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3 years ago

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A fuel cell vehicle draws 50 kW of power at 70 mph and is 40% efficient at rated power. You are asked to size the fuel cell syst

em so that a driver can go at least 400 miles at 70 mph before refueling. Specify the minimum volume and weight requirements for the fuel cell system (fuel cell fuel tank): Fuel cell power density is 1.5 kW/L and 1 kW/kg (note: the values taken into account efficiency already). Fuel tank energy density (compressed hydrogen with the tank mass taken into account): 4 MJ/L and 8 MJ/kg.

1 answer:

svetoff [14.1K]3 years ago

4 0

Answer:

a) fuel cell weight and volume: 50 kg, 33.3 L

b) fuel tank weight and volume: 321 kg, 643 L

Explanation:

<u>Fuel Cell</u>

Delivery of 50 kW from a source with a power density of 1.5 kW/L requires a source that has a volume of ...

(50 kW)/(1.5 kW/L) = 33.3 L

The weight of the power source is 1 kW/kg, so will be ...

(50 kW)/(1 kW/kg) = 50 kg

__

<u>Fuel Tank</u>

400 miles at 70 mph will take (400/70) h ≈ 5.71429 h. In that time, the energy used by the vehicle power plant is ...

(50 kW)(5.71429 h)(3600 s/h) = 1028.57 MJ

Since the power plant is 40% efficient, it must be supplied with 2.5 times that amount of energy, or 2571.43 MJ.

The tank volume and mass will then be ...

volume = 2571.43 MJ/(4 MJ/L) = 642.9 L

mass = 2571.43 MJ/(8 MJ/L) = 321.4 kg

_____

<em>Comment on fuel tank volume</em>

It appears the fuel tank would need to be equivalent in size to a sphere about 1.1 m in diameter.

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Explanation:

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$x_D = \text{bearing life in million revolution} \\ \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}$

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$C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}$

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