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Pachacha [2.7K]
3 years ago
11

The energy expenditure value of traveling by car is 3.6 mj/passenger-kilometer. The value for traveling by train is 1.1 mj/passe

nger-kilometer. What would be the best way to increase the efficiency of traveling by car?
Physics
1 answer:
andrezito [222]3 years ago
7 0

Answer:

Using lighter material in car construction, improving energy efficiency by enhancing engine design or replacing the engine with more efficient technologies.

Explanation:

Using lighter materials in the car construction, reducing the potential energy required to accelerate and to move the car, as well as energy losses due to rolling friction. There is evidence of such benefits by replacing steel and aluminium parts with components made of composite materials.  

Improving the design of internal combustion engines to minimize energy losses and accordingly, improving energy efficiency. A more radical approach is replacing internal combustion engines with electric engines, which offer higher efficiencies. Such conclusions can be easily inferred from model based on Work-Energy Theorem and Principle of Energy Conservation:

\eta_{engine} \cdot U_{engine} = \frac{1}{2} \cdot m_{car} \cdot v^{2} + \mu_{r} \cdot m_{car} \cdot g \cdot \Delta s

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The answer is "143.74^{\circ} \ C , 8.36\ g, and \ 2.77\ \frac{K}{J}"

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For point a:

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W=0\\\\Q=0\\\\m_e=0

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\frac{dU}{dt}=0-0+m_ih_i-0\\\\\Delta U=\int^{2}_{1}m_ih_idt\\\\

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T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}

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Follow the ideal gas table.

The u_2= 298.33\ \frac{kJ}{kg} and between temperature T =410 \ K \ and\  T=240\ K.

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For point b:

Consider the ideal gas equation.  therefore, p is pressure, V is the volume, m is mass of gas. \bar{R} \ is\  \frac{R}{M} (M is the molar mass of the  gas that is 28.97 \ \frac{kg}{mol} and R is gas constant), and T is the temperature.

n=\frac{pV}{TR}\\\\

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