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andrey2020 [161]
3 years ago
5

The intake and exhaust processes are not considered in the p-V diagram of Otto cycle. a) true b) false

Engineering
1 answer:
vovangra [49]3 years ago
6 0

Answer:

b) false

Explanation:

We know that Otto cycle is the ideal cycle for all petrol working engine.In Otto cycle all process are consider is ideal ,means there is no any ir-reversibility in the processes.

It consist four processes

1-2:Reversible adiabatic compression

2-3:Constant volume heat addition

3-4:Reversible adiabatic expansion

3-4:Constant volume heat rejection

Along with above 4 processes intake and exhaust processes are parallel to each other.From the P-v diagram we can see that all processes.

But actually in general we are not showing intake and exhaust line then it did not mean that in Otto cycle did not have intake and exhaust processes.

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Answer:

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Explanation:

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                           i (t) = 6*e^(-2*t)

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                                       dQ = i(t) . dt

                  Q = \int\limits^2_0 {i(t)} \, dt = \int\limits^2_0 {6*e^(-2t)} \, dt = 6*\int\limits^2_0 {e^(-2t)} \, dt

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                   = 6 * (-0.5)*e^-2t = - 3*( 1 / e^4 - 1) = 2.945 C

- The power can be calculated by using v(t) and i(t) as follows:

                 v(t) = 10* di / dt = 10*d(6*e^(-2*t)) /dt

                 v(t) = 10*(-12*e^(-2*t)) = -120*e^-2*t mV

                 P(t) = v(t)*i(t) = (-120*e^-2*t) * 6*e^(-2*t)

                 P(t) = -720*e^(-4t) uW

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                 W = \int\limits^3_0 {P(t)} \, dt = \int\limits^2_0 {-720*e^(-4t)} \, dt = -720*\int\limits^2_0 {e^(-4t)} \, dt

- Integrate and evaluate the on the interval:

                  W = -180*e^-4t = - 180*( 1 / e^12 - 1) = -180uJ

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