Posts

Showing posts with the label Numerical

previous post

RICH DAD POOR DAD BOOK REVIEW

-
  RICH  DAD  POOR  DAD  BOOK REVIEW This is amazing book.   • Explodes the myth that you need to earn a high income to become rich • Challenges the belief that your house is an asset • Shows parents why they can't rely on the school system to teach their kids about money • Defines once and for all an asset and a liability • Teaches you what to teach your kids about money for their future financial success It's been nearly 25 years since Robert Kiyosaki’s  Rich Dad Poor Dad  first made waves in the Personal Finance arena. It has since become the #1 Personal Finance book of all time... translated into dozens of languages and sold around the world. Rich Dad Poor Dad  is Robert's story of growing up with two dads — his real father and the father of his best friend, his rich dad — and the ways in which both men shaped his thoughts about money and investing. The book explodes the myth that you need to earn a high income to be rich and explains...
Post a Comment
Read more

Questions of Torque-Power of DC Machine

-
Problem: A Separately excited d.c motor has an armature resistance of 0.5 ohm. It runs off a 250 V d.c supply drawing an armature current of 20 A at 1,500 rpm. The torque developed for an armature current of 10 A will be______for the same field current. Solution: V t = 250 V; R a = 0.5 Ω; N = 1500 rpm; I a = 20 A Induced EMF, E a = V t - I a R a = 250 - 20*0.5 = 240 V Developed Power, P = E a I a = 240*20 = 4800 W Developed Torque, T = P / Ꞷ = 4800 / ( 2𝜋*1500 / 60 ) = 30.557 N-m Assuming flux remains constant,  T ∝ I a T 2 / T 1 = I a2 / I a1 T 2 / 30.557 = 10 /  20 T 2 = 0.5 * 30.557 = 15.278 N-m Problem: A 5 kw, 2000 V d.c shunt motor has an armature resistance of 1 ohm and shunt field resistance of 100 ohms. At no-load, the motor draws 6 A from a 200 V supply and runs at 1000 rpm. The rotational loss of the machine is ____W and the no load torque is _____N-m. Solution: Under No-Load conditions there is no shaft output of the DC Machine and hence the...
1 comment
Read more

Questions of DC Machine

-
Solved Examples Problem: A 4 pole generator supplies a current of 143 A. It has 492 conductors lap connected and delivering full load, brushes are given a shift of lead to 10°. calculate demagnetizing ampere turns per pole? The field winding is shunt connected and takes 10 A. Find no. of extra shunt turns necessary to neutralize this demagnetization. Solution: Current in armature, I a = 143 + 10 = 153 A ꞵ elect  = P/2 * 10° = 4/2 * 10° = 20° Z = 492 A = P = 4,  (A=P for lap winding) F ar (demagnetizing) = {(Z/2) / P} * {(2ꞵ elect ) / 180°} * {I a  / A}                                                                = {492 / (2*9)} * {(2*20°) / 180°} * {153 / 4}                                         ...
Post a Comment
Read more