Question

can you answer this question and explain why? thanks.

A 50-Hz 10-MVA, 66/22-kV distribution transformer has a full-load efficiency of 98% at 0.8 lagging power factor. The transformer core loss is 80 kW. The transformer supplies an industrial load that has a constant demand of 5000 kW at 0.65 lagging power factor. The cost of energy is $0.1 per kWh. 1. In the following questions, assume that the voltage at the load terminal is maintained constant at 22 kV. calculations. State any other assumptions made in your Calculate the capacitor rating (in kVAr) to improve the power factor at the (a) load terminal to 0.95 lagging. Also, determine the reduction in the average power loss of the transformer and the saving in energy costs per year when the capacitor is in service. Assume that one year has 365 days. (10 Marks) Note: Question No. 1 continues on page 2. (b) Suppose the transformer has a 20 % continuous overloading capability. Determine the additional load (in kW and KVA ) that can be supplied through the transformer following the installation of the capacitor if the additional load absorbs power at 0.8 lagging power factor. 9071623 92 (10 Marks) What is load shedding? Briefly discuss what are the factors to consider in load shedding. (5 Marks)

Answer 1

5oz, 10 MuA, 66/21kv Transformer details Efficieng 16. at o'8 lag. 1- full load PeCose loss of Transformer = Bokw fuu load copper loss Pat Industaial load = so0 o ku . 65 gpaver fator Cost of ener =$0.L per ku at load tenminal = 22Ku Valtge to have 3 a ssume Assumptions Configuration disti bu tion transformer 1s aounded neutral delta stas with (4) SO00 tan load Reative power of laad- power factor anglez Co3'0.65= 43.450 84 5.64 KVAL Qoad Sooo tan 49.45 mpoved pouwer factor Tequunad 42coso.35 = 18190 Sc ut load leminals wifh Capautor = Qod2 Reactve poues sequired - load So00tand Qload SODO tan 1g. 19 Ruoad = 1642.94 ku AT QLoad- d = 5845.641-16 42. 949 This muuh of eautie power should be sufplied by Capacitos Supply hees nactive powes wiy be drawn. qc= 4202.698 KVA S that from

Qutput Tuli load Efticienuy= 1 98v Qutputt PeotPef (104103)o. (10x18 x0-8) +o+ Pe¢ Pef = 8 3.26 53 Kul =fuu load Copper loss Tf fuoad Curen t toxio= 37 22 #10 Tf Cument loud I At given Load Soooku 0.65 Sooo03 J3k 22103 y0.65xI Pef Copper lass at given loadz Pe1 Pe1 832653 t 6243)2 Pe, = 49691 Kku Constant assumed to be Note: Ion losses aTe

improved to o- 351g Now when load power factor Lo ad teminals= L, at dauon Cumrent Sos103 J3 (221103) *o.JS I2 I, 138.122 A Peso Copper logg at given load & Impwed power fo c tor 83.9653 Pez RER.43)2 Pe 23. 0654 PL1 o.6S Pt (a- when load is at Totak power loss PL Total power loss when load is at o.gs PL,8+ *3.0654 103.0654 129.2691-103 o6r4-62043 Ku |P1-PL eduution inaerage powerloss Pe-Per = [129.2611-103-0 6| 29.2691 Parntege P4 Teduction in power lossa 2o »37 7.

Saving in Energ caste (P 2)365+Q4 L,-PL kwh 26.204)(365) Sauing in Energy Cast kass47.04) x(0H) - So Cost- Transfomes has ao1. capauity overload Supplied b tsans former= 101»2 maximum mVA ニ12MVA= 12000 KvA Let the adddional/oad be Prl >0 BPfla^ adCoso.8- 26-36° Reactive power taken by Load- Qa= Ptanb6 66 0.15P KVAY Presenent load th capaufor takes 5000vw of oDm ansfomer acive Power &- 164-2-94 I veactue paues p2+ 2 KuA KuA So (PeSe62-34 0ns (1200 .5625P 64.4235P - 116300718-6= 0 Quadratic equation Selving this Valid So P SS16.199 76kw

Can be SSI6.19g16 addiHonal load of So with o. Continous ovesloading Ca pacity. &upplied ) Load she dding Power System the Pass of removat of loads from that Can be S o res to ed to Noomal ualue load Shedding is n pouer ystem done Nlormalily importanu le ss On those feedars aYe which for erample sacuil feecders to Testorefrequeniy are than 0tne» fee ders. dwainy aoding ahudlug to nomal value. Cut