Question

This experiment will give information regarding the behavior of the MUT induction machine under full load conditions. In this experiment, rather than having a separate mechanical load for the MUT induction machine, the VSD controlled induction machine will act as a mechanical load. The VSD induction machine will act in torque control mode to ensure that a counteracting full load torque is applied to the MUT induction motor. It's rated at 5.5kW. When the motor deliver rated power output, if it connected to a 660V (VLL) voltage, it will draw 6.2A at 0.886 lagging power factor. If the supplied frequency is 50 Hz, its full load mechanical speed is 1460 r.p.m.

VLL = 422 V
IL = 13.6 A
pf = 0.830
Motor speed = -1470 rpm

a)What is the real power being delivered to the machine in this test?
b)What is the reactive power being delivered to the machine in this test?
c)Using the cantilever (approximate) equivalent circuit, determine the output power delivered by the MUT induction machine.
d) What is the machine efficiency?
e) Determine the electrical and mechanical torque.

Answer 1

a.) Real power being delivered to machine in the test=P =

Given, during test: Line voltage=422V; Line current =13.6A; pf= $\cos\phi$ =0.830

P = 8250.44 W or 8.25 kW

b.) Reactive power being delivered to machine in the test=Q = $\sqrt{3}V_LI_L\sin\phi$

Using other values same as in above part, $\sin\phi=\sqrt{1-\cos^2\phi}=\sqrt{1-(0.83)^2}=0.558$

Q = 5546.68 W or 5.55 kW

c.) The data is given for two voltages in the quesion:

1.Rated: Line voltage=660V; Line current =6.2A; pf= $\cos\phi$ =0.886, $n_1$ =1460 r.p.m., $P_{rated}$ =5.5kW, Torque, $T_1$

$T_1=\frac{P_{rated} }{2\pi n}=35.99 N-m$

2.Test: Line voltage=422V; Line current =13.6A; pf= $\cos\phi$ =0.830, $n_2$ =-1470r.p.m., $P_{out}$ , $T_2$

Since, the motor is being operated in torque control region and we know that: $T\propto V^2$ .

So, we can find Torque during test as follows:

$\frac{T_2}{T_1}= \frac{V_2^2}{V_1^2}\Rightarrow T_2=\frac{422^2}{660^2}\times 35.99=14.71Nm$

$P_{out}= T_2w_2=14.71\times \frac{1470}{60}\times 2\pi =2.263kW$

d.) $\eta=\frac{Output Power}{Input Power}\times 100=\frac{2.263}{8.25}\times 100=27.4\%$ (During test)

Under rated conditions:

$\eta=\frac{Output Power}{Input Power}\times 100=\frac{5500}{\sqrt{3}\times660\times 6.2\times 0.886}\times 100=89.6\%$ (Under rated conditions)

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