). A 15 wt% Na2SO4 solution is fed at the rate of 12 kg/min into a miser that in
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Question
). A 15 wt% Na2SO4 solution is fed at the rate of 12 kg/min into a miser that initially holds 100 kg of a 50-50 mixture (by weight) of Na2SO4 and water. The exit solution leaves at the rate of 10 kg/min. Assume uniform mixing, which means that the concentration of the exit solution is the same as the concentration in the mixer. (a) What is the total mass in the mixer at the end of 10 minutes? (b) What is the concentration of Na2SO4 in the mixer at the end of 10 minutes? (c) How long will it take for the mass fraction of Na2SO4 in the tank to reduce to 0.35?
Explanation / Answer
Initial tank mass, V0 = 100 kg
Initial concentration of tank, Ct0 = 0.5 [= 50/100]
Initial tank Na2SO4 content = 50% of 100 = 50 Kg
Inlet flow rate of Na2SO4 solution, Qi = 12 Kg/min
Concentration of inlet Na2SO4 solution, Ci = 0.15 [ =15/100]
Na2SO4 entering with the inlet = 15% of 12 Kg/min = 1.8Kg/min
Outlet flow rate of solution from tank, Qo = 10 Kg/min
Mass balance equation :
Let 'M' be tank volume, 'Ct' be tank concentration, 'Ci'be inlet concentration. 'Qi' be inlet mass flow rate, 'Qo' be outlet mass flow rate.
Mass balance :
dM/dt = Qi - Qo
Solute balance :
d(MCt)/dt = QiCi - QoCt
with initial condition: at t = 0 ; V0 = 100 ; Ct0 = 0.50
On integrating,
M = (Qi-Qo)t + 100 --(1)
ln( Ct - QiCi/Qo) = -Qot/M + ln(0.5) --(2)
The above equatons are coupled euations. That is, 'M' at any point to be used in equation 2, must be first calculated from equation 1.
a) Mass in mixer after 10 mins,
M = (Qi-Qo)t + 100
M = (12-10)*10 + 100 = 120 Kg
Total mass in mixer = 120 Kg
b)Concentration in the mixer after 10 mins
ln( Ct - QiCi/Qo) = -Qot/M + ln(0.5)
ln( Ct - 12*0.15/10) = -10*10/120 -0.693
Ct - 0.18 = 0.217
Ct = 0.397
Concentration of mixer at the end of 10 mns = 0.397 i.e 0.397*100 = 39.7%
c) Time requird for Ct to becoume 0.35.
M = (Qi-Qo)t + 100
M = 2t+100
ln( Ct - QiCi/Qo) = -Qot/M + ln(0.5)
ln( 0.35 - 12*0.15/10) = (-10t/(2t + 100) ) - 0.693
-1.772 + 0.693 = -10t/(2t + 100)
t = 14.38 mins
Time for concentration to drop to 0.35 = 14.38 mins
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