Question

1) A separately excited shunt generator develops 400V at 1900 RPM.
a) what will the voltage be at 1200 RPM
b) what will be the voltage if the field is reduced by 8%

2) 16.5 HP is needed to drive a generator rated at 12.3 KW, 230V, 50A at 1680 RPM. Ra=.32 ohm. Determine:

a) efficiency
b) torque ( use T= 9.55 P/n)
c) heat loss
d) armature voltage

3) At full load a self excited shunt generator delivers 24KW, 680V, 1750 RPM, Ra= 0.92 ohm Rf= 400 ohm, Rv is set at 20 ohm.
a) Find Ea and Ia
b) if Rv is increased to cause a 10% decrease in If find the new Eo and Io

Answer 1

the = ZNb or separately excited . 60A EAN E = kn = 400 = k1900- Ez = 5,1200 EL - 1200 400 1900- > E2= 252.63V Exo E = K20 → Ez = 0.92E, = 0.92 x 400 · - 368 V Pin: 16:5 HPS 12.31kW. - VA= 230V 1. Ia = 50A, Ra= 0.32.2 N=1680 RPM = E= 353.7V al y = fout = Pin- Pandos x 100% Pin Pin - = (1 - Tºda | X100 • -ЖА0 L 12.3x103) - = 93.57 CSScanned with CamScanner

—b) Torque : 9.55 € =9.55 x Esa = 100.53 Nm. c) Heat loss = Pcu, loss = I²Ra=800w (3) 2 Pout = 24kW = Vt la Ia = 35.3A Vt = 680V N=1750RPM Ra=0.922 Br = 4002 Ry=202 a) Ea = Vt + Ja ha = 912.50 - Iq = 35.3 A b) Rva . If & 10% Id': Vti Ea - 42.14 A Ra Scanned with CamScanner Ea' b = 0.9% Ea =641.23VEM travelclinic