2.3 Analyze the ideal Diesel cycle and cutoff ratio | University of Alberta MEC E 340: Applied Thermodynamics

University of Alberta MEC E 340: Applied Thermodynamics

Gas Power Cycles

Process constraints, state temperatures, and thermal efficiency

Model a fixed mass of air as a closed system completing a cycle through four numbered states. State 1 begins compression; state 2 ends compression; state 3 ends constant-pressure heat addition; and state 4 ends expansion. The ideal Diesel cycle has isentropic compression from 1 to 2, constant-pressure heat addition from 2 to 3, isentropic expansion from 3 to 4, and constant-volume heat rejection from 4 to 1. These idealizations do not describe every detail of an actual engine. Use heat added as positive and define heat rejected as a positive magnitude. Each example states the property model and data it uses.

What you will learn

Identify the four ideal Diesel-cycle processes and their heat-transfer directions.
Distinguish compression ratio from cutoff ratio and relate each to specific volume.
Use constant-specific-heat ideal-gas relations to determine temperatures and heat transfers.
Calculate ideal Diesel-cycle thermal efficiency and check whether results are physically reasonable.

1. Define the cycle and its ratios

During 1–2, air is compressed isentropically: specific volume falls and temperature rises. During 2–3, heat is added at constant pressure, so both specific volume and temperature increase. During 3–4, air expands isentropically. During 4–1, heat is rejected at constant volume, returning the air to its initial state.

The compression ratio describes the volume change during compression. The cutoff ratio describes the volume change during constant-pressure heat addition. A cutoff ratio greater than one means that specific volume increases from state 2 to state 3. Both ratios are dimensionless.

For a complete cycle, the working fluid returns to its initial state, so its net energy change is zero. Thus net work output per unit mass equals heat supplied minus the positive magnitude of heat rejected. In the ideal model, the two isentropic processes have no heat transfer.

r=\frac{v_1}{v_2},\qquad r_c=\frac{v_3}{v_2}

Process 2–3 is constant-pressure heat addition; process 4–1 is constant-volume heat rejection.
Compression ratio and cutoff ratio describe different processes.
Use absolute temperature in kelvins and consistent energy units.

2. Relate state temperatures using the ideal-gas model

For an ideal gas with constant specific heats, let the specific-heat ratio be

k=c_p/c_v

. Isentropic compression relates the temperatures at states 1 and 2 to the compression ratio. During constant-pressure heating, the ideal-gas relation makes temperature proportional to specific volume, so the temperature ratio from state 2 to state 3 equals the cutoff ratio.

Because process 4–1 is at constant volume,

v_4=v_1

. The specific-volume ratio during expansion is therefore

v_4/v_3=r/r_c

. Applying the isentropic relation to process 3–4 gives the temperature at state 4.

For heat transfer, use

c_p

during constant-pressure heating and

c_v

during constant-volume rejection. A specific heat in

\mathrm{kJ/(kg\cdot K)}

multiplied by a temperature difference in kelvins gives energy per unit mass in

\mathrm{kJ/kg}

. These relations apply here only when the problem specifies an ideal gas with constant specific heats.

T_2=T_1r^{k-1},\quad T_3=T_2r_c,\quad T_4=T_3\left(\frac{r_c}{r}\right)^{k-1}

Use $c_p$ for constant-pressure heat addition and $c_v$ for constant-volume heat rejection.
The expansion volume ratio is $r/r_c$ , not $r_c/r$ .
Use supplied property data; do not infer values from a schematic plot.

3. Apply the cycle energy balance and calculate efficiency

For the closed cycle, there is no net energy accumulation over a complete cycle. Heat enters only during 2–3 and leaves only during 4–1 in this ideal model. Calculate the heat transfers from the temperature changes. Subtract the positive heat-rejection magnitude from heat input to obtain net work per unit mass.

Thermal efficiency is the fraction of supplied heat converted to net work. Substituting the state relations and heat-transfer expressions gives a convenient relation using compression ratio, cutoff ratio, and

k

.

A P–v diagram shows the process sequence and constraints, but a schematic diagram does not give exact state values. Check that heat input and net work are positive, heat rejection is a positive magnitude, and efficiency is between zero and one. At fixed compression ratio and specific-heat ratio, increasing cutoff ratio lowers ideal Diesel-cycle efficiency.

w_{net}=q_{in}-q_{out}, \eta_{th}=1-

\frac{1}{r^{k-1}}\frac{r_c^k-1}{k(r_c-1)}

Keep heat rejection as a positive magnitude when finding net work.
Efficiency compares net work output with heat supplied.
A schematic plot shows process type and order, not measured property values.

Worked example

Find temperatures and efficiency from the ratios

An air-standard Diesel cycle has

T_1=300\ \mathrm{K}

, compression ratio

r=18

, and cutoff ratio

r_c=2.00

. Use the supplied constant properties

c_p=1.004\ \mathrm{kJ/(kg\cdot K)}

,

c_v=0.717\ \mathrm{kJ/(kg\cdot K)}

, and

k=1.40

. Find the state temperatures, heat transfers, and thermal efficiency.

Ideal Diesel cycle on P–v axes

Closed air-standard cycle; schematic and not to scale. The state order and process directions follow the stated ideal Diesel model.

Set the model
Treat the working fluid as a fixed mass of air in a closed cycle. Use the supplied constant-specific-heat ideal-gas model and the four ideal Diesel processes. Report heat rejection as a positive magnitude.
Find the temperatures
Use isentropic compression for 1–2, constant-pressure heating for 2–3, and isentropic expansion for 3–4. Since $v_4=v_1$ , the expansion volume ratio is $r/r_c$ .
$T_2=300(18)^{0.4}\approx953.6\ \mathrm{K},\quad T_3=2T_2\approx1907.2\ \mathrm{K},\quad T_4=T_3\left(\frac{2}{18}\right)^{0.4}\approx791.8\ \mathrm{K}$
Calculate heat and efficiency
Use the supplied $c_p$ for heat input on 2–3 and $c_v$ for the positive heat-rejection magnitude on 4–1. Net work is their difference. q_{in}=1.004(1907.2-953.6)\approx957.4\ kJ/kg, q_{out}=0.717(791.8-300)\approx352.7\ kJ/kg, \eta_{th}= $\frac{957.4-352.7}{957.4}$ \approx0.632

Answer: The state temperatures are approximately

T_2=953.6\ \mathrm{K}

,

T_3=1907.2\ \mathrm{K}

, and

T_4=791.8\ \mathrm{K}

. Heat input is about

957.4\ \mathrm{kJ/kg}

, heat rejection is about

352.7\ \mathrm{kJ/kg}

, and thermal efficiency is about 63.2%.

Check: Net work is about

604.7\ \mathrm{kJ/kg}

, equal to heat input minus heat rejection. It is positive. Temperature rises during compression and heat addition, then falls during expansion and heat rejection.

Worked example

Find cutoff ratio from end-of-heating temperature

An ideal air-standard Diesel cycle has

T_1=290\ \mathrm{K}

and compression ratio

r=16

. The temperature at state 3 is

1800\ \mathrm{K}

. Use the supplied constant properties

c_p=1.004\ \mathrm{kJ/(kg\cdot K)}

,

c_v=0.717\ \mathrm{kJ/(kg\cdot K)}

, and

k=1.40

. Find the cutoff ratio and heat added per kilogram.

Ideal Diesel cycle on P–v axes

Closed air-standard cycle; schematic and not to scale. The plot shows process order, not calculated state coordinates.

Find the end-of-compression temperature
Process 1–2 is isentropic compression. Use the specified compression ratio and initial temperature with the supplied value of $k$ .
$T_2=290(16)^{0.4}\approx879.1\ \mathrm{K}$
Determine the cutoff ratio
During constant-pressure heating, the temperature ratio equals the specific-volume ratio. That volume ratio is the cutoff ratio.
$r_c=\frac{T_3}{T_2}=\frac{1800}{879.1}\approx2.048$
Calculate heat input
Heat is added at constant pressure, so use the supplied $c_p$ and the temperature increase from state 2 to state 3. q_{in}=1.004(1800-879.1)\approx924.6\ kJ/kg

Answer: The cutoff ratio is approximately

2.05

, and heat added is approximately

924.6\ \mathrm{kJ/kg}

.

Check: The cutoff ratio exceeds one, consistent with increasing specific volume during constant-pressure heating. Heat input is positive because state 3 is hotter than state 2.

Worked example

Compare two cutoff ratios

For constant-specific-heat air with

k=1.40

, compare ideal Diesel-cycle thermal efficiency at the same compression ratio

r=16

for cutoff ratios

r_c=1.8

and

r_c=2.4

.

Ideal Diesel cycle on P–v axes

Closed air-standard cycle; schematic and not to scale. Shown to identify Diesel processes, not to compare exact state coordinates.

Use the efficiency relation
The compression ratio and specific-heat ratio are fixed. Evaluate the constant-specific-heat Diesel efficiency relation for each supplied cutoff ratio.
$\eta_{th}=1-\frac{1}{r^{k-1}}\frac{r_c^k-1}{k(r_c-1)}$
Compare the results
Substitution gives approximately 0.624 for cutoff ratio 1.8 and 0.595 for cutoff ratio 2.4. The larger cutoff ratio gives lower efficiency in this comparison.
$\eta_{th}(1.8)\approx0.624,\quad \eta_{th}(2.4)\approx0.595$

Answer: At

r_c=1.8

, efficiency is about 62.4%; at

r_c=2.4

, it is about 59.5%. For this model and fixed compression ratio, the larger cutoff ratio has lower efficiency.

Check: Both efficiencies lie between zero and one. The comparison is consistent with the efficiency relation at fixed

r

and

k

.

Common mistakes and how to avoid them

Using

v_1/v_2

as the cutoff ratio.

Correction: That ratio is the compression ratio. The cutoff ratio is

v_3/v_2

.

Using

c_v

to calculate heat added during 2–3.

Correction: Process 2–3 is at constant pressure, so use

c_p

in the constant-specific-heat model.

Reading exact properties from a schematic P–v plot.

Correction: Use supplied data and process relations for calculations; the sketch indicates process type and order only.

Subtracting a negative heat-rejection value after defining rejection as a positive magnitude.

Correction: With

q_{out}

defined as a positive magnitude, calculate net work as q_{in}-q_{out}.

Lesson summary

The ideal Diesel cycle consists of isentropic compression, constant-pressure heat addition, isentropic expansion, and constant-volume heat rejection.
Compression ratio is $r=v_1/v_2$ ; cutoff ratio is $r_c=v_3/v_2$ .
Use the stated constant-specific-heat relations and the specific heat appropriate to each heat-transfer process.
Thermal efficiency is net work divided by heat input; at fixed compression ratio, increasing cutoff ratio lowers ideal Diesel-cycle efficiency.

Check your understanding

Question 1

Which process defines the cutoff ratio?

Constant-pressure heat addition from state 2 to state 3.
Isentropic compression from state 1 to state 2.
Constant-volume heat rejection from state 4 to state 1.
Isentropic expansion from state 3 to state 4.

Show answer and explanation

Constant-pressure heat addition from state 2 to state 3.

The cutoff ratio is

v_3/v_2

, the specific-volume ratio across constant-pressure heat addition.

Question 2

At fixed compression ratio and

k

, what happens to ideal Diesel-cycle efficiency when cutoff ratio increases?

It decreases.
It increases.
It remains exactly constant.
It becomes greater than one.

Show answer and explanation

It decreases.

For the constant-specific-heat Diesel relation, increasing cutoff ratio at fixed compression ratio lowers efficiency.

Key terms

Air-standard cycle: An idealized cycle model that uses air as the working fluid and represents the cycle with specified ideal processes.
Compression ratio: The specific volume at state 1 divided by the specific volume at state 2, $v_1/v_2$ .
Cutoff ratio: The specific volume at the end of constant-pressure heat addition divided by that at its start, $v_3/v_2$ .
Isentropic process: A process with constant entropy; in this ideal-cycle model, compression and expansion are treated as internally reversible and adiabatic.
Thermal efficiency: Net work output divided by heat supplied to the cycle.

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2.3 · Analyze the ideal Diesel cycle and cutoff ratio

University of Alberta MEC E 340: Applied Thermodynamics

Process constraints, state temperatures, and thermal efficiency

What you will learn

1. Define the cycle and its ratios

2. Relate state temperatures using the ideal-gas model

3. Apply the cycle energy balance and calculate efficiency

Find temperatures and efficiency from the ratios

Find cutoff ratio from end-of-heating temperature

Compare two cutoff ratios

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Key terms

Continue through MEC E 340

About this lesson and its review