Fuel-cell bus range and efficiency claim
A transit agency is testing a 400-cell PEM fuel-cell bus. Each cell has 300 cm² of active area and operates at 0.65 V at a current density of 1.0 A/cm². The air compressor and other auxiliaries draw 12 kW from the stack's gross output. The bus requires 70 kW at the wheels on its test route, and the drivetrain efficiency from stack terminals to wheels is 90%. Hydrogen storage is 20 kg usable at 350 bar. The manufacturer claims a 350 km range at an average route speed of 40 km/h and a tank-to-wheel efficiency of 60%. Hydrogen LHV is 120 MJ/kg and the stack operates at a hydrogen stoichiometry of 1.5. The agency must verify the power, fuel consumption and range claims before accepting the vehicle.
1. Calculate the gross and net stack power at 1.0 A/cm², and determine whether the stack can deliver 70 kW at the wheels.
Read the worked answer
Cell current = 1.0 A/cm² × 300 cm² = 300 A.
Gross stack power = 400 cells × 0.65 V × 300 A = 78,000 W = 78 kW.
Net stack power = 78 − 12 = 66 kW.
Wheel power at 90% drivetrain efficiency = 66 × 0.90 = 59.4 kW.
This is below the 70 kW wheel requirement. To deliver 70 kW at the wheels, the stack must provide 70/0.90 = 77.8 kW net, or 89.8 kW gross. At 0.65 V per cell, that would require 89,800/(400 × 0.65) = 345 A, or 1.15 A/cm². At the higher current density the cell voltage will be lower than 0.65 V, so the actual current density would be even higher. The rated operating point is therefore inadequate for the stated 70 kW route demand; the claim only holds at a lower average power.
2. Estimate hydrogen consumption at 1.0 A/cm² and the range on 20 kg of usable hydrogen at 40 km/h.
Read the worked answer
Hydrogen molar flow = N × I/(2F) × stoichiometry = 400 × 300/(2 × 96,485) × 1.5 = 0.933 mol/s.
Mass flow = 0.933 mol/s × 2.016 g/mol = 1.88 g/s = 6.77 kg/h.
At 20 kg usable, operating time = 20/6.77 = 2.95 h.
Range at 40 km/h = 2.95 × 40 = 118 km.
This is far below the claimed 350 km. A 350 km range at 40 km/h would require 8.75 h of operation and about 59 kg of hydrogen at this consumption rate. The claim could only be approached at much lower average power, with hydrogen recirculation reducing the stoichiometry penalty, or with a much larger tank. The 20 kg, 350 bar storage is inconsistent with a 350 km claim for this stack.
3. Assess the 60% tank-to-wheel efficiency claim using the calculated hydrogen flow.
Read the worked answer
Hydrogen energy flow = 6.77 kg/h × 120 MJ/kg = 812.4 MJ/h = 225.7 kW.
Net stack electrical output = 66 kW, so tank-to-electric efficiency = 66/225.7 = 29.2%.
Wheel power = 59.4 kW, so tank-to-wheel efficiency = 59.4/225.7 = 26.3%.
The 60% claim is not supported. Even the gross stack power of 78 kW corresponds to only 34.6%, before auxiliaries. The cell voltage itself corresponds to about 0.65/1.25 = 52% on an LHV basis, so the system cannot reach 60% tank-to-wheel with 12 kW of auxiliaries and a stoichiometry of 1.5. If the unused hydrogen were recirculated and stoichiometry effectively reduced to 1.0, consumption would fall to about 4.51 kg/h and tank-to-wheel efficiency would rise to roughly 39%, still below 60%. The agency should reject the 60% claim and renegotiate the range guarantee.