A steam of 20.8 lb/min aqueous H2SO4 solution at 120 °F is to be diluted with a
ID: 1069493 • Letter: A
Question
A steam of 20.8 lb/min aqueous H2SO4 solution at 120 °F is to be diluted with a 5.94 lb/min stream of liquid water at 32 F. The incoming H2SO4 stream 0.450 mass fraction of H2SO4, and 0.550 mass fraction H20 The dilution happens in a mixer with heat exchanger coils inside. The temperature of diluted H2SO4 leaving the mixer is 120 °F 5.94 lb H2O/min m1 Mixer T1 32 °F m3 lb H2SO4 solution/min 120 OF T3 m2 20.8 lb H2SO4 solution/min m13H2so4 lb H2SO4/lb T2 120 °F m 3H20 lb H20/lb m2H2so4 0.450 lb H2SO4/lb Q btu/min m2H20 0.550 lb H20/lb What is m3, the flow rate of H2SO4 from the mixer? Number Ib H, SO, solution min What is m3H2so4, the mass fraction of H2SO4 in the exiting stream? (Scroll below for more questions.) Number lb H2So4 lbExplanation / Answer
Inlet aquesous feed flow rate= 20.8 lb/min. It contains 0.45 lb H2SO4/lb.
Hence its flow rate = 20.8*0.45 lb/min =9.36 lb/min
Rest of the flow rate is water which is =20.8-9.36= 11.44 lb/min
Flow rate of water entering = 5.94 lb/min
Total flow entering = 20.8+5.94= 26.74 lb /min solution of H2SO4.
Mass fraction H2SO4= 9.36/26.74 = 0.35 lb H2SO4/lb.
Water is enteting at 32 deg.F which is taken as reference temperaure, hence its enthalpy = 0
Speciifc enthalpy of H2SO4 at 77deg.F= 0.33 btu/lb.deg,F*(77-32)= 14.85 Btu/lb
Enthalpy of H2SO4 at 77 deg.F = Flow rate of pure H2SO4* specific heat* temperature difference + Flow rate of water*specific heat* temperature difference =
=9.36*0.33 btu/lb.deg.F*(120-77)+ 11.44*1*(120-77) =625Btu/min
Total inlet enthalpy = 625+0= 625 btu/min
Specific Enthalpy of exiting H2SO4 stream = 1btu/lb.deg.F* 0.35 (120-32)+0.65*0.33*(120-77) {H2SO4}=40 btu/lb
Enthalpy of exit H2SO4= 26.74*40 btu/min=1069.6 Btu/min
Heat flow into the mixture= 1069.6- 625 =444.6 Btu/min
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