Question

Cream and skim milk (1% fat wet basis) are produced from buffalo milk (6% fat wet basis) using centrifuge. Volumetric flow rates of buffalo milk and skimmed milk are 1000 L/h and 900 L/h, respectively. Determine the mass flow rate and fat content of cream in the process. Assume densities of cream and skimmed milk to be 1.05 kg/L.

Answer 1

Hope this will be useful

Volumetric flowrate× density= mass flow rate.

Wet basis means fraction of fat in total mass of milk.

Mass balance is applied.

At steady state we can equate the volumetric flowrate.

Skim milkt tream = Buffalos milk uating volumetice rate V t V₂ =V 90017h. + V volumetice - Ve = How rate of cream = 1000l/hu 100lm There-fore moes flow rate - , x denetley 21004/hox 1.05 kg/2 = 105 kg/hr w applyng now applying overall mass balance... → mass of skimmille + massiat of masshals rate / cream of Buffels I milk. 900x 1.05 + 105.: = (1000x density) = mass flowrako (rii crna) (m) " of Buffalo 1050 kg/h.= 900X1.05+105 a massflowrate of Buffalo T(m) o milk Fat balance. in, xx, tm ale z man (900X105) (6:01) + 105x2, = 10 50x (0.06) Content of fat -% = 1050X0-06 – 900X10580.01) - in cream = 51% wet basis 105