Six-car RGR trains running between two stations with a distance of 2000 m accele

Question

Six-car RGR trains running between two stations with a distance of 2000 m accelerate to Vmax = 100 km/h on a distance Sa = 600 m. From the moment Vmax is reached, two alternative regimes are possible: Maintain Vmas until braking must be applied b) Coasting is immediately used (c = 0.1 m/s, and maintained until braking must be applied. Assuming that in both cases braking rate is = 1.0 m/s2 and that energy consumption during Vmax travel is e-3 kWh/veh-km, compute the trade-off between the two regimes; energy saved per train, E (kWh), versus additional travel time, at(s), which case (b) involves in comparison with case (a)

Explanation / Answer

given distance , d = 2000 m
Vmax = 100 km/h = 27.778 m/s
accelerating distance Sa = 600 m
coasting , c = 0.1 m/s/s
braking rate, b = 1 m/s/s

energy consumption during Vmax travel e = 3000 Wh/km

case a) Vmax untill braking
   energy required to accelerate to Vmax = E1
   braking distance = Sb
   2*Sb*b = Vmax^2
   Sb = 385.50 m
   so, distance travelled at Vmax = 2000 - 600 - 385.50 = 1014.5 m
   Energy consumption in this distance = e*1014.5 /1000 = 3043.5 Wh
   energy consumed during braking = E2
   total energy consumption = E1 + E2 + 3043.5
   initial acceleration = a
   2aSa = Vmax^2
   s = 0.643 m/s/s
   time of travel = 27.778/0.643 + 1014.5/27.778 + 27.778/1 = 107.5 s

case b) Coasting untill breaking
   let assume the train coasted for distance a
   then final velocity = Vf
   2*0.1*a = Vf^2 - 27.77^2
   a = 5Vf^2 - 3855.8645
   breaking distnace= b
   2*1*b = Vf^2 = (a + 3855.8645)/5
   10b = a + 3855.8645
   but b + a = 1400 m
   so 10b = 1400 - b + 3855.8645
   b = 477.805
   and a = 922.194 m
   and Vf = 30.91 m/s

   so energy consumed = E1 + 3000*922.194/1000 + 0.5*m*(30.91^2 - 27.77^2) + E3 = E1 + 2766.582 + 91.93m + E3
   where E3 is energy required for breaking and m is mass of train
   as this train stops from higher speed, E3 > E2

   time of travel = 27.778/0.643 + (30.91 - 27.778)/0.1 + 30.91/1 = 105.43 s

comparing both cases
extra time taken in case 1 = 2.069 s
extra energy consumed in case 2 = -276.418 + 91.93m + (E3 - E2)
now E3 - E2 > 0
m > 1000 kg atleast
so energy consumed in case 2 is more than energy consumed in case 1

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.