CEE 496homework2sols

CEE 496homework2sols - CEE 496 —- ELECTRICAL BUILDlNG...

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Unformatted text preview: CEE 496 —- ELECTRICAL BUILDlNG SYSTEMS SOLUTION SET March 15, 2005 PROBLEM SET #2 1) A 120—volt, 20-amp circuit (circuit fused at 20 amps) feeds two convenience outlets in a kitchen. A coffee maker (625 watts) and a waffle iron (1100 watts) are plugged into one of the outlets. a) If both are running, will the circuit be overloaded? b) If a toaster rated at 1200 watts is plugged into the other outlet and turned on while the other two appliances are operating, will the fuse blow? Branch Circuit Rating: 20A Psource = V*| = 120V*20A = 24OOVA*1.OP,F, = 24OOW All loads are assumed to be non-continuous. All loads are purely resistive. a) 625W + 1100w = 1725w < 24OOW Not Overloaded b) 1725w + 1200w = 2925W > 24OOW Fuse will blo 2) A portable generator is needed at a construction site to provide temporary lighting for evening work. It has been determined that 40 fixtures, each equipped with a 100—watt incandescent lamp, are required. a) What is the total lighting load expressed in (kW)? b) if the generator is to be sized at a minimum of 125 percent of the lighting load, what should the minimum rated full load current (amps) of the generator be at 120 volts? The lighting is incandescent, so it is a purely resistive load. a) Calculate the total lighting load: Pload = (4Ofixtures)(1OOW/fixture) / 1000W/kW : b) Calculate the power needed by the generator: PSource = 1-25(4kW) = 5kw Calculate the minimum rated full load current: I: P/V = 5oooW/120V = 3) A 5 horsepower (1 horsepower equals 746 watts), three—phase motor with a power factor of 0.8 is being operated at 480 volts. What is the full load current? Calculate the real power: ' Preal = (746W/1HPNV) (5HP) = 3,730w Calculate the apparent power: 839 = P/P.F. = 3,730W/0.8P_F_ = 4,663VA Calculate the full load current: I0 = 839/\/3(VLL) = 4,663VA/\/3(480V) = 4) A 120—volt branch circuit serves three computers, each rated at 500 watts and operating for at least 8 hours a day, and one printer rated at 400 watts operating intermittently for a couple minutes at a time. a) What is the branch circuit rating, assuming a power factor of 0.9? b) What is the minimum conductor size that should be used? c) What is the maximum standard overcurrent protection device that should be used to protect the conductors? Continuous Load: 3(500W) = 1500w Non-continuous Load: 400W 3) |Branch circuit : 1-25 I (120V) (0.9p_F_) + 1.0 (400W) / (120V) (0.9m) i—= 21 .07A b) Satisfy integrity of terminals: 7 21.07A < 100A use 60°C column Select#12Awg, ampacity = 25A No derating necessary for temperature or # of current carrying conductors. c) Select 25A GB. or Fuse STOP! Section 240.4(D) says the maximum overcurrent device for #12AWG conductors is 20A. Therefore, you must use the next conductor size! Select #10Awg, ampacity = 30 Section 240.4(D) also says the maximum standard overcurrent device that can be used for #1 OAWG conductors is 30A. Therefore . . . elect 30A GB. or Fus 5) A branch circuit is made up of three phase conductors and an equipment grounding conductor enclosed in a common raceway. The branch circuit passes through a furnace room where the ambient temperature is 135 degrees F. If the branch circuit serves a 166-kVA, 480—volt, three-phase load, what is the minimum size copper conductor that can be used? Calculate the full load current: ' I9 = site/Java) = 166,000VA/x/3(480V) = 200A Satisfy integrity of terminals: 200A > 100A use 75°c column Select #3/0AWG, ampacity = 200A Satisfy integrity of insulation (derate for ambient temperature): loonducmr = 200A (0.58) = 116A 3 200A The minimum conductor ampacity = 200A/0.58 = 344A elect #500MCM, ampacity = 380 6) Nine current carrying No. 10 AWG type THWN copper conductors are installed in a single raceway running through a boiler room. The ambient temperature in the boiler room is 120 degrees F. a) What is the NEC allowable ampacity for each of these conductors? b) What is the minimum EMT conduit size that can be used? Ampacity of #1 OAWG using type THWN insulation is 35A a) Derate for ambient temperature and 9 current carrying conductors: 35A (0.75) (0.70) = b) From Table 5, Chapter 9 Area of #‘l OAWG, THWN = 0.021 ‘lsq_ in From Table 4, Chapter 9 Area of EMT conduit at 40% fill 2 9(0;0211'sq_in,) = 0.18998q_in_ Select %” EMT Area=0.213s_in. o 40% fill 7) What is the maximum number of 120-volt duplex receptacles that can be placed on a single 20-amp branch circuit? Conservativer assume the load will be continuous. (Hint: Use the receptacle loading specified in NEC Article 220-14 (I) as the basis for calculations.) From NEC 220—14(l), the minimum load computed for each receptacle can not be less than 180 VA. Calculate the maximum continuous load permitted for a 20A, 120V branch circuit: P = I*V/1.25 = (20A)(120V)/ 1.25 = 1920VA Maximum # of receptacles = 1920VA/ 180VA = ' 10.66 10 Receptacles 8) A certain piece of kitchen equipment requires a 40—amp branch circuit at 240 volts, single phase (line-to line, not line-to—neutral). The equipment grounding conductor is No. 10 AWG type THHN copper. Provide the following information about the branch circuit: a) Number of current carrying conductors comprising the branch circuit. b) Wire size assuming type THHN conductors. c) Minimum EMT raceway size. a) Per class notes, each phase conductor carries current and the equipment grounding conductor does not, therefore . . . wo current carrying conductors b) IBranch circuit 40A < 100A USG 60°C column Select#8AWG, ampacity = 40 c) From Table 5, Chapter 9 Area Of #BAWG, = 0.036631], in_ Area of #1 OAWG, THHN = 0.0211ASq_in_ Total conductor area = 2 (0.03663q_ in.) + 0.0211Sq_in_ = 0.09438q_in; From Table 4, Chapter 9 Area of EMT conduit at 40% fill 2 0.09438q_ in Select 1/2” EMT Area = 0.1228_in, o 40% fill ...
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CEE 496homework2sols - CEE 496 —- ELECTRICAL BUILDlNG...

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