Question 21 May 2021 Assume a four pole, lap wound, short shunt compound (cumulative) d.c, generator with an armature resistance of 0.020, field resistance of 0.040 and a shunt resistance of 100. The armature has 120 conductors in the slots and runs at 1000rpm. The generator supplies three loads connected as follows: i. a 72kW motor and a 2.4kW geyser connected in 21 May 2021 Assume a four pole, lap wound, short shunt compound (cumulative) d.c, generator with an armature resistance of 0.020, field resistance of 0.040 and a shunt resistance of 100. The armature has 120 conductors in the slots and runs at 1000rpm. The generator supplies three loads connected as follows: i. a 72kW motor and a 2.4kW geyser connected in parallel and a ii. a 1.8kW lighting load in series with the parallel combination in (1) above iii. the motor draws a current of 30A Neglect friction, rotational losses and assume a brush drop of 1.8V then: a. calculate the armature induced voltage [10 marks] b. calculate the flux per pole [05 marks) [05 marks] c. calculate the efficiency of the machine d. calculate the voltage across the lighting load [05 marks]

given data:-P=4 lap coound short shunt wmulative compound\begin{array}{l}r_{a}=0.02 \Omega \quad r_{f}=0.04 \Omega \quad R_{s h}=10 \Omega \\z=120 \text { conductors } \quad N=1000 \mathrm{rpm}\end{array}(i) 7.2 \mathrm{~kW} motor & 2.4 \mathrm{kr} geyser connected in parallel(ii) 1.8 \mathrm{~kW} lighting loads in series with parallel lombination (i)(iii) motor draws current =30 \mathrm{~A}solution:-since the motor & geyser connected in parallel both having same roltage somotor voltase V=\frac{P}{I}=\frac{7.2 \mathrm{~kW}}{30 \mathrm{~A}}=240 \mathrm{~V}V=240 \mathrm{~V} \rightarrow so voltage across geyser is also samecurrent flow in geyser Igey =\frac{P_{g y}}{Y}=\frac{2.4 \times 10^{3}}{240}=10 \mathrm{~A}I_{I}=30 A+10 A=40 A \quad I L=40 AVoltage drop alross the series connected coad\begin{array}{l}v_{\text {series }}=\frac{P_{L}}{I L}=\frac{18 \times 10^{3}}{I_{L}=40}=45 \mathrm{~V} \\\text { series }=45 \mathrm{~V}\end{array}The voltage drop across the seves field winding\begin{aligned}V_{S f}=I_{l} R_{f} & =40 \times 0.04=1.6 \\V_{S f} & =16 V\end{aligned}voltage drop across the armsture\begin{aligned}\left(V_{a}\right) & =V_{\text {gey }}+V_{\text {series }}+V_{s f} \\& =240 v+45 v+1.6 v \\V_{a} & =286.60 v\end{aligned}The current flow in shunt field winding Ishunt =\frac{Y_{a}}{R \text { shunt }}I_{\text {shunt }}=\frac{286.60}{10}=28.66 \mathrm{~A}So total armature current \sqrt{I_{a}}=I L+I Ishunt\text { IQ=8.66A }=40+28.66=6866 \mathrm{~A}a) ar mature induced vortage F a=Y a+I a r a\begin{array}{l}E_{a}=286.6+68.66 \times 0.02=287.97 \\E_{a}=287.97 \mathrm{~V}\end{array}(b)\begin{array}{l}\text { flux/pole }(\phi)=\frac{60 A E a}{P N z} \\\phi=\frac{60 \times 4 \times 287.97}{4 \times 1000 \times 120}=0.143 \mathrm{wb} \phi=0.143 \text { weter }\end{array}(c)\begin{array}{l}\text { Eftrieny }=\frac{\text { power detivexed }}{\text { power supplied }} \times 100=\frac{(7.2 \mathrm{kw}+2.4 \mathrm{~km}+1.8 \mathrm{kw})}{E_{a} \times \text { Ia }^{2}} \\=\frac{11.4 \times 10^{3}}{287.97 \times 6866} \times 100=57.65 \% \\\end{array}(d) yoltage drop across the l19nting loid\begin{array}{l}P=V_{i} \times I_{\text {line }} \Rightarrow V_{\text {lignting }}=\frac{P}{\text { line current }}=\frac{1.8 \mathrm{~kW}}{40 \mathrm{~A}}=45 \mathrm{~V} \\V=45 \mathrm{~V}\end{array} ...