Tractive Effort for Propulsion of a Train The tractive effort (Ft) is the force developed by the traction unit at the rim of the driving wheels for moving the unit itself and its train (trailing load). The tractive effort required for train propulsion on a level is track = + Ft Fa Fr If gradients are involved, the above expression becomes = + + for ascending gradient = + Ft Fa Fg Fr Fa Fg — − for descending gradient Fr — where = force required for giving linear acceleration to the train Fa = force required to overcome the effect of gravity Fg = force required to overcome resistance to train motion. Fr (a) Value of Fa If is the dead (or stationary) mass of the train and a its linear acceleration, then M = Fa Ma Since a train has rotating parts like wheels, axles, motor armatures and gearing etc., its (or effective accelerating) mass is more (about 8 15%) than its stationary mass. Me − These parts have to be given angular acceleration at the same time as the whole train is accelerated in the linear direction. Hence, = Fe Mea (ii) If is in tonne and in km/h/s, then converting them into absolute units, we have Me α = (1000 × (1000/3600) = 277.8 newton Fa Me) a Me a (b) Value of Fg As seen from Fig. 43.13, = = Fg W Mg sin θ sin θ In railway practice, gradient is expressed as the rise (in metres) a track distance of 100 m and is called percentage gradient. Therefore % = 100/100 G = = 100 Substituting the value of the above equation, we AC /AC sin θ sin θ in Get = = 9.8 × Fg Mg G/100 MG 10−2 When is in kg, = 9.8 × newton M Fg 10−2 MG When is given in tonne, then M = 9.8 × (1000 = 98 Newton Fg M) G MG 10−2
Tractive Effort for Propulsion of a Train The tractive effort (Ft) is the force developed by the traction unit at the rim of the driving wheels for moving the unit itself and its train (trailing load). The tractive effort required for train propulsion on a level track is Ft = Fa + Fr If gradients are involved, the above expression becomes Ft = Fa + Fg + Fr — for ascending gradient = Fa − Fg + Fr — for descending gradient where Fa = force required for giving linear acceleration to the train Fg = force required to overcome the effect of gravity Fr = force required to overcome resistance to train motion. (a) Value of Fa If M is the dead (or stationary) mass of the train and a its linear acceleration, then Fa = Ma Since a train has rotating parts like wheels, axles, motor armatures and gearing etc., its effective (or accelerating) mass Me is more (about 8 − 15%) than its stationary mass. These parts have to be given angular acceleration at the same time as the whole train is accelerated in the linear direction. Hence, Fe = Mea (ii) If Me is in tonne and α in km/h/s, then converting them into absolute units, we have Fa = (1000 Me) × (1000/3600) a = 277.8 Me a newton (b) Value of Fg As seen from Fig. 43.13, Fg = W sin θ = Mg sin θ In railway practice, gradient is expressed as the rise (in metres) a track distance of 100 m and is called percentage gradient. Therefore % G = 100/100 AC /AC = = 100 sin θ Substituting the value of sin θ in the above equation, we Get Fg = Mg G/100 = 9.8 × 10−2 MG When M is in kg, Fg = 9.8 × 10−2 MG newton When M is given in tonne, then Fg = 9.8 × 10−2 (1000 M) G = 98 MG Newton
25-Tractive-Effort.docx