Dairy Engineering

Lesson 16. SOLVING NUMERICAL

Module 4. Homogenizers

Lesson 16
SOLVING NUMERICAL

16.1 Problem

Calculate the power consumption of a homogenizer having capacity 10,000 lph, operated at 250 bar pressure.

1. Power Consumption = Pressure. Volumetric rate

E = P . V

V = 10,000lph =10/3600(m³/s). m³/s

1m³ = 1000lt

P = 250 bar =250 x 10⁵ kg/s²m

=0.69444 x 10⁵

≈70kw

16.1.2 In case the inlet pressure, efficiency of pump & motor are considered.

Calculate the power consumption of a homogenizer if the feed rate is 1800 lph and the operating pressure is 200 bar. The inlet pressure of milk is 2 bar, and efficiency of homogenizer is 95%.

Solution

Electrical Power

When Q_m = feeding rate l/h

P_h = Homogenisation pressure (bar)

P_in = Pressure at which milk enters Homogenizer (bar)

n _pump =Efficiency of pump

= 122.6 K.W

16.1.3 Find out the increase in temperature after homogenization for the given data below:

C milk = 3900J/kg K

ρ milk=1030kg/m³

P=250 x 10⁵kg/s².m

Solution

Increase in temperature of homogenized liquid is proportional to

E = P.V = V. ρ .C. ∆θ

Or ∆θ = ∆θ = P/C. ρ

∆θ =

16.1.4 Calculate the final outlet temperature of milk from the homogeniser if the inlet pressure is 2 bar, homogenising pressure 175 bar and milk inlet temperature is 50^oC.

Solution

Another assumption is that for every 40 bar pressure drop, the temperature of milk rises by 1ºC,

= 59. 325^oC