定义式:消除速率 14 mg/h,浓度 5 mg/L → CL = 2.8 L/h CL = 14 / 5 = 2.8 L/h。对应 70 kg 不吸烟成人茶碱的典型清除率。{"result":"CL = 2.8 L/h (definition: elimination rate 14 mg/h ÷ C 5 mg/L).","metadata":{"input":{"mode":"definition","eliminationRate":14,"concentration":5},"result":{"clearance":2.8,"clearanceUnit":"L/h","method":"Definition: CL = rate / C","model":"Definitional — applies to any pharmacokinetic model. Not medical advice."}}}
线性模式:Vd=31.5 L,t½=8 h → CL = 2.7288 L/h k = ln2/8 = 0.0866 h⁻¹;CL = k·Vd = 0.0866 × 31.5 = 2.7288 L/h。{"result":"CL = 2.7288 L/h (linear: k = 0.0866 h⁻¹ × Vd 31.5 L). t½ consistency: 0.693·31.5/2.7288 = 8 h.","metadata":{"input":{"mode":"linear","halfLife":8,"vd":31.5},"result":{"clearance":2.7288,"clearanceUnit":"L/h","k":0.0866,"halfLifeDerived":8,"method":"Linear: CL = k · Vd, k = ln2 / t½","model":"Linear first-order elimination. Not medical advice."}}}