Gas Power Cycles Questions and Answers

Answer: A

The correct statement is \(W_{AB} = -NRT_1 \ln\frac{V_2}{V_1}\), with \(W\) taken as the work done on the gas.

On a T-S diagram, horizontal lines are isotherms and vertical lines are adiabats (constant entropy). AB is the isothermal expansion at \(T_1\). For \(N\) moles of ideal gas the work done by the gas is \(\int P\,dV = NRT_1\ln\frac{V_2}{V_1}\), so the work done on it is

\[ W_{AB} = -\int_{V_1}^{V_2} P\,dV = -NRT_1 \ln\frac{V_2}{V_1} \]

The key follows this sign convention, so the expansion work is negative.

BC and DA are adiabats on which the temperature changes, so \(W = \Delta U \neq 0\) on both. CD is an isothermal compression at the cold temperature \(T_2\), so its work involves \(T_2\), not \(T_1\).

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: 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: D

A gas turbine works on the Brayton cycle.

The Brayton (or Joule) cycle has isentropic compression in the compressor, heat addition at constant pressure in the combustion chamber, isentropic expansion in the turbine and heat rejection at constant pressure. That matches the continuous-flow layout of a gas turbine.

The Otto cycle (heat added at constant volume) is the basis of the petrol engine, and the Diesel cycle (heat added at constant pressure in a piston engine) of the diesel engine. The Carnot cycle is an ideal reference that no practical engine follows.

Answer: A

The clearance ratio is clearance volume / swept volume of cylinder.

In a reciprocating compressor the clearance ratio is \(C = V_{c}/V_{s}\), the ratio of the volume left in the cylinder at the end of the delivery stroke to the volume swept by the piston. It typically lies between about 4 and 10 per cent, and it appears in the volumetric efficiency, \(\eta_{v} = 1 + C - C(p_{2}/p_{1})^{1/n}\).

Total volume over clearance volume is the compression ratio of an engine, and the other two options invert or change the denominator, so they are not the clearance ratio as defined.

Answer: C

The efficiency is \(\eta = 1 - \left(\frac{1}{r_{p}}\right)^{\frac{\gamma - 1}{\gamma}}\).

In the Brayton cycle the isentropic temperature ratio across the compressor is \(T_{2}/T_{1} = r_{p}^{(\gamma-1)/\gamma}\), and the efficiency is

\[ \begin{aligned} \eta &= 1 - \frac{T_{1}}{T_{2}} \\ &= 1 - \frac{1}{r_{p}^{(\gamma-1)/\gamma}} \end{aligned} \]

The options with the exponent \(\gamma - 1\) alone use the Otto-cycle form, which is written in terms of the compression ratio, not the pressure ratio. \(1 - r_{p}^{(\gamma-1)/\gamma}\) would be negative, since \(r_{p} \gt 1\).

Answer: B

The energy transfer is continuous in an axial compressor and intermittent in a reciprocating compressor.

In an axial compressor the rotor blades act on a steady stream of air passing through, so energy is transferred continuously. In a reciprocating compressor, work is done on the gas only during the compression stroke of each cycle, and the suction and delivery happen in separate parts of the cycle, so the transfer is intermittent.

The first option reverses the two. Calling both continuous or both intermittent ignores the basic difference between rotary (dynamic) and positive-displacement machines.

Answer: C

The thermal efficiency is 30%.

The useful output is the net work, turbine work minus compressor work:

\[ \begin{aligned} W_{net} &= 120 - 60 = 60\ \text{kJ} \\ \eta &= \frac{60}{200} = 0.30 \end{aligned} \]

Using the turbine work alone would give \(120/200 = 60\%\), which forgets that the compressor is driven by the turbine. 20%, 40% and 50% do not follow from the figures.

Answer: D

As the clearance volume increases, the amount of air delivered decreases.

At the end of delivery the air left in the clearance space expands again during the suction stroke. The larger the clearance, the further the piston must travel before the pressure falls below the suction pressure and fresh air can enter, so less air is drawn in and delivered per stroke.

For the same reason the volumetric efficiency falls, so the option saying it increases is wrong, as is 'amount of air delivered increases'. The work per kilogram of air delivered is, in theory, unaffected by clearance, because the energy spent compressing the clearance air is recovered as it re-expands.

Answer: D

Regeneration increases the efficiency of the cycle.

A regenerator uses the hot turbine exhaust to preheat the compressed air before it enters the combustion chamber. Less fuel is then needed to reach the same turbine inlet temperature, so the heat supplied falls while the turbine and compressor work stay the same.

Because the net work is unchanged, the options saying work output increases, decreases, or rises and then falls are all wrong. Only the efficiency changes, and it rises (provided the turbine exhaust is hotter than the compressor delivery).