Question

A Newtonian liquid food with a viscosity of 4 cP and a density of 1026 kg/m3 is to be pasteurized in a HTST system that heats the liquid food to 90°C followed by holding it in a 0.0275 m diameter pipe for a minimum time of 0.06 min. Calculate the maximum velocity and the length of the holding tube required, given that the flow rate is 30 L/min.

Answer 1

Sol:

Given,

The viscosity of liquid, μ = 4 cp = 0.04 Poise

Density of fluid, ρ = 1026 Kg/m3

Diameter of pipe, D = 0.0275 m

Minimum time, t =0.06 min = 3.6 sec

Flow rate, Q = 30 L/min = 30 x 10-3 m3/ 60 sec = 5 x 10-4  m3/s

Area of the tube, A = (π x D^2)/4 = 5.93 x 10-4 m2

Let maximum velocity be Vmax and avg veloctiy be Vavg and length of tube be L

As we have, Flow rate, Q = Vavg * Area

⇒ Vavg = Q/A

⇒ Vavg = 5.93 x 10-4 / 5 x 10-4  

⇒ Vavg = 1.186 m/s

Reynolds number, Re = ρDVavg/μ

⇒ Re = 836.574, the flow is Laminar (Re<2100)

For the laminar flow: Vmax = Vavg x 2

⇒ Vmax = 1.186 x 2

⇒ Vmax = 2.372 m/s

Required length of holding tube (in meter) = Vmax x Minimum holding time (in sec)

⇒ L = 2.372 x 3.6

⇒ L = 8.54 m

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