Unformatted text preview: From Eq. (567), ND = NV  NE = 14  4 = 10.
A typical specification for a single evaporator effect might be F, wF, TF, PF, wL, P, TS, PS,
TC , and PC .
For the total condenser, the variables are: (Into) V, T, P; (out of) V, PT , TT ; Q
Therefore, NV = 6.
The relationships among the variables are:
Energy balance: Q = V(HV in  HV out) Exercise 5.35 Analysis: (continued) (continued) Therefore, NE =1 and ND = 6  1 = 5.
Typical specs might be V, T, P, PT , TT
For the pump, the variables are L, wL, TL, PL, PP with no equations and 5 degrees of
freedom.
Now analyze the system. From Eq. (570), ND system = ND
all units u − interconnecting redundant variables
ND all units u (1) = 3(10) + 5 + 5 = 40 The number of interconnecting redundant variables are 4 (L, wL, T, P) for each of 3 liquid
streams and 3 (V, T, P) for each of 3 vapor streams, or 4(3) + 3(3) = 21
Therefore, from Eq. (1), ( N D )system = 40  21 = 19
The given specifications are as follows, where effects are numbered 1, 2, 3 from left to
right in Fig. 5.23: wF1 , w L3 , PS1 , TS1 . Therefore, additional specifications needed = 19  4
= 15,
which might be:
Feed rate to Effect 1 = F1
Temperature and pressure of feed to Effect 1 = TF1 , PF1
Pressure in each effect = P1, P2, P3
Temperature and pressure leaving condenser = TT, PT
Pressure leaving the pump = PP
Pressure of condensed steam leaving each effect = pressure of steam entering the
effect
Temperature of condensed steam leaving each effect = saturation for the exit
pressure Exercise 5.36
Subject: Degrees of freedom and additional specifications for a reboiled stripper. Given: Reboiled stripper with specifications in Fig. 5.28.
Find: (a)
(b)
(c)
(d) Number of variables
Number of relationships.
Number of degrees of freedom.
Additional specifications, if any. Analysis: (a) and (b) For an N 1 cascade and a partial reboiler from Table 5.3
NV
NE
Element
N 1 cascade
7(N1)+2C(N1)+2C+7
5(N1)+2C(N1)+2
Partial reboiler
3C+10
2C+6
Total
7N+2NC+3C+10
5N+2NC+3
From Eq. (568),
From Eq. (569), NV
NE unit = unit = NV
all elements NE
all elements e − NR C + 3 + N A e − NR (1)
(2) Since NR = 2 and NA = 0, Eqs. (1) and (2) become,
( NV )unit = (7N +2NC +3C +10) − 2 ( C + 3) = 7 N + 2 NC + C + 4 ( N E )unit = 5N +2NC +3 − 2 = 5N +2NC + 1 (c) From Eq. (567), ND = NV  NE = (7N+2NC+C+4)  (5N+2NC+1) = 2N + C + 3
(d) Specifications are:
All feed conditions
Number of stages Total C+2
1
C+3 Therefore, can make 2N additional specifications, which might be:
All stage and reboiler pressures
N
Adiabatic stages
N 1
Have one specification left. Make it the bottoms rate, distillate rate, or boilup rate or
ratio. Exercise 5.37
Subject: Degrees of freedom and specifications for thermally coupled distillation.
Given: Distillation system in Fig. 5.29.
Find: (a)
(b)
(c)
(d) Number of variables
Number of relationships.
Number of degrees of freedom.
Reasonable set of specifications. Analysis: Treat the system as one of two units with NR = 4 interconnecting...
View
Full Document
 Spring '11
 Levicky
 The Land

Click to edit the document details