Question 6.5: The following test data apply to a 7.5-hp, three-phase, 220-......

a. From Eq. 6.37, the rotational losses can be calculated as

P_{rot} = P_{\mathrm{nl}}  –  n_{ph} I_{1,\mathrm{nl}}^2 R_1 = 380  –  3×5.70^2×0.262=354  W

The line-to-neutral no-load voltage is equal to V_{1,\mathrm{nl}}= 219/ \sqrt{3} = 126.4  V and thus, from Eqs. 6.43 and 6.44,

Q_{\mathrm{nl}}=\sqrt{S_{\mathrm{nl}}^2  –  P_{\mathrm{nl}}^2} \quad \quad \quad (6.43)

S_{\mathrm{nl}}=n_{ph} V_{1,\mathrm{nl}} I_{1,\mathrm{nl}} \quad \quad \quad (6.44)

Q_{\mathrm{nl}}= \sqrt{(n_{ph}V_{1,\mathrm{nl}} I_{1,\mathrm{nl}})^2  –  P_{\mathrm{nl}}^2} = \sqrt{(3 × 126.4 × 5.7)^2  –  380^2}=  2128  W

and thus from Eq. 6.45

X_{\mathrm{nl}}=\frac{Q_{\mathrm{nl}}}{n_{ph}I_{1,\mathrm{nl}}^2}= \frac{2128}{3×5.7^2} = 21.8  Ω

We can assume that the blocked-rotor test at a reduced frequency of 15 Hz and rated current reproduces approximately normal running conditions in the rotor. Thus, from test 2 and Eqs. 6.47 and 6.48 with

Q_{\mathrm{bl}}=\sqrt{S_{\mathrm{bl}}^2  –  P_{\mathrm{bl}}^2} \quad \quad \quad (6.47)

S_{\mathrm{bl}}=n_{ph} V_{1,\mathrm{bl}} I_{1,\mathrm{bl}} \quad \quad \quad (6.48)

Q_{\mathrm{bl}} = \sqrt{(n_{ph}V_{1,\mathrm{bl}} I_{1,\mathrm{bl}})^2  –  P_{\mathrm{bl}}^2}= \sqrt{(3 × 15.3 × 18.57)^2 -675^2}= 520  VA

and thus from Eq. 6.49

X_{\mathrm{bl}}=\left(\frac{f_r}{f_{\mathrm{bl}}}\right) \left(\frac{Q_{\mathrm{bl}}}{n_{ph}I_{1,\mathrm{bl}}^2}\right) =\left(\frac{60}{15}\right) \left( \frac{520}{3× 18.57^2} \right) =2.01  Ω

Since we are told that this is a Class C motor, we can refer to Table 6.1 and assume that X_1= 0.3(X_1+ X_2)  \text{or}  X_1 = kX_2, where k = 0.429. Substituting into Eq. 6.57 results in a quadratic in X_2

X_2=(X_{\mathrm{bl}}  –  X_1) \left( \frac{X_{\mathrm{nl}}  –  X_1}{X_{\mathrm{nl}}  –  X_{\mathrm{bl}}}\right) \quad \quad \quad  (6.57)

k² X_2^2   +  (X_{\mathrm{bl}}(1  –  k)  –  X_{\mathrm{nl}}(1 + k))X_2 + X_{\mathrm{nl}}X_{\mathrm{bl}} = 0

or

(0.429)^2X_2 + (2.01(1  –  0.429)  –  22.0(1 + 0.429))X_2 + 22.0(2.01) \\ = 0.184X_2^2   –  30.29X_2 + 44.22 = 0

Solving gives two roots: 1.48 and 163.1. Clearly, X_2 must be less than X_{\mathrm{nl}} and hence it is easy to identify the proper solution as

X_2 = 1.48 Ω

and thus

X_1 = 0.633  Ω

From Eq. 6.58,

X_m = X_{\mathrm{nl}}  –  X_1 = 21.2  Ω

R_{\mathrm{bl}} can be found from Eq. 6.50 as

R_{\mathrm{bl}}=\frac{P_{\mathrm{bl}}}{n_{ph}I_{1,\mathrm{bl}}^2}=\frac{675}{3× 18.57^2}=0.652  Ω

and thus from Eq. 6.56

\begin{aligned}R_2 &= ( R_{\mathrm{bl}}  –  R_1 ) \left( \frac{X_2 +X_m}{X_m} \right) ^2 \\& = (0.652  –  0.262) \left( \frac{22.68}{21.2} \right) ^2 = 0.447  Ω\end{aligned}

The parameters of the equivalent circuit for small values of slip have now been calculated.

b. Although we could calculate the electromechanical starting torque from the equivalent- circuit parameters derived in part (a), we recognize that this is a double-squirrel-cage motor and hence these parameters (most specifically the rotor parameters) will differ significantly under starting conditions from their low-slip values calculated in part (a). Hence, we will calculate the electromechanical starting torque from the rated-frequency, blocked-rotor test measurements of test 4.

From the power input and stator I^2R losses, the air-gap power P_{\text{gap}} is

P_{\text{gap}} = P_{\mathrm{bl}}  –  n_{ph}I^2_{1,\mathrm{bl}}R_1 = 20,100  –  3 × 83.3^2 × 0.262 = 14,650  W

Since this is a four-pole machine, the synchronous speed can be found from Eq. 6.26 as

ω_s=\frac{4πf_e}{poles}=\left( \frac{2}{poles}\right) ω_e \quad \quad \quad (6.26)

 ω_s = 188.5  rad/sec. Thus, from Eq. 6.25 with s = 1

T_{mech} = \frac{P_{\text{mech}}}{ω_m}=\frac{P_{\text{gap}}}{ω_s}=\frac{n_{ph}I_2^2(R_2/s)}{ω_s} \quad \quad \quad (6.25)

T_{start} = \frac{P_{\text{gap}}}{ω_s}=\frac{14,650}{188.5} = 77.7  N  ·  m

The test value, T_{start}= 74.2  N · m is a few percent less than the calculated value because the calculations do not account for the power absorbed in the stator core loss or in stray-load losses.