Innovating Electrical Engineering Calculations

411.5 TT system In this system, all exposed-conductive-parts and extraneous-conductive-parts of the installation must be connected to a common earth electrode. The neutral point of the supply system is normally earthed at a point outside the influence area of the installation earth electrode but need not be so. The impedance of the earth fault loop therefore consists mainly in the two earth electrodes (i.e., the source and installation electrodes) in series, so that the magnitude of the earth fault current is generally too small to operate overcurrent relays or fuses, and the use of a residual current operated device is essential. "BS 7671, 411.5.l Every exposed-conductive-part which is to be protected by a single protective device shall be connected, via the main earthing terminal, to a common earth electrode. However, if two or more protective devices are in series, the exposed-conductive-parts may be connected to separate earth electrodes corresponding to each protective d...

Earth fault loop impedance test : Earth fault loop impedance testing is performed to measure the impedance of the earth fault loop in an electrical circuit. The purpose of this test is to ensure that the electrical installation is safe and can operate properly under fault conditions. During the test, the circuit is energized and a test current is injected into the circuit. The voltage drop across the circuit is measured, and the impedance of the earth fault loop is calculated using Ohm's Law. If the earth fault loop impedance is too high, it may result in a high fault current, which can cause the protective device, such as a circuit breaker, to trip. If the impedance is too low, it may not allow enough current to flow to operate the protective device, which can result in a dangerous situation. During the test, the circuit may be tested with both trip and no-trip conditions. The trip condition involves the activation of the protective device, which is designed to disconnect the circ...

  Breaker Size : Example: Design Current for the R-TPN Load of Max 10 KW value is 20.89 A We need to select the Next Higher size of Breaker available in the Market for Calculated Design Current (=20.89 A) Here In our Case 30/32 A breaker can be selected. It can be MCCB or MCB or RCBO depending upon requirements. Ib = 21 A In = 30 A Iz = 52.48 (From Cable Data) Note: The operating characteristics of a device protecting a conductor against overload shall satisfy the following two conditions: Ib <= In <= Iz I2 <= 1.45 Iz Ib = the current for which the circuit is designed, e.g. maximum demand In = the nominal current of the protective device Iz = the continuous current-carrying capacity of the conductor (see the AS/NZS 3008.1) I2 = the current ensuring the effective operation of the protective device and may be taken as equal to either— (a) the operating current in conventional time for circuit breakers (1.45 In); or (b) the fusing current in conventional time for fuses (1...

  Design Current : kVA and KW : kVA = KW / PF ; Let's say the power factor is 0.8 , which means KW is 80% of kVA For a 10 KW Load, kVA would be 12.5 kVA "Consequently, we can calculate the Current as well. "

  VBA ADDIN FOR ELECTRICAL CALCULATIONS Functions accessible in Excel : fx = CableTraySize( ) fx = CONTAINMENT( ) fx = EARTHING( ) fx = FAULTCURRENT( ) fx = kVA( ) fx = LLMF( ) fx = LMF( ) fx = LSF( ) fx = RSMF( ) fx = say( ) fx = TRCURRENT( ) fx = Z_CABLE( ) fx = Z_Transformer( ) fx = CABLE( ) fx = CURRENT( ) fx = MCCB( ) fx = SHORTCIR( ) fx = VDrop( )