Question 3.10: A steam locomotive is coupled with two single-cylinder, doub...
A steam locomotive is coupled with two single-cylinder, double-acting engine of 350 mm in diameter and 550 mm stroke. The driving wheels are 2.2 m in diameter. The pressure of steam supplied to the engine is 11 bar and dry saturated. The exhaust pressure of steam is 1.1 bar. The maximum cut-off is 80% of the stroke. Find out the tractive effort at 15 km/h speed with maximum cut-off. Assume the diagram factor 0.8 and mechanical efficiency 70%.
If the resistance to train is 12 N/1000 N at 90 km/h speed, determine the total train load that can be hauled at this speed if the cut-off is 20% of the stroke. Assume diagram factor at this speed as 0.70 and mechanical efficiency 80%.
Assume that the engine is directly coupled to the driving wheel.
Number of cylinder, n = 2
Diameter of a cylinder, D = 350 mm = 0.35 m
Stroke length, L = 550 mm = 0.55 m
Diameter of a driving wheel, D_{w} = 2.2 m
Pressure of steam supplied, p_{1} = 11 bar
Quality of steam, x_{1} = 1
Exhaust pressure of steam, p_{b} = 1.1 bar
Maximum cut-off = 0.8 V_{s}
Diagram factor, (D.F.) _{1} = 0.8
Mechanical efficiency = 70%
Tractive effort, P at 15 km/h :
Resistance to train = 12 N/1000 N
Speed = 90 km/h
Cut-off = 0.2 V_{s}
Diagram factor, (D.F.)_{2} = 0.7
Mechanical efficiency = 80%
(i) Tractive effort, P :
p_{m(th.)} = \frac{p_{1}}{r} (1 + \log_{e} r ) – p_{b} = \frac{11}{ 1 / 0.8} (1 + \log_{e} \frac{1}{0.8}) – 1.1 = 9.66 bar.
p_{m(act.)} = Diagram factor (D.F.)_{1} × p_{m(th.)}
= 0.8 × 9.66 = 7.73 bar
Speed of the train = \frac{π D_{w} N × 60}{1000}
= 15 (given)
∴ N = \frac{15 × 1000}{π × 2.2 × 60} = 36.2 r.p.m
As the engine is directly coupled to the driving wheel, the rotational speed of the engine is equal to the rotational speed of the driving wheel.
∴ I.P. = \frac{n × 10 p_{m} LAN }{3}
= \frac{2 × 10 × 7.73 × 0.55 × π / 4 × 0.35² × 36.2}{3}
[∵ n = 2 and the engine is double-acting]
= 98.7 kW
∴ B.P. = η_{mech.} × I.P. = 0.7 × 98.7 = 69.1 kW
B.P. = Tractive effort × speed in metres/sec.
69.1 = P × (\frac{15 × 1000}{ 60 × 60} ) where P is in kN
∴ P = \frac{69.1 × 3600}{15 × 1000} = 16.584 kN.
i.e., Tractive effort = 16.584 kN.
(ii) Total train load W_{t} :
= \frac{11}{5} (1 + \log_{e} 5 ) – 1.1 [ ∵ r = \frac{1}{0.2} = 5 ]
= 4.64 bar
p_{m(act.)} = Diagram factor (D.F.)_{2} × p_{m(th.)}= 0.7 × 4.64 = 3.25 bar
Speed of the engine = 36.2 × \frac{90}{15} = 217.2 r.p.m.
I.P. ∝ p_{m(act.)} N
∴ I.P. at net cut-off and new speed is given by
I.P. = 98.7 × \frac{3.25}{7.73} × \frac{217.2}{36.2} = 248.98 kW
B.P. = η_{mech.} × 248.98 = 0.8 × 248.98 = 199.2 kW
Tractive effort, P = \frac{B.P.}{Speed in metre /s} , where P is in kN
= \frac{199.2}{(\frac{90 × 1000}{60 × 60 })} = 7.968 kN
∴ Total train load, W_{t} = \frac{7.968 (kN) × 1000 (N)}{12 (N) } = 664 kN.