An ideal dual cycle has a compression ratio of 12 and uses...

An ideal dual cycle has a compression ratio of 12 and uses air as the working fluid. At the beginning of the compression process, air is at $14.7 \mathrm{psia}$ and $90^{\circ} \mathrm{F}$, and occupies a volume of $75 \mathrm{in}^3$. During the heat-addition process, $0.3 \mathrm{Btu}$ of heat is transferred to air at constant volume and 1.1 Btu at constant pressure. Using constant specific heats evaluated at room temperature, determine the thermal efficiency of the cycle.

Key Concepts

Compression Ratio

The compression ratio is defined as the ratio of the initial volume to the volume after compression in an engine cylinder. It is a critical parameter in engine cycles because higher compression ratios can lead to increased thermal efficiency by raising the thermal state of the working fluid before heat addition, although practical and material limitations exist.

Constant Specific Heats

The assumption of constant specific heats simplifies the thermodynamic analysis by treating the specific heat capacities of the working fluid as invariant with temperature. This means that the amount of energy needed to raise the temperature of the fluid remains constant throughout the process, allowing for easier calculation of energy transfers in the cycle.

Dual Cycle

The dual cycle is an ideal thermodynamic cycle that models internal combustion engines by combining two heat-addition processes: one at constant volume and one at constant pressure. This cycle integrates elements of both the Otto and Diesel cycles, enabling a more flexible representation of engine performance by accounting for heat additions in different ways during the power stroke.

Thermal Efficiency

Thermal efficiency is a key performance metric in thermodynamic cycles that quantifies the ratio of net work output to the total heat input. It provides a measure of how effectively the cycle converts supplied heat into useful work, and is influenced by the specific processes undergone during the cycle such as compression and heat addition.

Heat Addition Processes

In thermodynamic cycles, heat is typically added to the working fluid during the power stroke. The dual cycle specifically incorporates two heat addition processes: one at constant volume, which results in a rapid rise in temperature and pressure, and one at constant pressure, which allows the volume to increase while maintaining a steady pressure. Each of these processes has distinct effects on the cycle’s efficiency and behavior.

Transcript

00:01 We know that pv is equal to m r t this is general formula so p1 v1 is equal to m r t1 so we can calculate m from this formula so first we have p1 is equal to 14 .7 and t1 is equal to 120 degree fahrenheit so changing into r and v1 is equal to 98 iron cube change in into fit cube so putting the value in the formula of m and solving this so on calculating it we get m is equal to 0 .00388 i bm now t2 by t1 is equal to v1 x2 whole to the power k minus 1 so ongoing further we can put the value here so t2 by 580 is equal to 14 to the power 1 .4 minus 1.

01:12 So solving it we get t2 as this value.

01:17 Now going ahead and writing q223 is equal to mcv t3 minus t2.

01:24 This is formula.

01:26 So on putting all the data here in this formula and solving this, so first we will put all the data as i am putting here.

01:38 Now going forward and we can write this as this.

01:45 So this is coming as t3.

01:48 Now going forward, this is the formula for q3 to 4.

01:54 So just putting all the data here which are calculated earlier or given in the question.

02:00 So we get t4 as 3752 .37r.

02:07 Now going forward and solving further we can write t4 by t5 is equal to v5 by v4 hold to the power k minus 1.

02:19 So v1 by v2 into v2 by v4 hold to the power k minus 1 is equal to v1 by v2 into v3 by v4 whole to the power k minus 1.

02:31 As v2 is equal to v3.

02:34 So i just change the values.

02:38 So t4 by t5 is coming as v1 by v2 into t3 by t4 whole t3 the power k minus 1.

02:47 So on putting the calculated and given data here so that we can get the value of t5 from this equation...

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