A 30kw , 500 volt shunt generator has its driving machine uncoupled after it has reached normal operating temperature. It is then run as motor drawing an armature current of 4A from a 500 volt supply. The machine has an armature and shunt field resistance of 0,4 ohms and 100 ohms respectively. Calculate:
a.The full load efficiency as a generator, and
b. The efficiency of the machine when running as a motor supplying a load and drawing a current of 65A from the constant voltage busbars.
To calculate the full load efficiency of the shunt generator, we need to determine the input power and the output power.
Let's start with the input power. The input power to the generator can be calculated using the formula:
Input Power (Pg) = Voltage (V) * Current (I)
Given that the voltage is 500 volts and the current is 65 amps, the input power is:
Pg = 500 V * 65 A = 32,500 watts
Next, we need to calculate the output power. The output power is the product of the generated power and the generator efficiency.
Generated Power (Pg) = Output Power (Po) / Generator Efficiency (ηg)
However, to calculate the output power, we need to determine the generated power. The generated power is given by the formula:
Generated Power (Pg) = Voltage (V) * Current (I)
Given that the voltage is 500 volts and the current is 65 amps, the generated power is:
Pg = 500 V * 65 A = 32,500 watts
Now, we can calculate the generator efficiency using the formula:
Generator Efficiency (ηg) = Output Power (Po) / Input Power (Pg)
Substituting the known values, we have:
ηg = Po / Pg
Since we don't have the output power, we'll calculate it in the next step.
To calculate the output power, we need to find the losses in the generator. The losses include copper losses (I^2*R) and field losses (If^2*Rf).
Copper Losses (Pc) = Armature Current (I) ^2 * Armature Resistance (Ra)
Pc = 65 A^2 * 0.4 Ω = 1690 watts
Field Losses (Pf) = Shunt Field Current (If) ^2 * Shunt Field Resistance (Rf)
Pf = 65 A^2 * 100 Ω = 42,250 watts
Total Losses (Ploss) = Copper Losses (Pc) + Field Losses (Pf)
Ploss = 1690 W + 42,250 W = 43,940 watts
We can now calculate the output power:
Output Power (Po) = Generated Power (Pg) - Total Losses (Ploss)
Po = 32,500 W - 43,940 W = -11,440 watts
The output power appears to be negative, which indicates that the machine is not capable of supplying the load and is running at a loss. Hence, the generator is not able to maintain the desired output power at a load of 65A.
For Part (a), we are required to calculate the full load efficiency as a generator. However, since the generator is not operating efficiently at this load, we cannot calculate this value.
Moving on to Part (b), to calculate the efficiency of the machine when running as a motor supplying a load and drawing a current of 65A, we need to determine the losses and the input power.
Input Power (Pm) = Voltage (V) * Current (I)
Given that the voltage is 500 volts and the current is 65 amps, the input power is:
Pm = 500 V * 65 A = 32,500 watts
Next, we need to calculate the losses in the motor. The losses include copper losses (I^2*R) and field losses (If^2*Rf).
Copper Losses (Pc) = Armature Current (I) ^2 * Armature Resistance (Ra)
Pc = 65 A^2 * 0.4 Ω = 1690 watts
Field Losses (Pf) = Shunt Field Current (If) ^2 * Shunt Field Resistance (Rf)
Pf = 4 A^2 * 100 Ω = 1600 watts
Total Losses (Ploss) = Copper Losses (Pc) + Field Losses (Pf)
Ploss = 1690 W + 1600 W = 3290 watts
Now, we can calculate the output power. The output power is given by:
Output Power (Pout) = Input Power (Pm) - Total Losses (Ploss)
Pout = 32,500 W - 3290 W = 29,210 watts
Finally, we can calculate the efficiency of the motor using the formula:
Motor Efficiency (ηm) = Output Power (Pout) / Input Power (Pm)
ηm = Pout / Pm
Substituting the known values, we have:
ηm = 29,210 W / 32,500 W = 0.897 = 89.7%
Therefore, the efficiency of the machine when running as a motor and supplying a load of 65A is 89.7%.