Question

Consider the following blending process. A well-mixed tank initially contains 2000. L of a solution that is 30.0 wt% phosphoric acid. At some time we begin feeding two streams of acid of different concentration into the tank. One stream has a flowrate of 100.0 kg/min and an acid concentration of 80.0 wt %. The second stream has a flowrate of 200. kg/min with an acid concentration of 50.0 wt %. We also withdraw a stream at a rate of 5.0 kg/s. Assume the density of the streams is 1185. kg/m3 .

a. Derive the differential equation for the weight fraction acid as a function of time.

b. How much phosphoric acid is in the tank after 20.0 minutes?

Answer 1

let y(t) represent the % of salt in the tank at time t

y(0))= weight % of acid in the tank at zero time = 0.3

writing the acid balance around the tank

since the tank containd 2000 lits of acid of density 1185 kg/m3

weight of acid in the tank at any time t = (2000/1000)m3*1185 (kg/m3)* y/100 =2*1185/100 y =23.7y

d (23.7y/dt)= rate of mass in -rate of mass out

there are two masses entering , one at 100 kg/min ansd 80 wt % acid concentration and the other at 200 kg/min with an acid concentration of 50 wt%

d (y*23.7/dt)= 100*80/100 ( acid through stream 1)+ 200*50/100- 5*60 kg/min*y/100

d (23.7y/dt) =80+100-3y

dy/dt= (180-3y)/23.7 (1)

The differential equation for variation of weight fraction of acid with time is dy/dt= (180-3y)/23.7

integration equation (1)

dy/180-3y) = dt/23.7

-ln(180-3y)= t/23.7+ C , where C is integration constant

at t= 0 , y =0.3

-ln(180-3*y) = t/23.7 +C

where C is integration constant

t=0 y(0)= 0.3

-ln(180-0.3*3) =0 + C

C= -5.18794

-ln(180-3y) = t/23.7- 5.18794

t=20 minutes

-ln(180-3y)= 20/23.7- 5.18794

-ln(180-3y)= 0.8438332- 5.18794 =-4.34406

ln(180-3y)= 4.34406

180-3y= exp( 4.34406) = 77.01979

180-77.01979= 3y

102.9802=3y

y= 102.9802/3 =34.33 %

amount of salt in the tank after 20 minutes = 2m3*1185 kh/m3*34.33 =813.621 kg