Fault tree analysis was developed in 1962 for the U.S. Air Force by Bell Telephone Laboratories for use with the Minuteman system…was later adopted and extensively applied by the Boeing Comp…Full description
Fault Tree AnalysisDeskripsi lengkap
Three phase fault analysis with Matlab & Proteus circuitFull description
Fault tree analysis was developed in 1962 for the U.S. Air Force by Bell Telephone Laboratories for use with the Minuteman system…was later adopted and extensively applied by the Boeing Company…is ...
Reliability of safety instrumented systemsFull description
lecture note
Descripción: estimation of time pact analysis
With Rajiv Ranjan APPAREL STANDARDS SPECIFICATION AND QUALITY CONTROL DEFECT ANALYSIS OF 4 GARMENTS NOVEMBER 10, 2010 SHIRT DEFEC T NO. DEFECT NAME LOCATION CAUSE REMEDY DEPARTME NT PICTURE 1. ...Full description
Deskripsi lengkap
Bangladesh seismic fault modelingFull description
tenimiento de CiDescripción completa
about faults in the systemDescripción completa
Useful in Condition based monitering
about faults in the systemFull description
labFull description
analytical chemistryFull description
---Exp 13---------FAULT ANALYSIS-----HS P 427 -------Ex10.5------
PROBLEM: The one line diagram of a simple power system is shown in figure. The neutral of each generator is grounded through a current limiting reactor of 0.25/3 per unit on a 100MVA base. The system data expressed in per unit on a common 100 MVA base is tabulated below. The generators are running on no load at their rated voltage and rated frequency with their emf in phase. Determine the fault current for the following faults. (a) A balanced three phase fault at bus 3 through a fault impedance, Zf = j0.1 per unit. (b) A single line to ground fault at bus3 through a fault impedance, Zf = j0.1 per unit. (c) A line to line fault at bus3 through a fault impedance, Zf = j0.1 per unit. (d) A double line to ground fault at bus3 through a fault impedance, Zf = j0.1 per unit.
MATLAB PROGRAM: Z133 = j*0.22; Z033 = j*0.35; Zf = j*0.1; disp('(a) Balanced three-phase fault at bus 3') Ia3F = 1.0/(Z133+Zf) disp('(b) Single line-to-ground fault at bus 3') I03 = 1.0/(Z033 + 3*Zf + Z133 + Z133); I012=[I03; I03; I03] %sctm; global sctm a =cos(2*pi/3)+j*sin(2*pi/3); sctm = [1 1 1; 1 a^2 a; 1 a a^2]; Iabc3 = sctm*I012 disp('(c) Line-to-line fault at bus 3') I13 = 1.0/(Z133 + Z133 + Zf); I012 = [0; I13; -I13] Iabc3 = sctm*I012 disp('(d) Double line-to-ground fault at bus 3') I13 = 1/(Z133 + Z133*(Z033+3*Zf)/(Z133+Z033+3*Zf)); I23 = -(1.0 - Z133*I13)/Z133; I03 = -(1.0 - Z133*I13)/(Z033+3*Zf); I012 = [I03; I13; I23] Iabc3 = sctm*I012
