Question

Seu (Gl.213y GDabake the feed reach on (120% L 60 ottke-60 uta (m) is needed to prod rodu t mass ofl SG-L.13 SG-1.213 2) 40 DOF 0 if 3

All I need to calculate is the values of 44.4 kg/h and 144.4 Kg/h. Please show me how to get those numbers

Basically I need to know how to calculate m2 and m3

I know that m2=44.4 and m3=144.4

But how ???

Thank you

Answer 1

Part a

Basis - 1 m3 of product steam

Specific gravity = density of fluid / density of water

For the product stream

Specific gravity = 1.213

Density of product stream = specific gravity x density of water

= 1.213 x 1000 kg/m3

= 1213 kg/m3

Mass of solution = 1213 kg/m3 x 1m3 = 1213 kg

Molarity of product stream = 4 mol/L

Moles of H2SO4 / volume of solution = 4 mol/L

Mass of H2SO4 / (volume x molecular weight) = 4 mol/L

Mass of H2SO4 / (volume) = 4 mol/L x 98 g/mol x 1000L/m3

Mass of H2SO4 / (volume) = 392000 g/m3 x 1kg/1000g

Mass of H2SO4 / (volume) = 392 kg/m3

Mass of H2SO4 = 392 kg/m3 x 1m3 = 392 kg

Mass fraction of H2SO4 = mass of H2SO4 / mass of solution

= 392 / 1213 = 0.323 kg H2SO4 / kg solution

Part b

Mass of 20 wt% H2SO4 = 100 kg

Density of 20 wt% H2SO4 = 1.139 x 1000 = 1139 kg/m3

Volume of 20 wt% H2SO4 = mass/density

= 100 kg/(1139 kg/m3) = 0.0878 m3 x 1000L/m3

= 87.8 L

H2SO4 balance

H2SO4 in 20 wt% stream + H2SO4 in 60 wt% stream = H2SO4 in product stream

100 x (0.20) + m2 x (0.60) = m3 x (0.323)

20 + 0.6 x m2 = 0.323 x m3

m2 = 0.538 x m3 - 33.33 ....... Eq1

Water balance

Water in 20 wt% stream + Water in 60 wt% stream = Water in product stream

100 x (1 - 0.20) + m2 x (1 - 0.60) = m3 x (1‐0.323)

80 + 0.4 x m2 = 0.677 x m3

Put the value of m2 from eq1

80 + 0.4 x (0.538 x m3 - 33.33) = 0.677 x m3

80 + 0.2152 x m3 - 13.33 = 0.677 x m3

66.67 = 0.4618 x m3

m3 = 144 kg

V3 = m3/density of product stream

= 144 kg/(1213 kg/m3)

= 118.7 m³

m2 = 0.538 x 144 - 33.33

= 44.14 kg

V2 = 44 kg/(1498 kg/m3 )

= 0.0294 m3 x 1000L/m3

= 29.37 L

Feed notation = volume of 20% solution / volume of 60% solution

= 87.8 L /29.37 L

= 2.989 (L 20 wt% / L 60 wt%)

Part c

To produce 1250 kg/h of products

Volume requires V2 = (1250 kg/hr x 29.37 L) /144 kg

= 255 L/hr