Example 2 - Analysis of the Network Using entHC2 and entHCb

If we modify example one to include a change of phase of one of the streams, we can either follow the procedure used by Douglas and assume that the change in enthalpy occurs over 1°F or use the modified program entHC2. This program allows for discontinuous enthalpy vs temperature curves. Here is what help tells us:

>>help entHC2
 
  The enthalpy of one or more streams with different FCp + State changes .
   All streams must be either heated or cooled..
  function [H,Tcom]=entHC2(T,FC,Tj,dH)
  Argument  Holds                                  Size
     T      Temps. where FCp changes           1 row   N+1 columns
     FC     Value of FCp for T in an interval  1 row   N columns
     Tj     Temps where H is discontinuous     1 row   M columns
     dH     Changes in H at each Tj            1 row   M columns
  All Tjs must be inside the interval spanned by T.
  The returned row vector H gives the enthalpy relative to the value
     at T(1) for each of the Ts including each end of the 
     discontinuities.  The returned row vector Tcom is the combined 
     set of temperatures where the Hs have been set.  
  H(1) is 0.
  Example: >> Thot=[100 120 200 250];
           >> FChot=[4000 5000 1000];
           >> Tjmp=[110 212];
           >> dHs=[50000 100000];
           >> [Hhot,Tcom]=entHC2(Thot,FChot,Tjmp,dHs)
 

We will add a change of phase to the first cold stream in Example 1 to show how to use entHC2. Here is the revised diagram:

The first heat exchanger on the stream Cold #1, raises its temperature from 25°C to 125°C, the second one vaporizes the fluid and the third one increases the sensible heat. It is assumed that all change of state for the stream occurs at 125°C. Note that the heat capacity of the vapor is less than that of the liquid in Cold #1. The hot streams are the same as in our original problem. The following table shows how the input parameters for the flow*heat capacity used in entHC2 for the cold streams were determined.

In addition there is a jump of 5 MW in the enthalpy at 125°C. Thus our use of entHC2 proceeds by:

>>Tcold=[25 100 125 200 250];
>>FCcold=[0.25 0.55 0.45 0.3];
>>[Hcold,Tscold]=entHC2(Tcold,FCcold,125,5)
 
Hcold =
         0   18.7500   32.5000   37.5000   71.2500   86.2500
 
Tscold =
    25   100   125   125   200   250
 

A plot of these results shows the jump in enthalpy associated with the state change:

>>plot(Hcold,Tscold,'b')
>>grid
>>xlabel('Enthalpy MW')
>>ylabel('Temperature C')
 

The program entHCb allows us to see the effect of changes in state.

>>help entHCb
 
  Using entHC2 to produce composite curves and plot them with a horizontal shift
  function entHCb(Thot,FChot,Tjhot,dHhot,Tcold,FCcold,Tjcold,dHcold,Tunit,Hunit)
  Argument  Holds                                       Size
    Thot   Temps. where FCp changes for hot fluid  1 row  Nh+1 columns
    FChot  Value of FCp for Thot in an interval    1 row   Nh  columns
    Tjhot  Temps where Hhot is discontinuous       1 row   Mh  columns
    dHhot  Changes in H at each Tjhot              1 row   Mh  columns
    Tcold  Temps. where FCp changes for cold fluid 1 row  Nc+1 columns
    FCcold Value of Fcp for Tcold in an interval   1 row   Nc  columns
    Tjcold Temps where Hcold is discontinuous      1 row   Mc  columns
    dHcold Changes in H at each Tjcold             1 row   Mc  columns
    Tunit  Units for T: character data
    Hunit  Units for enthalpy: character data
  Uses ent HC1 and entHC2
  Example:
    >> Thot=[100 120 200 250];
    >> FChot=[4000 5000 1000];
    >> Tjhot=212;
    >> dHhot=100000;
    >> Tcold=[90 130 150 190];
    >> FCcold=[3000 9000 6000];
    >> Tjcold=110;
    >> dHcold=50000;
    >> entHCb(Thot,FChot,Tjhot,dHhot,Tcold,FCcold,Tjcold,dHcold,'F','Btu/hr')
 

Here is what we find using entHCb:

>>Thot=[25 50 200 275];
>>FChot=[0.15 0.46 0.15];
 

We get the unshifted curves first:

>>entHCb(Thot,FChot,[],[],Tcold,FCcold,125,5,'C','MW')

If you want to print the current graph, reply: y  

 
Give a new shift or 0 to stop.  14