This preview shows page 1. Sign up to view the full content.
Unformatted text preview: s for the 3 exiting streams
7
Mole fraction sums (one for each stream)
y
yi
2C
Phase equilibrium equations:
KiI = iI and KiII = II
xi
xi
Total number of equations = NE = 3C + 12
(c) Degrees of freedom = ND= NV  NE = (7C + 22)  (3C + 12) = 4C +10
(d) A possible set of specifications is:
For each entering stream: Total flow rate, temperature, pressure, and C1 mole fractions,
which totals 4(C + 2) = 4C + 8
For the remaining 2 variables, choose any combination of Q, temperature of one of the
three exiting streams, and/or pressure of one of the three exiting streams. Exercise 4.2
Subject: Determination of uniqueness of three different operations.
Given: (a) An adiabatic equilibrium stage with known vapor and liquid feed streams, and
known stage temperature and pressure.
(b) Same as (a), except that stage is not adiabatic.
(c) Partial condenser using cooling water, with known vapor feed (except for flow
rate), outlet pressure, and inlet cooling water flow rate.
Assumptions: Exiting streams in equilibrium.
Find: (a) Whether composition and amounts of exiting vapor and liquid can be computed.
(b) Same as part (a).
(c) Whether cooling water rate can be computed. Analysis: (a) With two steams in and two out, number of variables = NV = 4(C + 3) = 4C + 12
Equations are:
C
Component material balances
1
Energy balance
1
Pressure identity for 2 exiting streams
1
Temperature identity for 2 exiting streams
4
Mole fraction sums for 4 streams
C
Phase equilibrium equations
Therefore, number of equations = NE =2C + 7
Degrees of freedom = ND= NV  NE = (4C + 12)  (2C + 7) = 2C +5
Given specifications are: 2C + 4 variables for the two feed streams. Only one
specification left. Therefore can not specify both T and P for exiting streams.
(b) If stage in part (a) is not adiabatic, add Q as a variable to give NV = 4C + 13.
The number of equations stays the same, i. e. NE =2C + 7.
Thus, have one additional degree of freedom, giving, ND= NV  NE = 2C +6.
Can now specify both T and P for exiting streams.
(c) First, consider just the partial condensation of the vapor into two exiting streams
by heat transfer, Q, without considering the cooling water.
For three streams, NV =3(C + 3) + 1 (for Q) = 3C +10.
Equations are:
C
Component material balances
1
Energy balance
1
Pressure identity for 2 exiting streams
1
Temperature identity for 2 exiting streams
3
Mole fraction sums for 3 streams
C
Phase equilibrium equations
Therefore, NE = 2C + 6 and ND = NV  NE = C + 4 Exercise 4.2 (continued)
Analysis: (c) (continued)
Specified for the feed vapor are only C + 1 variables, because the feed rate is not
specified. Also the outlet pressure of the condenser is specified. This give C + 2. We are short
two variables from being able to compute Q.
An energy balance on the cooling water gives:
Q = mCP Tout − Tin But only Tin is given. Thus, with Q unknown and Tout not given, we are three variable short of
being able to compute the water rate, m. Thus, the problem can not be solved uniquely. We
could solve...
View
Full
Document
This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

Click to edit the document details