DC Machines Fundamentals: Solved Problems

Demonstrative Video

PROBLEM 01

Problem-1

Problem Statement

A long-shunt compound generator delivers a load current of 50 A at 500 V and has armature, series field and shunt field resistance of 0.05 \(\Omega\), 0.03 \(\Omega\), and 250 \(\Omega\) respectively. Calculate the generated voltage and the armature current. Allow 1 V per brush for contact drop.

Solution

Solution-1

\[\begin{aligned} I_{s h}&=500 / 250=2 \mathrm{~A}\\ I_a & =50+2=52 \mathrm{~A} \end{aligned}\]

\[=52 \times 0.03=1.56 \mathrm{~V}\]

\[\begin{aligned} I_{a} R_{a} &=52 \times 0.05=2.6 \mathrm{~V} \\ \text { Drop at brushes } &=2 \times 1=2 \mathrm{~V} \end{aligned}\]

Series winding Voltage drop

\[=500+2.6+1.56+2=506.16 \mathrm{~V}\]

\(+\)\(E_{g}=V+I_{a} R_{a}+\)

PROBLEM 02

Problem-2

Problem Statement

A short-shunt compound generator delivers a load current of 30 A at 220 V and has armature, series field and shunt field resistance of 0.05 \(\Omega\), 0.30 \(\Omega\), and 200 \(\Omega\) respectively. Calculate the induced emf and the armature current. Allow 1 V per brush for contact drop.

Solution

Solution-2

\[\begin{aligned} \text{Voltage drop in series winding}&=30 \times 0.3=9 \mathrm{~V} \\ \text{Voltage across shunt winding} &=220+9=229 \mathrm{~V}\\ I_{s h} &=229 / 200=1.145 \mathrm{~A} \\ I_{a} &=30+1.145=31.145 \mathrm{~A} \\ I_{a} R_{a} &=31.145 \times 0.05=1.56 \mathrm{~V} \\ \text { Brush drop } &=2 \times 1=2 \mathrm{~V} \end{aligned}\]

\[\begin{aligned} E_{g} &=V+\text { series drop }+\text { brush drop }+I_{a} R_{a} \\ &=220+9+2+1.56=232.56 \mathrm{~V} \end{aligned}\]

PROBLEM 03

Problem-3

Problem Statement

In a long-shunt compound generator, the terminal voltage is 230 V when generator delivers 150 A. Determine

induced emf
total power generated
distribution of this power.

Given the shunt field, series field, divertor and armature resistance are 92 \(\Omega\), 0.015 \(\Omega\), 0.03 \(\Omega\) and 0.032 \(\Omega\) respectively.

Solution

Solution-3

\[\begin{aligned} I_{sh} & = 230/92=2.5~\mathrm{A} \\ I_{a} & = 150+2.5=152.5~\mathrm{A} \end{aligned}\]

\[=0.03\times 0.015/0.045=0.01~\Omega\]

Combined resistance of series field and divertor resistance in parallel

PROBLEM 04

Problem-4

Problem Statement

The following information is given for a 300-kW, 600 V, long-shunt compound generator: Shunt field resistance = 75 \(\Omega\), armature resistance including brush resistance = 0.03 \(\Omega\), commutating field winding resistance = 0.011 \(\Omega\), series field resistance = 0.012 \(\Omega\), divertor resistance = 0.036 \(\Omega\). When the machine is delivering full load, calculate the voltage and power generated by the armature.

Solution

Solution-4

\[\begin{aligned} \text{Power output} & =300,000\mathrm{W}\\ \text{Output current} & =300,000/600 =500\mathrm{A}\\ I_{sh} & =600/75=8\mathrm{A}\\ I_{a} & =500+8=508\mathrm{A} \end{aligned}\]

\[=\dfrac{0.012\times 0.036}{0.048}=0.009~\Omega\]

combined resistance of series field and divertor resistance in parallel

PROBLEM 05

Problem-5

Problem Statement

A four-pole generator, having lap-wound armature winding has 51 slots, each slot containing 20 conductors. What will be the voltage generated in the machine when driven at 1500 rpm assuming the flux per pole to be 7 mWb?

Solution

Solution-5

\[\begin{aligned} Z & =51\times20=1020\\ A & =P=4\\ E_{g} & =\dfrac{\phi PN}{60}\times\dfrac{Z}{A}\\ & =\dfrac{7\times10^{-3}\times4\times1500}{60}\times\dfrac{1020}{4}\\ & =178.5\mathrm{V} \end{aligned}\]

PROBLEM 06

Problem-6

Problem Statement

An 8-pole dc generator has 500 armature conductors, and a useful flux of 0.05 Wb per pole. What will be the emf generated if it is lap-connected and runs at 1200 rpm? What must be the speed at which it is to be driven producing the same emf if its is wave-wound?

Solution

Solution-6

With Lap-winding:

\[\begin{aligned} P&=A=8\\ E & =\dfrac{\phi PN}{60}\times\dfrac{Z}{A}\\ & =\dfrac{0.05\times8\times1200}{60}\times\dfrac{500}{8}\\ & =500~\mathrm{V} \end{aligned}\]
\[\begin{aligned} E & =\dfrac{0.05\times8\times N}{60}\times\dfrac{500}{2}\\ \rightarrow500 & =\dfrac{5N}{3}\\ \Rightarrow N & =300~\text{rpm} \end{aligned}\]
If wave winding: