Question

Two 200 MVA alternators operate in parallel. The frequency drops in first machine from 50 Hz at no load to 48 Hz at full load, where as in case of other machine, the frequency drops from 50 Hz to 47 Hz under the same conditions. If the two machines has to share a total load of 300 MW, then the maximum load at unity p.f. which can be delivered by the two machines without overloading any of them is

• A. 380 MW
• B. 303.33 MW
• C. 333.33 MW
• D. 200 MW

Answer 1

The correct option is C 333.33 MW



New frequency of operation of 200 MW alternator,

$f_1=50-\frac{2}{200}{P}_1$

and $f_2=50-\frac{3}{200}{P}_2$

Total load, $P_1+P_2=300MW$ ...(i)

Units operated in parallel so,
$f_1=f_2=f$

$50-\frac{2}{200}{P}_1=50-\frac{3}{200}{P}_2$

$2P_1-3P_2=0$ ......(ii)

From equations (i) and (ii), we get

$P_1=180MW$

$P_2=120MW$

Machine (i) which has better speed regulation will be loaded first to its full load rating, so it will operate on maximum load of 200 MW

$2P_1-3P_2=0$ ... (iii)

$2\times 200-3P_2=0$

$P_2=\frac{400}{3}=133.33MW$

Total power delivered by two machine without overloading
$P_1+P_2=200+133.33=333.33MW$