 Pas
T= 82.2°C
d. = 0.034 m
n = 12
D82.2 = 0.0063 min
Q = 1.666X164 m3ls
rede
=
reduction = 12
Factor (n)
: F82.2Re = Pihinged
he
100g x0.18349 63 X 0.034
0.005
Re = 1259.0048
Re<2100
.. The flow is laminar flow
Umax = 2
2
lay
Umax = 2X0.

Your Answer

18 + 14 =