An ideal Otto cycle has a compression ratio of 9.2 and uses...

An ideal Otto cycle has a compression ratio of 9.2 and uses air as the working fluid. At the beginning of the compression process, air is at $98 \mathrm{kPa}$ and $27^{\circ} \mathrm{C}$. The pressure is doubled during the constant-volume heat-addition process. Accounting for the variation of specific heats with temperature, determine (a) the amount of heat transferred to the air, (b) the net work output, $(c)$ the thermal efficiency, and $(d)$ the mean effective pressure for the cycle.

Key Concepts

Mean Effective Pressure

Mean effective pressure (MEP) is a computed average pressure that, if acting on the piston during the entire power stroke, would produce the same amount of work as the actual varying pressures in the cycle. It serves as a useful parameter for comparing the performance of different engines independent of their displacement.

Net Work Output

The net work output is the difference between the work produced during expansion and the work consumed during compression in a thermodynamic cycle. It is a crucial measure of an engine's performance, directly impacting its power output and overall efficiency.

Thermal Efficiency

Thermal efficiency in the Otto cycle represents the fraction of the heat input that is converted into useful work. It is calculated based on the work done over the cycle and the heat added, and it depends on factors such as the compression ratio and the specific heat capacities of the working fluid.

Variable Specific Heats

In real thermodynamic processes, the specific heats of the working fluid can vary with temperature. Accounting for variable specific heats is important for obtaining accurate calculations of heat transfer, work output, and efficiency, especially when the temperature changes within the cycle are significant.

Compression Ratio

The compression ratio is the ratio of the maximum to minimum volume in the engine cylinder during the cycle. It is a key parameter in engine design, affecting the thermal efficiency and power output by influencing the pressure and temperature reached during the compression stroke.

Otto Cycle

The Otto cycle is the idealized thermodynamic cycle that models the functioning of spark-ignition internal combustion engines. It comprises two isentropic processes (compression and expansion) and two constant-volume processes (heat addition and rejection). This cycle is fundamental in understanding the energy conversion and efficiency of gasoline engines.

Constant-Volume Process

A constant-volume process, also known as an isochoric process, is one in which the volume of the working fluid remains fixed. In the Otto cycle, these processes occur during the heat addition and rejection steps, significantly determining the change in internal energy and the corresponding heat transfer.

Transcript

00:01 Our question is, an ideal auto cycle has a compression ratio of 8.

00:06 At the beginning of the compression process, air is at these values are given, heat is transferred to air during the constant volume heat addition process.

00:15 Taking into account the variation of specific heat with temperature, determine the pressure and temperature at the end of heat addition process.

00:22 This is a, b, the network output, c, the thermal efficiency, and d, the mean effective pressure for the cycle.

00:31 Have to calculate these values.

00:34 So as the properties of air are given, that is r is equal to 0 .287 kilojou per kg.

00:50 Here our diagram is left between b and u 750 jude.

01:32 The first process, we have to calculate the pressure in temperature and the end of heat addition process in the process one to two it is isentropic compression here t1 is equal to 300 kelvin has values of u1 and vr1 equals to at this temperature is 214 .0 .07 kilojou per kg is kilo jowl per kg and vrr1 equals 621 .2 vr 2 is equal to v2 over v1 u r1 equals to 1 over r1 u r1 which is 1 by 8 621 .2 which is equal to 77 .65 at this vr 2 u r2 we have temperature and u2 is equals to 6 .73 .1 kelvin and s 491 .2 kilojou per kg.

03:20 So p2 u2 over t2 is equals to p1 v1 this implies that p2 is equal to u1 over t1.

03:29 This implies that p2 is equal to u1 over v2 t2 over t1 p1.

03:37 Putting the value we get 8673 .1 over 300 to 95 is equals to 1705 kilopascal in the process 2 to 3 to 3 that is constant heat addition constant heat addition process we have q input heat is equals to u3 minus u2 that is u3 is equal to u2 plus q input from 2 to 3 putting the values 490 1 .2 plus 750 is equal to 1 to 4 1 .2 kilo jr per kg which is equals to t3 1539 kelvin and vr3 is equal to 6 .580...

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