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## Lesson 16. SOLVING NUMERICAL

Lesson 16

SOLVING NUMERICAL

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(m^{3}/s). m^{3}/s

1m^{3 }= 1000lt

P = 250 bar =250 x 10^{5}_{ }kg/s^{2}m

=0.69444 x 10^{5}_{ }

≈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/m^{3}

P=250 x 10^{5}kg/s^{2}.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^{o}C.

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

= 59. 325^{o}C