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luda_lava [24]
2 years ago
15

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.
Engineering
1 answer:
svetoff [14.1K]2 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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Leno4ka [110]

Answer:

ω=314.15 rad/s.

0.02 s.

Explanation:

Given that

Motor speed ,N= 3000 revolutions per minute

N= 3000 RPM

The speed of the motor in rad/s given as

\omega=\dfrac{2\pi N}{60}\ rad/s

Now by putting the values in the above equation

\omega=\dfrac{2\pi \times 3000}{60}\ rad/s

ω=314.15 rad/s

Therefore the speed in rad/s will be 314.15 rad/s.

The speed in rev/sec given as

\omega=\dfrac{ 3000}{60}\ rad/s

ω= 50 rev/s

It take 1 sec to cover 50 revolutions

That is why to cover 1 revolution it take

\dfrac{1}{50}=0.02\ s

4 0
2 years ago
Compare the use of a low-strength, ductile material (1018 CD) in which the stress-concentration factor can be ignored to a high-
kicyunya [14]

Answer:

Step 1 of 3

Case A:

AISI 1018 CD steel,

Fillet radius at wall=0.1 in,

Diameter of bar

From table deterministic ASTM minimum tensile and yield strengths for some hot rolled and cold drawn steels for 1018 CD steel

Tensile strength

Yield strength

The cross section at A experiences maximum bending moment at wall and constant torsion throughout the length. Due to reasonably high length to diameter ratio transverse shear will be very small compared to bending and torsion.

At the critical stress elements on the top and bottom surfaces transverse shear is zero

Explanation:

See the next steps in the attached image

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Consider a flat plate that is 25 mm long, 30 mm wide, and 1 mm thick and a 50 mm long cylinder with the same volume as the plate
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Answer:

Average heat transfer =42.448w/m^2k

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Explanation:

See attachment for step by step guide

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In subsea oil and natural gas production, hydrocarbon fluids may leave the reservoir with a temperature of 70°C and flow in subs
NeTakaya

Answer:

For detailed answer of "

In subsea oil and natural gas production, hydrocarbon fluids may leave the reservoir with a temperature of 70°C and flow in subsea surrounding of S°C. As a result of the temperature difference between the reservoir and the subsea surrounding, the knowledge of heat transfer is critical to prevent gas hydrate and wax deposition blockages. Consider a subsea pipeline with inner diameter of O.S m and wall thickness of 8 mm is used for transporting liquid hydrocarbon at an average temperature of 70°C, and the average convection heat transfer coefficient on the inner pipeline surface is estimated to be 2SO W/m2.K. The subsea surrounding has a temperature of soc and the average convection heat transfer coefficient on the outer pipeline surface is estimated to be ISO W /m2 .K. If the pipeline is made of material with thermal conductivity of 60 W/m.K, by using the heat conduction equation (a) obtain the temperature variation in the pipeline wall, (b) determine the inner surface temperature of the pipeline wall, (c) obtain the mathematical expression for the rate of heat loss from the liquid hydrocarbon in the pipeline, and (d) determine the heat flux through the outer pipeline surface."

see attachment.

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3 years ago
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uysha [10]

Answer:288 pm

Explanation:

Number of atoms(s) for face centered unit cell -

Lattice points: at corners and face centers of unit cell.

For face centered cubic (FCC), z=4.

- whereas

For an FCC lattices √2a =4r =2d

Therefore d = a/√2a = 408pm/√2a= 288pm

I think with this step by step procedure the, the answer was clearly stated.

8 0
2 years ago
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