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RICH DAD POOR DAD BOOK REVIEW

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  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...
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POWER FLOW, MAXIMUM POWER OUTPUT and MAXIMUM EFFICIENCY IN A DC MACHINE

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Power flow

 The flow of power from mechanical to electrical domain in a generator and electrical to mechanical domain in a motor occurs as shown in the diagram below:

Power Flow Diagram



 For Series and Separately excited machines: IL = Ia
For Shunt Generator: IL = Ia - If
For Shunt Motor:  IL = Ia + If  

 NOTE:- The amount of conductor and iron materials required for a Machine of given rating ∝ 1 / speed of machine     


 Maximum Power Output

 To achieve maximum power output in a DC Machine the following condition must be satisfied.

 Motor

 Developed Power, Pa = EaIa = (V - IaRa)Ia = VIa - Ia^2 Ra 
For maximum output,

dPa/dIa = V - 2IaRa = 0

Ia = V / 2Ra

 And  Ea = V - 2IaRa

 Ea = V/2

Efficiency, 𝜂 = Pa / VIa = EaIa / VIa = 1/2 = 50%

Suppose, Ra = 0.03pu
for maximum output

Ia = 100% / (2*3%) ≈ 17 times full load

Due to such high current, windings will be burnt immediately. Hence, maximum power output is undesirable condition.

Generator

 Output Power, Pa = VaIa
                                      = (Ea - IaRa)Ia
                                      = EaIa - Ia^2 Ra

 For maximum output power,  dPa / dIa = Ea - 2IaRa = 0

Ia = Ea / 2Ra

 Va = Ea - IaRa = Ea/2

Efficiency, 𝜂 = VaIa / EaIa = 1/2 = 50%

Here, also current will be high due to small Armature Resistance and it can damage the windings.

So, the desirable condition is maximum efficiency so that maximum part of Input Power is converted to Output Power.


Maximum Efficiency

 Assuming a Short Shunt Machine with the following parameters:




VBD = Brush Drop Voltage
Pk = No-Load Rotational Losses
Ra = Armature Resistance
Rf = Shunt Field Resistance
Rse = Series Field Resistance

If Machine operates as a generator,
Ia = IL + If

 Where IL is current delivered to load
 If is current in Shunt Field Winding
Ia is Armature Current

 𝜂 = Vt IL / (Vt Ia + Ia^2 Ra + VBD Ia + Pk + IL^2 Rse + If^2 Rf)

For Maximum Efficiency, d𝜂/dIL = 0

Vt(Vt Ia + Ia^2 Ra + VBD Ia + Pk + IL^2 Rse + If^2 Rf) - Vt IL(Vt + 2IaRa + VBD + 2ILRse) = 0

Substitute Ia = IL + If

 We get, IL^2 (Rse + Ra) = If^2 Rf + Pk + VBDIf + If^2 Ra

 which implies, Constant Losses = Variable Cu Losses

This can be applied to Shunt Machines by equating Rse to zero and for Series Machines by equating If and Rf to zero.



                     

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