A 400 Vshunt generator has full-load current of 200 A . Its armature resistance is 0.06 ohm, field resistance is 100 ohm and the stray losses are 2000 watt. Find the h.p. of prime-mover when it is delivering full load, and find the load for which the efficiency of the generator is maximum.

Question

Accepted Answer

[latex]\begin{array}{l} ext { The conventional circuit is shown in Fig. } 4.79 .\ ext { Shunt field current, } I_{s h}=\frac{V}{R_{s h}}=\frac{400}{100}=4 A\end{array}[/latex]

[latex]\begin{array}{l} ext { Armature current, } I_{a}=I_{L}+I_{s h}=200+4=204 A \ ext { Armature copper loss; }=I_{a}^{2} R_{a}=(204)^{2} imes 0.06=2497 W \ ext { Shunt field copper loss }=I_{s h}^{2} R_{s h}=(4)^{2} imes 100=1600 W \ ext { Total losses }=2497+1600+2000=6097 W \ ext { Output power }=V I_{L}=400 imes 200=80000 W \ ext { Input power }= ext { Output power }+ ext { losses } \=80000+6097=86097 W\end{array}[/latex]

[latex] ext { Horse power of prime-mover }=\frac{ ext { Input power }}{735.5}=\frac{86097}{735.5}=17.06 H.P. [/latex]

[latex] ext { Constant losses }= ext { stray losses }+ ext { shunt field copper loss }=2000+1600=3600 W[/latex]

Condition for maximum efficiency is,

Variable losses = constant losses.

[latex] ext { Let, } I_{L}^{\prime} ext { be the load current at which the efficiency is maximum and armature current is } I_{a}^{\prime}[/latex]

[latex] herefore \quad I_{a}^{\prime 2} R_{a}=3600 ext { or } \quad I_{a}^{\prime}=\sqrt{\frac{3600}{0.06}}=245 A[/latex]

[latex] ext { Load current, } I_{L}^{\prime}=I_{a}^{\prime}-I_{s h}=245-4=241 A[/latex]

[latex] ext { Load for which the efficiency is maximum }=I_{L}^{\prime} V=241 imes 400=96.4 kW[/latex]

Question 4.47: A 400 Vshunt generator has full-load current of 200 A . Its ...

Related Answered Questions