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Alecsey [184]
3 years ago
10

What is a Material in engineering

Engineering
1 answer:
NeX [460]3 years ago
6 0
Engineering material refers to the group of material that are used in the construction of manmade structures and components
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Answer:

FOR DARKEND1

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try copy and paste

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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 h
EastWind [94]

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.

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Explanation:

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Explanation:

Look at the photo

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