A six-pole lap-wound armature has 840 conductors and a flux per pole of 0.018 Wb Calculate the emf generated, when the machine is running at 600 rpm. \[\begin{aligned} P&=6 \\ \phi& =0.018~\text{ Wb} \\ N &=600\text{rpm} \\ Z&=840 \\ A & = P = 6~\quad \Leftarrow~\text{Lap wound} \\ E_g & =\frac{\phi ZNP}{60A}=\frac{0.018\times840\times600\times6}{60\times6}=151.2~\text{V} \end{aligned}\]
A six-pole, 2-pole wave connected armature has 300 conductors and runs at 1000 rpm. The emf generated on the open circuit is 400 V. Find the useful flux per pole \[\begin{aligned} P&=6 \\ A&=2 ~\quad \Leftarrow~\text{Wave wound} \\ Z&=300 \\ N& =1000\mathrm{rpm} \\ E_{g}&=400\mathrm{V} =\frac{\phi Z N P}{60A} \\ \Rightarrow ~ \phi &={\frac{60E_{g}A}{Z N P}}={\frac{60\times400\times2}{300\times1000\times6}}=0.0267{\mathrm{Wb}} \end{aligned}\]
A lap-wound dc shunt generator having 80 slots with 10 conductors per slot generates at no load an emf of 400 V running at 1000 rpm. At what speed should it be rotated to generate a voltage of 220 V on open circuit ?
\[\begin{aligned} \text{Number of slots on armature } & = 80\\ \text{ Conductors per slot} & =10 \end{aligned}\] \[\begin{aligned} E_g & =400~\operatorname{V} \\ N & =1000~\text{rpm} \end{aligned}\]
First part \[\begin{aligned} Z& =80\times10=800 \\ E_g & =\frac{\phi ZNP}{60A}\qquad \left(A =P\right)\\ \Rightarrow~400 & =\frac{\phi\times1000\times800}{60}\\ \Rightarrow~\phi &= 0.03 ~\text{Wb} \end{aligned}\] Second part
\[\begin{aligned} E_g& =220~\text{V} \\ \phi & =0.03~\text{Wb} \\ \Rightarrow~ 220& =\frac{0.03\times N\times800}{60} \\ \Rightarrow~N& =550~\text{rpm} \end{aligned}\]
A 4-pole dc shunt generator with lap-connected armature supplies a load of \(100 \mathrm{~A}\) at \(200 \mathrm{~V}\). The armature resistance is \(0.1 \Omega\) and the shunt-field resistance is \(80 \Omega\). Find (i) total armature current, (ii) current per armature path, and (iii) emf generated. Assume a brush contact drop of \(2 \mathrm{~V}\).
Given data:
\[\begin{aligned} P & =4 \\ V &=200 \mathrm{~V} \\ R_f &=80 \Omega \end{aligned}\] \[\begin{aligned} R_a & =0.1 \Omega \\ I_L &=100 \mathrm{~A} \end{aligned}\]
(i) Total armature current \[\begin{aligned} & I_f=\frac{V}{R_f}=\frac{200}{80}=2.5 \mathrm{~A} \\ & I_a=I_L+I_f=100+2.5=102.5 \mathrm{~A} \end{aligned}\]
(ii) Current per armature path
\[\begin{aligned} \text{For lap-wound} ~&A=P=4 \\ I_a &=102.5 \mathrm{~A}\\ \text { Current per armature path } & =\frac{102.5}{4}=25.625 \mathrm{~A} \end{aligned}\]
(iii) EMF generated
\[\begin{aligned} E_g & =V+I_a R_a+\text { Brush Drop }\\ &=200+(102.5\times 0.1)+2\\ &=212.25 \mathrm{~V} \end{aligned}\]
The armature of a four-pole, lap-wound shunt generator has 120 slots with 4 conductors per slot. The flux per pole is \(0.05 \mathrm{~Wb}\). The armature resistance is \(0.05 \Omega\) and the shunt-field resistance is \(50 \Omega\). Find the speed of the machine when supplying \(450 \mathrm{~A}\) at a terminal voltage of \(250 \mathrm{~V}\).
\[\begin{aligned} V & =250 \mathrm{~V} \\ I_L & =450 \mathrm{~A}\\ R_f &=50 \Omega \\ \end{aligned}\] \[\begin{aligned} R_a & =0.05 \Omega \\ P & =4 \\ \phi & =0.05 \mathrm{~Wb} \end{aligned}\]
\[\begin{aligned} I_f &={\frac{V}{R_{f}}}={\frac{250}{50}}=5\mathrm{A} \\ I_a &=I_L+I_f=450+5=455\text{A} \\ E_g &=V+I_a R_a=250+455\times0.05=272.75\text{V} \end{aligned}\]
\[\begin{aligned} \text{Number of slots on armature} & = 120\\ \text{Conductors per slot} & =4\\ \text{Total number armature conductors}~Z&=120\times4=480\\ \text{For lap wound generator}~A & =P=4 \\ E_g & =\frac{\phi ZNP}{60\mathrm{A}}\\ \Rightarrow ~272.75 & =\frac{0.05\times480\times N\times4}{60\times4}\\ \Rightarrow ~N & =682\text{rpm} \end{aligned}\]
A 230 V dc shunt machine has an armature resistance of 0.5 \(\Omega\) and a field resistance of 115 \(\Omega\). If this machine is connected to 230 V supply mains, find the ratio of speed as generator to the speed as a motor: The line current in each is 40 A. \[\begin{aligned} V& =230\text{V} \qquad I_L =40\text{A} \qquad R_a =0.5 \Omega \qquad R_f =115 \Omega \end{aligned}\]
Generator operation \[\begin{aligned} I_L& =40\text{A} \\ I_{f}& =\frac{V}{R_f}=\frac{230}{115}=2\text{A} \\ I_a &=I_L+I_f\\ &=40+2=42~\text{A} \\ E_g &=V+I_a R_a\\ &=230+42\times0.5=251~\text{V} \end{aligned}\] Motor operation \[\begin{aligned} I_L &=40\text{A} \\ I_f &=2\mathrm A \\ I_a &=I_L-I_f\\ &=40-2=38~\text{A} \\ E_b &=V-I_a R_a\\ &=230-38\times0.5=211~\text{V} \end{aligned}\]
\[\begin{aligned} E & =\frac{\phi ZNP}{60} \\ E& \propto N \\ \frac{E_b}{E_g}& =\frac{N_2}{N_1} \\ \frac{211}{251}& =\frac{N_2}{N_1} \\ \frac{N_2}{N_1}& =1.1896 \end{aligned}\]
A short-shunt compound generator supplies 200 A at 100 V. The resistance of armature, series field and shunt field is respectively, 0.04,0.03 and 60 \(\Omega\). Find the emf generated.
\[\begin{aligned} V&=100\mathrm{V} \\ R_f&=60 \Omega \\ I_L&=200\mathrm A \end{aligned}\] \[\begin{aligned} R_a&=0.04 \Omega \\ R_s &= 0.03 \Omega \end{aligned}\]
\(\text{Voltage drop in series field winding}=I_LR_s=200\times0.03=6~\text{V}\) \[\begin{aligned} I_f &={\frac{V+I_{L}R_{s}}{R_{f}}}={\frac{100+6}{60}} =1.77\mathrm A \\ I_a &=I_L+I_f =200+1.77=201.77~\text{A}\\ E_g& =V+I_L R_s+I_a R_a \\ &=100+6+201.77\times0.04=114.07~\text{V} \end{aligned}\]
A 120 V dc shunt motor draws a current of 200 A.The armature resistance is 0.02 \(\Omega\) and shunt field resistance 30 \(\Omega\). Find the back emf. If the lap wound armature has 90 slots with 4 conductors per slot at what speed will the motor run when the flux per pole is 0.04 Wb ?
\[\begin{aligned} V&=120~\text{V} \\ R_a&=0.02 ~\Omega \\ I_L&=200~\mathrm{A} \end{aligned}\] \[\begin{aligned} \phi&=0.04~\text{Wb} \\ R_{f}&=30~\Omega \end{aligned}\]
First part
\[\begin{aligned} I_f &=\frac{V}{R_f}=\frac{120}{30}=4\mathrm{A} \\ I_a& =I_L-I_f=200-4=196\text{A} \\ E_b &=V-I_a R_a=120-196\times0.02=116.08~\text{V} \end{aligned}\]
Second part
\[\begin{aligned} \text{ For-lap wound armature}~A&=P\\ \text{Number of slots on the armature}& =90\\ \text{Total armature conductors}~& Z=90\times4=360 \\ E_b & =\frac{\phi Z N P}{60A} \\ 116.08& =\frac{0.04\times360\times N}{60} \\ N & =\frac{116.08\times60}{0.04\times360}=484\text{rpm} \end{aligned}\]