Optimizing Performance: Solved Problems on DC Motor Efficiency

Problem-1

A shunt generator delivers 195 A at terminal potential difference of 250 V. the armature resistance and shunt field resistance are 0.02 \(\Omega\) and 50 \(\Omega\), respectively. The iron and friction losses equal 950 W. Find

E.M.F generated
Cu losses
output of the prime mover
commercial, mechanical and electrical efficiencies.

Solution-1

\(\quad I_{s h}=250 / 50=5 \mathrm{~A}\)
\(I_{a}=195+5=200 \mathrm{~A}\)
Armature voltage drop \(=I R_{a}=200 \times 0.02=4 \mathrm{~V}\)
Generated e.m.f. =250+4=254 V
Armature Cu loss = \(I_{a}^{2} R_{a}=200^{2} \times 0.02=800 \mathrm{~W}\)
Shunt Cu loss \(=V . I_{\text {sh }}=250 \times 5=1250 \mathrm{~W}\)
Total Cu loss \(=1250+800=2050 \mathrm{~W}\)
Stray losses =950 W
Total losses =2050+950=3000 W
Output =\(250 \times 195=48.750 \mathrm{~W}\)
Input =48,750+3000=51750 W
Output of prime mover =51,750 W
Generator input =51,750 W
Stray losses =950 W

Electrical power produced in armature \(=51,750-950=50,800\)
Mechanical efficiency \(\eta_{m} =(50,800 / 51.750) \times 100=98.2 \%\)
Electrical or Cu losses \(=2050 \mathrm{~W}\)
Electrical efficiency\(\eta_{e} =\frac{48,750}{48,750+2,050} \times 100=95.9 \%\)
Commercial efficiency \(\eta_{c} =(48.750 / 51,750) \times 100=94.2 \%\)

Problem-2

A shunt generator has a F.L current of 196 A at 220 V. The stray losses are 720 W and the shunt field coil resistance is 55 \(\Omega\). If it has a F.L efficiency of 88%, find the armature resistance. Also, find the load current corresponding to maximum efficiency.

Solution-2

Output = \(220 \times 196=43120~\mathrm{W}\)
\(\eta=88\%\) (overall efficiency)
Total losses = 49000-43120=5880 W
Shunt field current = 220/55=4 A
\(I_a=196+4=200~\mathrm{A}\)
Shunt Cu loss = \(220\times 4=880~\mathrm{W}\)
Stray losses = 720 W
Constant losses = 880+720=1600

Armature Cu loss = 5880-1600=4280 W
Therefore, \[\begin{aligned} I_a^2R_a & = 4280\\ 200^2R_a & =4280\\ R_a & = 4280/200^2=0.107~\Omega \end{aligned}\]
For maximum efficiency, \[\begin{aligned} I^2 R_{a}&=\text { constant losses }=1,600 \mathrm{~W}\\ I&=\sqrt{1,600 / 0.107}=122.34 \mathrm{~A} \end{aligned}\]

Problem-3

A long-shunt generator running at 1000 rpm supplies 22 KW at a terminal voltage of 220 V. The resistance of armature, shunt field and the series field are 0.05, 110 and 0.06 \(\Omega\), respectively. The overall efficiency at the above load is 88%. Find

Cu losses
iron and friction losses
the torque exerted by the prime mover.

Solution-3

\(I_{s h} =220 / 110=2 \mathrm{~A}\)
\(I =22,000 / 220=100 \mathrm{~A}\)
\(I_{a} =102 \mathrm{~A}\)
Drop in series field winding \(=102 \times 0.06=6.12 \mathrm{~V}\)
\(I_{a}^{2} R_{a}=102^{2} \times 0.05=520.2 \mathrm{~W}\)

Series field loss \(=102^{2} \times 0.06=624.3 \mathrm{~W}\)
Shunt field loss \(=4 \times 110=440 \mathrm{~W}\)
Total Cu losses \(=520.2+624.3+440=1584.5 \mathrm{~W}\)
Output \(=22,000 \mathrm{~W}\)
Input \(=22,000 / 0.88=25,000 \mathrm{~W}\)
Total losses \(=25,000-22,000=3,000 \mathrm{~W}\)

Iron and friction losses \(=3,000-1,584.5=1,415.5 \mathrm{~W}\)
\[\begin{aligned} T \times \dfrac{2 \pi N}{60}&=25,000 \\ T&=\dfrac{25,000 \times 60}{1,000 \times 6.284}=238.74 \mathrm{~N}-\mathrm{m} \end{aligned}\]

Problem-4

A 4-pole dc generator is delivering 20 A to a load of 10 \(\Omega\). If the armature resistance is 0.5 \(\Omega\) and the shunt field resistance is 50 \(\Omega\), calculate the induced emf and the efficiency of the machine. Allow a drop of 1 V per brush

Solution-4

Terminal voltage =\(20 \times 10=200~ \mathrm{V}\)
\(I_{\mathrm{sh}}=200 / 50=4 \mathrm{~A}\)
\(I_{a}=20+4=24 \mathrm{~A}\)
\(I_{a} R_{a}=24 \times 0.5=12 \mathrm{~V}\)
Brush drop =\(2 \times 1=2 \mathrm{~V}\)
\(E_{g}=200+12+2=214 \mathrm{~V}\)

Since iron and friction losses are not given, only electrical efficiency of the machine can be found out.
Total power generated in the armature =\(214 \times 24=5.136 \mathrm{~W}\)
Useful output =\(200 \times 20=4.000 \mathrm{~W}\)
\(\eta_{e} =4,000 / 5,136=0.779 \text { or } 77.9 \%\)

Problem-5

A long-shunt compound-wound generator gives 240 volts at F.L output of 100 A. The resistance of various windings of the machine are: armature (including brush contact) 0.1 \(\Omega\) , series field 0.02 \(\Omega\), inter-pole field 0.025 \(\Omega\), shunt field (including regulating resistance) 100 \(\Omega\). The iron loss at F.L is 1000 W; windage and friction losses total 500 W. Calculate F.L efficiency of the machine.

Solution-5

Output =\(240 \times 100=24,000 \mathrm{~W}\)
Total armature circuit resistance \(=0.1+0.02+0.025=0.145 \Omega\)
\(I_{\text {sh }}=240 / 100=2.4 \mathrm{~A}\)
\(I_{a}=100+2.4=102.4 \mathrm{~A}\)
Armature circuit copper loss \(=102.4^{2} \times 0.145=1,521 \mathrm{~W}\)
Shunt field copper loss \(=2.4 \times 240=576 \mathrm{~W}\)
Iron loss \(=1000 \mathrm{~W}\)
Friction loss \(=500 \mathrm{~W}\)
Total loss \(=1,521+1.500+576=3.597 \mathrm{~W}\)
\(\eta=\dfrac{24,000}{24.000+3.597}=0.87=87 \%\)