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author | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
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committer | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
commit | 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (patch) | |
tree | dbb9e3ddb5fc829e7c5c7e6be99b2c4ba356132c /3718/CH15/EX15.1/Ex15_1.sce | |
parent | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff) | |
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initial commit / add all books
Diffstat (limited to '3718/CH15/EX15.1/Ex15_1.sce')
-rw-r--r-- | 3718/CH15/EX15.1/Ex15_1.sce | 20 |
1 files changed, 20 insertions, 0 deletions
diff --git a/3718/CH15/EX15.1/Ex15_1.sce b/3718/CH15/EX15.1/Ex15_1.sce new file mode 100644 index 000000000..64758ed74 --- /dev/null +++ b/3718/CH15/EX15.1/Ex15_1.sce @@ -0,0 +1,20 @@ +//Chapter 15: Environmental Pollution and Control
+//Problem: 1
+clc;
+
+MM = 294// Molar mass, K2Cr2O7
+
+//Declaration of Variables
+v_eff = 25 // cm cube,
+v = 8.3 // cm cube, K2Cr2O7
+M = 0.001 // M, K2Cr2O7
+
+// Solution
+w = v * 8 * 6 * M / 1000.
+
+mprintf("8.3 cm cube of 0.006 N K2Cr2O7 =%.2e g of O2\n",w)
+mprintf(" 25 ml of the effluent requires %.2e g of O2\n",w)
+
+cod = w * 10 ** 6 / 25.
+mprintf(" 1l of the effluent requires %.2fg of O2\n",cod)
+mprintf(" COD of the effluent sample is %.2f ppm or mg/L",cod)
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