Question

An automotive fuel cell consumes fuel at a rate of 28m3/h and delivers 80kW of power to the wheels. If the hydrogen fuel has a heating value of 141,790 kJ/kg and a density of 0.0899 kg/m3, determine the efficiency of this fuel cell.

Answer 1

Answer:

The efficiency of this fuel cell is 80.69 percent.

Explanation:

From Physics we define the efficiency of the automotive fuel cell ($\eta$), dimensionless, as:

$\eta = \frac{\dot W_{out}}{\dot W_{in}}$ (Eq. 1)

Where:

$\dot W_{in}$ - Maximum power possible from hydrogen flow, measured in kilowatts.

$\dot W_{out}$ - Output power of the automotive fuel cell, measured in kilowatts.

The maximum power possible from hydrogen flow is:

$\dot W_{in} = \dot V\cdot \rho \cdot L_{c}$ (Eq. 2)

Where:

$\dot V$ - Volume flow rate, measured in cubic meters per second.

$\rho$ - Density of hydrogen, measured in kilograms per cubic meter.

$L_{c}$ - Heating value of hydrogen, measured in kilojoules per kilogram.

If we know that $\dot V = \frac{28}{3600}\,\frac{m^{3}}{s}$, $\rho = 0.0899\,\frac{kg}{m^{3}}$, $L_{c} = 141790\,\frac{kJ}{kg}$ and $\dot W_{out} = 80\,kW$, then the efficiency of this fuel cell is:

(Eq. 1)

$\dot W_{in} = \left(\frac{28}{3600}\,\frac{m^{3}}{s}\right)\cdot \left(0.0899\,\frac{kg}{m^{3}} \right)\cdot \left(141790\,\frac{kJ}{kg} \right)$

$\dot W_{in} = 99.143\,kW$

(Eq. 2)

$\eta = \frac{80\,kW}{99.143\,kW}$

$\eta = 0.807$

The efficiency of this fuel cell is 80.69 percent.