Question

It is desired to use a rotary vacuum filter to separate slurry containing 20 kg of water per kilogram of solid material. Tests on the rotary filter at the conditions to be used for the filtration have indicated that the dimensionless ratio α/β' is 0.6 and 19 kg of filtrate (not including wash water) is obtained for each 27.8 kg of slurry. The temperature and pressure of the surroundings are 25 oC and 1 atm, respectively. The pressure drop to be maintained by the vacuum pump is 45 kPa. The fraction of the drum area submerged in the slurry is 0.3 and the fraction of the drum area available for suction is 0.1. On the basis of the following assumptions, estimate the kilowatt rating of the motor necessary to operate the vacuum pump if the filter handles 16000 kg/h of slurry. Assumptions are as follows:

Resistance of the filter medium is negligible.

Effects of air leakage are included in the value given for α/β'.

The value of β' is based on the temperature and pressure of the ambient air.

The filter removes all the solids from the slurry.

The vacuum pump operates isentropically with an overall efficiency of 50 percent for the pump and motor.

The ratio of Cp/Cv remains constant at a value of 1.4.

Answer 1

Data provided in the problem statement

Mass of slurry = 20 kg water + 1 kg solids = 21 kg slurry

Ratio α/β = 0.6

Fraction of drum available for suction (a) = 0.1

Fraction of drum area submerged in slurry  (f) = 0.3

Properties of water and air at surrounding temp = 25 C

Viscosity of water at 25 C ( w ) = 0.0008891 pa.s = 0.031399 Kpa. m / hr

Viscosity of air at 25 C ( a ) = 18.37 * 10-6 pa.s = 0.0006487 Kpa. m / hr

Pressure drop ( P ) = 45 kPa

Density of water at 25 C ( ) = 997 kg/ m3

Mass filtrate obtained from 27.8 kg of slurry = 19 kg

Volume of filtrate = mass of filtrate / density of filtrate = 19 / 997 = 0.01905 m3

Dry solids are 1 kg per 21 kg of slurry

so dry solids in 27.8 kg slurry = 27.8 / 21 = 1.324 kg dry solids

Dry solids per unit volume of filtrate ( w) = 1.324 kg dry solids / 0.01905 m3 = 69.5 kg / m3

Now filter handle slurry of 16000 kg /hr which contains 1 kg dry solids per 21 kg of slurry

Weight of dry cake / hr = 16000 kg /hr * ( 1 / 21 ) = 761.9 kg dry solids / hr

Now volume of dry air can be determined as

So

Rating of the motor necessary to operate the vacuum pump = 2 kW would be sufficient.