A steam of 29.2 lb/min aqueous H2SO4 solution at 60 F is to be diluted with a 29

Question

A steam of 29.2 lb/min aqueous H2SO4 solution at 60 F is to be diluted with a 29.2 lb/min stream of liquid water at 32 F. The in H2SO4 stream 0.600 mass fraction of H2SO4, and 0.400 mass fraction H20. coming The dilution happens in a mixer with heat exchanger coils inside. The temperature of diluted H2SO4 leaving the mixer is 60 29.2 lb H20/min 32 FT Mixer m3 lb H2SO4 solution/min 60 °F m2 29.2 lb H2SO4 solution/min m3H2so4 lb H2SO4/lb 60 °F m3H20 b H20/lb m2H2so4 0.600 lb H2SO4/b Q btu/min m2H20 0.400 lb H20/lb What is m3, the flow rate of H2SO4 from the mixer? Number lb H,So, solution min What is m3H2so4, the mass fraction of H2SO4 in the exiting stream? (Scroll below for more questions.) Number

Explanation / Answer

-m1 = 29.2 lb/min liquid water stream mixes with m2=29.2 lb/min AQ. H2SO4 soln.

-but, m2 H2SO4 =0.6 lb H2SO4/lb

And m2 H2O = 0.4 lb H2O/lb

So, flow in m2 stream is ,

= flow of H2SO4 = 29.2*0.6 = 17.52 lb/min

& flow of H2O = 29.2. * 0.4 = 11.68 lb/min

m3 = m1+m2 = 29.2+29.2. = 58.4 lb H2SO4 soln/min

Now, molecular weight of H2O= 18 gm=0.0396 lb

So, = 58.4 lb/(1 mol H2O*0.0396 lb H2O) = 1474.7 mol H2O

Now,molecular weight of H2SO4= 98 gm= 0.2156 lb

So,= 58.4 lb/(1 mol H2SO4*0.2156 lb H2SO4)= 270.87 mol H2SO4

Total mol= 1474.7 + 270.87 mol of soln

So,mass fraction of H20 in soln = 1474.7/1745.57= 0.845

& mass fraction of H2SO4 in soln = 270.87/1745.57= 0.155

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