CHAPTER 4.5: ELECTRIC TRACTION-SPEED TIME CURVES AND MECHANICS OF TRAIN MOVEMENT

Example 4.5.1, Page number 778

In [1]:
#Variable declaration
speed = 45.0      #Scheduled speed(kmph)
D = 1.5           #Distance between 2 stops(km)
t = 20.0          #Time of stop(sec)
alpha = 2.4       #Acceleration(km phps)
beta = 3.2        #Retardation(km phps)

#Calculation
t_total = D*3600/speed                   #Total time(sec)
T = t_total-t                            #Actual time for run(sec)
k = (alpha+beta)/(alpha*beta)            #Constant
V_m = (T/k)-((T/k)**2-(7200*D/k))**0.5   #Maximum speed over the run(kmph)

#Result
print('Maximum speed over the run, V_m = %.f kmph' %V_m)

Maximum speed over the run, V_m = 74 kmph

Example 4.5.2, Page number 778

In [1]:
#Variable declaration
V_m = 65.0        #Maximum speed(kmph)
t = 30.0          #Time of stop(sec)
speed = 43.5      #Scheduled speed(kmph)
alpha = 1.3       #Acceleration(km phps)
D = 3.0           #Distance between 2 stops(km)

#Calculation
t_total = D*3600/speed                                 #Total time of run including stop(sec)
T = t_total-t                                          #Actual time for run(sec)
V_a = D/T*3600                                         #Average speed(kmph)
beta = 1/((7200.0*D/V_m**2*((V_m/V_a)-1))-(1/alpha))   #Value of retardation(km phps)

#Result
print('Value of retardation, β = %.3f km phps' %beta)
print('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here')
print('      ERROR: β unit is km phps & not km phps as mentioned in textbook solution')

Value of retardation, β = 1.198 km phps

NOTE: Changes in the obtained answer from that of textbook is due to more precision here
      ERROR: β unit is km phps & not km phps as mentioned in textbook solution

Example 4.5.3, Page number 778-779

In [1]:
#Variable declaration
speed = 25.0       #Scheduled speed(kmph)
D = 800.0/1000     #Distance between 2 stations(km)
t = 20.0           #Time of stop(sec)
V_m_per = 20.0     #Maximum speed higher than(%)
beta = 3.0         #Retardation(km phps)

#Calculation
t_total = D*3600/speed                                 #Total time of run including stop(sec)
T = t_total-t                                          #Actual time for run(sec)
V_a = D/T*3600                                         #Average speed(kmph)
V_m = (100+V_m_per)*V_a/100                            #Maximum speed(kmph)
alpha = 1/((7200.0*D/V_m**2*((V_m/V_a)-1))-(1/beta))   #Value of acceleration(km phps)

#Result
print('Rate of acceleration required to operate this service, α = %.2f km phps' %alpha)

Rate of acceleration required to operate this service, α = 1.85 km phps

Example 4.5.4, Page number 779

In [1]:
#Variable declaration
D = 2.0          #Distance between 2 stations(km)
V_a = 40.0       #Average speed(kmph)
V_1 = 60.0       #Maximum speed limitation(kph)
alpha = 2.0      #Acceleration(km phps)
beta_c = 0.15    #Coasting retardation(km phps)
beta = 3.0       #Braking retardation(km phps)

#Calculation
t_1 = V_1/alpha                                        #Time for acceleration(sec)
T = 3600*D/V_a                                         #Actual time of run(sec)
V_2 = (T-t_1-(V_1/beta_c))*beta*beta_c/(beta_c-beta)   #Speed at the end of coasting period(kmph)
t_2 = (V_1-V_2)/beta_c                                 #Coasting period(sec)
t_3 = V_2/beta                                         #Braking period(sec)

#Result
print('Duration of acceleration, t_1 = %.f sec' %t_1)
print('Duration of coasting, t_2 = %.f sec' %t_2)
print('Duration of braking, t_3 = %.f sec' %t_3)

Duration of acceleration, t_1 = 30 sec
Duration of coasting, t_2 = 137 sec
Duration of braking, t_3 = 13 sec

Example 4.5.5, Page number 781-782

In [1]:
#Variable declaration
r = 1.0    #Tractive resistance(N/tonne)

#Calculation
tractive_res_i = 0.278*r     #Tractive resistance(N/tonne) = Energy consumption(Wh/tonne-km)
beta = 1/277.8               #Tractive resistance(N/tonne) = Retardation(km kmps/tonne)
energy = 98.1*1000/3600      #1% gradient = energy(Wh per tonne km)

#Result
print('Case(i)  : Tractive resistance of 1 N per tonne = %.3f Wh per tonne-km' %tractive_res_i)
print('Case(ii) : Tractive resistance of 1 N per tonne = %.5f km phps per tonne' %beta)
print('Case(iii): 1 percent gradient = %.2f Wh per tonne km' %energy)
print('\nNOTE: Slight change in the obtained answer from that of textbook is due to more precision here')

Case(i)  : Tractive resistance of 1 N per tonne = 0.278 Wh per tonne-km
Case(ii) : Tractive resistance of 1 N per tonne = 0.00360 km phps per tonne
Case(iii): 1 percent gradient = 27.25 Wh per tonne km

NOTE: Slight change in the obtained answer from that of textbook is due to more precision here

Example 4.5.6, Page number 782

In [1]:
#Variable declaration
W = 254.0      #Weight of motor-coach train(tonne)
no = 4.0       #Number of motor
t_1 = 20.0     #Time(sec)
V_m = 40.25    #Maximum speed(kmph)
G = 1.0        #Gradient(%)
gamma = 3.5    #Gear ratio
n = 0.95       #Gear efficiency
D = 91.5/100   #Wheel diameter(m)
r = 44.0       #Train resistance(N/tonne)
I = 10.0       #Rotational inertia(%)

#Calculation
W_e = W*(100+I)/100                  #Accelerating weight of train(tonne)
alpha = V_m/t_1                      #Acceleration(km phps)
F_t = 277.8*W_e*alpha+W*r+98.1*W*G   #Tractive effort(N)
T = F_t*D/(2*n*gamma)                #Torque developed(N-m)
T_each = T/no                        #Torque developed by each motor(N-m)

#Result
print('Torque developed by each motor = %.f N-m' %T_each)
print('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here & more approximation in textbook')
print('      ERROR: W = 254 tonne, not 256 tonne as mentioned in textbook problem statement')

Torque developed by each motor = 6615 N-m

NOTE: Changes in the obtained answer from that of textbook is due to more precision here & more approximation in textbook
      ERROR: W = 254 tonne, not 256 tonne as mentioned in textbook problem statement

Example 4.5.7, Page number 782

In [1]:
#Variable declaration
W = 203.0      #Weight of motor-coach train(tonne)
no = 4.0       #Number of motors
T = 5130.0     #Shaft torque(N-m)
V_m = 42.0     #Maximum speed(kmph)
G = 100.0/250  #Gradient
gamma = 3.5    #Gear ratio
n = 0.93       #Gear efficiency
D = 91.5/100   #Wheel diameter(m)
r = 45.0       #Train resistance(N/tonne)
I = 10.0       #Rotational inertia(%)

#Calculation
W_e = W*(100+I)/100                     #Accelerating weight of train(tonne)
F_t = n*4*T*2*gamma/D                   #Tractive effort(N)
alpha = (F_t-W*r-98.1*W*G)/(277.8*W_e)  #Acceleration(km phps)
t_1 = V_m/alpha                         #Time taken by train to attain speed(sec)

#Result
print('Time taken by train to attain speed, t_1 = %.1f sec' %t_1)

Time taken by train to attain speed, t_1 = 20.2 sec

Example 4.5.8, Page number 782-783

In [1]:
import matplotlib.pyplot as plt

#Variable declaration
V_a = 42.0          #Average speed of train(kmph)
D = 1400.0/1000     #Distance(km)
alpha = 1.7         #Acceleration(km phps)
beta = 3.3          #Retardation(km phps)
r = 50.0            #Tractive resistance(N/tonne)
I = 10.0            #Rotational inertia(%)

#Calculation
T = D*3600/V_a                                       #Time for run(sec)
k = (alpha+beta)/(alpha*beta)                        #Constant
V_m = (T/k)-((T/k)**2-(7200*D/k))**0.5               #Maximum speed over the run(kmph)
t_1 = V_m/alpha                                      #Time of acceleration(sec)
t_3 = V_m/beta                                       #Time(sec)
t_2 = T-(t_1+t_3)                                    #Time(sec)
D_1 = D-(V_a*t_1/(2*3600))                           #Distance(km)
We_W = (100+I)/100                                   #W_e/W
energy = (0.0107*V_m**2*We_W/D)+(0.278*r*D_1/D)      #Energy consumption(Wh per tonne-km)
fig = plt.figure(1)
plt.plot(
          [0,t_1,t_1,(t_1+t_2),(t_1+t_2),(t_1+t_2+t_3)],
          [0,V_m,V_m,V_m,V_m,0],
          color='b'
          )                                           #Plotting speed-time curve
plt.plot(
         [t_1,t_1],
         [0,V_m],
         color='r',ls='--'
         )
plt.plot(
         [t_1+t_2,t_1+t_2],
         [0,V_m],
         color='r',ls='--'
         )
plt.axis([0,125,0,54])
plt.xlabel('Time(seconds)')
plt.ylabel('Speed (km/h)')
plt.show()

#Result
print('Speed-time curve for the run is shown in figure 1')
print('Energy consumption at the axles of train = %.1f Wh per tonne-km' %energy)

Speed-time curve for the run is shown in figure 1
Energy consumption at the axles of train = 34.9 Wh per tonne-km

Example 4.5.9, Page number 783

In [1]:
#Variable declaration
V_A = 48.0       #Speed(kmph)
t_1 = 24.0       #Time taken to accelerate from rest to speed(sec)
t_2 = 69.0       #Coasting time(sec)
r = 58.0         #Constant resistance(N/tonne)
beta = 3.3       #Retardation(km phps)
t_3 = 11.0       #Retardation time(sec)
t_iii_a = 20.0   #Station stop time(sec)
t_iii_b = 15.0   #Station stop time(sec)
I = 10.0         #Rotational inertia(%)

#Calculation
alpha = V_A/t_1                                                   #Acceleration(km phps)
V_B = beta*t_3                                                    #Speed at B(km phps)
beta_c = (V_A-V_B)/t_2                                            #Retardation during coasting(km phps)
distance_acc = 1.0/2*t_1*V_A/3600                                 #Distance covered during acceleration(km)
distance_coasting = (V_A**2-V_B**2)/(2*beta_c*3600)               #Distance covered during coasting(km)
distance_braking = t_3*V_B/(3600*2)                               #Distance covered during braking(km)
distance_total = distance_acc+distance_coasting+distance_braking  #Total distance(km)
speed_iii_a = distance_total*3600/(t_1+t_2+t_3+t_iii_a)           #Scheduled speed with a stop of 20 sec(kmph)
speed_iii_b = distance_total*3600/(t_1+t_2+t_3+t_iii_b)           #Scheduled speed with a stop of 15 sec(kmph)

#Result
print('Case(i)  : Acceleration, α = %.f km phps' %alpha)
print('Case(ii) : Coasting retardation, β_c = %.2f km phps' %beta_c)
print('Case(iii): Scheduled speed with a stop of 20 seconds = %.2f kmph' %speed_iii_a)
print('           Scheduled speed with a stop of 15 seconds = %.2f kmph' %speed_iii_b)
print('\nNOTE: ERROR: Calculation mistakes in the textbook solution')

Case(i)  : Acceleration, α = 2 km phps
Case(ii) : Coasting retardation, β_c = 0.17 km phps
Case(iii): Scheduled speed with a stop of 20 seconds = 29.71 kmph
           Scheduled speed with a stop of 15 seconds = 30.96 kmph

NOTE: ERROR: Calculation mistakes in the textbook solution

Example 4.5.10, Page number 784

In [1]:
#Variable declaration
W = 350.0      #Weight of train(tonne)
G = 1.0        #Gradient
alpha = 0.8    #Acceleration(km phps)
u = 0.25       #Co-efficient of adhesion
r = 44.5       #Train resistance(N/tonne)
I = 10.0       #Rotational inertia(%)

#Calculation
W_e = W*(100+I)/100                  #Accelerating weight of train(tonne)
F_t = 277.8*W_e*alpha+W*r+98.1*W*G   #Tractive effort(N)
adhesive_weight = F_t/(u*9.81*1000)  #Adhesive weight(tonnes)

#Result
print('Minimum adhesive weight of the locomotive = %.1f tonnes' %adhesive_weight)
print('\nNOTE: ERROR: Train resistance is 44.5 N per tonne & not 45 N per tonne as mentioned in textbook problem statement')

Minimum adhesive weight of the locomotive = 55.2 tonnes

NOTE: ERROR: Train resistance is 44.5 N per tonne & not 45 N per tonne as mentioned in textbook problem statement

Example 4.5.11, Page number 784

In [1]:
#Variable declaration
W = 400.0       #Weight of train(tonne)
G = 100.0/75    #Gradient
alpha = 1.6     #Acceleration(km phps)
r = 66.75       #Train resistance(N/tonne)
I = 10.0        #Rotational inertia(%)
V = 48.0        #Speed(kmph)
n = 0.7         #Overall efficiency of equipment

#Calculation
W_e = W*(100+I)/100                  #Accelerating weight of train(tonne)
F_t = 277.8*W_e*alpha+W*r+98.1*W*G   #Tractive effort(N)
t = V/alpha                          #Time(sec)
energy_a = F_t*V*t/(2*3600**2)       #Energy usefully employed(kWh)
G_r = 98.1*G+r                       #Force(N)
work_tonne_km = G_r*1000             #Work done per tonne per km(Nw-m)
energy_b = work_tonne_km/(n*3600)    #Energy consumption(Wh per tonne-km)

#Result
print('Case(a): Energy usefully employed in attaining speed = %.2f kWh' %energy_a)
print('Case(b): Specific energy consumption at steady state speed = %.1f Wh per tonne-km' %energy_b)

Case(a): Energy usefully employed in attaining speed = 15.26 kWh
Case(b): Specific energy consumption at steady state speed = 78.4 Wh per tonne-km

Example 4.5.12, Page number 784-785

In [1]:
#Variable declaration
W = 200.0       #Trailing weight(tonne)
G = 1.0         #Gradient(%)
alpha = 1.0     #Acceleration(km phps)
u = 0.2         #Co-efficient of adhesion
r = 50.0        #Train resistance(N/tonne)
I = 10.0        #Rotational inertia(%)

#Calculation
W_L = ((277.8*(100+I)/100*alpha)+98.1*G+r)*W/(u*9.81*1000-((277.8*(100+I)/100*alpha)+98.1*G+r))  #Weight of locomotive(tonnes)

#Result
print('Minimum adhesive weight of a locomotive, W_L = %.1f tonnes' %W_L)
print('\nNOTE: ERROR: Calculation mistake in textbook solution in calculating W_L')

Minimum adhesive weight of a locomotive, W_L = 60.2 tonnes

NOTE: ERROR: Calculation mistake in textbook solution in calculating W_L