Physics, asked by Bhaijan4094, 1 year ago

With increase in engine speed the ignition of spark in petrol engine has to be

Answers

Answered by sahilarora199587
0

Explanation:

spark ignition engine on the whole requires very little modification to run on producer gas. Generally depending upon the make of engine(compression ratio and rpm), theignition timing has to be advanced by about 30–40°C. This is done because of low flame speed of producer gas as compared to gasoline.

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Answered by harshitapandey19
0

Answer:

A spark ignition engine on the whole requires very little modification to run on producer gas. Generally depending upon the make of engine(compression ratio and rpm), theignition timing has to be advanced by about 30–40°C. This is done because of low flame speed of producer gas as compared to gasoline.

A spark-ignition engine operates on a 10% rich mixture of carbon monoxide and air. The conditions at the end of compression are 8.5 bar and 600 K, and it can be assumed that the combustion is adiabatic at constant volume. Calculate the maximum pressure and temperature achieved if dissociation occurs.

(There are two different approaches for solving this type of problem; both of these will be outlined below. The first approach, which develops the chemical equations from the degrees of dissociation, is often the easier method for hand calculations because it is usually possible to estimate the degree of dissociation with reasonable accuracy, and it can also be assumed that the degree of dissociation of the water vapour is less than that of the carbon dioxide. The second approach is more appropriate for computer programs because it enables a set of simultaneous (usually nonlinear) equations to be defined.)

General considerations

The products of combustion with dissociation have to obey all of the laws which define the conditions of the products of combustion without dissociation, described in Chapter 10, plus the ratios of constituents defined by the equilibrium constant, Kpr. This means that the problem becomes one with two iterative loops: it is necessary to evaluate the degree of dissociation from the chemical equation and the equilibrium constant, and to then ensure that this obeys the energy equation(i.e. the First Law of Thermodynamics). The following examples show how this can be achieved.

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