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2.1 · State air-standard and cold-air-standard assumptions

Learn to state air-standard and cold-air-standard assumptions through clear examples and targeted practice.

University of Alberta MEC E 340: Applied Thermodynamics

Gas Power Cycles

How idealizations simplify the analysis of gas power cycles

Begin with a closed cycle containing a fixed mass of air. Label its states in order as 1,2,…,n1,2,\ldots,n, with the final state returning to state 11. For this lesson, the cycle’s detailed processes are not specified; no particular cycle shape or performance value is assumed. The purpose is to state which idealizations make a gas-cycle model easier to analyze. The first law still applies: for a closed system, changes in energy are accounted for by heat transfer and work. Air-standard assumptions change how the working fluid and energy transfers are represented; they do not replace that balance.

What you will learn

  • State the air-standard assumptions and explain what each one changes in a cycle model.
  • Distinguish air-standard analysis from cold-air-standard analysis.
  • Use supplied specific-heat data to calculate a simple energy change and identify the model being used.
  • Recognize when a stated assumption is not enough to determine a cycle’s performance.

1. Start with the system and the baseline assumptions

The system is a closed cycle with air as its working fluid. In a real engine, fuel and air react, and combustion products leave through the exhaust. In an air-standard model, the working fluid instead completes a cycle: it returns to its starting state. This makes it possible to analyze the cycle without following fresh reactants and exhaust gases through the engine.
The usual air-standard assumptions are: (1) the working fluid is air circulating in a closed loop, and air is treated as an ideal gas; (2) combustion is replaced by heat transfer into the air from an external source; (3) exhaust is replaced by heat transfer from the air that restores it to its initial state; and (4) the processes within the cycle are internally reversible. These are modeling assumptions, not claims that a real engine contains only air or has no irreversibilities.
The ideal-gas assumption lets properties be related using an ideal-gas model. It does not, by itself, mean that specific heats are constant. The heat-addition and heat-rejection replacements are also important: the model accounts for energy crossing the system boundary as heat rather than calculating a chemical reaction or exchanging mass with the surroundings. No reaction calculation is part of this topic.
ΔE=Q−W\Delta E=Q-W
  • System: a closed cycle containing air; the cycle returns to its starting state.
  • Combustion and exhaust are represented by heat transfer, not by modelling fuel reactions or exhaust flow.
  • Air-standard assumptions do not automatically fix specific heats at one value.

2. Add the cold-air-standard assumption only when stated

Cold-air-standard analysis includes the air-standard assumptions and adds one more: the specific heats of air are taken to be constant at a reference temperature, commonly room temperature. Use the reference temperature and specific-heat values supplied by the problem or its stated source; do not silently choose property values.
With this added approximation, sensible energy changes can be calculated using constant specific heats. For a fixed mass of ideal-gas air, a constant-volume specific-heat model gives ΔU=mcv(T2−T1)\Delta U=m c_v(T_2-T_1), while a constant-pressure specific-heat model gives ΔH=mcp(T2−T1)\Delta H=m c_p(T_2-T_1). These formulas apply to the stated property model and the relevant energy change; they do not imply that an entire cycle is constant-volume or constant-pressure.
The distinction matters because specific heats of air can vary with temperature. Air-standard analysis can account for that variation if appropriate property data or relations are provided. Cold-air-standard analysis deliberately suppresses the variation to simplify calculations. The word “cold” names the constant-specific-heat approximation; it does not mean the cycle must operate only at low temperature.
Δu=cv(T2−T1),Δh=cp(T2−T1)\Delta u=c_v(T_2-T_1),\qquad \Delta h=c_p(T_2-T_1)
  • Cold-air-standard means air-standard plus constant specific heats evaluated at a stated reference temperature.
  • Use cvc_v for internal-energy change and cpc_p for enthalpy change in the corresponding ideal-gas model.
  • A constant-specific-heat assumption is a simplification, not a new cycle process.

3. Choose the model, then keep its assumptions consistent

Before calculating, identify the system, working fluid, numbered states, and which model the problem specifies. Write down any supplied property values and their units. If the problem says only “air-standard,” do not assume constant specific heats unless that is also stated or clearly established by the problem’s method.
For a closed-cycle energy balance, heat and work are energy transfers across the boundary. Use the sign convention in the equation above: QQ is positive into the system and WW is positive out of it. Over a complete cycle, the working fluid returns to its initial state, so its net change in energy is zero. That fact does not tell you the separate heat and work amounts; further process information or property data are needed.
A schematic cycle diagram can help organize states and processes, but air-standard assumptions alone do not specify which cycle is being analyzed or the path between states. Therefore, do not infer a cycle shape, state values, or performance from these assumptions alone. Check that a calculation uses the given temperature scale, specific-heat basis, mass basis, and sign convention consistently.
ΔEcycle=0\Delta E_{\mathrm{cycle}}=0
  • Do not infer constant specific heats from air-standard wording alone.
  • A completed cycle has zero net change in the working fluid’s energy, but this alone does not determine heat or work.
  • Only calculate a numerical property change when the required data and model are supplied.

Worked example

Identify the assumptions in a model statement

A problem says: “Model the working fluid as air in a closed cycle. Treat air as an ideal gas. Replace combustion by heat supplied to the air and replace exhaust by heat rejected from the air.” It does not say that specific heats are constant. Which assumptions are stated, and can you call this cold-air-standard analysis?
  1. List the model features
    The statement specifies a closed cycle with air as the working fluid, ideal-gas behaviour, and heat-transfer replacements for combustion and exhaust.
  2. Check for the added cold-air condition
    Cold-air-standard analysis also requires constant specific heats evaluated at a reference temperature. That condition is not stated, so the wording does not establish cold-air-standard analysis.
Answer: The stated features are air in a closed cycle, ideal-gas behaviour, and heat-transfer replacements for combustion and exhaust. Constant specific heats are not stated, so it is not justified to label the model cold-air-standard.
Check: The distinction depends on whether specific heats are fixed at a reference temperature—not on whether the working fluid is air.

Worked example

Calculate a supplied cold-air-standard energy change

For a closed mass of air, a process takes the air from T1=300 KT_1=300\ \mathrm{K} to T2=700 KT_2=700\ \mathrm{K} at constant volume. For this example only, use the supplied cold-air-standard value cv=0.718 kJ/(kg⋅K)c_v=0.718\ \mathrm{kJ/(kg\cdot K)} and mass m=0.50 kgm=0.50\ \mathrm{kg}. Find the internal-energy change.
  1. Select the property relation
    The process is at constant volume and the problem supplies a constant cvc_v, so the ideal-gas internal-energy change is calculated from the temperature change using that value.
    ΔU=mcv(T2−T1)\Delta U=m c_v(T_2-T_1)
  2. Substitute the supplied data
    The temperature rise is positive. The units of mass, specific heat, and temperature difference combine to give energy in kilojoules.
    ΔU=(0.50 kg)(0.718 kJ/(kg⋅K))(700−300) K\Delta U=(0.50\ \mathrm{kg})(0.718\ \mathrm{kJ/(kg\cdot K)})(700-300)\ \mathrm{K}
Answer: ΔU=143.6 kJ\Delta U=143.6\ \mathrm{kJ}, or approximately 144 kJ144\ \mathrm{kJ}.
Check: The result is positive because the air’s temperature increases. The specific-heat value was supplied in the problem; it is not a universal value for every temperature range.

Worked example

Spot an unsupported numerical conclusion

A student is given an air-standard cycle description but no temperatures, specific-heat data, or process details. The student assumes cold-air-standard properties and reports a numerical cycle work. Is the result supported by the stated information?
  1. Check which property model is specified
    Air-standard assumptions alone do not establish constant specific heats. The student has added a cold-air-standard assumption without support from the problem statement.
  2. Check what determines cycle work
    A numerical cycle-work calculation needs enough information to determine energy changes and transfers through the cycle, such as appropriate state or process information and property data. None is supplied here.
Answer: No. The assumptions given do not justify cold-air-standard properties, and the missing state, process, and property information does not determine a numerical cycle work.
Check: A plausible-looking number is not a result unless both the model and the data needed for it are stated.

Common mistakes and how to avoid them

Treating air-standard and cold-air-standard as interchangeable.
Correction: Cold-air-standard analysis adds the constant-specific-heat assumption to the air-standard assumptions.
Assuming ideal-gas behaviour automatically means constant specific heats.
Correction: Ideal-gas behaviour and constant specific heats are separate modelling choices.
Treating combustion and exhaust as actual heat-transfer devices in the engine.
Correction: They are replacements in the model: combustion is represented by heat supplied, and exhaust by heat rejected to restore the cycle.
Calculating cycle work from the assumptions alone.
Correction: The assumptions define a model but do not provide the state changes or property data needed for a numerical result.

Lesson summary

  • Air-standard analysis models a closed cycle with air as the ideal-gas working fluid; combustion and exhaust are replaced by heat transfer, and cycle processes are treated as internally reversible.
  • Cold-air-standard analysis adds the approximation that specific heats are constant at a stated reference temperature.
  • Use only supplied or identified property data, and do not infer a particular cycle or numerical performance from the assumptions alone.

Check your understanding

Question 1

Which additional assumption distinguishes cold-air-standard analysis from air-standard analysis?
  1. Air is treated as an ideal gas.
  2. The working fluid completes a closed cycle.
  3. Specific heats are constant at a reference temperature.
  4. Combustion is represented by heat supplied to the air.
Show answer and explanation
Specific heats are constant at a reference temperature.
Cold-air-standard analysis includes the air-standard assumptions and adds constant specific heats evaluated at a reference temperature.

Question 2

A closed-cycle model specifies air-standard assumptions but says nothing about specific heats. What is the safest conclusion?
  1. Specific heats must be constant.
  2. The model is necessarily cold-air-standard.
  3. The specific-heat treatment is not established by the stated assumptions.
  4. The cycle has zero heat transfer.
Show answer and explanation
The specific-heat treatment is not established by the stated assumptions.
Air-standard assumptions do not, by themselves, specify constant specific heats. They also do not imply zero heat transfer.

Question 3

In a cold-air-standard calculation, a supplied constant-volume specific heat is used for an internal-energy change. What temperature difference belongs in the relation?
  1. The final temperature minus the initial temperature.
  2. The initial temperature minus the final temperature in every case.
  3. The reference temperature minus the final temperature.
  4. No temperature difference is needed.
Show answer and explanation
The final temperature minus the initial temperature.
The relation is Δu=cv(T2−T1)\Delta u=c_v(T_2-T_1), with the sign determined by the direction of the temperature change.

Key terms

Air-standard assumption
A set of cycle-model assumptions that treats air as the ideal-gas working fluid in a closed cycle and represents combustion and exhaust by heat transfer.
Cold-air-standard assumption
Air-standard analysis with the additional approximation that specific heats are constant at a stated reference temperature.
Specific heat
A property that relates a temperature change to an internal-energy or enthalpy change in the stated model; use cvc_v for internal energy and cpc_p for enthalpy.

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