Electrical Machines Test 1

Concept:

In a DC generator, E_g ∝ Nϕ_g

In a DC generator, E_m ∝ Nϕ_m

In a DC generator, generated emf E_g = V_t + I_aR_a

In a DC motor, back emf E_b = V_t - I_aR_a

For the same speed, terminal voltage and current

\(\frac{{{\phi }_{m}}}{{{\phi }_{g}}}=\frac{{{V}_{t}}-{{I}_{a}}{{R}_{a}}}{{{V}_{t}}+{{I}_{a}}{{R}_{a}}}\)

Calculation:

Given that, terminal voltage (V_t) = 220 V

Armature current (I­_a) = 50 A

Armature resistance (R_a) = 0.2 Ω

\(\frac{{{\phi }_{m}}}{{{\phi }_{g}}}=\frac{220-50\times 0.2}{220+50\times 0.2}=\frac{210}{230}\)

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

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

Armature current (I_a) = 15 A

Armature resistance (R_a) = 0.7 Ω

Voltage (V) = 200 V

Field resistance (R_sh) = 0.3 Ω

Back emf, E_b1 = V – I_a(R_a + R_sh)

= 200 – 15(0.7 + 0.3)

= 185 V

When a series resistance is added,

E_b2 = V – I_a(R_a + R_sh + R_se)

= 200 – 15(0.7 + 0.3 + 5)

= 110 V

E_b ∝ Nϕ and ϕ ∝ I_a

\(\Rightarrow \frac{{{E}_{b1}}}{{{E}_{b2}}}=\frac{{{N}_{1}}}{{{N}_{2}}}\)

\(\Rightarrow {{N}_{2}}=\frac{{{E}_{b2}}}{{{E}_{b1}}}\times {{N}_{1}}\)

\(=\frac{110}{185}\times 800\)

Z₁ = (0.01 + j0.04) pu, kvA₁ = 1000 kVA

Z₂ = (0.02 + j0.06) pu, kvA₂ = 500 kVA

Load shared by transformer 2,

\(\begin{array}{l}{S_2} = \frac{{{z_1}}}{{{z_2}}} \times kV{A_2}\\ \Rightarrow {S_2} = \frac{{\sqrt {{{\left( {0.01} \right)}^2} + {{\left( {0.04} \right)}^2}} }}{{\sqrt {{{\left( {0.02} \right)}^2} + {{\left( {0.06} \right)}^2}} }} \times 500\end{array}\) 

\(= \frac{{0.0412}}{{0.0632}} \times 500 = 326.2\;kVA\) 

Largest kVA load on both transformers,

When torque slip relation is linear,

\(T \propto \frac{{s{V^2}}}{{{\omega _s}{r_2}}}\)

Given that, power = 7.5 kW

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

Slip (s) = 0.04

Rotor speed (N_r) = N_s (1 – s) = 960 rpm

Torque at full load \(= {T_{fl}} = \frac{P}{\omega }\)

\(= \frac{P}{{\frac{{2\pi {N_r}}}{{60}}}}\)

\(= \frac{{7.5 \times {{10}^3} \times 60}}{{2\pi \times 960}} = 74.6\;N.m\)

No load torque = T_nL = 6 Nm

\({T_{fL}} \propto \frac{{{s_{fl}}\;V_{fl}^2}}{{{\omega _s}{r_2}}} \propto \frac{{0.04 \times {{400}^2}}}{{{\omega _s}{r_2}}}\)

\({T_{nL}} \propto \frac{{{s_{nL}}\;V_{nl}^2}}{{{\omega _s}{r_2}}} \propto \frac{{{s_{nl}} \times {{200}^2}}}{{{\omega _s}{r_2}}}\)

\(\Rightarrow \frac{{74.6}}{6} = \frac{{0.04 \times {{400}^2}}}{{{s_{nl}} \times {{200}^2}}}\)

⇒ s_nl = 0.01287

The slip at which maximum torque occurs without internal resistance added to the rotor,

\({{S}_{m1}}=\frac{{{r}_{2}}}{{{x}_{2}}}=0.2\) 

Slip at with maximum torque occurs with an external resistance added to the rotor

\({{S}_{m2}}=\frac{{{r}_{2}}+{{r}_{ext}}}{{{x}_{2}}}=\frac{{{R}_{2}}}{{{x}_{2}}}\) 

The relation between, starting torque and maximum torque is

\(\frac{{{T}_{st}}}{{{T}_{max}}}=\frac{2{{s}_{m2}}}{s_{m}^{2}+1}\) 

\(\Rightarrow 0.75=\frac{2{{s}_{m2}}}{1+s_{m2}^{2}}\) 

⇒ s_m2 = 0.4514

\(\Rightarrow \frac{{{r}_{2}}+{{r}_{ext}}}{{{x}_{2}}}=0.4514\) 

\(\Rightarrow \frac{{{r}_{2}}}{{{x}_{2}}}+\frac{{{r}_{ext}}}{{{x}_{2}}}=0.4514\) 

\(\Rightarrow 0.2+\frac{0.5}{{{x}_{2}}}=0.4514\) 

Synchronous speed \(=\frac{120f}{P}\)

\(=\frac{120\times 50}{4}=1500~rpm\) 

Rated full load current per phase

\({{I}_{ph}}=\frac{1500\times {{10}^{3}}}{\sqrt{3}\times 6.6\times {{10}^{3}}}=131.2~A\) 

At no load, phase current = 0

\(Phase~voltage\left( E \right)=V=\frac{6.6\times {{10}^{3}}}{\sqrt{3}}=3810.5~V\) 

δ = 0°

Synchronizing power co-efficient for 3-phases.

\(=3\left( \frac{EV}{{{x}_{s}}} \right)\cos \delta\) 

\(=3\left( \frac{3810.5\times 3810.5}{45} \right)\cos 0{}^\circ\) 

= 967.99 kW

With 4 poles, 1 mech-rad = 2 elec.rad

Synchronizing power co-efficient = 2 × 967.99

= 1935.98 kW

But one mech rad \(=\frac{180}{\pi }\) mech. Degree

Synchronizing power per mechanical degree of rotor displacement \(=\frac{1935.98}{\left( \frac{180}{\pi } \right)}=33.78~kW\)

Synchronizing torque \(\left( \tau \right)=\frac{P}{\omega }\)

\(=\frac{P}{\frac{2\pi {{N}_{s}}}{60}}\) 

\(=\frac{60\times 33.78\times {{10}^{3}}}{2\pi \times 1500}\)