Electric Machinery Fundamentals Power & Energy_12 pptx

A MATLAB program to calculate the torque-speed characteristic of this motor is shown below: % M-file: prob9_19.m % M-file to create a plot of the torque-speed curve of the % a series dc

(a) At full load, I A=I L−I F=100 A − 0.75 A = 99.25 A, and

F

V I

This field current would produce a voltage EAo of 290 V at a speed of no = 1200 r/min The actual EA

is 240 V, so the actual speed at full load will be

% M-file to create a plot of the torque-speed curve of the

% a cumulatively compounded dc motor

% Get the magnetization curve

r_f = 200; % Field resistance (ohms)

r_adj = 120; % Adjustable resistance (ohms)

r_a = 0.19; % Armature + series resistance (ohms)

i_l = 0:2:100; % Line currents (A)

n_f = 1500; % Number of turns on shunt field

n_se = 12; % Number of turns on series field

% Calculate the armature current for each load

i_a = i_l - v_t / (r_f + r_adj);

% Now calculate the internal generated voltage for

% each armature current

e_a = v_t - i_a * r_a;

% Calculate the effective field current for each armature

% current

i_f = v_t / (r_f + r_adj) + (n_se / n_f) * i_a;

% Calculate the resulting internal generated voltage at

% 1800 r/min by interpolating the motor's magnetization

% curve

e_a0 = interp1(if_values,ea_values,i_f);

% Calculate the resulting speed from Equation (9-13)

n = ( e_a / e_a0 ) * n_0;

% Calculate the induced torque corresponding to each

% speed from Equations (8-55) and (8-56)

t_ind = e_a * i_a / (n * 2 * pi / 60);

% Plot the torque-speed curves

The resulting torque-speed characteristic is shown below:

(c) The no-load speed of this machine is the same as the no-load speed of the corresponding shunt dc

motor with Radj = 120 Ω, which is 1125 r/min The speed regulation of this motor is thus

% a differentially compounded dc motor

% Get the magnetization curve

r_f = 200; % Field resistance (ohms)

r_adj = 120; % Adjustable resistance (ohms)

r_a = 0.19; % Armature + series resistance (ohms)

i_l = 0:2:40; % Line currents (A)

n_f = 1500; % Number of turns on shunt field

n_se = 12; % Number of turns on series field

% Calculate the armature current for each load

i_a = i_l - v_t / (r_f + r_adj);

% Now calculate the internal generated voltage for

% each armature current

e_a = v_t - i_a * r_a;

% Calculate the effective field current for each armature

% current

i_f = v_t / (r_f + r_adj) - (n_se / n_f) * i_a;

% Calculate the resulting internal generated voltage at

% 1800 r/min by interpolating the motor's magnetization

% curve

e_a0 = interp1(if_values,ea_values,i_f);

% Calculate the resulting speed from Equation (9-13)

n = ( e_a / e_a0 ) * n_0;

% Calculate the induced torque corresponding to each

% speed from Equations (8-55) and (8-56)

t_ind = e_a * i_a / (n * 2 * pi / 60);

% Plot the torque-speed curves

The resulting torque-speed characteristic is shown below:

This curve is plotted on the same scale as the torque-speed curve in Problem 6-17 Compare the two curves

9-19 A series motor is now constructed from this machine by leaving the shunt field out entirely Derive the

torque-speed characteristic of the resulting motor

SOLUTION This motor will have extremely high speeds, since there are only a few series turns, and the flux

in the motor will be very small A MATLAB program to calculate the torque-speed characteristic of this motor is shown below:

% M-file: prob9_19.m

% M-file to create a plot of the torque-speed curve of the

% a series dc motor This motor was formed by removing

% the shunt field from the cumulatively-compounded machine

r_a = 0.19; % Armature + series resistance (ohms)

i_l = 20:1:45; % Line currents (A)

n_f = 1500; % Number of turns on shunt field

n_se = 12; % Number of turns on series field

% Calculate the armature current for each load

i_a = i_l;

% Now calculate the internal generated voltage for

% each armature current

e_a = v_t - i_a * r_a;

% Calculate the effective field current for each armature

% current (Note that the magnetization curve is defined

% in terms of shunt field current, so we will have to

% translate the series field current into an equivalent

% shunt field current

i_f = (n_se / n_f) * i_a;

% Calculate the resulting internal generated voltage at

% 1800 r/min by interpolating the motor's magnetization

% curve

e_a0 = interp1(if_values,ea_values,i_f);

% Calculate the resulting speed from Equation (9-13)

n = ( e_a / e_a0 ) * n_0;

% Calculate the induced torque corresponding to each

% speed from Equations (8-55) and (8-56)

t_ind = e_a * i_a / (n * 2 * pi / 60);

% Plot the torque-speed curves

The extreme speeds in this characteristic are due to the very light flux in the machine To make a practical series motor out of this machine, it would be necessary to include 20 to 30 series turns instead of 12

9-20 An automatic starter circuit is to be designed for a shunt motor rated at 15 hp, 240 V, and 60 A The

armature resistance of the motor is 0.15 Ω, and the shunt field resistance is 40 Ω The motor is to start with no more than 250 percent of its rated armature current, and as soon as the current falls to rated value,

a starting resistor stage is to be cut out How many stages of starting resistance are needed, and how big should each one be?

SOLUTION The rated line current of this motor is 60 A, and the rated armature current is I A=I L− = 60 I F

A – 6 A = 54 A The maximum desired starting current is (2.5)(54 A) = 135 A Therefore, the total initial starting resistance must be

9-21 A 15-hp 120-V 1800 r/min shunt dc motor has a full-load armature current of 60 A when operating at rated

conditions The armature resistance of the motor is R = 0.15 A Ω, and the field resistance R is 80 F Ω The adjustable resistance in the field circuit Radj may be varied over the range from 0 to 200 Ω and is currently set to 90 Ω Armature reaction may be ignored in this machine The magnetization curve for this motor, taken at a speed of 1800 r/min, is given in tabular form below:

Note: An electronic version of this magnetization curve can be found in file

prob9_21_mag.dat, which can be used with MATLAB programs Column

1 contains field current in amps, and column 2 contains the internal generated

voltage EA in volts

(a) What is the speed of this motor when it is running at the rated conditions specified above?

(b) The output power from the motor is 7.5 hp at rated conditions What is the output torque of the motor? (c) What are the copper losses and rotational losses in the motor at full load (ignore stray losses)?

(d) What is the efficiency of the motor at full load?

(e) If the motor is now unloaded with no changes in terminal voltage or Radj, what is the no-load speed of the motor?

(f) Suppose that the motor is running at the no-load conditions described in part (e) What would happen

to the motor if its field circuit were to open? Ignoring armature reaction, what would the final state speed of the motor be under those conditions?

steady-(g) What range of no-load speeds is possible in this motor, given the range of field resistance adjustments

available with R ? adj

F

V I

(c) The copper losses are

( ) (2 ) ( )( )2

Therefore, the rotational losses are

rot conv OUT 13, 260 W 11,190 W 2070 W

11,190 W

14,100 W

P P

A o Ao

E

E

(In reality, the motor speed would be limited by rotational losses, or else the motor will destroy itself first.)

(g) The maximum value of R = 200 Ω, so adj

F

V I

This field current would produce a voltage E Ao of 153 V at a speed of n o = 1800 r/min The actual E is A

230 V, so the actual speed will be

F

V I

E

9-22 The magnetization curve for a separately excited dc generator is shown in Figure P9-7 The generator is

rated at 6 kW, 120 V, 50 A, and 1800 r/min and is shown in Figure P9-8 Its field circuit is rated at 5A The following data are known about the machine:

Note: An electronic version of this magnetization curve can be found in file

p97_mag.dat, which can be used with MATLAB programs Column 1 contains field current in amps, and column 2 contains the internal generated

voltage EA in volts

R A=0.18 Ω V F =120 V

adj 0 to 30

N F =1000 turns per pole

Answer the following questions about this generator, assuming no armature reaction

(a) If this generator is operating at no load, what is the range of voltage adjustments that can be achieved

by changing Radj?

(b) If the field rheostat is allowed to vary from 0 to 30 Ω and the generator’s speed is allowed to vary from

1500 to 2000 r/min, what are the maximum and minimum no-load voltages in the generator?

SOLUTION

(a) If the generator is operating with no load at 1800 r/min, then the terminal voltage will equal the

internal generated voltage EA The maximum possible field current occurs when Radj = 0 Ω The current

F

V I

F

V I

(b) The maximum voltage will occur at the highest current and speed, and the minimum voltage will

occur at the lowest current and speed The maximum possible field current occurs when R = 0 adj Ω The current is

F

V I

V I

9-23 If the armature current of the generator in Problem 9-22 is 50 A, the speed of the generator is 1700 r/min,

and the terminal voltage is 106 V, how much field current must be flowing in the generator?

SOLUTION The internal generated voltage of this generator is

From the magnetization curve, this value of E Ao requires a field current of 4.2 A

9-24 Assuming that the generator in Problem 9-22 has an armature reaction at full load equivalent to 400

A⋅turns of magnetomotive force, what will the terminal voltage of the generator be when I = 5 A, F n = m

9-25 The machine in Problem 9-22 is reconnected as a shunt generator and is shown in Figure P9-9 The shunt

field resistor Radj is adjusted to 10 Ω, and the generator’s speed is 1800 r/min

(a) What is the no-load terminal voltage of the generator?

(b) Assuming no armature reaction, what is the terminal voltage of the generator with an armature current

(b) At an armature current of 20 A, the internal voltage drop in the armature resistance is

( 20 A )( 0.18 Ω ) = 3 6 V As shown in the figure below, there is a difference of 3.6 V between EA and

T

V at a terminal voltage of about 106 V

A MATLAB program to locate the position where the triangle exactly fits between the E A and V T lines is shown below This program created the plot shown above Note that there are actually two places where the difference between the EA and VT lines is 3.6 volts, but the low-voltage one of them is unstable The code shown in bold face below prevents the program from reporting that first (unstable) point

% M-file: prob9_25b.m

% M-file to create a plot of the magnetization curve and the

% field current curve of a shunt dc generator, determining

% the point where the difference between them is 3.6 V

% Get the magnetization curve This file contains the

% three variables if_values, ea_values, and n_0

% First, initialize the values needed in this program

r_f = 24; % Field resistance (ohms)

r_adj = 10; % Adjustable resistance (ohms)

r_a = 0.19; % Armature + series resistance (ohms)

i_f = 0:0.02:6; % Field current (A)

n = 1800; % Generator speed (r/min)

% Calculate Ea versus If

Ea = interp1(if_values,ea_values,i_f);

% Calculate Vt versus If

Vt = (r_f + r_adj) * i_f;

% Find the point where the difference between the two

% lines is 3.6 V This will be the point where the line

% line "Ea - Vt - 3.6" goes negative That will be a

% close enough estimate of Vt

diff = Ea - Vt - 3.6;

% This code prevents us from reporting the first (unstable)

% location satisfying the criterion

% We have the intersection Tell user

disp (['Ea = ' num2str(Ea(ii)) ' V']);

disp (['Vt = ' num2str(Vt(ii)) ' V']);

disp (['If = ' num2str(i_f(ii)) ' A']);

% Plot the curves

plot([i_f(ii) i_f(ii)], [0 Ea(ii)], 'k-');

plot([0 i_f(ii)], [Vt(ii) Vt(ii)],'k-');

plot([0 i_f(ii)], [Ea(ii) Ea(ii)],'k-');

(c) The rated current of this generated is 50 A, so 20 A is 40% of full load If the full load armature

reaction is 200 A⋅turns, and if the armature reaction is assumed to change linearly with armature current, then the armature reaction will be 80 A⋅turns The figure below shows that a triangle consisting of 3.6 V and (80 A⋅turns)/(1000 turns) = 0.08 A fits exactly between the E A and V T lines at a terminal voltage of

103 V

The rated current of this generated is 50 A, so 40 A is 80% of full load If the full load armature reaction

is 200 A⋅turns, and if the armature reaction is assumed to change linearly with armature current, then the armature reaction will be 160 A⋅turns There is no point where a triangle consisting of 3.6 V and (80

A⋅turns)/(1000 turns) = 0.16 A fits exactly between the E A and V T lines, so this is not a stable operating condition

(c) A MATLAB program to calculate the terminal characteristic of this generator without armature

reaction is shown below:

% M-file: prob9_25d.m

% M-file to calculate the terminal characteristic of a shunt

% dc generator without armature reaction

% Get the magnetization curve This file contains the

% three variables if_values, ea_values, and n_0

load p97_mag.dat;

if_values = p97_mag(:,1);

ea_values = p97_mag(:,2);

n_0 = 1800;

% First, initialize the values needed in this program

r_f = 24; % Field resistance (ohms)

r_adj = 10; % Adjustable resistance (ohms)

r_a = 0.18; % Armature + series resistance (ohms)

i_f = 0:0.005:6; % Field current (A)

n = 1800; % Generator speed (r/min)

% Calculate Ea versus If

Ea = interp1(if_values,ea_values,i_f);

% Calculate Vt versus If

Vt = (r_f + r_adj) * i_f;

% Find the point where the difference between the two

% lines is exactly equal to i_a*r_a This will be the

% point where the line line "Ea - Vt - i_a*r_a" goes

% This code prevents us from reporting the first (unstable)

% location satisfying the criterion

The resulting terminal characteristic is shown below:

A MATLAB program to calculate the terminal characteristic of this generator with armature reaction is shown below:

% M-file: prob9_25d2.m

% M-file to calculate the terminal characteristic of a shunt

% dc generator with armature reaction

% Get the magnetization curve This file contains the

% three variables if_values, ea_values, and n_0

% First, initialize the values needed in this program

r_f = 24; % Field resistance (ohms)

r_adj = 10; % Adjustable resistance (ohms)

r_a = 0.18; % Armature + series resistance (ohms)

i_f = 0:0.005:6; % Field current (A)

n = 1800; % Generator speed (r/min)

n_f = 1000; % Number of field turns

% Calculate Ea versus If

Ea = interp1(if_values,ea_values,i_f);

% Calculate Vt versus If

Vt = (r_f + r_adj) * i_f;

% Find the point where the difference between the Ea

% armature reaction line and the Vt line is exactly

% equal to i_a*r_a This will be the point where