Question

(4) A 2.5 KW electrical motor with 0.9 power factor and 75% efficiency, is supplied with single core general purpose PVC cable passing through thermally insulated wall and 20m length. The heater is protected via BS3036 fuse with ambient temperature of 25°C. The ratings of BS3036 are shown in Table 1. The maximum permissible voltage drop is 6% of the rated voltage of 240 V. BS3036 Current ratings (A) 5, 15, 20, 30, 45,60 Table 1 Find: a) The design current b) Nominal current c) Correction factor d) Voltage drop e) The fuse rating f) The cable size

Answer 1

Given Power(P)= 2.5 kw = 2.5X103w Power faction=og Voltage lut 240V power = VB VL ILCOS & @) Design Current =/ з V Cosф = 5x103 V3x 240x0,9 = 13.36 A (e). The fuse rating depends, which type of protection is given There are two types protecting 1 Corse excurs current protection 2) close excess current protection * In cese of course excess crorrent proteching fure Iching = 0.81 x Normel corrent Cerrying Cepecity = 0.81% 20 = 16.24 . . So choose 16.2A fuse

Pin Cere of close cace 18 cortest protection fuse relics=123x Nosmel current celrying Фра. # = 1.2320 = 296 A As per the IE rules for powel Circuit the minimum nominel cross sectional aree is 3.5 mmv. Current reling of 3.5 mmv coppel conductor =20A (Refer to Is 684 pelt a -1964) (60080) e do not know the concept of option bredd questions. plz like my answer.