Calculation of Protection of Generators and Transformers Against Ground Faults

I. D. Kutyavin
Submitted 1945 | SovietRxiv: ru-194501.08442 | Translated from Russian

Abstract Generated abstract

This paper develops a revised analytical method for calculating relay protection of generators and transformers against ground faults, arguing that graphical analytical procedures are unnecessarily laborious at the design stage when theoretical relay characteristics are used. It derives design equations for wattmetric protection schemes, introducing simplified calculated currents, selectivity and sensitivity conditions, dead zone estimates, relay pickup power relations, and adjustments for multiple grounded neutral points. The method is then extended to transformer protection and illustrated with a numerical generator station example, showing how grounding resistance, relay settings, and permissible arc current constrain the choice of protection schemes. The author concludes that the analytical formulation provides a simpler and more intuitive calculation procedure with substantially less labor than the method recommended in contemporary relay protection guidelines.

Full Text

Calculation of Protection of Generators and Transformers Against Ground Faults

I. D. Kutyavin
Associate Professor, Candidate of Technical Sciences

The Guidelines for Relay Protection (R. U.) in the new edition (Vol. II, Section G—4, § 5) recommend using, for calculating the protection of generators against faults to the frame, only the graphical-analytical method of calculation, considering it more advanced. This method of calculation should certainly be used for selecting the basic protection parameters according to the actual relay characteristic when setting up protective devices; however, using it in the design of protection, when the theoretical relay characteristic is used, is pointless, since the analytical method requires considerably less labor for the calculation and gives results that are no less accurate.

The analytical method for calculating protection, given in the first edition of the R. U., has by now become obsolete and therefore is evidently not recommended for use. In the present work this method is given in a revised form, taking into account the basic requirements of the R. U. However, the article gives only the derivation of the basic equations for calculation, since the method for selecting individual calculation data can be taken from the R. U.

Determination of Calculation Data

A major drawback of the calculation method recommended by the R. U. is the introduction of a whole series of current concepts \((I_{\mathrm{zgt}}, I^{(t)}_{\mathrm{zgp}}, I_{\mathrm{zst}}, I_{\mathrm{zrasch}}\ \text{and}\ I_{\mathrm{z p}})\), which greatly complicates the protection calculation and often leads to an incorrect understanding of the question. In the present work only two calculated currents are introduced:

\(I_{\mathrm{zm}}\)—the maximum ground-fault current flowing through the external current transformers when a generator terminal is faulted to the frame, if only one grounded neutral point is connected to the generator-voltage network.

\(I_{\mathrm{dm}}\)—the maximum permissible current in the arc at the point of insulation breakdown to the frame when a generator terminal is faulted, if all grounded neutral points are connected to the generator-voltage network. This current is determined according to the R. U. (Vol. II, Section G—1, § 5).

Both currents are related by the equation

\[ I_{\mathrm{dm}} = n_{\mathrm{m}} I_{\mathrm{zm}}, \tag{1} \]

where \(n_{\mathrm{m}}\) is the maximum number of grounded neutral points in the generator-voltage system, connected simultaneously to the buses and having identical grounding resistances.

For different grounding resistances, the maximum number of grounded neutral points is determined from the following:

\[ n_{\mathrm{m}}=R_1\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\ldots\right), \tag{2} \]

where \(R_1\) is the grounding resistance of the neutral point under consideration,
\(R_2, R_3 \ldots\) are the grounding resistances of the other neutral points.

The voltage across the grounding resistance when a short circuit occurs to the frame in \(W_x\) turns of the stator winding through some resistance at the fault point is

\[ U_{\mathrm{z}}=\frac{U_{\mathrm{f}}}{W_{\mathrm{f}}} W_x K_{\mathrm{p}}=U_{\mathrm{f}}K_{\mathrm{p}}x, \tag{3} \]

where \(U_{\mathrm{f}}\) is the phase voltage of the generator,
\(W_{\mathrm{f}}\) is the total number of turns in a phase of the stator winding,
\(x\) is the relative number of turns short-circuited to the frame,
\(K_{\mathrm{p}}\) is the coefficient of completeness of the ground fault.

The voltage at the terminals of the wattmetric relay is

\[ U_{\mathrm{r}}=\frac{U_{\mathrm{z}}}{n_{\mathrm{n}}}= \frac{U_{\mathrm{f}}K_{\mathrm{p}}x}{n_{\mathrm{n}}} =U_{\mathrm{rm}}K_{\mathrm{p}}x, \tag{4} \]

where \(n_{\mathrm{n}}\) is the transformation ratio of the zero-sequence voltage transformer.

The maximum voltage at the relay terminals for \(x=1\) and \(K_{\mathrm{p}}=1\) is:

\[ U_{\mathrm{rm}}=\frac{U_{\mathrm{f}}}{n_{\mathrm{n}}}. \tag{5} \]

The current flowing into the current circuit of the protection from the side of the zero-sequence voltage transformer, on the basis of Fig. 1, is

\[ i_{\mathrm{n}}= \frac{U_{\mathrm{r}}}{ r_{\mathrm{d}}+\dfrac{0.33Z_x-Z_{\mathrm{r}}}{0.33Z_x+Z_{\mathrm{r}}} } \simeq \frac{U_{\mathrm{r}}}{r_{\mathrm{d}}} = \frac{U_{\mathrm{rm}}}{r_{\mathrm{d}}}K_{\mathrm{p}}x =i_{\mathrm{nm}}K_{\mathrm{p}}x, \tag{6} \]

where \(r_{\mathrm{d}}\) is the additional resistance in the circuit of the zero-sequence voltage transformer,
\(Z_x\) is the no-load impedance of the external transformers,
\(Z_{\mathrm{r}}\) is the resistance of the current coil of the relay.

The maximum value of this current, on the basis of (6), for \(x=1\) and \(K_{\mathrm{p}}=1\) is

\[ i_{\mathrm{nm}}\simeq\frac{U_{\mathrm{rm}}}{r_{\mathrm{d}}}. \tag{7} \]

The current flowing into the current coil of the relay from the side of the zero-sequence voltage transformer is

\[ i_{\mathrm{nr}}=i_{\mathrm{n}}\frac{0.33Z_x}{0.33Z_x+Z_{\mathrm{r}}} =i_{\mathrm{nm}}K_{\mathrm{p}}K_0x, \tag{8} \]

where

\[ K_0=\frac{0.33Z_x}{0.33Z_x+Z_{\mathrm{r}}} \]

is the tap coefficient.

The current in the grounding resistance under consideration is

\[ I_{\mathrm{z}}=\frac{U_{\mathrm{z}}}{R_1} =\frac{U_{\mathrm{f}}K_{\mathrm{p}}x}{R_1} =I_{\mathrm{zm}}K_{\mathrm{p}}x; \tag{9} \]

the maximum value of this current on the basis of (9), for \(K_{\mathrm{p}}=1\) and \(x=1\):

\[ I_{zm}=\frac{U_{\phi}}{R_1}. \tag{10} \]

The current flowing into the current coil of the relay from the external current transformers and caused by the flow in the primary circuit of the current \(I_z\):

\[ i_{vr}=\frac{I_z}{n_t}K_0K_{\mathrm{p}} =\frac{I_{zm}}{n_t}K_0K_{\mathrm{p}}x =i_{vm}K_0K_{\mathrm{p}}x. \tag{11} \]

The maximum value of this current for \(x=1\) and \(K_{\mathrm{p}}=1\)

\[ i_{vm}=\frac{I_{zm}}{n_t}, \tag{12} \]

where \(n_t\) is the transformation ratio of the external current transformers.

Calculation of Protection According to the Circuit of Fig. 1

a) Ground fault outside the protection zone of the generator under consideration (for example, in the stator winding of a neighboring generator) through some transient resistance.

The current in the current winding of the wattmeter relay is

\[ i'_r=i_{nr}-K_{\tau}i_{vr}+i_{nb} =(i_{nm}-K_{\tau}i_{vm})K_0K_{\mathrm{p}}x+i_{nb}, \tag{13} \]

where \(K_{\tau}\) is a coefficient taking into account the heating of the grounding resistance by the current. For cast-iron spirals this coefficient, according to the Operating Rules, is equal to 0.8, while for alloys with a small temperature coefficient of resistance it is equal to unity.

In order that the protection of the generator under consideration should not operate, the current \(i'_r\), according to (13), must for all values of \(x\) be less than the pickup current of the wattmeter relay (the selectivity condition)

\[ i'_r \leq \beta i_{rt}=\beta\frac{P_{rt}}{U_{rm}x}, \tag{14} \]

where \(\beta\) is the reserve coefficient.

On the basis of (13) and (14)

\[ \beta=\frac{i'_r}{i_{rt}} =\frac{\left[(i_{nm}-K_{\tau}i_{vm})K_0K_{\mathrm{p}}x+i_{nb}\right]U_{rm}x}{P_{rt}}. \tag{15} \]

In Fig. 2 the following curves are plotted as functions of \(x\):

1) the selectivity straight line on the basis of (13);

2) the pickup-current curve of the wattmeter relay

\[ i_{rt}=\frac{P_{rt}}{U_{rm}x}; \]

3) the curve of variation of the reserve coefficient on the basis of (15). As is seen from Fig. 2, the reserve coefficient \(\beta\), at a certain value of \(x\), reaches a maximum. On the basis of the selectivity condition (14), the maximum value of the reserve coefficient must be less than unity.

Let us determine the \(x\) at which \(\beta\), according to expression (15), reaches a maximum:

\[ \frac{d\beta}{dx} = \frac{\left[2(i_{nm}-K_{\tau}i_{vm})K_0K_{\mathrm{p}}x+i_{nb}\right]U_{rm}}{P_{rt}} =0, \]

whence

\[ x_{\beta m}=-\frac{i_{nb}}{2(i_{nm}-K_{\tau}i_{vm})K_0K_{\mathrm{p}}}. \tag{16} \]

On the basis of (15) and (16)

\[ i_{\mathrm{vm}}= \frac{i_{\mathrm{nb}}^{2} U_{\mathrm{rm}}}{4\beta_{\mathrm{m}} K_{0} K_{\mathrm{p}} K_{\mathrm{t}} P_{\mathrm{rt}}} + \frac{i_{\mathrm{nm}}}{K_{\mathrm{t}}}. \tag{17} \]

b) A ground fault in the winding of the generator under consideration through some transient resistance, with only one grounded neutral point in operation (the neutral point of the generator under consideration).

Fig. 1

Fig. 1

Fig. 2

Fig. 2

The current in the current winding of the wattmetric relay for a ground fault at the boundary of the dead zone is

\[ i''_{\mathrm{p}}=i_{\mathrm{nm}} K_{0} K_{\mathrm{p}} x-i_{\mathrm{nb}} . \tag{18} \]

In order for the protection to operate, the sensitivity condition must be satisfied:

\[ i''_{\mathrm{p}} \geqslant i_{\mathrm{rt}}= \frac{P_{\mathrm{rt}}}{U_{\mathrm{rm}} x}. \tag{19} \]

On the basis of (18) and (19)

\[ i_{\mathrm{nm}}= \frac{1}{K_{0}K_{\mathrm{p}}} \left( \frac{P_{\mathrm{rt}}}{x_{0}^{2} U_{\mathrm{rm}}} + \frac{i_{\mathrm{nb}}}{x_{0}} \right). \tag{20} \]

On the basis of (12), (17), and (20)

\[ I_{\mathrm{zm}}= \frac{n_{\mathrm{t}}}{K_{\mathrm{p}}K_{0}K_{\mathrm{t}}} \left( \frac{i_{\mathrm{nb}}^{2} U_{\mathrm{rm}}}{4\beta P_{\mathrm{rt}}} + \frac{P_{\mathrm{rt}}}{x_{0}^{2} U_{\mathrm{rm}}} + \frac{i_{\mathrm{nb}}}{x_{0}} \right). \tag{21} \]

From expression (21), for a known current \(I_{\mathrm{zm}}\), the dead zone \(x_{0}\) or the relay operating power \(P_{\mathrm{rt}}\) can be determined.

Expressions (20) and (21), which relate the main protection parameters to one another, are the design equations.

The structure of expression (21) makes it possible to determine the optimal operating power at which the current \(I_{\mathrm{zm}}\) has a minimum value. Opti-

the minimum pickup power is determined by differentiating expression (21):

\[ \frac{d I_{\mathrm{zm}}}{d P_{\mathrm{rt}}} = -\frac{i_{\mathrm{nb}}^{2} U_{\mathrm{rm}}}{4 \beta_{\mathrm{m}} P_{\mathrm{rt}}^{2}} + \frac{1}{x_{0}^{2} U_{\mathrm{rm}}} =0, \]

whence

\[ P_{\mathrm{rt}}=\frac{i_{\mathrm{nb}} \cdot U_{\mathrm{rm}} x_{0}}{2\sqrt{\beta_{\mathrm{m}}}} . \tag{22} \]

On the basis of (20) and (22),

\[ i_{\mathrm{nm}} = \frac{i_{\mathrm{nb}}\left(0.5+\sqrt{\beta_{\mathrm{m}}}\right)} {K_{0}K_{\mathrm{p}}x_{0}\sqrt{\beta_{\mathrm{m}}}} . \tag{23} \]

On the basis of (21) and (22),

\[ I_{\mathrm{zm}} = \frac{n_{\mathrm{t}} i_{\mathrm{nb}}\left(1+\sqrt{\beta_{\mathrm{m}}}\right)} {K_{0}K_{\mathrm{p}}K_{\mathrm{t}}x_{0}\sqrt{\beta_{\mathrm{m}}}} . \tag{24} \]

Unfortunately, equations (22), (23), and (24) cannot be used for calculation, since the pickup power according to (22) is obtained within the limits of 0.1 W, whereas the minimum pickup power for the IM-143 relay of the KhEMZ plant is 0.15 W.

c) Ground fault in the protection zone of the generator under consideration, with the number of grounded neutral points equal to \(n\).

The sensitivity equation is

\[ \left[ i_{\mathrm{nm}}+(n-1)\frac{I_{\mathrm{zm}}}{n_{\mathrm{t}}} \right] K_{0}K_{\mathrm{p}}x_{\mathrm{op}} - i_{\mathrm{nb}} = \frac{P_{\mathrm{rt}}}{x_{\mathrm{op}}U_{\mathrm{rm}}}, \tag{25} \]

where \(n\), for unequal grounding resistances of the neutral points, is determined according to (2).

Solving equation (25) with respect to \(x_{\mathrm{op}}\) gives the expression for the magnitude of the dead zone for any number of grounded neutral points:

\[ x_{\mathrm{op}} = \frac{ i_{\mathrm{nb}} + \sqrt{ i_{\mathrm{nb}}^{2} + 4\frac{P_{\mathrm{rt}}}{U_{\mathrm{rm}}} K_{0}K_{\mathrm{p}} \left[ i_{\mathrm{nm}}+(n-1)\frac{I_{\mathrm{m}}}{n_{\mathrm{t}}} \right] } }{ 2K_{0}K_{\mathrm{p}} \left[ i_{\mathrm{nm}}+(n-1)\frac{I_{\mathrm{zm}}}{n_{\mathrm{t}}} \right] }. \tag{26} \]

Calculation of the protection according to the circuit of Fig. 3

a) Ground fault in the winding of an adjacent generator through a certain transition resistance, taken into account by the fault-completeness coefficient \(K_{\mathrm{p}}\).

The braking current flowing into the current circuit of the protection from the side of the zero-sequence voltage transformer:

\[ i_{\mathrm{t}} \cong \frac{U_{\mathrm{p}}}{r_{\mathrm{d}}} = \frac{U_{\mathrm{rm}}}{r_{\mathrm{d}}}K_{\mathrm{p}}x = i_{\mathrm{tm}}K_{\mathrm{p}}x . \tag{27} \]

The maximum value of this current at \(x=1\) and \(K_{\mathrm{p}}=1\) is

\[ i_{\mathrm{tm}}\cong\frac{U_{\mathrm{rm}}}{r_{\mathrm{d}}}. \tag{28} \]

The current flowing into the current coil of the relay is

\[ i'_{\mathrm{p}} = -i_{\mathrm{tm}}K_{\mathrm{p}}K_{0}x + i_{\mathrm{nb}} . \tag{29} \]

In order that the protection of the generator under consideration not operate, it is necessary that \(i'_p\) from (29), for all values of \(x\), be less than the pickup current of the wattmetric relay (selectivity condition):

\[ i'_p \ll \beta i_{pt}=\beta \frac{P_{pt}}{U_{pm}x}. \tag{30} \]

Solving the problem analogously to the preceding case, we obtain the following expression for the maximum braking current:

\[ i_{tm}=\frac{i_{kb}^{2}U_{pm}}{4\beta_m K_0K_nP_{pt}}. \tag{31} \]

b) Fault to the frame in the winding of the generator under consideration through some transition resistance, with one grounded neutral point in operation.

The current in the current coil of the wattmetric relay during a ground fault at the boundary of the dead zone is

\[ i''_p=(-i_{tm}+i_{zm})K_0K_nx_0-i_{nb} \tag{32} \]

In order for the protection to operate, the sensitivity condition must be satisfied:

\[ i''_p \gg i_{pt}=\frac{P_{pt}}{U_{pm}x_0}. \tag{33} \]

Fig. 3

Fig. 3

On the basis of (12), (13), (32), and (33),

\[ I_{zm}=\frac{n_t}{K_0K_n} \left( \frac{i_{nb}^{2}U_{pm}}{4\beta_mP_{pt}} +\frac{P_{pt}}{U_{pm}x_0^{2}} +\frac{i_{nb}}{x_0} \right). \tag{34} \]

Equations (31) and (34) are the calculation equations for this scheme. Analogously to the preceding case, the optimum pickup power is

\[ P_{pt}=\frac{x_0U_{pm}i_{nb}}{2\sqrt{\beta_m}}. \tag{35} \]

The maximum braking current at the optimum pickup power is

\[ i_{tm}=\frac{i_{nb}}{2K_0K_nx_0\sqrt{\beta_m}}. \tag{36} \]

The maximum ground-fault current for the neutral point under consideration is

\[ I_{zm}=\frac{n_t i_{nb}(1+\sqrt{\beta_m})}{K_0K_nx_0\sqrt{\beta_m}}. \tag{37} \]

The value of the dead zone when the calculated number of grounded neutral points is equal to \(n\) is

\[ x_{0n}= \frac{ i_{nb}+ \sqrt{ i_{nb}^{2} +4\frac{P_{pt}}{U_{pm}}K_0K_n \left( -i_{tm}+n\frac{I_{zm}}{n_t} \right) } }{ 2K_0K_n \left( -i_{tm}+n\frac{I_{zm}}{n_t} \right) }. \tag{38} \]

Calculation of Transformer Protection Against Ground Faults

Figure 4 shows a scheme for protecting transformers against ground faults, recommended by the PUE (Vol. II, Section T-1). This scheme is exactly the same as the generator protection scheme in Fig. 3; therefore equations (31) and (34) are also the calculation equations for the present case.

In the event of a fault to the frame in the winding of a transformer connected in delta, the minimum voltage at the relay terminals is \(0.5 U_{\mathrm{pm}}\); therefore, in order to increase reliability, it is desirable in calculating the protection to take the dead-zone value \(x_0 < 0.4\).

Fig. 4

Fig. 4

Example. Carry out the calculation of generator protection against faults to the frame, if it is known that the station has 3 generators installed, each with a capacity of 12 MW. The rated current of the generators is 1375 A and the rated voltage is 6.3 kV.

Calculation data (determined on the basis of the PUE).

The transformation ratio of the external current transformers is

\[ n_{\mathrm{t}}=\frac{1500}{10}\ \mathrm{A} \]

and that of the zero-sequence voltage transformer is

\[ n_{\mathrm{n}}=\frac{6000}{100}\ \mathrm{V}. \]

The tapping coefficient of the external current transformers is \(K_0 = 0.8\). The safety factor is \(\beta_{\mathrm{m}} = 0.5\). The coefficient of completeness of the ground fault is \(K_{\mathrm{p}} = 1\). The calculated unbalance current is \(0.02\) A. The operating power of the wattmetric relay is \(P_{\mathrm{rt}} = 0.2\) W. The permissible dead-zone value is \(x_0 = 0.13\). The maximum permissible arc current is \(I_{\mathrm{dm}} = 180\) A.

The grounding resistances are assumed to be from rheostat fields. The maximum voltage at the relay terminals, on the basis of (5), is

\[ U_{\mathrm{pm}}=\frac{6300}{\sqrt{3}\cdot 60}=61\ \mathrm{V}. \]

The maximum ground-fault current, when one generator is operating alone, on the basis of (21) and when using the scheme of Fig. 1, is

\[ I_{\mathrm{zm}}=\frac{150}{0.8} \left( \frac{0.02\cdot 61}{4.05\cdot 0.2} + \frac{0.2}{0.13^2\cdot 61} + \frac{0.02}{0.13} \right) =77\ \mathrm{A}. \]

The actual value of the current in the arc, on the basis of (1), when all three generators with grounded neutral points are operating in parallel, is

\[ I_{\mathrm{d}}=3\cdot 77=231>I_{\mathrm{dm}}=180\ \mathrm{A}. \]

Consequently, the use of the protection scheme according to Fig. 1 is possible only for two generators (Nos. 1 and 2); therefore, for the third generator the protection scheme according to Fig. 3 is adopted. Then the actual current in the arc is

\[ I_{\mathrm{d}}=2\cdot 77=154\ \mathrm{A}. \]

The grounding resistance of the neutral points of generators Nos. 1 and 2, on the basis of (10), is

\[ R_1=R_2=\frac{6300}{\sqrt{3}\cdot 77}=47.3\ \Omega. \]

The maximum value of the zero-sequence current in the secondary circuit of the protection, on the basis of (20),

\[ i_{\mathrm{nm}}=\frac{1}{0.8}\left(\frac{0.2}{0.13^2\cdot 61}+\frac{0.02}{0.13}\right)\simeq 0.435\ \mathrm{A}. \]

The additional resistance in the secondary circuit of the protection for generators No. 1 and 2, on the basis of (7), is

\[ r_{\mathrm{d}}=\frac{61}{0.435}=140\ \Omega. \]

The magnitude of the dead zone of the protection of generators No. 1 and 2 when two grounded neutral points are in operation, on the basis of (26), is

\[ x_{02}= \frac{ 0.02+\sqrt{0.02^2+4\,\frac{0.2\cdot 0.8}{61}\left[0.435+(2-1)\frac{77}{150}\right]} }{ 2\cdot 0.8\left[0.435+(2-1)\frac{77}{150}\right] } =0.08. \]

When generator No. 3 operates alone, the current required for protection against a ground fault is produced by grounding the neutral point of the auxiliary-service transformer through a resistance of 47.3 ohms.

The maximum restraining current in the current circuit of the protection of generator No. 3, on the basis of (31), is

\[ i_{\mathrm{k}}=\frac{0.02^2\cdot 61}{4\cdot 0.5\cdot 0.2\cdot 0.8}=0.0763\ \mathrm{A}. \]

The additional resistance in the secondary circuit of the protection of generator No. 3, on the basis of (26), is

\[ r_{\mathrm{d}}=\frac{61}{0.0736}=800\ \Omega. \]

The magnitude of the dead zone of the protection of generator No. 3 when two grounded neutral points are in operation, on the basis of (38), is

\[ x_{02}= \frac{ 0.02+\sqrt{0.02^2+4\,\frac{0.2\cdot 0.8}{61}\left(-0.0763+2\frac{77}{150}\right)} }{ 2\cdot 0.8\left(-0.0763+2\frac{77}{150}\right) } =0.08. \]

Conclusion

The analytical calculation method presented makes it possible to calculate the protection with a considerably smaller expenditure of labor. This method is simpler and more intuitive in comparison with that recommended by the R. U. of the 1939 edition.

Noticed Misprints

Page Line Printed Should be
89 Fig. 3 (Figure inverted)
92 7 from bottom $0_2$ $-i_{02}$
95 19 from bottom $Z$ $Z_{\mathrm{T}}$
97 In equation (27), $\alpha$ is an exponent
98 19 from bottom $Z_m\Delta$ $Z_mY$
101 15 from bottom surrounding circles
178 17 from bottom $$s=\frac{n}{n}$$ $$s=\frac{n}{n_0}$$
193 4 from top $$I_2^2=\frac{E}{1+Z^2}$$ $$I_2^2=\frac{\varepsilon}{1+Z^2}$$
203 12 from top $$I=I_{\mathrm{M}}\left(1-e^{-\frac{r_{\mathrm{M}}}{L_{\mathrm{M}}}}\right)=\ldots$$ $$I=I_{\mathrm{M}}\left(1-e^{-\frac{r_{\mathrm{M}}}{L_{\mathrm{M}}}t}\right)=\ldots$$
203 17 from bottom concluding concluding
214 8 from bottom (28) (26)
221 4 from top $$\ldots\frac{i_0^2-i''_a}{i_{01}-i''_a}$$ $$\ldots\frac{i_{02}-i''_a}{i_{01}-i''_a}$$
318 Caption to Fig. 5 Heat losses due to incomplete combustion. Heat losses from incomplete combustion.
318 14 from bottom (Fig. 3) (Fig. 4)
323 1 from top ature perature
323 1 from top $r=58^\circ$ $r=587^\circ$
325 In Fig. 6 $q'_{\mathrm{water}}$ and $q''_{\mathrm{water}}$ $q'_{\mathrm{water}}$ and $q''_{\mathrm{water}}$
326 33 from top $/2\ 3+25=293\ \mathrm{K},$ $/273+25=293\ \mathrm{K}/,$

Submission history

Calculation of Protection of Generators and Transformers Against Ground Faults