130 lines
3.1 KiB
Plaintext
130 lines
3.1 KiB
Plaintext
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/*
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**********************************************************************
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*
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* CLP(R) Version 2.0 (Example Programs Release)
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* (C) Copyright, March 1986, Monash University
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*
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**********************************************************************
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*/
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%
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% Calculate excess air needed for a chemical reaction to proceed in some way
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% The goal
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% ?- go.
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% is a sample reaction.
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%
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mol_comp(n2, [f(n, 2)]).
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mol_comp(o2, [f(o, 2)]).
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mol_comp(co2, [f(c, 1),f(o, 2)]).
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mol_comp(co, [f(c, 1),f(o, 1)]).
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mol_comp(c, [f(c, 1)]).
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mol_comp(h2o, [f(h, 2),f(o, 1)]).
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comp(air, vol, [f(n2, 0.79),f(o2, 0.21)]).
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comp(flue, vol, [f(co2, 0.125),f(co, 0.025),f(o2, 0.03),f(n2, 0.82)]).
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comp(full_comb, vol, [f(co2, 1),f(n2, T)]).
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comp(fuel, vol, [f(c, 1)]).
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basis(1.0).
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bal(LEFT, RIGHT, VARS) :-
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bal_side(LEFT, VARS, NEW_VARS),
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bal_side(RIGHT, NEW_VARS, N_VARS),
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zero(N_VARS).
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zero([]).
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zero([f(X, 0)|L]) :-
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zero(L).
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bal_side([], [], []).
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bal_side([], [f(A, X)| L1], [f(A, MX)| L2]) :-
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MX = -X,
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bal_side([], L1, L2).
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bal_side([f(S,X)| L], VARS, NEW_VARS) :-
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comp(S, T, COMP_LIST),
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printf("species %s \n",[S]),
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add_species(X, COMP_LIST, VARS, N_VARS),
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bal_side(L,N_VARS, NEW_VARS).
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add_species(X, [f(SPEC, PROP)| L], VARS, NEW_VARS) :-
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printf("molecule %s \n",[SPEC]),
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mol_comp(SPEC, ELEMENTS),
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add_elements(X, PROP, ELEMENTS, VARS, N_VARS),
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add_species(X, L, N_VARS, NEW_VARS).
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add_species(X, [], VARS, VARS).
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add_elements(X, PROP, [f(n, NUM)| L], VARS, NEW_VARS) :-
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add_elements(X, PROP, L, VARS, NEW_VARS).
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add_elements(X, PROP, [f(E, NUM)| L], [f(E, Z)| VARS], [f(E, Z1)| NEW_VARS]):-
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Z = X*PROP*NUM + Z1 ,
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printf("atom %s \n",[E]),
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add_elements(X, PROP, L, VARS, NEW_VARS).
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add_elements(X, PROP, [f(F, NUM)| L], [f(E, Z)| VARS], [f(E, Z)| NEW_VARS]):-
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add_elements(X, PROP, [f(F, NUM)| L], VARS, NEW_VARS).
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add_elements(X, PROP, [], VARS, VARS).
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excess(A, A1) :-
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B = A - A1,
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B = ANS * A1,
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PERC = ANS * 100 ,
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printf("Excess = %f%% \n",[PERC]).
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go:- basis(BASE),
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bal([f(air, A),f(fuel, BASE)], [f(flue, F)], [f(c,0),f(o,0)]),
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printf("With base of %f mol of fuel\n",[BASE]),
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printf("air (just to burn the carbon) = %f mol\n",[A]),
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printf("flue (mol) :%f\n",[F]),
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nl,
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bal([f(air, A1),f(fuel, BASE)], [f(full_comb, F1)], [f(c,0),f(o,0)]),
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printf("Compared with theoretical air (for complete combustion):\n",[]),
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printf(" %f mol\n",[A1]),
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excess(A, A1).
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% Answer:
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% species air
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% molecule n2
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% molecule o2
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% atom o
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% species fuel
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% molecule c
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% atom c
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% species flue
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% molecule co2
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% atom c
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% atom o
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% molecule co
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% atom c
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% atom o
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% molecule o2
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% atom o
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% molecule n2
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% With base of 1.000000 mol of fuel
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% air (just to burn the carbon) = 5.317460 mol
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% flue (mol) :6.666667
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%
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% species air
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% molecule n2
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% molecule o2
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% atom o
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% species fuel
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% molecule c
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% atom c
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% species full_comb
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% molecule co2
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% atom c
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% atom o
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% molecule n2
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% Compared with theoretical air (for complete combustion):
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% 4.761905 mol
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% Excess = 11.666667%
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?- printf("\n>>> Sample goal: go/0\n", []).
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