6.3.1 Weighted-Harmonic Factor for Single-Phase Transformers

Figure 6.1 illustrates the equivalent circuit of a single-phase transformer for the fundamental (h = 1) and harmonic voltages of hth order.

The ohmic losses of the hth harmonic expressed in terms of the fundamental ohmic losses are

$\frac{W_{\mathrm{ohmic},h}}{W_{\mathrm{ohmic},1}}={\left(\frac{V_{\mathit{ph}}}{V_{p1}}\right)}^2\left(\frac{R_{\mathit{ph}}+R_{sh}^{\prime }}{R_{p1}+R_{s1}^{\prime }}\right)\left(\mathrm{RATIO}\right)\text{,}$

where

$\mathrm{RATIO}=\frac{{\left(R_{p1}+R_{s1}^{\prime }+R_{\mathrm{load},1}^{\prime }\right)}^2+{\omega }^2{\left(L_{p1}+L_{s1}^{\prime }+L_{\mathrm{load},1}^{\prime }\right)}^2}{{\left(R_{\mathit{ph}}+R_{sh}^{\prime }+R_{\mathrm{load},h}^{\prime }\right)}^2+{\left(h\omega \right)}^2{\left(L_{\mathit{ph}}+L_{sh}^{\prime }+L_{\mathrm{load},h}^{\prime }\right)}^2}\text{.}$

In most residential power transformers the influence of the skin effect can be neglected. However, the skin effect of an ohmic load cannot be ignored, because $R_{\mathrm{load},h}^{\prime },R_{\mathrm{load},l}^{\prime }\gg R_{\mathit{ph}},R_{p1},R_{sh}^{\prime },R_{s1}^{\prime }\text{.}$ Therefore, one obtains

$\frac{W_{\mathrm{ohmic},h}}{W_{\mathrm{ohmic},1}}\approx {\left(\frac{V_{\mathit{ph}}}{V_{p1}}\right)}^2\frac{{\left(R_{\mathrm{load},1}^{\prime }\right)}^2+{\omega }^2{\left(L_{p1}+L_{s1}^{\prime }+L_{\mathrm{load},1}^{\prime }\right)}^2}{{\left(R_{\mathit{load},h}^{\prime }\right)}^2+{\left(h\omega \right)}^2{\left(L_{\mathit{ph}}+L_{sh}^{\prime }+L_{\mathit{load},h}^{\prime }\right)}^2}\text{.}$

Due to $L_{\mathit{ph}}\approx L_{\mathrm{p}1},L_{sh}^{\prime }\approx L_{s1}^{\prime },L_{\mathrm{load},h}^{\prime }\approx L_{\mathrm{load},1}^{\prime }\text{,}$ and the fact that the load resistance R_load, h′ increases less than linearly with frequency [18], the following simplified expression is considered:

$\frac{W_{\mathrm{ohmic},h}}{W_{\mathrm{ohmic},1}}\approx \frac{1}{h^n}{\left(\frac{V_{\mathit{ph}}}{V_{p1}}\right)}^2\text{,}$

where 0 < n < 2 depends on the resistance and inductance values of the equivalent circuit of a load.

The iron-core losses of a single-phase transformer due to a voltage harmonic of order h are expressed in terms of the corresponding losses caused by the fundamental as

$\frac{W_{\mathrm{iron},h}}{W_{\mathrm{iron},1}}={\left(\frac{B_{\max ,h}}{B_{\max ,1}}\right)}^2\frac{\left\{\sigma {\left(\frac{fh}{100}\right)}^2{k}_m^2+\varepsilon \left(\frac{fh}{100}\right)\right\}}{\left\{\sigma {\left(\frac{f}{100}\right)}^2+\varepsilon \left(\frac{f}{100}\right)\right\}}\text{,}$

where B_max,1 and B_max,h define fundamental and harmonic maximum flux densities, respectively, and σ and ɛ are [26] coefficients related to eddy-current and hysteresis losses, respectively. The term k_m takes the reaction of the eddy currents within the laminations and between the laminations on the original field into account [18]. With (h B_max,h) ∝ E_ph and B_max,1 ∝ E_p1, the relation becomes

$\frac{W_{\mathrm{iron},h}}{W_{\mathrm{iron},1}}={\left(\frac{E_{\mathit{ph}}}{h{E}_{p1}}\right)}^2\left(\frac{R_{fe,1}}{R_{fe,h}}\right)\text{,}$

where R_fe,1 and R_fe,h are the iron-core loss resistances for the fundamental (h = 1) and harmonic of order h, respectively.

The total losses for a harmonic of order h are then determined from the sum of Eqs. 6-6 and 6-8:

$\frac{W_{\mathrm{total},h}}{W_{\mathrm{total},1}}\approx \frac{1}{h^n}{\left(\frac{V_{\mathit{ph}}}{V_{p1}}\right)}^2+\frac{1}{h^2}{\left(\frac{E_{\mathit{ph}}}{E_{p1}}\right)}^2\left(\frac{R_{fe,1}}{R_{fe,h}}\right)\text{.}$

According to [27], the core-loss resistance ratio is R_fe,h/R_fe,1 = h^0.6 and because of the leakage reactance being proportional to h, one obtains the inequality (E_ph/E_p1) < V_ph/V_p1) or

${\left(\frac{E_{\mathit{ph}}}{E_{p1}}\right)}^2={\left(\frac{V_{\mathit{ph}}}{V_{p1}}\right)}^m\text{,}$

where the exponent m is less than 2. Therefore, for single-phase transformers Eq. 6-9 becomes

$\frac{W_{\mathrm{total},h}}{W_{\mathrm{total},1}}\approx \frac{1}{h^n}{\left(\frac{V_{\mathit{ph}}}{V_{p1}}\right)}^2+\frac{1}{h^{2.6}}{\left(\frac{V_{\mathit{ph}}}{V_{p1}}\right)}^m\text{.}$

Depending on the relative sizes of the ohmic and iron-core losses at full load, Eq. 6-11 can be rewritten for all occurring harmonics and is called the weighted harmonic factor [20]:

$\frac{W_{\mathrm{total},h\Sigma }}{W_{\mathrm{total},1}}=K_1{\displaystyle \sum _{h=2}^{\infty }\frac{1}{{\left(h\right)}^k}{\left(\frac{V_{\mathit{ph}}}{V_{p1}}\right)}^{\ell }\text{,}}$

where W_total,hΣ and W_total,1 are the total harmonic losses and the total fundamental losses, respectively. Also 0 ≤ k ≤ 2.0 and 0 ≤ ℓ ≤ 2.0. Average values for k and ℓ are obtained from measurements as discussed in a later section.

6.3.2 Measured Temperature Increases of Transformers

Measured temperature rises in transformers and induction machines as they occur in a residential/commercial power system – together with calculated loss increases due to voltage harmonics – represent a base for the estimation of the lifetime reduction due to voltage harmonics. It is thereby assumed that for small additional temperature rises – as compared with the rated temperature rise – the additional losses are proportional to the additional temperature rise.

6.3.2.1 Single-Phase Transformers

A single-phase 150 VA transformer was tested [18,19] and Fig. 6.2a,b shows the measured temperature rises in percent of the rated temperature rises at full load for moderate (e.g., T_amb = 23 °C) and high (e.g., T_amb = 40 °C) ambient temperatures.

Note that if the third harmonic voltage v₃(t) superposed with the fundamental voltage v₁(t) produces a peak-to-peak voltage {v₁(t) + v₃(t)} that is maximum, then the peak-to-peak flux density in the core {B₁(t) + B₃(t)} is minimum. That is, the total harmonic losses are smaller than those when the peak-to-peak voltage is minimum, resulting in a maximum peak-to-peak value of the flux density. This “alternating” behavior of the harmonic losses and associated temperatures is discussed in detail in [27,28].

6.3.2.2 Three-Phase Transformers

Three 150 VA single-phase transformers were assembled to Δ/Y-grounded and Y-grounded/Y-grounded configurations and they were subjected to the same harmonic voltage conditions as the single-phase transformer of the preceding section. The temperatures of the transformer windings were measured at the same location as for the single-phase transformer. The tests were limited to balanced operation. However, no phase-lock loop was available and no stable relation between fundamental and harmonic voltage systems could be maintained. Therefore, the temperature data of Fig. 6.3a,b represent average values and the alternating behavior could not be observed, although it also exists in three-phase transformers.

6.3.3 Weighted-Harmonic Factor for Three-Phase Induction Machines

The per-phase equivalent circuits of a three-phase induction motor are shown for the fundamental (h = 1) and a (integer, sub-, inter-) harmonic of hth order in Fig. 6.4. In Fig. 6.4 the fundamental slip is

$s_1=\frac{n_{s1}-n_m}{n_{s1}}\text{,}$

and the harmonic slip is

$s_h=\frac{n_{sh}-n_m}{n_{sh}}\text{.}$

The ohmic losses caused by either the fundamental or a harmonic of order h are proportional to the square of the respective currents. Therefore, one can write for the ohmic losses caused by voltage harmonics:

$\frac{W_{\mathrm{ohmic},h}}{W_{\mathrm{ohmic},1}}=\left(\frac{R_{\mathit{ph}}+R_{sh}^{\prime }}{\left(R_{p1}+R_{s1}^{\prime }\right)}\right){\left(\frac{V_{\mathit{ph}}}{V_{p1}}\right)}^2{\left(\frac{Z_{p1}}{Z_{\mathit{ph}}}\right)}^2\text{,}$

where

$Z_{p1}=\sqrt{{\left(R_{p1}+R_{s1}^{\prime }/s_1\right)}^2+{\left({\omega }_1{L}_{p1}+{\omega }_1{L}_{s1}^{\prime }\right)}^2}$

and

$Z_{\mathit{ph}}=\sqrt{{\left(R_{\mathit{ph}}+R_{sh}^{\prime }/s_h\right)}^2+{\left(h{\omega }_1{L}_{\mathit{ph}}+h{\omega }_1{L}_{sh}^{\prime }\right)}^2}\text{.}$

For all time harmonics (h = 2, 3, 4, …) an induction machine operating at rated speed can be considered being at standstill because s₁< <s_h. With R_ph, Rʹ_sh/s_h < < hω₁L_ph ≈ hω₁L_p1, hω₁Lʹ_sh ≈ hω₁Lʹ_s1, the per-unit starting impedance

$Z_{\mathrm{start}pu}=\frac{\sqrt{{\left({\omega }_1{L}_{p1}+{\omega }_1{L}_{s1}^{\prime }\right)}^2}}{\sqrt{{\left(R_{p1}+R_{s1}^{\prime }/s_1\right)}^2+{\left({\omega }_1{L}_{p1}+{\omega }_1{L}_{s1}^{\prime }\right)}^2}}\text{,}$

and the slip $s_h=\left\{h\mp \left(1–s_1\right)\right\}/h\approx \left(h\mp 1\right)/h\text{,}$ Eq. 6-14 can be rewritten as

$\frac{W_{\mathrm{ohmic},h}}{W_{\mathrm{ohmic},1}}=\left(\frac{R_{\mathit{ph}}+R_{sh}^{\prime }}{\left(R_{p1}+R_{s1}^{\prime }\right)}\right){\left(\frac{V_{\mathit{ph}}}{V_{p1}}\right)}^2\frac{1}{h^2{Z^2}_{\mathrm{start}pu}}$

The additional iron-core losses due to a voltage harmonic of order h referred to the iron-core losses at the fundamental frequency are (see Fig. 6.4)

$\frac{W_{\mathrm{iron},h}}{W_{\mathrm{iron},1}}={\left(\frac{V_{\mathit{ph}}}{V_{p1}}\right)}^2{\left(\frac{Z_{fe,1}}{Z_{fe,h}}\right)}^2\frac{R_{fe,h}}{R_{fe,1}}={\left(\frac{E_{\mathit{ph}}}{E_{p1}}\right)}^2\frac{R_{fe,1}}{R_{fe,h}}\text{,}$

where

$Z_{fe,1}=\frac{\left(j{\omega }_1{L}_{m1}+R_{fe,1}\right)\left(R_{p1}+j{\omega }_1{L}_{p1}\right)}{j{\omega }_1{L}_{m1}}+R_{fe,1}\text{,}$

$Z_{fe,h}=\frac{\left(jh{\omega }_1{L}_{mh}+R_{fe,h}\right)\left(R_{\mathit{ph}}+jh{\omega }_1{L}_{\mathit{ph}}\right)}{jh{\omega }_1{L}_{mh}}+R_{fe,h}\text{,}$

where L_mh ≈ L_m1, L_ph ≈ L_p1. From [31] follows $\frac{R_{fe,h}}{R_{fe,1}}\approx h^{0.6}\text{,}$ and therefore the iron-core losses are

$\frac{W_{\mathrm{iron},h}}{W_{\mathrm{iron},1}}\approx {\left(\frac{E_{\mathit{ph}}}{E_{p1}}\right)}^2\left(\frac{1}{h^{0.6}}\right)\text{.}$

Summing Eqs. 6-16 and 6-18, the total losses for a harmonic of order h of a three-phase induction machine become

$\frac{W_{\mathrm{total},h}}{W_{\mathrm{total},1}}\approx \left(\frac{R_{\mathit{ph}}+R_{sh}^{\prime }}{R_{p1}+R_{s1}^{\prime }}\right){\left(\frac{V_{\mathit{ph}}}{V_{p1}}\right)}^2\frac{1}{h^2{Z}_{\mathrm{start}pu}^2}+{\left(\frac{E_{\mathit{ph}}}{E_{p1}}\right)}^2\left(\frac{1}{h^{0.6}}\right)\text{,}$

or for all occurring harmonics, one can show that the total harmonic losses W_total,hΣ relate to the total fundamental losses, as defined by the weighted harmonic factor [32]:

$\frac{W_{\mathrm{total},h\Sigma }}{W_{\mathrm{total},1}}=K_2{{\displaystyle \sum _{h=2}^{\infty }\frac{1}{{\left(h\right)}^k}\left(\frac{V_{\mathit{ph}}}{V_{p1}}\right)}}^{\ell }\text{,}$

where 0 ≤ k ≤ 2 and 0 ≤ ℓ ≤ 2.

Note that the relations for single-phase transformers (Eq. 6-12) and that for a three-phase induction motor (Eq. 6-20) are identical in structure, though the exponents k and ℓ may assume different values for single-phase transformers and three-phase induction motors. These exponents will be identified in the next section and it will be shown that the same structure of the weighted harmonic factor will also be valid for three-phase transformers and single-phase induction machines as well as universal machines.

Depending on the values of k and ℓ one obtains different loss or temperature dependencies as a function of the frequency. Unfortunately, these dependencies can vary within wide ranges from constant to hyperbolic functions. This is so because the iron-core losses depend on the electric steel (e.g., conductivity, lamination thickness, permeability) used and the winding configurations in primary or stator and secondary or rotor. These can exhibit different conductivities (e.g., Cu, Al) or skin and proximity effects. In order to study these dependencies various k and ℓ combinations will be assumed, as is depicted in Figs. 6.5 to 6.7:

• k = 0 and ℓ = 1 results in a linear frequency-independent characteristic (Fig. 6.5).

• k = 1 and ℓ = 1 results in an inverse frequency-dependent characteristic (Fig. 6.6).

• k = 2 and ℓ = 2 results in a quadratic inverse frequency-dependent characteristic (Fig. 6.7).

In real transformers and induction machines combinations of these characteristics will be possible.

6.5.1 Application Example 6.1: Temperature Rise of a Single-Phase Transformer due to Single Harmonic Voltage

Determine the temperature rise ΔT_h of a single-phase transformer provided V₃ = 0.10 pu = 10%, T_amb = 40 °C, and T_rated = 100 °C. Assume k_avg = 0.90 and ℓ_avg = 1.75.

Solution to Application Example 6.1

The rated temperature rise is ΔT_rated = T_rated – T_amb = T_{rise rat} = 60 °C. For the harmonic with order h = 3 and amplitude V₃ = 0.10 pu ≡ 10%, one obtains (through linear extrapolation) the weighted-harmonic factor by replacing $V_{\mathit{ph}}/V_{p1}$ by 10:

${\displaystyle \sum _{h=3}\frac{1}{{\left(3\right)}^{0.9}}{\left(10\right)}^{1.75}}=20.98\text{,}$

resulting with Fig. 6.14 in the additional temperature increase due to the 3rd harmonic (average) ΔT_h = 3 = 3.3% or ΔT_h = 3 = 1.98 °C.

6.5.2 Application Example 6.2: Temperature Rise of a Single-Phase Induction Motor due to Single Harmonic Voltage

Determine the temperature rise ΔT_h of a single-phase induction motor provided V₃ = 0.10 pu ≡ 10%, T_amb = 40 °C, and T_rated = 100 °C. Assume k_avg = 0.85 and ℓ_avg = 1.40.

Solution to Application Example 6.2

The rated temperature rise is ΔT_rated = T_rated – T_amb = T_{rise rat} = 60 °C. For the harmonic with order h = 3 and amplitude V₃ = 0.10 pu ≡ 10% one obtains the weighted-harmonic factor by replacing $V_{\mathit{ph}}/V_{p1}$ by 10 as follows ${\displaystyle \sum _{h=3}\frac{1}{{\left(3\right)}^{0.85}}{\left(10\right)}^{1.40}}=9.88\text{,}$ resulting with Fig. 6.14 in the additional temperature increase due to the 3rd harmonic (average) ΔT_h = 3 = 11% or ΔT_h = 3 = 6.6 °C.

Transformers, induction machines, and universal machines have different loss or temperature sensitivities with respect to voltage harmonics and, therefore, the additional temperature rises are different for all five types of devices. The additional temperature rise (loss) versus weighted-harmonic factor function indicates that the most sensitive components are single-phase induction machines, whereas the least sensitive devices are transformers with resistive load and universal machines. This is so because for any harmonic terminal voltage, a single-phase machine (and to some degree an unbalanced three-phase induction machine) develops forward-and backward-rotating harmonic fields and responds like being under short-circuit conditions due to the large slip s_h of the harmonic field with respect to the rotating rotor. In transformers the resistive and inductive loads can never have zero impedance due to the nature of the frequency dependency of such loads.

6.8.1 Application Example 6.3: Aging of a Single-Phase Induction Motor with E = 0.74 eV Due to a Single Harmonic Voltage

Determine for an activation energy of E = 0.74 eV the slope E/K and for the additional temperature rise ΔT_h = 3 = 6.6 °C (see Application Example 6.2 with T_amb = 40°, T₂ = T_rat = 100 °C) the reduced lifetime of a single-phase induction motor.

Solution to Application Example 6.3

The slope (E/K) is E/K = 8.58·10³ kelvin. Thus one obtains from $t_{\mathsf{1}}=40e^{-8.58\cdot {10}^3\mathrm{▵}{T}_h/373\left(373+\mathrm{▵}{T}_h\right)}$, with ΔT_h = 3 = 6.6 °C, a reduced lifetime of $t_1|{}_{\mathrm{▵}{T}_{h=3}=6.6°\mathrm{C}}=26.81\mathrm{years}$.

6.8.2 Application Example 6.4: Aging of a Single-Phase Induction Motor with E = 0.51 eV Due to a Single Harmonic Voltage

Determine for an activation energy of E = 0.51 eV the slope E/K and for the additional temperature rise ΔT_h = 3 = 6.6 °C (see Application Example 6.2 with T_amb = 40°, T₂ = T_rat = 100 °C) the reduced lifetime of a single-phase induction motor.

Solution to Application Example 6.4

The slope (E/K) is E/K = 5.91·10³ kelvin. Thus one obtains from $t_1=40e^{-5.91\cdot {10}^3\mathrm{▵}{T}_h/373\left(373+\mathrm{▵}{T}_h\right)}\text{,}$ with ΔT_h = 6.6 °C, a reduced lifetime of $t_1|{}_{\mathrm{▵}{T}_{h=3}=6.6°\mathrm{C}}=30.37\mathrm{years}$.

From Application Examples 6.3 and 6.4 one concludes that lower values of the activation energy E result in lower decreases in lifetime.

6.10.1 Application Example 6.5: Estimation of Lifetime Reduction for Given Single-Phase and Three-Phase Voltage Spectra with High Harmonic Penetration with Activation Energy E = 1.1 eV

Estimate the lifetime reductions of induction machines, transformers, and universal machines for the single- and three-phase voltage spectra of Table E6.5.1 and their associated lifetime reduction for an activation energy of E = 1.1 eV. The ambient temperature is T_amb = 23 °C, the rated temperature is T₂ = 85 °C, and the rated lifetime of t₂ = 40 years can be assumed.

Table E6.5.1

Possible Voltage Spectra with High-Harmonic Penetration

h	${\left({\mathit{V}}_{\mathit{h}},/,{\mathit{V}}_{\mathbf{60}\mathbf{Hz}}\right)}_{\mathbf{1}\mathbf{\Phi }}\left(\mathbf{\%}\right)$	${\left({\mathit{V}}_{\mathit{h}},/,{\mathit{V}}_{\mathbf{60}\mathbf{Hz}}\right)}_{\mathbf{3}\mathbf{\Phi }}\left(\mathbf{\%}\right)$
1	100	100
2	2.5	0.5
3	5.71	1.0
4	1.6	0.5
5	1.25	7.0
6	0.88	0.2
7	1.25	5.0
8	0.62	0.2
9	0.96	0.3
10	0.66	0.1
11	0.30	2.5
12	0.18	0.1
13	0.57	2.0
14	0.10	0.05
15	0.10	0.1
16	0.13	0.05
17	0.23	1.5
18	0.22	0.01
19	1.03	1.0
	All higher harmonics < 0.2%

Solution to Application Example 6.5

Calculation of weighted harmonic factor for single-phase spectrum of Table E6.5.1 based on the values for k_avg = 0.85 and ℓ_avg = 1.4 for single-phase induction motors:

$\begin{array}{l}\frac{1}{2^{0.85}}{\left(2.5\right)}^{1.4}=2.000,\frac{1}{3^{0.85}}{\left(5.71\right)}^{1.4}=4.506,\\ \frac{1}{4^{0.85}}{\left(1.6\right)}^{1.4}=0.594,\dots ,\frac{1}{{17}^{0.85}}{\left(0.23\right)}^{1.4}=0.0115,\\ \frac{1}{{18}^{0.85}}{\left(0.22\right)}^{1.4}=0.0103,\frac{1}{{19}^{0.85}}{\left(1.03\right)}^{1.4}=0.0853\text{.}\end{array}$

Summing all contributions results in the weighted harmonic-voltage factor for single-phase induction motors ${\displaystyle \sum _{h=2}^{h\max }\frac{1}{h^k}}{\left(\frac{V_{\mathit{ph}}}{V_{p1}}\right)}^{\ell }\approx 8.5$.

Calculation of weighted harmonic factor for three-phase spectrum of Table E6.5.1 using the values for k_avg = 0.95 and ℓ_avg = 1.6 for three-phase induction motors:

$\begin{array}{l}\frac{1}{2^{0.95}}{\left(0.5\right)}^{1.6}=0.1707,\frac{1}{3^{0.95}}{\left(1.0\right)}^{1.6}=0.3522,\\ \frac{1}{4^{0.95}}{\left(0.5\right)}^{1.6}=0.0884,\dots ,\frac{1}{{17}^{0.95}}{\left(1.5\right)}^{1.6}=0.1297,\\ \frac{1}{{18}^{0.95}}{\left(0.01\right)}^{1.6}=0.00004,\frac{1}{{19}^{0.95}}{\left(1.0\right)}^{1.6}=0.0609\text{.}\end{array}$

Summing all contributions results in the weighted harmonic-voltage factor for three-phase induction motors ${\displaystyle \sum _{h=2}^{h\max }\frac{1}{h^k}}{\left(\frac{V_{\mathit{ph}}}{V_{p1}}\right)}^{\ell }\approx 8.6$.

Calculation of weighted harmonic factor for single-phase spectrum of Table E6.5.1 based on the values for k_avg = 0.90 and ℓ_avg = 1.75 for single-phase transformers:

$\begin{array}{l}\frac{1}{2^{0.9}}{\left(2.5\right)}^{1.75}=2.664,\frac{1}{3^{0.9}}{\left(5.71\right)}^{1.75}=7.847,\\ \frac{1}{4^{0.9}}{\left(1.6\right)}^{1.75}=0.6536,\dots ,\frac{1}{{17}^{0.9}}{\left(0.23\right)}^{1.75}=0.00597,\\ \frac{1}{{18}^{0.9}}{\left(0.22\right)}^{1.75}=0.00524,\frac{1}{{19}^{0.9}}{\left(1.03\right)}^{1.75}=0.744\text{.}\end{array}$

Summing all contributions results in the weighted harmonic-voltage factor for single-phase transformers ${\displaystyle \sum _{h=2}^{h\max }\frac{1}{h^k}}{\left(\frac{V_{\mathit{ph}}}{V_{p1}}\right)}^{\ell }\approx 12.3\text{.}$

Calculation of weighted harmonic factor for three-phase spectrum of Table E6.5.1 based on the values for k_avg = 0.90 and ℓ_avg = 1.75 for three-phase transformers:

$\begin{array}{l}\frac{1}{2^{0.9}}{\left(0.5\right)}^{1.75}=0.1593,\frac{1}{3^{0.9}}{\left(1\right)}^{1.75}=0.37203,\\ \frac{1}{4^{0.9}}{\left(0.5\right)}^{1.75}=0.0854,\dots ,\frac{1}{{17}^{0.9}}{\left(1.5\right)}^{1.75}=0.15876,\\ \frac{1}{{18}^{0.9}}{\left(0.01\right)}^{1.75}=0.0000234,\frac{1}{{19}^{0.9}}{\left(1\right)}^{1.75}=0.07065\text{.}\end{array}$

Summing all above contributions results in the weighted harmonic-voltage factor for three-phase transformers ${\displaystyle \sum _{h=2}^{h\max }}\frac{1}{h^k}{\left(\frac{V_{\mathit{ph}}}{V_{p1}}\right)}^{\ell }\approx 11.8\text{.}$

Calculation of weighted harmonic factor for single-phase spectrum of Table E6.5.1 based on the values for k_avg = 1.0 and ℓ_avg = 2.0 for universal motors:

$\begin{array}{c}\frac{1}{2}{\left(2.5\right)}^2=3.125,\frac{1}{3}{\left(5.71\right)}^2=10.868,\\ \frac{1}{4}{\left(1.6\right)}^2=0.640,....,\frac{1}{17}{\left(0.23\right)}^2=0.00311,\\ \frac{1}{18}{\left(0.22\right)}^2=0.0027,\frac{1}{19}{\left(1.03\right)}^2=0.0558\text{.}\end{array}$

Summing all contributions results in the weighted harmonic-voltage factor for universal motors ${\displaystyle \sum _{h=2}^{h\max }}\frac{1}{h^k}{\left(\frac{V_{\mathit{ph}}}{V_{p1}}\right)}^{\ell }\approx 15.6\text{.}$

Based on Fig. 6.14, T₂ = 85 °C, T_amb = 23 °C, E = 1.1 eV, and rated lifetime of t₂ = 40 years the above harmonic factors result in the additional temperature rises and lifetime reductions of Table E6.5.2.

6.10.2 Application Example 6.6: Estimation of Lifetime Reduction for Given Single-Phase and Three-Phase Voltage Spectra with Moderate Harmonic Penetration with Activation Energy E = 1.1 eV

Estimate the lifetime reductions for induction machines, transformers, and universal machines for the single- and three-phase voltage spectra of Table E6.6.1 and their associated lifetime reduction for an activation energy of E = 1.1 eV. The ambient temperature is T_amb = 23 °C, the rated temperature is T₂ = 85 °C, and the rated lifetime of t₂ = 40 years can be assumed.

Table E6.6.1

Possible Voltage Spectra with Moderate-Harmonic Penetration

h	${\left({\mathit{V}}_{\mathit{h}},/,{\mathit{V}}_{\mathbf{60}\mathbf{Hz}}\right)}_{\mathbf{1}\mathbf{\Phi }}\left(\mathbf{\%}\right)$	${\left({\mathit{V}}_{\mathit{h}},/,{\mathit{V}}_{\mathbf{60}\mathbf{Hz}}\right)}_{\mathbf{3}\mathbf{\Phi }}\left(\mathbf{\%}\right)$
1	100	100
2	0.5	0.5
3	3.0	0.5
4	0.3	0.5
5	2.0	3.0
6	0.2	0.2
7	1.0	2.5
8	0.2	0.2
9	0.75	0.3
10	0.1	0.1
11	1.0	1.0
12	0.1	0.1
13	0.9	0.85
14	0.1	0.05
15	0.3	0.1
16	0.05	0.05
17	0.5	0.3
18	0.05	0.01
19	0.4	0.2
	All higher harmonics < 0.2%

Solution to Application Example 6.6

Calculation of weighted harmonic factor for single-phase spectrum of Table E6.6.1 based on the values for k_avg = 0.85 and ℓ_avg = 1.4 for single-phase induction motors:

$\begin{array}{l}\frac{1}{2^{0.85}}{\left(0.5\right)}^{1.4}=0.2103,\frac{1}{3^{0.85}}{\left(3\right)}^{1.4}=1.8297,\\ \frac{1}{4^{0.85}}{\left(0.3\right)}^{1.4}=0.05703,....,\frac{1}{{17}^{0.85}}{\left(0.5\right)}^{1.4}=0.0341,\\ \frac{1}{{18}^{0.85}}{\left(0.05\right)}^{1.4}=0.0013,\frac{1}{{19}^{0.85}}{\left(0.4\right)}^{1.4}=0.0227\text{.}\end{array}$

Summing all contributions results in the weighted harmonic-voltage factor for single-phase induction motors ${\displaystyle \sum _{h=2}^{h\max }}\frac{1}{h^k}{\left(\frac{V_{\mathit{ph}}}{V_{p1}}\right)}^{\ell }\approx 3.4$.

Calculation of weighted harmonic factor for three-phase spectrum of Table E6.6.1 using the values for k_avg = 0.95 and ℓ_avg = 1.6 for three-phase induction motors:

$\begin{array}{l}\frac{1}{2^{0.95}}{\left(0.5\right)}^{1.6}=0.1707,\frac{1}{3^{0.95}}{\left(0.5\right)}^{1.6}=0.1162,\\ \frac{1}{4^{0.95}}{\left(0.5\right)}^{1.6}=0.0884,....,\frac{1}{{17}^{0.95}}{\left(0.3\right)}^{1.6}=0.00986,\\ \frac{1}{{18}^{0.95}}{\left(0.01\right)}^{1.6}=0.00004,\frac{1}{{19}^{0.95}}{\left(0.2\right)}^{1.6}=0.00464\text{.}\end{array}$

Summing all contributions results in the weighted harmonic-voltage factor for three-phase induction motors ${\displaystyle \sum _{h=2}^{h\max }}\frac{1}{h^k}{\left(\frac{V_{\mathit{ph}}}{V_{p1}}\right)}^{\ell }\approx 2.6\text{.}$

Calculation of weighted harmonic factor for single-phase spectrum of Table E6.6.1 based on the values for k_avg = 0.90 and ℓ_avg = 1.75 for single-phase transformers:

$\begin{array}{l}\frac{1}{2^{0.90}}{\left(0.5\right)}^{1.75}=0.1593,\frac{1}{3^{0.90}}{\left(0.3\right)}^{1.75}=2.544,\\ \frac{1}{4^{0.90}}{\left(0.3\right)}^{1.75}=0.0349,....,\frac{1}{{17}^{0.90}}{\left(0.5\right)}^{1.75}=0.0232,\\ \frac{1}{{18}^{0.90}}{\left(0.05\right)}^{1.75}=0.000392,\frac{1}{{19}^{0.90}}{\left(0.4\right)}^{1.75}=0.01421\text{.}\end{array}$

Summing all contributions results in the weighted harmonic-voltage factor for single-phase transformers ${\displaystyle \sum _{h=2}^{h\max }}\frac{1}{h^k}{\left(\frac{V_{\mathit{ph}}}{V_{p1}}\right)}^{\ell }\approx 4.1\text{.}$

Calculation of weighted harmonic factor for three-phase spectrum of Table E6.6.1 based on the values for k_avg = 0.90 and ℓ_avg = 1.75 for three-phase transformers:

$\begin{array}{l}\frac{1}{2^{0.90}}{\left(0.5\right)}^{1.75}=0.1593,\frac{1}{3^{0.90}}{\left(0.5\right)}^{1.75}=0.1106,\\ \frac{1}{4^{0.90}}{\left(0.5\right)}^{1.75}=0.0854,....,\frac{1}{{17}^{0.90}}{\left(0.3\right)}^{1.75}=0.00949,\\ \frac{1}{{18}^{0.90}}{\left(0.05\right)}^{1.75}=0.0000234,\frac{1}{{19}^{0.90}}{\left(0.2\right)}^{1.75}=0.00423\text{.}\end{array}$

Summing all contributions results in the weighted harmonic-voltage factor for three-phase transformers ${\displaystyle \sum _{h=2}^{h\max }}\frac{1}{h^k}{\left(\frac{V_{\mathit{ph}}}{V_{p1}}\right)}^{\ell }\approx 3.1\text{.}$

Calculation of weighted harmonic factor for single-phase spectrum of Table E6.6.1 based on the values for k_avg = 1.0 and ℓ_avg = 2.0 for universal motors:

$\begin{array}{l}\frac{1}{2}{\left(0.5\right)}^2=0.125,\frac{1}{3}{\left(3\right)}^2=3.0,\\ \frac{1}{4}{\left(0.3\right)}^2=0.0225,\dots ,\frac{1}{17}{\left(0.5\right)}^2=0.0147,\\ \frac{1}{18}{\left(0.05\right)}^2=0.000139,\frac{1}{19}{\left(0.4\right)}^2=0.00842\text{.}\end{array}$

Summing all contributions results in the weighted harmonic-voltage factor for universal motors:

${\displaystyle \sum _{h=2}^{h\max }}\frac{1}{h^k}{\left(\frac{V_{\mathit{ph}}}{V_{p1}}\right)}^{\ell }\approx 4.4\text{.}$

Based on Fig. 6.14 the above harmonic factors, T₂ = 85 °C, T_amb = 23 °C, E = 1.1 eV, and rated lifetime of t₂ = 40 years result in the additional temperature rises and lifetime reductions of Table E6.6.2.

6.13.1 Application Example 6.7: Temperature Increase of Rotating Machine with a Step Load

An enclosed fan-cooled induction motor has a thermal time constant of τ_θ = 3 h and a steady-state rated temperature of θ_f = 120 °C at an ambient temperature of θ_amb = 40 °C. The motor has at time t = 0_- a temperature of θ = θ_amb, and the motor is fully loaded at t = 0₊. Calculate the time t_95% when the fully loaded motor reaches 95% of its final temperature θ_f = 120 °C.

Solution to Application Example 6.7

With Eq. 6-39 one obtains

$0.95\cdot {\theta }_f={\theta }_f-\left({\theta }_f-{\theta }_{amb}\right)e^{\left(\frac{-t_{95\%}}{{\tau }_{\theta }}\right)}\text{,}$

$0.95\cdot 120=\left(120-40\right)e^{\left(\frac{-t_{95\%}}{{\tau }_{\theta }}\right)}\text{,}$

or $t_{95\%}=2.59\cdot {\tau }_{\theta }=7.77h\text{.}$

6.14.1 Steady-State Operation

Steady-state temperature is reached when the operating time of the machine t_oper is large as compared with the time constant τ_θ of the machine. That is,

$t_{\mathrm{oper}}\ge \left(3-4\right)\cdot {\tau }_{\theta }\text{.}$

According to experience the thermal time constant for openly ventilated machines is

${\tau }_{\theta }\approx 1h$

and for enclosed but ventilated machines

${\tau }_{\theta }\approx \left(3-4\right)h\text{.}$

During steady-state operation the rated output power must be delivered by the machine.

6.14.2 Short-Term Operation

For short-term operation one assumes that the machine cools down to the ambient temperature and its temperature rise (initial temperature rise) is θ_o = 0. Fig. 6.20 illustrates the transient temperature of this operating mode.

The absence of any ventilation during standstill (time t_standstill) requires that the time constant during standstill be longer than that during operation (time t_oper), that is, τ_standstul > τ_oper. For short-term operation the times t_oper and t_standstill must relate to the time constants as follows:

$t_{\mathrm{oper}}<\left(3-4\right){\tau }_{\mathrm{oper}}\text{,}$

$t_{\mathrm{standstill}}<\left(3-4\right){\tau }_{\mathrm{standstill}}\text{.}$

The transient temperature is for short-term operation (see Fig. 6.20):

${\theta }_{\mathrm{short}-\mathrm{term}}={\theta }_{\mathrm{rated}}={\theta }_{\mathrm{short}-\mathrm{term}\_\mathrm{final}}\left[1-e^{\left(-\frac{t_{\mathrm{oper}}}{{\tau }_{\mathrm{oper}}}\right)}\right]\text{.}$

The final short-term temperature is obtained from Eq. 6-45:

${\theta }_{\mathrm{short}-\mathrm{term}\_\mathrm{final}}={\theta }_{\mathrm{rated}}/\left[1-e^{\left(-\frac{t_{\mathrm{oper}}}{{\tau }_{\mathrm{oper}}}\right)}\right]\text{,}$

with

${\theta }_{\mathrm{short}–\mathrm{term}\_\mathrm{final}}\propto P_{\mathrm{loss}\_\mathrm{short}-\mathrm{term}}$

and

${\theta }_{\mathrm{rated}}\propto P_{\mathrm{loss}\_\mathrm{rated}}\text{.}$

It follows for the losses during short-term operation:

$P_{\mathrm{loss}\_\mathrm{short}-\mathrm{term}}=P_{\mathrm{loss}\_\mathrm{rated}}/\left[1-e^{\left(-,\frac{t_{\mathrm{oper}}}{{\tau }_{\mathrm{oper}}}\right)}\right]\text{.}$

For machines whose losses consist of iron-core and copper losses (e.g., induction machines) the total losses at rated operation are

$P_{\mathrm{loss}\_\mathrm{rated}}=p·P_{\mathrm{rated}}+q·P_{\mathrm{rated}}\text{.}$

The first term of Eq. 6-50 pertains to the iron-core and the second term to the copper losses. The losses during short-term operation are

$P_{\mathrm{loss}\_\mathrm{short}-\mathrm{term}}=p\cdot P_{\mathrm{rated}}+q\cdot P_{\mathrm{rated}}{\left(\frac{P_{\mathrm{loss}\_\mathrm{short}-\mathrm{term}}}{P_{\mathrm{rated}}}\right)}^2\text{.}$

Introducing Eqs. 6-50 and 6-51 into Eq. 6-49 yields the ratio

$\frac{P_{\mathrm{loss}\_\mathrm{short}-\mathrm{term}}}{P_{\mathrm{rated}}}=\sqrt{\left[1+\frac{p}{q}\right]\left\{1/\left[1-e^{\left(\frac{t_{\mathrm{oper}}}{{\tau }_{\mathrm{oper}}}\right)}\right]\right\}-\frac{p}{q}}\text{,}$

where the ratio p/q is available from the manufacturer of the machine.

6.14.3 Steady State with Short-Term Operation

In this case the short-term load is superposed with the steady-state load (see Fig. 6.21). The operating time t_oper and the pause time t_pause relate to the time constant as follows:

$t_{\mathrm{oper}}<\left(3-4\right)\cdot {\tau }_{\theta }\text{,}$

$t_{\mathrm{pause}}>\left(3-4\right)\cdot {\tau }_{\theta }\text{,}$

and the ratio between the required power during steady-state with short-term load to the rated power is

$\frac{P_{\mathrm{steady}-\mathrm{state}+\mathrm{short}-\mathrm{time}}}{P_{\mathrm{rated}}}=\sqrt{\left\{1/\left[1-e^{\left(\frac{t_{\mathrm{oper}}}{{\tau }_{\theta }}\right)}\right]\right\}}\text{.}$

In Eq. 6-55 the iron-core losses are neglected (p = 0).

6.14.4 Intermittent Operation

The mode of intermittent operation occurs most frequently. One can differentiate between two cases:

• irregular load steps (see Fig. 6.22), and

• regular load steps.

The case with irregular load steps is discussed in Application Example 6.8.

The case with regular load steps can be approximated as follows based on Fig. 6.23:

$\frac{P_{\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{ermittent}}}{P_{\mathrm{rated}}}=\sqrt{\left(1+\frac{p}{q}\right)\left(1+\frac{t_{\mathrm{pause}}}{t_{\mathrm{oper}}}\cdot \frac{{\tau }_{\theta }}{{\tau }_{\mathrm{standstill}}}\right)-\frac{p}{q}}\text{.}$

6.14.5 Steady State with Intermittent Operation

Steady state with superimposed periodic intermittent operation (see Fig. 6.24) does not occur frequently. The ratio between the power required for this case and the rated power is

$\frac{P_{\mathrm{steady}-\mathrm{state}+\mathrm{periodic}\mathrm{intermittent}}}{P_{\mathrm{rated}}}=\sqrt{\left(1+\frac{t_{\mathrm{pause}}}{t_{\mathrm{oper}}}\right)}$

for t_oper < (3–4) τ_θ and t_pause < (3–4) τ_θ. In summary one can state that for the motor sizing the maximum temperature (and not the torque) is important.

Intermittent operation occurs in drives such as used in hybrid cars, wind-power plants, air-conditioning and refrigeration systems, and others. For this reason a detailed analysis of steady state with superposed intermittent operation will be presented in Application Example 6.8.

6.14.6 Application Example 6.8: Steady State with Superimposed Periodic Intermittent Operation with Irregular Load Steps

A three-phase, squirrel-cage induction motor is fed by a voltage-source inverter. The induction motor is operated at variable frequency and constant (rated) flux (that is, $\left|\tilde{E}\right|/f=\mathrm{constant}$) and has the following nameplate data: V_L-L = 460 V, f = 60 Hz, p = 4 poles, n_{m_rat} = 1720 rpm, P_{out_rat} = 29.594 kW or 39.67 hp. The stator winding is Y-connected, and the parameters per phase are R_s = 0.5 Ω, Rʹ_r = 0.2 Ω, X_s = Xʹ_r = 1 Ω, X_m = 30 Ω at f = 60 Hz, and R_fe → ∞. The axial moment of inertia of the motor is J_m = 0.234 kgm², the viscous damping coefficient B can be assumed to be zero, and the inertia of the load referred to the motor shaft is J_load = 4.766 kgm². The steady-state/intermittent load cycle is shown in Fig. E6.8.1.

The induction motor is operated at 60 Hz (natural torque-speed characteristic).

a) Determine the rated synchronous speed, rated synchronous angular velocity, rated angular velocity, rated slip, rated stator current $\left|{\tilde{I}}_s\right|$, rated rotor current |Ĩ_r′|, and rated induced voltage |Ẽ|.

b) Calculate the maximum torque T_max at s_m and the fictitious maximum torque T_{max_fict} at s = 1.

c) Derive from the equation of motion the slip s(t) as a function of the initial slip s(0), the load torque T_L(t), the maximum fictitious torque T_{max_fict}, the time constant τ_m, and the time t.

d) Apply the solution for s(t) to the different regions of Fig. E6.8.1 and plot s(t) from t = 0 to the maximum time t = 5 s of the load-cycle time period of T_period = 5 s.

e) Calculate the torque T(t) and plot it from t = 0 to t = 5 s.

f) Based on the slip function compute pointwise the stator current |Ĩ_s(t)| and plot |Ĩ_s(t)| from t = 0 to t = 5 s.

g) Plot |Ĩ_s(t)|² as a function of time from t = 0 to t = 5 s.

h) Determine the rms value of the motor (stator) current i_s(t) for the entire load cycle as specified in Fig. E6.8.1; that is, determine

$I_{s\_rms}=\sqrt{\frac{1}{T}{\displaystyle \underset{t}{\overset{t+T}{\int }}\left|{\tilde{I}}_s\left(t\right)\right|{}^{2}dt}}\text{.}$

i) Is this induction motor over- or underdesigned? If it is underdesigned, what is the reduction of lifetime?