Tutorial Sheet - 5
1) A 1-phase transformer has 400 primary and 1000 secondary turns. The net cross-sectional area of the 2
core is 60cm .If the primary winding be connected to a 50-Hz at 500V, calculate (a) the peak value of the flux density in the core, and (b) the voltage induced in the secondary winding. 2
[ (a) 0.94Wb/m ; (b) 1250V ] 2) The required no-load ratio in a 1- phase, 50-Hz, core-type transformer is 6000/250V. Find the number of turns in each winding if the flux is to be about 0.06 Wb. [ 480,20 ] 3) A 1-phase, 50Hz, core-type transformer has square cores of 20-cm side. The permissible maximum 2
density is 1Wb/m .Calculate the numbers of turns per limb on the high- and low- voltage sides for a 3000/220-V ratio. [191,1] 4) A single –phase auto transformer is to transmit a power of WkW from the primary network at E 1 volts to the secondary E2 volts. Determine the rating of each section of the winding and compare it with the rating of a two – winding winding transformer, assuming an efficiency of ŋ in both cases. cases. Evaluate for E2=90% e1, W=100kW and ŋ=0.97. Ignore magnetizing current. [Primary 103.2/10.32, Secondary 100/7.22.] 5) An 11500/2300-V transformer is rated at 100kVA as a 2-winding transformer .If the two winding are connected in series to form an auto-transformer, what will be the voltage ratio and output. [ 13.8/11.5 kV,600 kVA or 13.8/2.8 kV,120kVA] 6) The efficiency, at unity power factor, of a 6600/884-V, 200-kVA, 1-phase transformer is98% both at full load and half load. The power factor on no load is 0.2 and the full load regulation at a lagging power factor of 0.8is 4%. Draw the equivalent circuit referred to the l.v side, and insert all values. values. [ R 0= 108 Ω; X0=22.1 Ω;X=0.036 Ω] 7) A 100-kVA, 6600/330-V 50-Hz, single-phase transformer took 10A and 436W at 100V in a shortcircuit test,, the figures referring to the high-voltage side. Calculate the voltage to be applied to the high-voltage side on full load at power factor 0.8 lagging when the secondary terminal voltage is 330 V. [6784 V.] 8) A 110-kVA, 1-phase transformer has a ratio of 11000/440 V. The iron loss measured on open-circuit is 1100 W. With the secondary secondary winding short-circuited, short-circuited, a voltage voltage of 500V at normal frequency applied to the primary produces full-load current, the wattmeter reading 1000 W. Calculate (a) the
secondary terminal voltage,(b) the efficiency, when a current of 250 V at a lagging power factor of 0.8 is taken by a load connected to the low-voltage terminals, the primary voltage being 11000 V. [ (a) 425 V ; (b) 97.6% ] 9) Find the efficiency of a 150-kVa transformer at 25%, 33% and 100% full load, (a)at unity power factor ,(b) at power factor 0.8 lagging ,if the copper loss is 1600 W at full load and the iron loss is 1400 W . Ignore the effects of temperature rise and magnetizing current. [(a) 96.15, 96.94, 98.04 %; (b) 95.23, 96.21, 97.56%.] 10) In a 25-kVA , 2000/200 –V transformer the iron and copper losses are 350 and 400 W respectively. Calculate the efficiency on unity power factor at (a)full load and (b)half load , (c) Determine the load for maximum efficiency and the iron and the copper loss in this case. [ (a) 97.1% ; (b) 96.5% ; (c) 23.4kVA,350 W each.] 11) A transformer has its maximum efficiency of 0.98 at 15 kVA at unity power factor. During the day it is loaded as following :-12 hours- 2 kW at power factor 0.5 6
hours-12 kW at power factor 0.8
6
hours-18 kW at power factor 0.9
Find the “all-day” efficiency (output in kWh/inpu t in kWh). [0.97] 12) A 100-kVA, 1000/10000-V, 50-c/s transformer has an iron loss is 1200 W. The copper loss with 6 A in the high-voltage winding is 500 W. Calculate the efficiency at (i)25% (ii)50% (iii)100 % of normal load, for power factor of (a)1.0 and (b)0.8, the output terminal voltage being maintained at 10000 V . Find also the load for maximum efficiency at both power factors. [ (a)(i) 95.10%,(ii)97.00 %,(iii) 97.47%; 93kVA; (b)(i) 93.95%,(ii) 96.27%,(iii) 96.86%; 93kVA ]
13) Calculate the values of R 0, X0, R t and Xt in the diagram for the equivalent circuit of a 1-phase, 2 – kVA, 200/400-V,50-Hz transformer of which the following are test results: Open circuit :200 V 0.7 A, 70W on low –voltage (primary) side. Short-circuit: 15 V, 10 A, 80W on high-voltage (secondary) side.
[571, 330, 0.2 and 0.317Ω ]
14) The diagram shows the equivalent circuit for a 1-phase transformer. Figures give resistance and reactances in ohms in terms of the primary side. The ratio of secondary to primary turns is 10 and the load is inductive. Find (a) the secondary terminal voltage; (b) the primary current; (c) the efficiency. [(a)1860 V; (b) 25.9 V; (c) 94.9%.]
15) The equivalent circuit shown refers to a 200/400- V 1-phase, 50-Hz, 4-kVA, transformer, the values given being reduced to the low-voltage side. For a high-voltage current of 10 A at a lagging power factor of 0.8, calculate (a) the low–voltage input current; (b) the efficiency; (c) the voltage at the terminals of the high-voltage side.
[(a)20.65 A ; (b) 96.08% ;(c)386 V.]
