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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /527/CH2/EX2.20/2_20exam.sce | |
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diff --git a/527/CH2/EX2.20/2_20exam.sce b/527/CH2/EX2.20/2_20exam.sce new file mode 100755 index 000000000..78508cc93 --- /dev/null +++ b/527/CH2/EX2.20/2_20exam.sce @@ -0,0 +1,79 @@ +//Engineering and Chemical Thermodynamics
+//Example 2.20
+//Page no :89
+
+clear ; clc ;
+//solution (a)
+//Given data
+Cv = 3/2 * 8.314 ;
+Cp = 5/2 * 8.314 ;
+n = 1;
+R = 8.314 ;
+T1 = 1000 ; //[K]
+P1 = 10 ; //[bar]
+T2 = 1000 ; //[K]
+P2 = 0.1 ; //[bar]
+T3 = 300 ; //[K]
+T4 = 300 ; //[K]
+
+k = Cp / Cv ;
+P3 = P2 * (T3 / T2)^(k/(k-1)); //[bar]
+P4 = P1 * (T4 / T1)^(k/(k-1)) ; //[bar]
+
+// (1)
+ del_U_12 = 0 ; // As process 1-2 is isothermal
+ W_12 = n * R * T1 * log(P2 / P1);
+ Q_h_12 = W_12 ;
+ disp(" Example: 2.20 Page no : 89") ;
+printf('(a)\n (1)\n del_U = %d J',del_U_12) ;
+printf('\n Work = %d J',W_12) ;
+printf('\n Heat = %d J',Q_h_12) ;
+
+//(2)
+ Q_23 = 0 ; // As adiabatic process
+ del_U_23 = n * Cv *(T3 - T2) ;
+ W_23 = del_U_23 ;
+printf('\n (2)\n del_U = %g J',del_U_23) ;
+printf('\n Work (J) = %d J',W_23) ;
+printf('\n Heat (J) = %d J',Q_23) ;
+
+//(3)
+ del_U_34 = 0 ; // As isothermal process
+ W_34 = n * R * T3 * log(P4 / P3) ; // Eqn E2.20.A
+ Q_c_34 = del_U_34 - W_34 ;
+printf('\n (3)\n del_U = %g J',del_U_34) ;
+printf('\n Work = %d J',W_34) ;
+printf('\n Heat = %d J',Q_c_34) ;
+
+//(4)
+ Q_41 = 0 ; // As adiabatic process
+ del_U_41 = n * Cv * (T1 - T4) ;
+ W_41 = del_U_41 ;
+printf('\n (4)\n del_U = %g J',del_U_41) ;
+printf('\n Work = %d J',W_41) ;
+printf('\n Heat = %d J',Q_41) ;
+
+//Solution (b)
+//Users can refer figure E2.20
+
+//Solution (c)
+ W_total = W_12 + W_23 + W_34 + W_41 ;
+ Q_absor = Q_h_12 ;
+ effi = W_total / Q_absor ;
+printf('\n\n(c) efficiency = %g',effi)
+
+//Solution (d)
+x = 1 - T3 / T1 ;
+printf('\n\n(d) 1 - Tc/Th = %g',x);
+disp(" i.e Efficiency = 1 - Tc/Th");
+
+//Solution (e)
+disp("(e) The process can be made more efficient by raising Th or by lowering Tc .");
+disp("Table E2.20B") ;
+disp(" T(K) P(bar) v(m^3/mol)") ;
+P = [P1 , P2 , P3 , P4 ] ;
+T = [T1 , T2 , T3 , T4 ] ;
+for i = 1:4
+ v(i) = R * T(i) * 10^-5/ P(i) ;
+ printf("\n %d %.4f %f \n",T(i) ,P(i) ,v(i)) ;
+end
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