diff options
| author | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
|---|---|---|
| committer | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
| commit | 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (patch) | |
| tree | dbb9e3ddb5fc829e7c5c7e6be99b2c4ba356132c /3472/CH18/EX18.3 | |
| parent | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff) | |
| download | Scilab-TBC-Uploads-7f60ea012dd2524dae921a2a35adbf7ef21f2bb6.tar.gz Scilab-TBC-Uploads-7f60ea012dd2524dae921a2a35adbf7ef21f2bb6.tar.bz2 Scilab-TBC-Uploads-7f60ea012dd2524dae921a2a35adbf7ef21f2bb6.zip | |
initial commit / add all books
Diffstat (limited to '3472/CH18/EX18.3')
| -rw-r--r-- | 3472/CH18/EX18.3/Example18_3.sce | 38 |
1 files changed, 38 insertions, 0 deletions
diff --git a/3472/CH18/EX18.3/Example18_3.sce b/3472/CH18/EX18.3/Example18_3.sce new file mode 100644 index 000000000..31876c981 --- /dev/null +++ b/3472/CH18/EX18.3/Example18_3.sce @@ -0,0 +1,38 @@ +// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES
+
+// EXAMPLE : 11.3 :
+// Page number 331-332
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+V = 6600.0 // Voltage(V)
+R = 0.045 // Resistance(ohm)
+X = 0.45 // Reactance(ohm)
+Load = 10000.0*10**3 // Total load(W)
+PF = 0.8 // Lagging power factor
+I_a = 437.5 // Armature current(A)
+
+// Calculations
+I = Load/(3**0.5*V*PF) // Load current(A)
+I_working = PF*I // Working component of current(A)
+I_watless = (1-PF**2)**0.5*I // Watless component of current(A)
+I_second = (I_a**2+I_watless**2)**0.5 // Load current supplied by second alternator(A)
+PF_second = I_a/I_second // Lagging power factor of second alternator
+V_ph = V/3**0.5 // Terminal voltage per phase(V)
+I_R = I_second*R // Voltage drop due to resistance(V)
+I_X = I_second*X // Voltage drop due to reactance(V)
+sin_phi_second = (1-PF_second**2)**0.5
+E = ((V_ph+I_R*PF_second+I_X*sin_phi_second)**2+(I_X*PF_second-I_R*sin_phi_second)**2)**0.5 // EMF of the alternator(V/phase)
+E_ll = 3**0.5*E // Line-to-line EMF of the alternator(V)
+
+// Results
+disp("PART II - EXAMPLE : 11.3 : SOLUTION :-")
+printf("\nArmature current of other alternator = %.1f A", I_second)
+printf("\ne.m.f of other alternator = %.f V (line-to-line)", E_ll)
+printf("\nPower factor of other alternator = %.3f (lagging)", PF_second)
|