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Tatiana [17]
3 years ago
6

In the last 5 meters of braking, you lose ___ of your speed.

Engineering
1 answer:
expeople1 [14]3 years ago
5 0

Answer:

answer is

a)

3/4

Explanation:

In the last 5 meters of braking, you lose 3/4 of your speed.

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b. equal to the specific entropy of the gas at the inlet.

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If a process is completely reversible, without the need to provide energy in the form of heat, then the process is isentropic.

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2 years ago
An ideal Otto cycle has a compression ratio of 9.2 and uses air as the working fluid. At the beginning of the compression proces
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(a) The amount of heat transferred to the air, q_{out} is 215.5077 kJ/kg

(b) The net work output, W_{net}, is 308.07 kJ/kg

(c) The thermal efficiency is 58.8%

(d) The Mean Effective Pressure, MEP, is 393.209 kPa

Explanation:

(a) The assumptions made are;

c_p = 1.005 kJ/(kg·K), c_v = 0.718 kJ/(kg·K), R = 0.287 kJ/(kg·K),

Process 1 to 2 is isentropic compression, therefore;

T_{2}= T_{1}\left (\dfrac{v_{1}}{v_{2}}  \right )^{k-1} = 300.15\times 9.2^{0.4} = 729.21 \, K

From;

\dfrac{p_{1}\times v_{1}}{T_{1}} = \dfrac{p_{2}\times v_{2}}{T_{2} }

We have;

p_{2} = \dfrac{p_{1}\times v_{1}\times T_{2}}{T_{1} \times v_{2}} = \dfrac{98\times 9.2\times 729.21}{300.15 } = 2190.43 \, kPa

Process 2 to 3 is reversible constant volume heating, therefore;

\dfrac{p_3}{T_3} =\dfrac{p_2}{T_2}

p₃ = 2 × p₂ = 2 × 2190.43 = 4380.86 kPa

T_3 = \dfrac{p_3 \times T_2}{p_2} =\dfrac{4380.86  \times 729.21}{2190.43} = 1458.42 \, K

Process 3 to 4 is isentropic expansion, therefore;

T_{3}= T_{4}\left (\dfrac{v_{4}}{v_{3}}  \right )^{k-1}

1458.42= T_{4} \times \left (9.2 \right )^{0.4}

T_4 = \dfrac{1458.42}{(9.2)^{0.4}}  = 600.3 \, K

q_{out} = m \times c_v \times (T_4 - T_1) = 0.718  \times (600.3 - 300.15) = 215.5077 \, kJ/kg

The amount of heat transferred to the air, q_{out} = 215.5077 kJ/kg

(b) The net work output, W_{net}, is found as follows;

W_{net} = q_{in} - q_{out}

q_{in} = m \times c_v \times (T_3 - T_2) = 0.718  \times (1458.42 - 729.21) = 523.574 \, kJ/kg

\therefore W_{net} = 523.574 - 215.5077 = 308.07 \, kJ/kg

(c) The thermal efficiency is given by the relation;

\eta_{th} = \dfrac{W_{net}}{q_{in}} \times 100=  \dfrac{308.07}{523.574} \times 100= 58.8\%

(d) From the general gas equation, we have;

V_{1} = \dfrac{m\times R\times T_{1}}{p_{1}} = \dfrac{1\times 0.287\times 300.15}{98} =0.897\, m^{3}/kg

The Mean Effective Pressure, MEP, is given as follows;

MEP =\dfrac{W_{net}}{V_1 - V_2} = \dfrac{W_{net}}{V_1 \times (1- 1/r)}= \dfrac{308.07}{0.897\times (1- 1/9.2)} = 393.209 \, kPa

The Mean Effective Pressure, MEP = 393.209 kPa.

3 0
3 years ago
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