列车采用韶山3型电力机车牵引,机车质量P=138t,列车牵引质量G=2620t;车辆均采用滚动轴承;计算当列车以最低计算速度运行时,列车基本阻力与列车平均单位基本阻力?
正确答案:
Vjmin=48km/h
=(2.25+0.019V+0.00032V2)g
=(0.92+0.0048V+0.000125V2)g
Vjmin=48km/h
=(2.25+0.019V+0.00032V2)g
=(2.25+0.019×48+0.00032×482)×9.81=38.3(N/t)
车辆采用滚动轴承;当考虑列车牵引质量时,即列车满载,所以为重车:
=(0.92+0.0048V+0.000125V2)g
=(0.92+0.0048×48+0.000125×482)×9.81=14.1(N/t)
列车平均单位基本阻力为:
=(2.25+0.019V+0.00032V2)g
=(0.92+0.0048V+0.000125V2)g
Vjmin=48km/h
=(2.25+0.019V+0.00032V2)g
=(2.25+0.019×48+0.00032×482)×9.81=38.3(N/t)
车辆采用滚动轴承;当考虑列车牵引质量时,即列车满载,所以为重车:
=(0.92+0.0048V+0.000125V2)g
=(0.92+0.0048×48+0.000125×482)×9.81=14.1(N/t)
列车平均单位基本阻力为:
答案解析:有
微信扫一扫手机做题