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pashok25 [27]
3 years ago
15

Write the following statements as Prolog clauses, in the order given: If it is raining or snowing, then there is precipitation.

If it is freezing and there is precipitation, then it is snowing. If it is not freezing and there is precipitation, then it is raining. It is snowing. What answer does Prolog give to the following queries: Is it freezing
Engineering
1 answer:
Anika [276]3 years ago
8 0

Answer:

a. See Explanation Below

b. No

Explanation:

a.

Interpreting the statements line by line.

First, the prolog clauses need to be initialised, as follows:

:- dynamic freezing/0, precipitation/0, snowing/0, raining/0.

If it is raining or snowing, then there is precipitation.

The statement is splited in two.

1st, if it is raining then there is precipitation

Or

2nd, if it is raining, then there is precipitation.

So, both statements are given as:

precipitation :- raining.

precipitation :- snowing.

If it is freezing and there is precipitation, then it is snowing.

This is a conditional statement that checks if it is freezing AND there's is precipitation.

If true, then it is snowing.

The above statement is given as:

snowing :- freezing, precipitation.

If it is not freezing and there is precipitation, then it is raining.

This is a conditional statement that checks if it is not freezing AND there's is precipitation.

If true, then it is raining.

The not clause is represented as \+

The above statement is the given as:

raining :- \+ freezing, precipitation.

It is snowing.

This is represented as follows

snowing.

So, the full prolog clause is as follows

:- dynamic freezing/0, precipitation/0, snowing/0, raining/0.

precipitation :- raining.

precipitation :- snowing.

snowing :- freezing, precipitation.

raining :- \+ freezing, precipitation.

snowing.

b. What answer does Prolog give to the following queries:

Is it freezing

No

Reason;

All of the conditions above point to raining or snowing; none points to freezing.

Hence, it is no.

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Answer:

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c) W_cycle = 400 KW , n_th = 66.67 %  , Reversible

Explanation:

Given:

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  TL = 400 K

  TH = 1200 K

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For each case W_cycle , n_th ( Thermal Efficiency ) :

(a) QH = 600 kW, QC = 400 kW

(b) QH = 600 kW, QC = 0 kW

(c) QH = 600 kW, QC = 200kW

- Determine whether the cycle operates reversibly, operates irreversibly, or is impossible.

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- The work done by the cycle is given by first law of thermodynamics:

                                 W_cycle = QH - QC

- For categorization of cycle is given by second law of thermodynamics which states that:

                                 n_th < n_max     ...... irreversible

                                 n_th = n_max     ...... reversible

                                 n_th > n_max     ...... impossible

- Where n_max is the maximum efficiency that could be achieved by a cycle with Hot and cold reservoirs as follows:

                                n_max = 1 - TL / TH = 1 - 400/1200 = 66.67 %

And,                         n_th = W_cycle / QH

a) QH = 600 kW, QC = 400 kW

   - The work done by cycle according to First Law is:

                                W_cycle = 600 - 400 = 200 KW

   - The thermal efficiency of the cycle is given by n_th:

                                n_th = W_cycle / QH

                                n_th = 200 / 600 = 33.33 %

   - The type of process according to second Law of thermodynamics:

               n_th = 33.333 %                n_max = 66.67 %

                                       n_th < n_max  

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b) QH = 600 kW, QC = 0 kW

   - The work done by cycle according to First Law is:

                                W_cycle = 600 - 0 = 600 KW

   - The thermal efficiency of the cycle is given by n_th:

                                n_th = W_cycle / QH

                                n_th = 600 / 600 = 100 %

   - The type of process according to second Law of thermodynamics:

                 n_th = 100 %                 n_max = 66.67 %

                                     n_th > n_max  

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c) QH = 600 kW, QC = 200 kW

   - The work done by cycle according to First Law is:

                                W_cycle = 600 - 200 = 400 KW

   - The thermal efficiency of the cycle is given by n_th:

                                n_th = W_cycle / QH

                                n_th = 400 / 600 = 66.67 %

   - The type of process according to second Law of thermodynamics:

               n_th = 66.67 %                 n_max = 66.67 %

                                     n_th = n_max  

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