Electrical Machines Test 2

V_t = 230 V

I_a = 36 A, R_a = 0.278 Ω

DC generator:

E_g = V_t + I_a R_a

= 230 + (36) (0.278) = 240 V

\({\omega _m} = \frac{E}{{k\;{\phi _p}}} = \frac{{240}}{{\left( {212.21} \right)\left( {0.091} \right)}}\) 

= 113.09 rad/s = 1080 rpm

DC Motor:

E_b = V_t - I_aR_a

= 230 – (36) (0.278) = 220 V

\({\omega _m} = \frac{E}{{k\;{\phi _p}}} = \frac{{220}}{{\left( {212.21} \right)\left( {0.01} \right)}}\) 

= 103.67 = 990 rpm

The operating principles of electrical to mechanical and mechanical to electrical conversion devices are similar, hence, the common name electro-mechanical device. However, their structural details differ depending on their function.

Answer: C Electrical to mechanical and mechanical to electrical

As the energy storage capacity of the magnetic field is higher, it is most commonly used as coupling medium in electro-mechanical energy conversion devices.

Either electric field or magnetic field can be used, however most commonly we use magnetic field because of its greater energy storage capacity.

Energy stored is kinetic energy, since the system is of linear motion.

At maximum power condition, δ = 90°,

\({I_a} = \frac{{{{\bar E}_f} - {{\bar V}_{an}}}}{{j{X_S}}}\)

\( = \frac{{11,500\angle 90^\circ - \frac{{13,800}}{{\sqrt 3 }}\angle 0^\circ }}{{20\angle 90^\circ }}\)

Developed torque \(\left( T \right)=\left( \frac{PZ}{2\pi A} \right)\phi {{I}_{a}}\)

P is number of poles

Z is number of conductors

ϕ is flux per pole

I_a is armature current

A is number of parallel paths

Total flux = 0.14 wb

⇒ Pϕ = 0.14 wb

Current in each conductor \(=\frac{{{I}_{a}}}{A}\)

Armature ampere conductors \(=Z\left( \frac{{{I}_{a}}}{A} \right)\)

= 4500

\(\Rightarrow T=\frac{0.14\times 4500}{2\pi }=100.26~N-m/r\)

Copper loss = I²_a(R_a + R_s) = 50² × (0.1 + 0.3) = 1000 W

Power loss in brush contacts = 2 × I_a = 2 × 50 = 100 W

Friction and windage losses = 200 W

Total power loss = 1000 + 100 + 200 = 1300 W

\(efficiency = \frac{{10000}}{{10000 + 1300}} = 88.49\%\)

 therefore machine is acting as a generator and operating at lagging Pf.

\(syn.\;speed\;{N_S} = \frac{{120 \times 50}}{4} = 1500\;rpm\)

Full load speed given as 1440 rpm

\(\therefore \% slip = \frac{{{N_S} - N}}{{{N_s}}} \times 100 = \left( {\frac{{1500 - 1440}}{{1500}}} \right) \times 100 = 4\% \)

Rotor voltage frequency = s × f = 0.04 × 50 = 5%

2 cycles/sec

Synchronous speed, \({N_S} = \frac{{120f}}{p}\)

\(= \frac{{120 \times 50}}{6}\)

= 1000 rpm

Slip, \(s = \frac{{{N_S} - N}}{{{N_S}}} = \frac{{1000 - 960}}{{1000}}\)

= 0.04

Resistance of backward field \(= \frac{{\frac{{10}}{2}}}{{2 - s}}\)

\(= \frac{5}{{2 - s}} = 2.55{\rm{\Omega }}\)

Concept:

Active power is directly proportional to the load angle i.e. P ∝ δ

Reactive power is directly proportional to the voltage ∝ V

Application:

Machine A is supplying an active power to Machine B i.e.

⇒ δ_A > δ_B

Machine B is supplying reactive power to Machine A

For a synchronous motor, N = Ns for all value of torque.

1) The speed is constant, hence the load does not know that the line voltage has dropped. Therefore, the mechanical power will remain unchanged.

2) \(P = \frac{{{E_f}{V_t}}}{{{X_S}}}\sin \delta \)

P, E_f, and X_S are the same but V_t has fallen consequently sin δ must increase, which means δ increases.

3) As δ increases, the poles fall slightly behind their former position.

4) The terminal voltage is smaller than before, the motor internal voltage is bigger than the terminal voltage and as a result, the power factor will be less than unity and leading.

S_P = 2.4 ∠cos^-1 0.707

= P_P + jQ_P = 1.697 + j 1.697 MVA

The reactive power rating of synchronous motor.

Q_p = 1.697 MVAR

The motor is used to drive a new load of 0.405 MW and the motor is 91% efficient.

\({S_m} = \frac{{{P_m}}}{{0.91}} + j{Q_m}\) 

\(= \frac{{0.45}}{{0.91}} + j{Q_m}\) 

= 0.5 – j 1.697

MVA rating = 1.769 MVA

Mmf method is called optimistic method because it gives voltage regulation lower than the actual value.

Motor becomes over-excited and starts operating at leading pf with the increase in field current.

Here, I_a will increase.

A potential transformer is a device that transfers electrical power through inductive coupling. This means it uses electromagnetic fields to transfer energy from one circuit to another without direct electrical connection.

• A potential transformer is primarily used for measurement and protection in power systems.
• It provides isolation between the high-voltage circuit and the measuring instruments.
• Power transfer occurs inductively rather than conductively.

In contrast, an auto transformer can transfer power both conductively and inductively, as it has common windings between the input and output.

Concept:

The efficiency of the Transformer \((\eta ) = \frac{{XS\cos \phi }}{{XS\cos \phi + {P_i} + {X^2}{P_e}}}\)

Where, X = Fraction of load

S = Apparent power in kVA

Pi = Iron losses

Pcu = Copper losses

Maximum efficiency of transformer occurred at a fraction of load, \(X = \sqrt {\frac{{{P_i}}}{{{P_{cu}}}}}\)

In a transformer, a short circuit test is used to find copper losses and the open circuit is used to find core losses.

Calculation:

From the given table, iron losses (W_i) = 120 W

Copper losses (W_cu) = 600 W

kVA rating of the transformer at maximum efficiency \( = \sqrt {\frac{{{W_i}}}{{{W_{cu}}}}} \times kVA = \sqrt {\frac{1}{5}} \times 48 = 21.467\;kVA\)

Maximum efficiency, \(\eta = \frac{{21.467 \times {{10}^3} \times 0.8}}{{21.467 \times {{10}^3} \times 0.8 + 120 + 600{{\left( {\frac{1}{{\sqrt 5 }}} \right)}^2}}} \times 100 = 98.621\;\% \)

The leading p.f. has negative v.r. and lagging p.f. has major portion of positive voltage regulation.

Synchronous speed, \({N_S} = \frac{{120\;f}}{P}\)

\(\Rightarrow {N_s} = \frac{{120\; \times \;50}}{6} = 1000\;rpm\)

N_r = (1 - s) N_s = (1 – 0.04) 1000 = 960 rpm

Output power \(= {T_{sh}} \times \frac{{2\pi N}}{{60}}\)

\(= 2\pi \times \frac{{960}}{{60}} \times 149.3 = 15\;kW\)

Friction and windage losses = 200 W

Rotor gross output = 15,200 W

Rotor input \(= 15,200 \times \frac{{1000}}{{960}} = 15,833\;W\)

Stator cu and iron losses = 1620 W

Stator input = 15,833 + 1620 = 17,453 W

Efficiency \(= \frac{{Output}}{{input}} = \frac{{15000}}{{17453}} \times 100\)

= 85.94%

Per phase power \(=\frac{30}{3}=10~kW\)

\({{I}^{2}}\left( {{r}_{1}}+r_{2}^{'} \right)=10\times {{10}^{3}}\)

\(\Rightarrow {{\left( 100 \right)}^{2}}\left( {{r}_{1}}+r_{2}^{'} \right)={{10}^{4}}\)

\(\Rightarrow {{r}_{1}}+r_{2}^{'}=1\)

Given, \({{r}_{1}}=r_{2}^{'}\)

\({{r}_{1}}=r_{2}^{'}=0.5~\text{ }\!\!\Omega\!\!\text{ }\)

\(Starting~torque~\left( {{T}_{s}} \right)=\frac{180}{2\pi {{N}_{s}}}\times {{I}^{2}}\times \frac{r_{2}^{'}}{s}\) 

At starting, slip(s) = 1

Synchronous speed (N_s) \(=\frac{120f}{P}=\frac{120\times 50}{4}=1500~rpm\)

\({{T}_{s}}=\frac{180}{2\pi \times 1500}\times {{\left( 100 \right)}^{2}}\times \frac{0.5}{1}\) 

starting capacitor is added to produce 90° phase difference b/w the I_a, I_m

\(\begin{array}{l} {{\rm{\Phi }}_{\rm{m}}} + {ϕ _{\rm{a}}} = 90\\ ta{n^{ - 1}}\left( {\frac{{{X_a} - {X_c}}}{{{R_a}}}} \right) + ta{n^{ - 1}}\left( {\frac{{{X_m}}}{{{R_m}}}} \right) = 90\\ ta{n^{ - 1}}\left( {\frac{{5 - {X_c}}}{7}} \right) + ta{n^{ - 1}}\left( {\frac{4}{{2.3}}} \right) = 90\\ {X_c} = 0.975 \end{array}\)

\({\tan ^ - }^1\left( {\frac{{5 - {X_c}}}{7}} \right) = 90^\circ - 60.10^\circ \)

\(\frac{{5 - {X_c}}}{7} = tan 29.89^\circ \)

Solving this we get

∴ C = 3.264 mf

Note:

To obtain the maximum torque at starting itself. Then

ϕ_m+ 2ϕ_a = 90⁰