Question 5.8: Data are given in Fig. 5.19 with respect to the losses of th......

For the specified operating conditions, the armature current is

$I_a=\frac{45 × 10^3}{\sqrt{3}×230}= 113  A$

The I²R losses must be computed on the basis of the dc resistances of the windings at 75°C. Correcting the winding resistances by means of Eq. 5.32 gives

$\frac{R_T}{R_t}=\frac{234.5+T}{234.5+t}$           (5.32)

Field-winding resistance$R_f$ at 75°C = 35.5 Ω

Armature dc resistance$R_a$ at 75°C = 0.0399 Ω/phase

The field I²R loss is therefore

$I_f^2R_f = 5.50^2 × 35.5 = 1.07  kW$

According to ANSI standards, losses in the excitation system, including those in any field- rheostat, are not charged against the machine.

The armature I²R loss is

$3I_a^2R_a = 3 × 113^2 × 0.0399 = 1.53  kW$

and from Fig. 5.19 at $I_a = 113  A$ the stray-load loss = 0.37 kW. The stray-load loss is considered to account for the losses caused by the armature leakage flux. According to ANSI standards, no temperature correction is to be applied to the stray load loss.

Core loss under load is primarily a function of the main core flux in the motor. As is discussed in Chapter 2, the voltage across the magnetizing branch in a transformer (corresponding to the transformer core flux) is calculated by subtracting the leakage impedance drop from the terminal voltage. In a directly analogous fashion, the main core flux in a synchronous machine (i.e., the air-gap flux) can be calculated as the voltage behind the leakage impedance of the machine. Typically the armature resistance is small, and hence it is common to ignore the resistance and to calculate the voltage behind the leakage reactance. The core loss can then be estimated from the open-circuit core-loss curve at the voltage behind leakage reactance.

In this case, we do not know the machine leakage reactance. Thus, one approach would be simply to assume that the air-gap voltage is equal to the terminal voltage and to determine the core-loss under load from the core-loss curve at the value equal to terminal voltage.³ In this case, the motor terminal voltage is 230 V line-to-line and thus from Fig. 5.19, the open-circuit core loss is 1.30 kW.

To estimate the effect of ignoring the leakage reactance drop, let us assume that the leakage reactance of this motor is 0.20 per unit or

$X_{\mathrm{al}}= 0.2 \left(\frac{220²}{45 × 10^3}\right) = 0.215  Ω$

Under this assumption, the air-gap voltage is equal to

which corresponds to a line-to-line voltage of $\sqrt{3} (153) = 265  V$. From Fig. 5.19, the corresponding core-loss is 1.8 kW, 500 W higher than the value determined using the terminal voltage. We will use this value for the purposes of this example.

Including the friction and windage loss of 0.91 kW, all losses have now been found:

$\text{Total losses} = 1.07 + 1.53 + 0.37 + 1.80 + 0.91 = 5.68  kW$

The total motor input power is the input power to the armature plus that to the field.

$\text{Input power} = 0.8 × 45 + 1.07 = 37.1  kW$

and the output power is equal to the total input power minus the total losses

$\text{Output power} = 37.1  –  5.68 = 31.4  kW$

Therefore

$\text{Efficiency}=\frac{\text{Output power}}{\text{Input power}}=1-\frac{31.4}{37.1}  =  0.846 = 84.6\%$