Show all work and answers are: (a) DOF (25 pts) For the non-reacting process bel
ID: 702872 • Letter: S
Question
Show all work and answers are:
(a) DOF
(25 pts) For the non-reacting process below, Stream F has a mass flow rate of 20,000 kg/hr and contains 20 wt% KNO3 and 80 wt% water. Stream T is 100 wt% water. Stream M is 50 wt% KNO3. Stream C is 96 wt% KNO3. Stream R is a saturated solution of 0.6 kg KNO3 per kg of water 5. (a) Perform the degree of freedom analysis. Include the DOF for the overall system (b) Calculate the mass flow rate of stream C c) Calculate the mass flow rate of stream R (d) Calculate the mass fraction of KNOs in the stream entering the evaporator in stream E.Explanation / Answer
All the individual balance will be denoted by the stream letter followed 'k' for KNO3 and 'w' for water.
In the above system ,the number of unknown are :
E,Ew,Ek,T,M,Mw,Mk,C,Cw,Ck,R,Rw,Rk
Number of independent equations :
F = T + C ( overall balance)
E = F + R
Ew = Fw + Rw
Ek = Fk + Fw
M = E - T
Mw = 0.5(E-T)
Mk = 0.5*(E-T)
Ck = Fk
C = Ck / wt frac of KNO3 in C
Cw = C - Ck
R = M - C
Rk = 0.6/1.6 R
Rw = 1/1.6R
As number equations and number of of unknowns are equal, the degree of freedom for the whole system is zero.
For evaporator : Unknowns = E, Ew, Ek, T, M,Mw,Mk
M = E - T
Mk = Ek
Mw = Ew - T
Ew = E - Ek
Mw = 0.5M
Number of unknowns = 7 ; Number of equations = 5. DOF = 7 - 5 = 2 ;
Similarly for crystallizer, Number of unknowns = 6 ; Number of equations = 5. DOF = 6 - 5 = 1 ;
for mixer, Number of unknowns = 6 ; Number of equations = 5. DOF = 6 - 5 = 1 ;
b) Mass flow rate of F = 20,000 Kg/hr with 20 wt% KNO3 and 80% Water
Fk = KNO3 = 0.2*20000 = 4000 Kg/hr
Fw = Water = 0.8 * 20000 = 16000 Kg/hr
Lets assume that Stream E entering the evaporator has flow rate of 'E' with Ek and Ew being flow rate of KNO3 and water respectively.
E = R + F = R + 20000
Given that R has 0.6 Kg of KNO3 per Kg of water. Fraction of KNO3 = 0.6/(1+0.6) = 0.6/1.6 = 3/8. Similarly fraction of water = 5/8R
Ek = 3/8R + 4000
Ew = 5/8 + 16000
Given that T is totally water. Thus all the KNO3 entering the evaporator exits to stream M.
Given that stream M contains 50 wt% KNO3 . The indicidual and overall balance around M is given as,
M = E - T
Mk = Ek = 4000 + 3/8R
Mw = Ew - T = 16000 + 5/8R - T
In crystallizer, the product has 96% wt KNO3. This is the only point where KNO3 laves the system. For continuous system, input should be equal. That is all the KNO3 entering the system should leave the system.
Thus, KNO3 in C = KNO3 in F = 4000 Kg/hr
C = KNO3 flow rate/ wt fraction of KNO3 = 4000/0.96 = 4166.66 Kg/hr
Water leaving in C (Cw) = 166.66 Kg/hr
c) Using overall material balance for water,
Fw = T + Cw
T = 16000 - 166.66 = 15833.34 Kg/hr
Rewriting equations of M stream,
Mk = Ek = 4000 + 3/8R
Mw = Ew - T = 16000 + 5/8R - 15833.34 = 166.66 + 5/8R
Given that 50% M is KNO3 and 50% is water, i.e. Mk = Mw
4000 + 3/8R = 166.66 + 5/8R
R = 15333.36 Kg/hr
d) Calculating E ,
E = F + R = 20000 + 15333.36 = 35333.36 Kg/hr
Ek = 4000 + 3/8R = 9750.01 Kg/hr
Ew = 16000 + 5/8R = 25583.35 Kg/hr
Fraction of KNO3 in E = Ek/E = 9750.01/35333.36 = 0.276