Gas Power Cycles Questions and Answers

Answer: C

For these conditions \(\eta_{O} \lt \eta_{D}\).

Draw both cycles on a T-s diagram. They start from the same state 1 and reject heat along the same constant-volume line back to it. Since they reject the same amount of heat, they leave that line at the same state 4, so they share the same entropy at the end of expansion and therefore the same state 3, the state of maximum temperature.

They differ only in how heat is added to reach state 3: the Otto cycle along a constant-volume line, the diesel cycle along a constant-pressure line. Traced back from state 3, a constant-pressure line on the T-s diagram is flatter than a constant-volume line, so it stays at higher temperatures and encloses more area beneath it. The diesel cycle therefore takes in more heat for the same heat rejected, and \(\eta = 1 - Q_{out}/Q_{in}\) is larger.

This is the reverse of the familiar comparison at the same compression ratio, where the Otto cycle is more efficient; here the fixed maximum temperature makes the diesel cycle use the higher compression ratio. Equal efficiencies, or \(\eta_{O} = 1 - \eta_{D}\), have no basis.

Answer: A

The ratio is the cut-off ratio.

In the Diesel cycle heat is added at constant pressure while the piston moves out, so the cylinder volume grows during combustion. The cut-off ratio is \(r_{c} = V_{3}/V_{2}\), the volume at the end of combustion over the volume at its start. A larger cut-off ratio lowers the Diesel cycle efficiency.

The compression ratio is the total volume over the clearance volume, \(V_{1}/V_{2}\). The pressure ratio relates pressures, as in the Brayton cycle, and the back work ratio is the fraction of turbine work used by the compressor, also a gas-turbine term.

Answer: D

The Otto cycle is the most efficient of the cycles for the same compression ratio. Its efficiency is \[ \eta = 1 - \frac{1}{r^{\gamma - 1}} \] and for the same compression ratio the Otto cycle rejects the least heat, since all the heat is added at constant volume. The Diesel cycle has the lowest efficiency for the same compression ratio, and the Dual and Trinkler cycles come between them.

The Diesel cycle adds heat at constant pressure during expansion, so at the same compression ratio it is less efficient than the Otto cycle.

Answer: C

The ideal Stirling cycle is two reversible isotherms and two reversible isochores.

The working gas is compressed isothermally at the low temperature, heated at constant volume, expanded isothermally at the high temperature and cooled at constant volume. With an ideal regenerator, the heat given up in the constant-volume cooling is stored and returned in the constant-volume heating, so the cycle reaches the Carnot efficiency.

Two isobars and two adiabatics make the Brayton (Joule) cycle. Two isotherms and two isobars make the Ericsson cycle. Isentropic compression, constant-volume heat addition, isentropic expansion and constant-volume heat rejection describe the Otto cycle.

Answer: A

The correct T-S diagram is (A).

The Otto cycle is 1 to 2 adiabatic compression, 2 to 3 heat addition at constant volume, 3 to 4 adiabatic expansion and 4 to 1 heat rejection at constant volume. A quasistatic adiabatic process is isentropic, so 1 to 2 and 3 to 4 are vertical lines. At constant volume \(dS = C_{V}\,dT/T\), so T rises exponentially with S along 2 to 3 and falls along 4 to 1: curved lines rising to the right. An engine runs clockwise on a T-S diagram, taking in heat at the higher temperature.

Diagram (A) has exactly this shape and direction. (B) has the same shape traversed anticlockwise, which is a refrigerator, not an engine. (C) and (D) are rectangles, with horizontal isothermal sides, which describe the Carnot cycle, not the Otto cycle.

Answer: A

The result is \(T_{4a} \lt T_{4}\).

With constant-volume heat rejection, expansion 3-4 stops at the volume \(v_{1}\). For constant-pressure rejection to return to state 1, expansion 3-4a must continue down to the pressure \(p_{1}\), which is below \(p_{4}\) (\(p_{4} \gt p_{1}\) because state 4 is hotter than state 1 at the same volume). Further isentropic expansion lowers the temperature, since along 3-4a

\[ \begin{aligned} \frac{T_{4a}}{T_{4}} &= (p_{1}/p_{4})^{(\gamma - 1)/\gamma} \\ &\lt 1 \end{aligned} \]

So \(T_{4a}\) is lower than \(T_{4}\), not equal to it, higher, or double.

Answer: C

The optimum pressure ratio is \(\left(\frac{T_{3}}{T_{1}}\right)^{\frac{\gamma}{2(\gamma-1)}}\).

Write \(x = r_{p}^{(\gamma-1)/\gamma}\). Then \(T_{2} = T_{1}x\) and \(T_{4} = T_{3}/x\), and the net work per unit mass is

\[ \begin{aligned} w &= c_{p}\left(T_{3} - \frac{T_{3}}{x} - T_{1}x + T_{1}\right) \\ \frac{dw}{dx} &= c_{p}\left(\frac{T_{3}}{x^{2}} - T_{1}\right) = 0 \Rightarrow x = \sqrt{\frac{T_{3}}{T_{1}}} \end{aligned} \]

So \(r_{p} = (T_{3}/T_{1})^{\gamma/(2(\gamma-1))}\); at this ratio \(T_{2} = T_{4} = \sqrt{T_{1}T_{3}}\). The exponent \(\gamma/(\gamma-1)\) is the largest ratio the cycle can have, where the net work falls to zero, and the other two exponents are not stationary points.

Answer: B

For the same compression ratio, \(\eta_{th,Otto} \gt \eta_{th,Diesel}\).

\[ \begin{aligned} \eta_{Otto} &= 1 - \frac{1}{r^{\gamma - 1}} \\ \eta_{Diesel} &= 1 - \frac{1}{r^{\gamma - 1}}\left[\frac{r_{c}^{\gamma} - 1}{\gamma(r_{c} - 1)}\right] \end{aligned} \]

For any cut-off ratio \(r_{c} \gt 1\), the bracketed factor exceeds 1, so the Diesel cycle loses more. Physically, the Otto cycle adds all its heat at the highest temperature, at constant volume, while the Diesel adds part of its heat at constant pressure as the piston moves away, at a lower effective temperature.

The two would be equal only in the limit \(r_{c} \to 1\), and no heat engine reaches an efficiency of 1. In practice Diesel engines are more efficient because they run at much higher compression ratios, which this comparison excludes.

Answer: A

The net work output is about 335 kJ/kg.

With \(c_{v} = 1.005/1.4 = 0.7179\) kJ/kg-K, follow the cycle:

\[ \begin{aligned} T_{2} &= 300 \times 20^{0.4} = 994.3\ \text{K} \\ T_{3} &= 1.5\,T_{2} = 1491.5\ \text{K} \\ T_{4} &= T_{3}\left(\tfrac{1.5}{20}\right)^{0.4} = 529.2\ \text{K} \end{aligned} \]

Heat is added at constant pressure and rejected at constant volume:

\[ \begin{aligned} q_{in} &= 1.005(1491.5 - 994.3) = 499.7 \\ q_{out} &= 0.7179(529.2 - 300) = 164.6 \\ w_{net} &= 499.7 - 164.6 \approx 335\ \text{kJ/kg} \end{aligned} \]

The distractors are the intermediate quantities: 500 kJ/kg is the heat added and 165 kJ/kg the heat rejected. The thermal efficiency, \(335/500 = 0.67\), is a check.

Answer: C

The efficiency is \(1 - 1/(\text{Compression ratio})^{(\gamma - 1)}\).

In the air-standard Otto cycle heat is added and rejected at constant volume, and the two other processes are isentropic between the same two volumes. With compression ratio \(r = V_{1}/V_{2}\):

\[ \begin{aligned} \eta &= 1 - \frac{T_{4} - T_{1}}{T_{3} - T_{2}} = 1 - \frac{T_{1}}{T_{2}} \\ \frac{T_{2}}{T_{1}} &= r^{\gamma - 1} \Rightarrow \eta = 1 - \frac{1}{r^{\gamma - 1}} \end{aligned} \]

The exponent is \(\gamma - 1\), not \(\gamma\), because the isentropic temperature-volume relation is \(TV^{\gamma - 1}\) = constant. The forms in terms of a pressure ratio raised to \((\gamma - 1)/\gamma\) describe the Brayton (gas turbine) cycle, where heat is added at constant pressure; the pressure ratio with exponent \(\gamma - 1\) matches neither cycle.