Question

A mixer is used to mix 3 different streams. Stream A is an acid solution containing HNO3 = 2% weight, H2SO4 = 57% weight. Stream B contains nitric acid solution with composition of 90% weight HNO3. Stream C is a concentrated sulfuric acid solution containing 93% weight H2SO4. The mixture of those 3 streams flown out from the mixer with a flowrate of 1000 kg/hour and composition of: HNO3 = 27% weight, H2SO4 = 60% weight. a) Draw a block diagram for the process, completed with all the known variables. b) Flowrate of stream A (kg/hour) into the mixer c) Flowrate of stream B (kg/hour) into the mixer d) Flowrate of stream C (kg/hour) into the mixer e) Composition of the mixture in %moles coming out from the mixer

Answer 1

Let the flow rate of stream A = X kg/hr

flow rate of stream B = Y kg/hr

flow rate of stream C = Z kg/hr

Part a

Block diagram

Part b, c and d

Overall mass balance

X + Y + Z = 1000

Y = 1000 - X - Z......... Eq1

HNO3 Balance

X*0.02 + Y*0.90 + Z*0 = 1000*0.27

0.02X + 0.9Y = 270........ Eq2

Put the value of Y from eq1 into eq2

0.02X + 0.9*(1000 - X - Z) = 270

0.02X + 900 - 0.9X - 0.9Z = 270

0.88X + 0.9Z = 630

Multiply by 0.57

0.5016X + 0.513Z = 359.1............ Eq3

H2SO4 Balance

X*0.57 + Y*0 + Z*0.93 = 1000*0.60

0.57X + 0.93Z = 600

Multiply by 0.88

0.5016X + 0.8184Z = 528............... Eq4

Solve eq3 and eq4 simultaneously

Eq4 - eq3

0.3054Z = 168.9

Z = 533 kg/hr

0.5016X + 0.8184Z = 528

0.5016X + 0.8184*533 = 528

0.5016X = 91.8

X = 183 kg/hr

Y = 1000 - 533 - 183 = 284 kg/hr

Z = 533 kg/hr

Part e

At the output of mixer

Mass of HNO3 = 1000 x 0.27 = 270 kg/hr

Moles of HNO3 = mass/molecular weight

= ( 270kg/hr) / (63kg/kmol)

= 4.286 kmol/hr

Mass of H2SO4 = 1000 x 0.60 = 600 kg/hr

Moles of H2SO4 = mass/molecular weight

= ( 600kg/hr) / (98kg/kmol)

= 6.122 kmol/hr

Mass of water = 1000 - 600 - 270 = 130 kg/hr

Moles of water = mass/molecular weight

= ( 130kg/hr) / (18kg/kmol)

= 7.222 kmol/hr

Total Moles at output = 4.286 + 6.122 + 7.222

= 17.630 kmol/hr

Mol% of HNO3 = moles of HNO3 x 100 / total moles

= 4.286 x 100 / 17.630

= 24.3 %

Mol% of H2SO4 = moles of H2SO4 x 100 / total moles

= 6.122 x 100 / 17.630

= 34.7 %

Mol% of water = 100 - 24.3 - 34.7 = 41 %