# CHAPTER 13: RATINGS, SELECTION, AND MAINTENANCE OF ELECTRIC MACHINERY¶

## Example 13.1, Page number 460¶

In [1]:
#Variable declaration
T = 125.0         #Recorded temperature by the embedded detectors(°C)
life_orig = 10.0  #Standard life of the motor(years)

#Calculation
delta_T = T-105            #Positive temp diff b/w the given max hottest spot temp of its insulation and ambient temp recorded
R = 2**(delta_T/10.0)      #Life reduction factor
Life_calc = life_orig/R    #Reduced life expectancy of the motor(years)

#Result
print('Life reduction factor , R = %.f ' %R)
print('Reduced life expectancy of the motor , Life_calc = %.1f years' %Life_calc)

Life reduction factor , R = 4
Reduced life expectancy of the motor , Life_calc = 2.5 years


## Example 13.2, Page number 460¶

In [1]:
#Variable declaration
T = 75.0            #Recorded temperature by the embedded detectors(°C)
life_orig = 10.0    #Standard life of the motor(years)

#Calculation
delta_T = 105-T            #Positive temp diff b/w the given max hottest spot temp of its insulation and ambient temp recorded
E = 2**(delta_T/10.0)      #Life extension factor
Life_calc = life_orig*E    #Increased life expectancy of the motor(years)

#Result
print('Life extension factor , E = %.f ' %E)
print('Increased life expectancy of the motor , Life_calc = %.f years' %Life_calc)

Life extension factor , E = 8
Increased life expectancy of the motor , Life_calc = 80 years


## Example 13.3, Page number 461¶

In [1]:
#Variable declaration
T_A = 105.0            #Recorded temperature by the embedded detectors at hottest spot limiting temp by class A(°C)
T_B = 130.0            #Recorded temperature by the embedded detectors at hottest spot limiting temp by class B(°C)
life_orig = 5.0        #Standard life of the motor(years)

#Calculation
delta_T = T_B-T_A          #Positive temp diff b/w the given max hottest spot temp of its insulation and ambient temp recorded
E = 2**(delta_T/10.0)      #Life extension factor
Life_new = life_orig*E    #Increased life expectancy of the motor(years)

#Result
print('New life if wound with class B insulation , Life_new = %.1f years' %Life_new)

New life if wound with class B insulation , Life_new = 28.3 years


## Example 13.4, Page number 463¶

In [1]:
#Variable declaration
P_o = 25.0          #Rated power of SCIM(hp)
T_ambient = 40.0    #Standard ambient temperature recorded by the embedded hot-spot detectors(°C)
T_hottest = 115.0   #Hottest-spot winding temperature recorded by the embedded hot-spot detectors(°C)

#Calculation
delta_Tf = 90.0                #Allowable temperature rise for the insulation type used(°C)
T_rise = T_hottest-T_ambient   #Actual temperature rise of the SCIM(°C)
P_f = P_o*(delta_Tf/T_rise)    #Approximate power the motor can deliver at temperature rise(hp)
power = 30.0                   #Power rating that may be stamped on the name plate(hp)
delta_Tf_e = 90.0              #Temperature rise that must be stamped on the name plate(°C)

#Result
print('Case(a): Allowable temperature rise for the insulation type used = %.f°C' %delta_Tf)
print('Case(b): Actual temperature rise of the SCIM = %.f°C' %T_rise)
print('Case(c): Approximate power the motor can deliver at the temperature rise , P_f = %.f hp' %P_f)
print('Case(d): Power rating that may be stamped on the name plate = %.f hp' %power)
print('Case(e): Temperature rise that must be stamped on the name plate = %.f°C' %delta_Tf_e)

Case(a): Allowable temperature rise for the insulation type used = 90°C
Case(b): Actual temperature rise of the SCIM = 75°C
Case(c): Approximate power the motor can deliver at the temperature rise , P_f = 30 hp
Case(d): Power rating that may be stamped on the name plate = 30 hp
Case(e): Temperature rise that must be stamped on the name plate = 90°C


## Example 13.5, Page number 464¶

In [1]:
#Variable declaration
P_o = 50.0            #Power rating of the WRIM(hp)
T_hottest = 160.0     #Hottest-spot winding temperature recorded by the embedded hot-spot detectors(°C)
T_ambient = 40.0      #Standard ambient temperature recorded by the embedded hot-spot detectors(°C)
P_f_a = 40.0          #Power rating of load(hp)
P_f_b = 55.0          #Power rating of load(hp)

#Calculation
#Case(a)
delta_T_o = T_hottest-T_ambient         #Temperature rise for the insulation type used(°C)
delta_T_f_a = (P_f_a/P_o)*delta_T_o     #Final temperature rise(°C)
T_f_a = delta_T_f_a+T_ambient           #Approximate final hot-spot temperature(°C)
#Case(b)
delta_T_f_b = (P_f_b/P_o)*delta_T_o     #Final temperature rise(°C)
T_f_b = delta_T_f_b+T_ambient           #Approximate final hot-spot temperature(°C)

#Result
print('Case(a): Approximate final hot-spot temperature at a continuous output load of 40 hp , T_f = %.f°C ' %T_f_a)
print('Case(b): Approximate final hot-spot temperature at a continuous output load of 55 hp , T_f = %.f°C ' %T_f_b)
print('         Yes, Motor life is reduced at 110 percent motor load because allowable maximum hot-spot motor temperature is 155°C')

Case(a): Approximate final hot-spot temperature at a continuous output load of 40 hp , T_f = 136°C
Case(b): Approximate final hot-spot temperature at a continuous output load of 55 hp , T_f = 172°C
Yes, Motor life is reduced at 110 percent motor load because allowable maximum hot-spot motor temperature is 155°C


## Example 13.6, Page number 464¶

In [1]:
#Variable declaration
P_o = 55.0             #Power rating of the WRIM(hp)
T_ambient = 40.0       #Standard ambient temperature recorded by the embedded hot-spot detectors(°C)
life_orig = 10.0       #Standard life of the motor(years)
T_f = 172.0            #Approximate final hot-spot temperature(°C)

#Calculation
delta_T = T_f-155.0       #Positive temp diff b/w the given max hottest spot temp of its insulation and ambient temp recorded
R = 2**(delta_T/10.0)     #Life reduction factor
Life_calc = life_orig/R   #Reduced life expectancy of the motor(years)

#Result
print('Reduced life expectancy of the motor , Life_calc = %.2f years' %Life_calc)

Reduced life expectancy of the motor , Life_calc = 3.08 years


## Example 13.7, Page number 466¶

In [1]:
#Variable declaration
P_o = 200.0     #Power rating of the test motor(hp)
P1 = 200.0      #Operating power of test motor at t1 duration(hp)
t1 = 5.0        #Time duration for which test motor is operated(min)
P2 = 20.0       #Operating power of test motor at t2 duration(hp)
t2 = 5.0        #Time duration for which test motor is operated(min)
P3 = 0.0        #Operating power of test motor at t3 duration at rest(hp)
t3 = 10.0       #Time duration for which test motor is operated(min)
P4 = 100.0      #Operating power of test motor at t4 duration(hp)
t4 = 10.0       #Time duration for which test motor is operated(min)

#Calculation
rms_hp = ((P1**2*t1+P2**2*t2+P3**2*t3+P4**2*t4)/(t1+t2+t3+t4/3))**0.5   #HP required for intermittent varying load

#Result
print('Horsepower required for such an intermittent varying load , rms hp = %.f hp' %rms_hp)
print('125 hp motor would be selected because that is the nearest larger commercial standard rating')

Horsepower required for such an intermittent varying load , rms hp = 114 hp
125 hp motor would be selected because that is the nearest larger commercial standard rating


## Example 13.8, Page number 472¶

In [1]:
#Variable declaration
V = 120.0   #Rated output voltage of separately excited dc generator(V)
I = 100.0   #Rated output current of separately excited dc generator(A)
R = 0.1     #Armature resistance(ohm)

#Calculation
V_b = V          #Base voltage(V)
I_b = I          #Base current(A)
R_b = V_b/I_b    #Base resistance(ohm)
R_pu = R/R_b     #Per-unit value of armature resistance

#Result
print('Case(a): Base voltage , V_b = %.f V' %V_b)
print('Case(b): Base current , I_b = %.f A' %I_b)
print('Case(b): Base resistance , R_b = %.1f Ω' %R_b)
print('Case(d): Per-unit value of armature resistance , R_p.u = %.3f p.u' %R_pu)

Case(a): Base voltage , V_b = 120 V
Case(b): Base current , I_b = 100 A
Case(b): Base resistance , R_b = 1.2 Ω
Case(d): Per-unit value of armature resistance , R_p.u = 0.083 p.u


## Example 13.9, Page number 473¶

In [1]:
import math
import cmath

#Variable declaration
V = 500.0     #Rated voltage of the alternator(V)
P = 20.0      #Rated power of the alternator(kVA)
I = 40.0      #Rated current of the alternator(A)
R = 2.0       #Armature resistance(ohm)
X = 15.0      #Armature reactance(ohm)

#Calculation
V_b = V                             #Base voltage(V)
I_b = I                             #Base current(A)
R_pu = R*I_b/V_b                    #Per-unit value of armature resistance
jX_pu = X*I_b/V_b                   #Per-unit value of armature reactance
Z_pu1 = complex(R_pu,jX_pu)         #Per-unit value of armature impedance
Z_pu2 = complex(R,X)*(I/V)          #Per-unit value of armature impedance

#Result
print('Case(a): Per unit value of armature resistance , R_p.u = %.2f p.u' %R_pu)
print('Case(b): Per unit value of armature reactance , jX_p.u = j%.1f p.u' %jX_pu)
print('Case(c): Per-unit value of armature impedance , Z_p.u = %.3f∠%.1f° p.u (Method 1)' %(abs(Z_pu1),cmath.phase(Z_pu1)*180/math.pi))
print('         Per-unit value of armature impedance , Z_p.u = %.3f∠%.1f° p.u (Method 2)' %(abs(Z_pu2),cmath.phase(Z_pu2)*180/math.pi))

Case(a): Per unit value of armature resistance , R_p.u = 0.16 p.u
Case(b): Per unit value of armature reactance , jX_p.u = j1.2 p.u
Case(c): Per-unit value of armature impedance , Z_p.u = 1.211∠82.4° p.u (Method 1)
Per-unit value of armature impedance , Z_p.u = 1.211∠82.4° p.u (Method 2)


## Example 13.10, Page number 473¶

In [1]:
#Variable declaration
V_orig = 500.0      #Rated voltage of the alternator(V)
kVA_orig = 20.0     #Rated power of the alternator(kVA)
I = 40.0            #Rated current of the alternator(A)
R = 2.0             #Armature resistance(ohm)
X = 15.0            #Armature reactance(ohm)
V_new = 5000.0      #New voltage of the alternator(V)
kVA_new = 100.0     #New power of the alternator(kVA)
Z_pu_orig = 1.211   #Original per-unit value of armature impedance

#Calculation
Z_pu_new = Z_pu_orig*(kVA_new/kVA_orig)*(V_orig/V_new)**2   #New per-unit impedance

#Result
print('New per-unit impedance , Z_pu(new) = %.5f p.u' %Z_pu_new)

New per-unit impedance , Z_pu(new) = 0.06055 p.u


## Example 13.11, Page number 474¶

In [1]:
#Variable declaration
V = 2300.0        #Line voltage of 3-phase distribution system(V)
V_p = 1328.0      #Phase voltage of 3-phase distribution system(V)
V_b = 69000.0     #Common base line voltage(V)
V_pb = 39840.0    #Common base phase voltage(V)

#Calculation
V_pu_line = V/V_b          #Distribution system p.u line voltage
V_pu_phase = V_p/V_pb      #Distribution system p.u phase voltage

#Result
print('Case(a): Distribution system p.u line voltage , V_pu = %.2f p.u' %V_pu_line)
print('Case(b): Distribution system p.u phase voltage , V_pu = %.2f p.u' %V_pu_phase)

Case(a): Distribution system p.u line voltage , V_pu = 0.03 p.u
Case(b): Distribution system p.u phase voltage , V_pu = 0.03 p.u


## Example 13.12, Page number 474¶

In [1]:
import math
import cmath

#Variable declaration
VA_b = 50.0    #Base power rating of the 3-phase Y-connected alternator(MVA)
V_b = 25.0     #Base voltage of the 3-phase Y-connected alternator(kV)
X_pu = 1.3     #Per unit value of synchronous reactance
R_pu = 0.05    #Per unit value of resistance

#Calculation
#Case(a)
Z_b1 = V_b**2/VA_b      #Base impedance(ohm)
S_b = VA_b              #Base power rating of the 3-phase Y-connected alternator(MVA)
I_b = S_b/V_b           #Base current(kA)
Z_b2 = V_b/I_b          #Base impedance(ohm)
#Case(b)
Z_b = Z_b1              #Base impedance(ohm)
X_s = X_pu*Z_b          #Actual value of synchronous reactance per phase(ohm)
#Case(c)
R_a = R_pu*Z_b          #Actual value of armature stator resistance per phase(ohm)
#Case(d)
Z_s1 = complex(R_a,X_s)     #Synchronous impedance per phase(ohm)
Z_pu = complex(R_pu,X_pu)   #Per unit value of impedance
Z_s2 = Z_pu*Z_b             #Synchronous impedance per phase(ohm)
#Case(d)
S = S_b                     #Base power rating of the 3-phase Y-connected alternator(MVA)
P = S*R_pu                  #Full-load copper losses for all three phases(MW)

#Result
print('Case(a): Base impedance , Z_b = %.1f Ω' %Z_b1)
print('         Base impedance , Z_b = %.1f Ω' %Z_b2)
print('Case(b): Actual value of synchronous reactance per phase , X_s = j%.2f Ω' %X_s)
print('Case(c): Actual value of armature stator resistance per phase , R_a = %.3f Ω' %R_a)
print('Case(d): Synchronous impedance per phase , Z_s = %.2f∠%.1f° Ω (Method 1)' %(abs(Z_s1),cmath.phase(Z_s1)*180/math.pi))
print('         Synchronous impedance per phase , Z_s = %.2f∠%.1f° Ω (Method 2)' %(abs(Z_s2),cmath.phase(Z_s2)*180/math.pi))
print('Case(e): Full-load copper losses for all three phases , P = %.1f MW' %P)

Case(a): Base impedance , Z_b = 12.5 Ω
Base impedance , Z_b = 12.5 Ω
Case(b): Actual value of synchronous reactance per phase , X_s = j16.25 Ω
Case(c): Actual value of armature stator resistance per phase , R_a = 0.625 Ω
Case(d): Synchronous impedance per phase , Z_s = 16.26∠87.8° Ω (Method 1)
Synchronous impedance per phase , Z_s = 16.26∠87.8° Ω (Method 2)
Case(e): Full-load copper losses for all three phases , P = 2.5 MW