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:
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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.
Temp. Interval (°C) |
Fcp Cold #1 (MW/°C) |
Fcp Cold #2 (MW/°C) |
Total Cold Fcp (MW/°C) |
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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')
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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
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You should note that with the added heat required for the phase change, we now need 2.5 MW more heat for heating the cold streams than we need in the way of cooling the hot streams. Here is a comparison of the two cases--with and without the change of phase:
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No Phase Change |
With Phase Change and different Fcp | ||
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DTmin°C4 |
Cooling Utility (MW) |
Heating Utility (MW) |
Cooling Utility (MW) |
Heating Utility (MW) |
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4All DTmin were read from the graphs except the middle one with no phase change.
The minimum heating and cooling requirements almost reverse roles as a result of the phase change in the cold stream.