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author | prashantsinalkar | 2017-10-10 12:38:01 +0530 |
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committer | prashantsinalkar | 2017-10-10 12:38:01 +0530 |
commit | f35ea80659b6a49d1bb2ce1d7d002583f3f40947 (patch) | |
tree | eb72842d800ac1233e9d890e020eac5fd41b0b1b /3269/CH4 | |
parent | 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (diff) | |
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updated the code
Diffstat (limited to '3269/CH4')
-rwxr-xr-x | 3269/CH4/EX4.5/Ex4_5.sce | 28 | ||||
-rw-r--r-- | 3269/CH4/EX4.6/Ex4_6.sce | 65 |
2 files changed, 46 insertions, 47 deletions
diff --git a/3269/CH4/EX4.5/Ex4_5.sce b/3269/CH4/EX4.5/Ex4_5.sce index 8de1ba2af..5e5951cc2 100755 --- a/3269/CH4/EX4.5/Ex4_5.sce +++ b/3269/CH4/EX4.5/Ex4_5.sce @@ -1,14 +1,14 @@ -// Example 4.5
-clear all;
-clc;
-
-// Given data
-t = 30; // Time of uranium sufficiency in years
-// Assuming once through Light Water Reactor (LWR)fuel cycle
-U_LWR = 0.0055; // Uranium Utilization factor for LWR
-// Assuming once through Liquid Metal cooled Fast Breeder Reactor (LMFBR) fuel cycle
-U_LMFBR = 0.67; // Uranium Utilization factor for LMFBR
-// Estimation
-est_time = 30*U_LMFBR/U_LWR;
-// Result
-printf("The time for which Uranium would fuel LMFBR = %d years \n",ceil(est_time));
+// Example 4.5 +clear; +clc; + +// Given data +t = 30; // Time of uranium sufficiency in years +// Assuming once through Light Water Reactor (LWR)fuel cycle +U_LWR = 0.0055; // Uranium Utilization factor for LWR +// Assuming once through Liquid Metal cooled Fast Breeder Reactor (LMFBR) fuel cycle +U_LMFBR = 0.67; // Uranium Utilization factor for LMFBR +// Estimation +est_time = 30*U_LMFBR/U_LWR; +// Result +printf("The time for which Uranium would fuel LMFBR = %d years \n",ceil(est_time));
\ No newline at end of file diff --git a/3269/CH4/EX4.6/Ex4_6.sce b/3269/CH4/EX4.6/Ex4_6.sce index e2f3078b2..3b82cb570 100644 --- a/3269/CH4/EX4.6/Ex4_6.sce +++ b/3269/CH4/EX4.6/Ex4_6.sce @@ -1,33 +1,32 @@ -// Example 4.6
-clear all;
-clc;
-
-// Given data
-A_U = 238; // Atomic Mass number of Uranium
-A_O = 16; // Atomic Mass number of Oxygen
-amt_UO2 = 33000; // Amount of Uranium dioxide (UO2) present in kilogram(kg)
-x_P = 0.032; // Enrichment of 3.2 w/o uranium product
-x_T = 0.002; // Enrichemnt of 0.2 w/o residual tails
-// From Figure 4.45
-x_F = 0.00711; // Enrichemnt of 0.711 w/o feed
-
-// 1.
-// Estimation of enriched uranium in kg
-M_P = A_U*amt_UO2/(A_U+2*A_O);
-// Estimation of amount of Uranium feed in kg
-M_F = ((x_P-x_T)/(x_F-x_T))*M_P;
-// Result
-printf(" \n The amount of uranium feed required per reload = %d kg \n",ceil(M_F));
-
-// 2.
-V_x_P = (1-2*x_P)*log((1-x_P)/x_P); // Value function of uranium product with enrichemnt of 3.2 w/o
-V_x_F = (1-2*x_F)*log((1-x_F)/x_F); // Value function of feed with enrichemnt of 0.711 w/o
-V_x_T = (1-2*x_T)*log((1-x_T)/x_T); // Value function of tallings with enrichemnt of 0.2 w/o
-rate_SWU = 130.75; // Enrichment cost in dollars per SWU
-// Calculation
-SWU = M_P*(V_x_P-V_x_T)-M_F*(V_x_F-V_x_T); // Separative Work (SWU) in kg
-enrich_cost = ceil(SWU)*rate_SWU; // Enrichment cost in dollars
-// Result
-printf("\n The enrichment cost = $ %d \n",ceil(enrich_cost));
-// Due to approximation of Separative Work Unit(SWU), there is a difference in the value of enrichment cost on comparison with the textbook value.
-
+// Example 4.6 +clear; +clc; + +// Given data +A_U = 238; // Atomic Mass number of Uranium +A_O = 16; // Atomic Mass number of Oxygen +amt_UO2 = 33000; // Amount of Uranium dioxide (UO2) present in kilogram(kg) +x_P = 0.032; // Enrichment of 3.2 w/o uranium product +x_T = 0.002; // Enrichemnt of 0.2 w/o residual tails +// From Figure 4.45 +x_F = 0.00711; // Enrichemnt of 0.711 w/o feed + +// 1. +// Estimation of enriched uranium in kg +M_P = A_U*amt_UO2/(A_U+2*A_O); +// Estimation of amount of Uranium feed in kg +M_F = ((x_P-x_T)/(x_F-x_T))*M_P; +// Result +printf(" \n The amount of uranium feed required per reload = %d kg \n",ceil(M_F)); + +// 2. +V_x_P = (1-2*x_P)*log((1-x_P)/x_P); // Value function of uranium product with enrichemnt of 3.2 w/o +V_x_F = (1-2*x_F)*log((1-x_F)/x_F); // Value function of feed with enrichemnt of 0.711 w/o +V_x_T = (1-2*x_T)*log((1-x_T)/x_T); // Value function of tallings with enrichemnt of 0.2 w/o +rate_SWU = 130.75; // Enrichment cost in dollars per SWU +// Calculation +SWU = M_P*(V_x_P-V_x_T)-M_F*(V_x_F-V_x_T); // Separative Work (SWU) in kg +enrich_cost = ceil(SWU)*rate_SWU; // Enrichment cost in dollars +// Result +printf("\n The enrichment cost = $ %d \n",ceil(enrich_cost)); +// Due to approximation of Separative Work Unit(SWU), there is a difference in the value of enrichment cost on comparison with the textbook value.
\ No newline at end of file |