Answer:
The efficiency of this fuel cell is 80.69 percent.
Explanation:
From Physics we define the efficiency of the automotive fuel cell ([tex]\eta[/tex]), dimensionless, as:
[tex]\eta = \frac{\dot W_{out}}{\dot W_{in}}[/tex] (Eq. 1)
Where:
[tex]\dot W_{in}[/tex] - Maximum power possible from hydrogen flow, measured in kilowatts.
[tex]\dot W_{out}[/tex] - Output power of the automotive fuel cell, measured in kilowatts.
The maximum power possible from hydrogen flow is:
[tex]\dot W_{in} = \dot V\cdot \rho \cdot L_{c}[/tex] (Eq. 2)
Where:
[tex]\dot V[/tex] - Volume flow rate, measured in cubic meters per second.
[tex]\rho[/tex] - Density of hydrogen, measured in kilograms per cubic meter.
[tex]L_{c}[/tex] - Heating value of hydrogen, measured in kilojoules per kilogram.
If we know that [tex]\dot V = \frac{28}{3600}\,\frac{m^{3}}{s}[/tex], [tex]\rho = 0.0899\,\frac{kg}{m^{3}}[/tex], [tex]L_{c} = 141790\,\frac{kJ}{kg}[/tex] and [tex]\dot W_{out} = 80\,kW[/tex], then the efficiency of this fuel cell is:
(Eq. 1)
[tex]\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)[/tex]
[tex]\dot W_{in} = 99.143\,kW[/tex]
(Eq. 2)
[tex]\eta = \frac{80\,kW}{99.143\,kW}[/tex]
[tex]\eta = 0.807[/tex]
The efficiency of this fuel cell is 80.69 percent.
